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# TheoryofElasticityI MEM660

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THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E ua ons ofElastieit Chapter 4 StressStrain Relations and Governing Equations of Elasticity When a deformable body is subjected to certain loadings deformation will occur and stresses will be developed within the body In general the level of stress developed in the body depends not only on the final state of deformation but also on the history and rate of the applied loading the temperature diange etc and the deformed body usually does not return to its original configuration after the loading is removed However if the deformation is infinitesimal and the deformation process is adiabatic then it is generally acceptable to assume that the stress and strain are independent of the loading rate and history and that the body does return to its original configuration after the loading is removed Sudi a deformation is called elastic and can be studies by using the Theory of Elasticity 41 Generalized Hooke39s Law The stress in an elastic body if depending only on the deformation can be expressed as1 017 f1 em 4 1 Expanding the equation into a Taylor39s series in the neighborhood of em 0 yields 0 62 0 6 75k1i fix eklemnquot39 0 017 M aek magma If the deformation is linear elastic then the higherorder nonlinear terms in the above equation can be neglected Consequently we have 03917 A17 C17klekl 42 where A17 f170 and CW 6f1706ekl are constants It is easy to see from 42 that A17 represents the state of stress at ekl 0 namely the initial stress which can be neglected without loss of generality in the context of linear elastic deformation Therefore we have 03917 CWekl 43a or in matrix form all C1111 C1122 C1133 C1123 C1131 C1112 C1132 C1113 C1121 ell 0 22 C2211 C2222 C2233 C2223 C2231 C2212 C2232 C2213 C221 822 033 C3311 C3322 C3333 C3323 C3331 C3312 C3332 C3313 C3321 833 023 C2311 C2322 C2333 C2323 C2331 C2312 C2332 C2313 C2321 823 031 C3111 C3122 C3133 C3123 C3131 C3112 C3132 C3113 C3121 831 0 12 C1211 C1222 C1233 C1223 C1231 C1212 C1232 C1213 C1221 ell 0 32 C3211 C3222 C3233 C3223 C3231 C3212 C3232 C3213 C321 832 013 C1311 C1322 C1333 C1323 C1331 C1312 C1332 C1313 C1321 813 all C2111 C2122 C2133 C2123 C2131 C2112 C2132 C2113 C2121 821 Equation 43a or 43b is called the generalized Hooke39s law in which CW are elastic constants Since both the stress and strain are secondorder tensors according to the Quotient Rule C must be a 17 kl 1 We shall use e1 7 to denote the Cauchy s infinitesimal strain tensor in this and subsequent chapters TM Tan Drexel University 4 1 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uations ofElastieit fourthorder tensor with 81 components representing 81 elastic constants as can be seen in 43b However due to the symmetry of stress and strain tensors ie 03917 0397 and ek elk we have can CW lelk Therefore the number of independent elastic constants reduces to 36 Furthermore we shall show later that for an elastic deformation there exists a strain energy density inetion Wkly such that 03917 w 44 Bel Consequently C1714 Ckll 45 and the number of independent elastic constants reduces to 21 for generalized anisotropic linear elastic materials By introducing the following abbreviated notations 01011 02022 03033 04023 05031 06012 51511 52522 53533 542523 552531 562512 we can avoid carrying the double sums and reduce the number of subscripts of eadi elastic constant to two Thus the generalized Hooke39s law can be expressed in the following matrix form 71 C11 C12 C13 C14 C15 C16 51 0392 C12 C22 C23 C24 C5 C26 52 73 C13 C23 C33 C34 C35 C36 53 4 6 74 C14 C24 C34 C44 C45 C46 54 75 C15 C25 C35 C45 C55 C56 55 76 C16 C26 C36 C46 C56 C66 56 or simply 039 Cl 4631 It should be noted that in the above equation 54 2523 723 55 2531 731 56 2512 712 are the socalled engineering shearing strain components 42 Elastic Symmetry Most engineering materials possess some types of symmetry in their crystals whidi can be observed in their elastic properties By making the reference coordinates coincide with these planes of symmetry and by showing that the constants are invariant with respect to the coordinate transformation we can further reduce the number of independent elastic constants a Monoelinie materials A monoclinic material has one plane of symmetry If we set up two coordinate systems X and Xl39 sudi that XX1 X39ZXZ X 3 7X3 and the common X1 7X2 plane coincides with the plane of symmetry as shown in Fig 41 then the elastic constants become invariant with respect to the coordinate TM Tan Drexel University 4 2 October 29 2007 THEORY OF ELASTICITY 4 StressStrain Relations and Governing Equations of Elasticity transformation between X and that is the material properties at points P and P39 shown in Fig 41 should be identical for the two coordinate systems X2 X Plane of X3 Symmetry gt X5 Fig41 A monoclinic material having X12 plane as the plane of symmetry The direction cosines for such a transformation are given by 391 0 039 aijz iE O 1 O 0 0 1 and the components of stresses and strains can be shown to satisfy the following relationship 7 01 e 61 fori l 2 3 6 47a I I I I The elastic constants on the other hand remain unchanged upon such transformation ie C39 The generalized Hooke s law of 46a thus can be written in the coordinates as 039 C 39lle39 Clle39l Upon expanding this expression we have 7139 C11ei C126 C136s C146 C15e C166 48 plus five similar equations for 0 0 0g Substituting 47 in 48 gives 71 C1161 C1262 C1363 C1464 C1565 C1666 From 46 we have 01 C1161 C1262 C1363 C1464 C1565 C1666 plus five similar equations for 02 03 06 Comparing the two equations above shows that C14 2 C15 2 0 Similarly by examining the other five equations for 02 03 06 and 0 0 0g we have C24 2 C25 2 C34 2 C35 2 C46 2 C56 2 0 Thus the matrix of elastic constants for a monoclinic material becomes Tll Tan Drexel University 4 3 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uations ofElastieit and the number of independent elastic constants reduces to 13 l7 Orthotropie materials An orthotropic material has three planes of symmetry that are mutually orthogonal to one another By making the coordinate planes coincide with the planes of symmetry and following a procedure similar to that for the monoclinic materials we can show that the number of independent elastic constants reduces to nine and the matrix of elastic constants becomes C11 C12 C13 0 0 0 C12 C22 C23 0 0 C13 C23 C33 0 0 0 0 C41 0 0 0 0 0 C55 0 0 0 0 0 C66 e Hexagonal transversely isotropic materials A hexagonal or transversely isotropic material has one plane of symmetry and one axis of symmetry perpendicular to the plane of symmetry By making the X3 axis the axis of symmetry hence the X1 7 X2 plane the plane of symmetry we can show that the number of independent elastic constants reduces to five and the matrix of elastic constants becomes C12 C13 0 0 0 C12 C11 C13 0 0 0 C13 C13 C33 0 0 0 0 0 0 C44 0 0 0 0 0 41 0 0 0 0 0 0 C11 7 C122 a Isotropic materials For isotropic materials the material properties are independent of the coordinate systems The number of independent elastic constants reduces to two and the matrix of elastic constants becomes C11 C12 C12 0 0 0 C12 C11 C12 0 0 0 C C12 C12 C11 0 0 0 0 0 0 curd2 0 0 0 0 0 0 eyed2 0 0 0 0 0 0 eyed2 which can be rewritten in the following form TM Tan Drexel University 4 4 October 29 2007 THEORY OF ELASTI CI TY 4 StnessStmin Relations and Governing E uutions ofElusticit it2u it it 0 0 0 it it2u it 0 0 0 it it it2u 0 0 0 C 49 0 0 0 u 0 0 0 0 0 0 u 0 0 0 0 0 0 u in whidi H12 y C11C12 l and Ll are called the Lum constants and the Hooke s law in terms of the Lame constants becomes 03917 it lzekk 2ue 410 The inverse relation is given by 2m a 60 411 322u 43 Engineering Elastic Constants tn the previous section we have shown that for an isotropic material the number of independent elastic constants reduces to two ie the Lame constants it and u While these constants are obtained mathematically their physical meanings are difficult to interpret and hence are rarely used directly in solving engineering problems Instead the socalled engineering elastic constants including the bulk modulus K the Young39s modulus E the shear modulus G and the Poisson39s ratio 1 are more commonly used in practice We shall illustrate the physical significance of these engineering elastic constants and derive the relationship among all the constants using the following examples u Uniform pressure iBulk Modulus When a body is subjected to a uniform pressure the state of stress is given by 03911 03922 03933 P 012 U23 03931 0 where p is the magnitude of the pressure applied on the surface Since the pressure is always acting in the direction perpendicular to the surface the traction thus can be expressed as t1 7pm 711517117 where n is the unit normal to the surface Furthermore since t 717117 we have 017 P r Performing a contraction on the above equation and 410 the generalized Hooke s law yields a 1 5 3r a 039 xt llekk 2ue 31 2102 b A comparison of a and b results in TM Tan Drexel University 4 5 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uations ofElastieit p NZ jg NZ 1g 3 3 where e e11 e22 e33 If the first invariant of the strain tensor represents the rate of volume d1ange The Bulk modulus or Modulus ofCompression whid r measures the amount of volume dqange of a material under uniform pressure load or the rigidity of the material in a dilatation deformation is defined as KL12 412 e 3 u l7 Simple tension 7 Young s Modulus and Poisson s Ratio A state of simple tension in the X1 direction is dqaracterized by 011 00 03922 03933 U12 03923 03931 0 where 0390 is the applied uniform tensile stress Substituting this state of stress in 411 yields il 39 2 it 2511 011 711 M00 31 211 31 211 The Young 39s modulus or the modulus of elasticity which measures the rigidity of the material in a unidirectional extension deformation is defined as 5 342 l 413 511 A I It is noted from the generalized Hooke39s law 411 that a simple tension in the X1 direction also produces strains in the X2 and X3 directions even though there is no traction applied in those directions From 411 and using 413 we have 22533 03911 A e mus12 201 11 where the negative sign indicates that contractions will occur in both the X2 and X3 directions when the material is subjected to a simple tension in the X1 direction The Poisson s ratio whid measures the ratio of lateral contraction due to the axial extension for a material under a simple tension condition is defined as V2f224 414 511 511 2a 1 e Simple shear 7 Shear Modulus When an elastic body is subjected to a simple shear of To in the X1 7 X2 plane we have 71210 03911 03922 03933 03923 03931 0 Substituting this state of stress in 410 yields 012 To 2mm 712 where 712 is the engineering shearing strain The shear modulus whid r measures the rigidity of the material in a pure shear is defined as TM Tan Drexel University 4 6 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uutions ofElustieit G u 415 12 The relationship among the Lam constants it and y and the commonly used engineering elastic constants E v G and K are given in Table 41 3H2 2 it1v1 72v 9K1lt71 gtlt MHZ4 E 241 1721 K2 3 Ev v 72v 313K7E 3Kv 31172v 1v v Consider the following relations from Table 41 7 E and K E 7 21v 3172v Since physically the values of E G and K cannot be negative we have from the above two expressions 1v20 and 172120 ISVSOE For most of engineering materials the values of Poisson s ratio do not deviate mud from 13 For highly incompressible materials rubber for example where the volume remains almost und ranged regardless of the level of stress applied K gt oo then 1 gt 05 and G u E3 Furthermore a negative Poisson39s ratio although mathematically possible has never been observed in isotropic materials For some anisotropic materials however there has been evidence of effective Poisson 395 ratios with values greater than 05 For isotropic materials with 0 E V g 05 the Lam constant 2 must be positive as can be seen from the equation TM Tan Drexel University 4 7 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElusticit Ev 1v172v By using table 41 we can show that the Hooke s law in terms of the engineering elastic constants E G and 1 can be written in the following alternative forms v v 039171 1217 172V 572 2G e 1721 572 416 1 1 v 51E1V07 V51 Ukk E017 m5170kkj 417 In unabridged notation 417 becomes 1 l 71 0 0 0 l i E i 0 0 0 511 E 1E E 03911 E 1E E 03911 v v v v 522 7E E if 0 0 0 03922 7E E 7E 0 0 0 03922 v v 1 v v 1 7 7 0 0 0 7 0 0 0 833 E E E a E E E 1 a 418 823 0 0 0 11 0 0 023 0 0 0 0 0 03923 2G 531 0 0 0 1V 0 03931 0 0 0 0 i 0 031 E 2G 2 039 039 12 0 0 0 0 0 H V 12 0 0 0 0 0 i 12 E 2G in whida the Poisson s effect on the normal deformation can be clearly seen 44 SumJnary of the Governing Equations in the Theory of Linear Elasticity So far we have derived the following equations for linear elastic materials I Three equilibrium equations 0W 12 0 419 I Six straindisplacement relations for an infinitesimal deformation 1 e 50 M 420 I Six stressstrain relations Hooke s law 03917 it 39lzekk 2ue17 421a 2 7 1 a 4 21b W17 017 3A2 170 10 When solving an elasticity problem we consider the deformation of an elastic body of volume V enclosed by a surface S subjected to the actions of body force l7 and surface traction t1 We seek solutions of the six stress components 039 six strain components 2 and three displacement components 141 that satisfy 17 419 420 and 421 at every point in Vand the traction boundary condition on S given by t 039 n 422 TM Tan Drexel University 4 8 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElastieit where n is the unit normal to the surface It is noted that b and t cannot be prescribed arbitrarily as they must satisfy the overall equilibrium Instead of the prescribed surface traction an elasticity problem may also involve a prescribed displacement u on a portion or the entire surface ie u u 423 In fact most of the realworld problems involve both prescribed traction and displacement over its surface In sud a case the surface S is divided into two separated regions namely SL and 8 over which the displacements and surface tractions are prescribed respectively They are referred to as the displacement and traction boundary conditions The solution to an elasticity problem is the one in which the components of displacement stress and strain satisfy the governing equations 419 421 and the boundary conditions 422 and 423 Finally the components of strain must satisfy the compatibility equations 517kzekz17 T 51qu 7 gym 0 424 to ensure the uniqueness of the displacement solution Sometimes it is more convenient to combine some of the governing equations and express them in terms of only one set of variables sud as those of displacement or stress This can be accomplished rather easily using the procedures described below a Governing equations in terms ofaisplacements Differentiating 421a with respect to x and using 420 we have 0 1 Him 141 425 Substituting 425 in 419 gives x1uuW tau 12 0 426 or in vector form x1uVl euVZuli0 427 where If e1 um is the first invariant of the strain tensor Equation 426 or 427 called the Navier39s equation is the governing equation of elasticity in terms of the displacements Solving this set of equations subject to the boundary conditions yields the displacement solution of the elasticity problem The strain components can be obtained by dii 39 39 O the quot J J according to 420 and the stress components can be subsequently computed using the Hooke39s law given in 421 It is noted that as the solution of the Navier39s equation is in terms of the components of displacement the compatibility conditions are satisfied automatically 1 Governing equations in terms of stresses Substituting 421b in 424 and using 419 we obtain wk V m 171 4771 428 0 kk 1v 17 where v it21u is the Poisson39s ratio Equation 428 called the BeltramieMiehell compatibility equation by Mid ell in 1900 and by Beltrami in 1892 without body force terms is the governing equation TM Tan Drexel University 4 9 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElusticit of elasticity in terms of stresses Solving this set of equations and the equilibrium equations 419 subject to the boundary conditions yields the solution to the elasticity problem If the body force field is conservative then there exists a potential function 1 such that 171a V 6x 1 er 429 Substituting 429 in 428 gives v 171 1 Ulchk makk17 5 Wick 2W 430 If the body forces are constant eg gravitational force or zero 1417 yrkk 0 430 becomes 1 0W 1V 0km 0 431 Applying tensor contraction to 431 yields v211 0 432 where 11 01 is the first invariant of the stress tensor Equation 432 implies that 11 is a harmonic mction Furthermore by taking the Laplace of 431 and using 432 we have v40 0 433 Equation 433 is called biharmonic equation whose solutions are called biharmonic functions Thus when body forces are constant or absent the solution of an elasticity problem reduces to that of finding the appropriate biharmonic functions that satisfy the boundary conditions 45 Strain Energy Density Function Consider an elastic body in a state of equilibrium If the body undergoes a further stable incremental deformation denoted by V114 under an adiabatic condition ie there is no heat gained or lost then according to the first law of thermodynamics we have dWe jVdudV 434 where dWe jvbyzuyzw jst u s 435 is the incremental work done by body force l7 and surface traction t1 on the incremental displacements d141 and dU is the incremental increase of the intrinsic energy per unit volume of the body during this process By using the relation t1 717117 and the Theorem of Gauss 435 can be written as dWe Maw b 1udv j V adudv 436 The first integral on the righthand side of 436 vanishes due to the equilibrium conditions The integrant of the second integral can be written as alzdulyz 017d217 dw Harlem 437 TM Tan Drexel University 4 10 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElusticit Combining 434 through 437 gives jVdudV Ivawdequ 438 Since the volume of the integral in 438 can be taken arbitrarily we have du Jade 439 Since the lefthand side of 439 is an exact differential there exists a function Wkly sud that 039 ale de 17 17 e 17 17 or 6W 03917 E 440 The function W called the strain energy density function is the potential energy per unit volume stored in the body The existence of the function Wkly when the deformation in the body takes place isothermally can be proven using the second law of thermodynamics For an elastic body that obeys the generalized Hooke39s law given in 43 the strain energy density function can be easily identified as 1 1 WEC17kleiyekl aneg 441 For isotropic material we have by using 410 W 15172kk 2ue17 gt17 gaujeuekk 7142127 7 217217 3441155 27 21125 442 where I and I are the first and second invariants of the strain tensor respectively Equation 442 can also be written in the following form 7 4 2 2 2 2 2 2 2 W 7 3211 222 233 u 211 222 233 2212 2223 2231 443 Since both it and u are nonnegative quantities we conclude that w 2 0 444 46 Uniqueness of Solution For a given elasticity problem in whid the body force 171 in V surface traction t1 on 8 and displacement 14 on SL are prescribed we seek a solution in which the components of stress strain and displacement satisfy the governing equations 419 420 and 421 and boundary conditions 422 and 423 One may ask if sud a solution is unique in the sense that the components of stress and displacement are determined without ambiguity To show that the solution if existing is indeed unique let us assume that there are two possible solutions denoted by ufl0391 21 and 149042 22 respectively to the same problem Since both 1717 1717 TM Tan Drexel University 4 11 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElusticit solutions satisfy all the governing equations and the boundary conditions it is easy to see that the difference of these two solutions defined as 7 1 Z 7 1 Z 7 1 Z Auliul 7141 Aggie 7037 Ae ie iew will satisfy the following conditions A017 0 in V 445a A017 leekk 2yAelZ in V 445b A141 0 on SL 445c Aalznz At 0 on S 445d Defining W 7 1A0 A3 i W 7 W 4 46 T 2 17 17 we can show that 2 WW IVAaWAerV IVAa Aulyde AW Aew 7 Am IV AL393917A141 7 dV 7quot AalzyzAuldV Integration by parts ISAaleulnzdS Theorem of Gauss A017 0 447 mamas Mm Ag jg Agmms jg AtlAuldS s s 5 Since A141 0 on SL At 0 on S and the volume integral in 447 is arbitrary we have W1A039Ae 0 2 17 17 In view of 445b we conclude that A037 0 and A217 0 The only possible difference between the two solutions is a rigid body motion whid does not produce any strain or stress Finally since the displacements are prescribed on S the rigidbody motion must vanish Consequently the solution if exists must be unique 47 Principle of Superposition Recall the governing equations of linear elasticity 419 421 and the boundary conditions 422 and 423 Since all these equations are linear ie they contain only terms of up to the first degree in the dependent variables and their derivatives therefore the components of stress strain and displacement due to some combined sets of external loads can be obtained by superposing the results due to ead set of load acting separately This is known as the Principle UfSuperposition The proof of this principle is straightforward and can be found in any textbook It should be noted that in deriving 419 421 we have assumed that the deformation is infinitesimal and the distinction between the initial and the deformed configurations of the body is negligible There are however certain situations in whid the deformation may be infinitesimal but the difference between the initial and deformed configurations may not be ignored In such cases the TM Tan Drexel University 4 12 October 29 2007 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uutions ofElusticit bars of a solid cross section In contrast stresses in the bars of a thinwalled cross section depend largely on the end conditions 10 I xx I a 17 c fwltxgt I x 39 fmax0 I x l x 17 I i O xL 10 Fig 43 Normal stresses due to torsion along bars of various cross section Figure 44 shows the second example by Hoff where a statically determinate truss is subjected to a set of four selfequilibrating concentrated forces at one end The numerical values written on some of the truss elements represent the forces acting in those members It can be seen that the effect of this set of forces decays very slowly over the length of the truss indicating that the SaintVenant s principle does not work well in this case Fig 44 Atruss subjected to a selfequilibrated force system Problems 41 A thin plate made of a homogeneous isotropic material is subjected to a proportional biaxial loading as shown Two strain gages are being placed at the center of the plate to measure the strain components 211 and 222 As the load readies 039 35 gtlt103 psi the readings from the two gages are 211 35x10 3 and 222 10gtlt10 3 respectively Determine the Young s modulus E and Poisson s ratio 1 of this material TM Tan Drexel University 4 14 October 29 2007 THEORY OF ELASTI CI TY 4 StnessStmin Relations and Govemin E nations 0 Elusticit 7220 20 in 14140 psi A lt B 822 45 7 all 20 Le 011 20 14140 ps1 uo psi 11 A 45 l 14140 psi 2 Problem 41 Problem 42 A parallelogramshaped thin plate as shown in the figure is made of a material with elastic constants E 30gtlt106 psi and v 03 The plate is subject to a uniform shearing stress of 14140 psi as shown If the deformation is infinitesimal determine a the dqange in length of line E and b the principal strains and their directions Show that for isotropic materials the principal directions of stresses coincide with those of principal strains Show that if a material possesses two planes of symmetry that are mutually orthogonal to each other then the third orthogonal plane must be a plane of symmetry Show that the number of independent elastic constants reduces to nine for an orthotropic material and to five for a transversely isotropic material Given the following state of stress cx vx127x 725mg 0 0391 725mg cx vx 7x12 0 0 0 cvxf x a Do the components satisfy the equilibrium equations b Are they a possible solution of a problem in elasticity Why TM Tan Drexel University October 29 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Chapter 5 Extension Torsion and Bending ofPrismatic Bars In this diapter we shall study the behavior of prismatic bars subjected to various types of loading conditions that cause the bars to extend bend or twist The primary solution method we shall employ is the socalled semifinverse method in which certain assumptions are made regarding the components of stress strain or displacement yet leaving enough freedom in those quantities so that the governing equations and boundary conditions can be satisfied It should be noted that in many cases it is often difficult if not impossible to specify the displacement andor traction boundary conditions exactly In sudi cases statically equivalent conditions based on the SaintVenant s Principle must be used instead In deriving the governing equations of elasticity we used the tensor or the abridged notation to take the advantage of its compactness When solving specific problems however no particular benefit can be gained by using sudi a notation Therefore in this diapter we shall use the variables x y and z to replace x1 x2 and x3 and denote the components of displacement in the X1 X2 and X3 directions by 14 v and w respectively Thus the governing equations derived in the previous chapters will be rewritten in the following forms a Equilibrium Equations From 419 we have 6039 Law e Lam bx 0 M 6y 62 6039 6039 6039 quoty W 92 b 0 51b 6x 6y 62 y 6039 Law W Lam H72 0 Sic 6x 6y 62 b Hooke s Law Substituting 420 in 421b and using the relations in Table 41 we obtain the following stressdisplacement relations in terms of Young s modulus E shear modulus G and Poisson s ratio 1 039m 7 1039W 039ZZ 52a 039W 7103922 03 52b 03922 7 1039m 039W 52C ZONE 52d 6y 62 E 91 G 2 ZONE quotzx 52e 62 6x E m C Q 21V A 520 6x 6y E 9 G c Compatibility Equations From 430 we have 1 621 v 62w V2 1 V2 2 53 0 1v 6x2 171 W 6962 a TM Tan Drexel University 5 1 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Z 2 via La 121 V Viv26 531 W 1v By 171 6y 2 Z V2022 v2w2 f 53c Z Z via y LQ a V 53d 1v ayaz Oyaz Z Z VZUZXLa 11 2 a V 53e 1v 626x 626x 621 62 via W 1 e V 530 1v may 5 My where 11 is the first invariant of the stress tensor and 11 is a potential function from which the components of body force can be derived iie 6y 6y 6y I 7 b 7 1727 54 quot 6x y ax y 62 d Traction Boundary Conditions From 422 we have t awn axyny runZ 55a ty awn aWny ayznz 55b tZ o xznx ayzny 72211 55c where 71 11 my 712 is the unit vector normal to the surface of the boundary 51 Extension of a Prismatic Bar by an Axial Force Fig 51 Extension of a prismatic bar by an axial force Consider a prismatic bar of length L with one end fixed at the xy plane and the other end subjected to an axial force P as shown in Fig 51 Force P is directed along the z axis Which coincides with the centerline of the bar The elementary ie Mechanics of Materials theory assumes that on cross sections perpendicular to the zaxis shearing stress vanishes and normal stress distributes uniformly ie 03922 PA a 7W cryZ an o xy 0 56 TM Tan Drexel University 5 2 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs where A is the crosssectional area It is easy to show that in the absence of body force the state of stress given in 56 satisfies the equilibrium equations 51 and the compatibility equations 53 The tractionfree boundary condition ie t1 0 on the lateral surface of the bar is also satisfied as can be shown by substituting the components of the unit normal to the surface 111 15119 0 and 56 into 55 On the z L surface 11 0 01 the exact traction boundary condition should be 0 to 0 t039 y yz z 22 t1 717117 03912 or tx a In order for the stress components given in 56 to be the solution of this problem the normal stress 039ZZ must be uniformly distributed over the entire cross section at the free end This condition however is rarely satisfied exactly The concentrated force in Fig 51 for instance would have to be represented by an infinitely large stress at the point of application and zero stress everywhere else on the cross section For a slender bar whose length is much larger than the typical dimension of the cross section a set of statically equivalent condition based on the SaintVenant s principle see Chapter 4 is often used to replace the exact condition For the particular example shown in Fig 51 the statically equivalent conditions at 2 L are the resultant forces in the x and y directions and the resultant moments about the three coordinate axes must vanish and the resultant force in the z direction must be equal to the applied concentrated force P These conditions can be expressed in the following equations 2F L crudl 0 ZFy L 0 9sz 0 2132 L crudi P 57a 2M L auydA 0 ZMy 7 A 0 2sz 0 2M2 L iyam x039yz7lA 0 571 Since the state of stress given in 56 satisfies 57 it is therefore the solution of the problem It should be noted however that the first two equations in 57b are satisfied only if P is applied at the centroid of the cross section The LUL r J39 o J39 J ie the Hooke s law and the boundary conditions at z 0 are properly specified For instance by constraining the displacements and the rotations with respect to the three coordinate axes at 000 we obtain t r can be determined by substituting 56 into 52 n integrating the resulting equations provided that the displacement va wg 14 AE AE AE 58 Further discussion on the determination of quot 1 r will be given in Section 52 The total amount of extension of the bar is characterized by the zdisplacement at the free end From 58 we have wz L PLAE which agrees with the solution of elementary theory 52 A Prismatic Bar Stretched by Its Own Weight A slender prismatic bar of length L supported at its top surface is being stretdied by its own weight as shown in Fig 52 We set up a coordinate system so that the origin is located at the centroid of the bottom surface and the z axis is directed along the centerline of the bar The components of body force are given by bxl7 0 bz7pg 59 where pg is the weight density From 59 and 54 we obtain the potential function 1 as being wng 510 TM Tan Drexel University 5 3 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars The coefficients of eadi individual terms in the above equation must vanish ie 0140xyavoxy0 517 By 6x 2 6 w x y 0 518 ix Equations 516a and 517 imply that 140 is a linear function of y 00 is a linear function of x and the coefficients of the two linear terms must be equal in value but opposite in sign ie 140 ayl7 519 00 fax 5 520 in whidi a b and c are arbitrary constants From 516b and 518 we conclude that m0 is a quadratic function of x and y and that the coefficients of the xy term must vanish ie wox2y2dxeyf 521 where d e andf are unknown constants Substituting 519 520 and 521 in 514 and 513 yields u7xzidz2yb 522 v7yziezinyc 523 wzzvx2y2dxeyf 524 Constants in 522 7 524 can be determined using the displacement boundary conditions at 2 L the top surface of the bar Since the exact condition ie 14 v w 0 everywhere on the cross section is impossible to satisfy results can only be obtained for certain approximate boundary conditions specified based on the SaintVenant s principle Consider for instance the following set of displacement boundary conditions at 0 0 L ie at the centroid of the cross section on the top surface 14 v w 0 to prevent translations in x y and 2 directions 525a 6 0 or 0 to prevent rotation of cross section about the x axis 525b z 60 6w a 0 or a 0 to prevent rotation of cross section about the y ax1s 525c z 60 614 a 0 or a 0 to prevent rotation of cross section about the z ax1s 525d x It is easy to show that these conditions would yield 11 b c d e 0 and f ingZ2E Thus the components of displacement are given by u7xz v7yz wZZLZlx2yz 526 TM Tan Drexel University 5 5 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 00L lt L 000 max Fig 53 Deformation of a bar stretched by its own weight Figure 53 shows the cross section y 0 of the bar before dashed lines and after solid lines deformation has occurred Points along the z axis would displace vertically only as can be seen by substituting x y 0 in 526 ie u 0 0 v 0 0 aux0 wi 527 y 0 y 0 90 2E and the maximum vertical displacement wmax ingZ2E occurs at 0 0 0 Other points on the cross section however would displace both vertically and horizontally It can be seen from 526 that 14 and v are linear functions of x and y respectively Therefore longitudinal fibers that were parallel to the z axis before deformation would deform into straight lines inclined to the z axis Furthermore points on the cross section 2 c where 0 g c S L would displace to ZCw6CZLZlx2yz whidi represents a surface of paraboloid to which all longitudinal fibers would be perpendicular after deformation Therefore there is no shearing strain associated with this deformation Due to the assumed uniformly distributed normal stress given in 511 and the displacement boundary condition given in 525 the top surface of the bar would deform concavely as shown in Fig 53 This of course would not have happened if the entire top surface were completely constrained In the latter case even though the stress distribution on the top surface would be very different from being uniform the resultant force must still be equal to the total weight of the bar in view of the overall equilibrium Thus based on the SaintVenant s principle the difference between the two solutions based on different displacement boundary conditions would be appreciable only in a small region near the top surface 53 Torsion of a Prismatic Bar of Circular Cross Section Consider a prismatic bar of circular cross section and of length L with one end fixed on the xy plane and the other end subjected to a torque MZ as shown in Fig 54a Under the action of the torque the bar would be twisted and a longitudinal line on the lateral surface eg the dashed line in Fig 54a would deform into a helical curve the solid line in Fig 54a The elementary theory of torsion of circular bars assumes that w 0 ie plane cross sections would remain plane after deformation and that any TM Tan Drexel University 5 6 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs cross section at a distance 2 from the xed end would simply rotate by an angle 6 Consider for instance the cross section shown in Fig 54b showing only the first quadrant A point originally located at P would move by an angle 6 to P after deformation The inplane displacement components are given by a A generator deforms b Displacement and stress at into a helical curve point P on a cross section Fig54 Torsion of a circular prismatic bar 14rcos667rcos xcos6717ysin6 528a vrsins 67rsin xsin6ycos671 528b where x cos and y sin If 6 is infinitesimal then cos6 m 1 and sin6 m 6 528 becomes 147y6 vx6 529 If we further assume that 6 varies linearly in z ie 6 20 where or is known as the angle of twist per unit length of the bar then 529 can be written as 147yzot vxzo 530 The corresponding stress components can be readily derived from 52 0x2 iGozy cry Gooc 531a am039W crmcrxy 0 531b in which the assumption that w 0 has been employed It is easy to show that these stress components satisfy equilibrium 51 and compatibility 53 To deck the traction boundary conditions we first examine the lateral surface of the bar For a circular cross section as shown in Fig 54b the unit normal has the following components 71 n n2 0 532 Substituting 531 and 532 in 55 we have t awn axyny 03211Z 0 533a ty 03911 799119 79211 0 533b TM Tan Drexel University 5 7 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs circular bars The same pattern of deformations has also been observed in prismatic bars of varying cross sections Based on these observations SaintVenant in 1855 hypothesized the following displacement components for prismatic bars of general cross sections subjected to end torques 14 foxyz v oocz w byway 535 The corresponding stress components can be derived by substituting 535 in 52 ie am 7W 03922 7xy 0 536a 039CZ GaKa Zi y 039 2 Go a Zx 536b 6x 9 0y Substituting 536 in 51 yields the following Laplace s equation for 1xy 62 62 V2 2 5 6x 0y Thus warping functions x are harmonic functions Furthermore by substituting 536 into 55 0 537 it can be shown that the tractionfree boundary condition on the lateral surface of the bar is satisfied if the warping function satisfies the following equation on the boundary 7yjnxxny 0 538a or ll Z ynx 7 amy 538b in whid 71 and my are the x and y components of the unit normal At the two ends of the bar the statically equivalent traction conditions based on the SaintVenant s principle will be used It is easy to show that the stress components given in 536 would yield FZ Mx My 0 The resultant force in the x direction is given by2 F Laud4 Leggy Gangpg j Hizw j dv GaJLxEi ynx xny st 0 in which 537 538b and the Theorem of Gauss see Chapter 1 have been used Similar result can be obtained for the resultant force in the y direction ie ry j 039 dA j Gaa WxdA 0 A 91 A W Finally the resultant moment about the z axis must be equal to the applied torque MZ Thus M2 IJ039xzyo39yzxdAGog x2y2xiydfl 53 2 LS Sokolnikoff Mathematical Theory of Elasticity 2 d Ed McGrawHill 1956 p 112 TM Tan Drexel University 5 9 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs or M2 Do 5395 where 7 2 2 62 6 DiGHEx y xaiyajdl 540 is the torsional rigidity of the bar From 539b we conclude that the angle of twist per unit length 0 must be proportional to the applied torque M2 The torsion problem is thus reduced to that of finding a harmonic function 1x y satisfying the boundary condition 538 associated with the particular cross section It is noted that since any arbitrary constant satisfies both 537 and 538 the warping function therefore can be determined only to within an arbitrary constant This constant however will have no effect on the stresses as can be seen from 537 55 Prandtl39s Stress Function for Torsion Problems Consider the function Ga77xzyz 541 where 7 is the conjugate harmonic function of the warping function shy so that they satisfy the CauchyRiemann equations 6 756ZI 6 757675 542 6x 6y 0y 6x Differentiating 541 with respect to x and y respectively and using 542 we have sza Zixisz alx 543a 6x 6x 0y aw a Ziy Goia Ziy 5435 0y 0y 6x Comparing 543 with 536b we have a a 7 544 6y 9 x The function xy introduced by L Prandtl in 1903 is called the stress function3 for torsion problems It is easy to show that 544 satisfies the equilibrium equations Eliminating the warping function I from the two equations in 543 yields the following Poisson s equation for the stress function xy 62 62 Vz mf6yfizca 545 3 A rigorous derivation of the stress functions for torsion can be found in LS Sokolnikoff s Mathematical Theory of Elasticity 2 01 ed McGrawHill 1956 pp 114119 TM Tan Drexel University 5 10 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Fig56 A unit normal on the boundary of an arbitrary cross section As for the traction boundary condition on the lateral surface of the bar let us consider an arbitrary cross section as shown in Fig 56 showing only the first quadrant The components of unit normal to the surface can be expressed as d dx y n n x dc 9 dc 546 where dc is an infinitesimal arc element counterclockwise being positive along the boundary of the cross section Substituting 546 in the boundary condition 538a and using 543 we have dijiw 547 By dc 6x dc dc Equation 547 implies that the stress function xy must remain constant along the boundary of the cross section For bars with simpleconnected cross sections eg solid bars this constant can be set to zero ie 0 as the value of the constant has no effect on the stresses Thus eadi torsion problem is reduced to that of finding a stress function xy satisfying the governing Poisson s equation 545 and the boundary condition 5 0 For bars with multipleconnected cross sections eg hollow bars the value of 5 can still be set to zero along its outer boundary Along the inner boundaries however the values of 5 can no longer be chosen arbitrarily We shall discuss this in more detail in Section 58 The torque applied at the two ends of the bar can be obtained by taking the moment of the stress components on a cross section about the z axis ie M2jjiaxzyayzxiA7 yxm2jj m 548 in which 0 along the boundary has been observed Therefore the torque is numerically equal to twice the volume enclosed by the surface representing function xy and the xy plane It is noted that each term in the integrand of 548 contributes to one half of the total torque Thus one half of the torque is due to the stress component 039CZ and the other half due to cry Example 51 Torsion ofan Elliptic Prismatic Bar Consider a prismatic bar of elliptic cross section as shown in Fig 57 The boundary of the cross section is described by the equation TM Tan Drexel University 5 11 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs w Fig 57 The cross section of an elliptic prismatic bar 2 Z y 2710 a NIX N Q in whidi 11 and b are the major and minor axes respectively of the ellipse By assuming x2 y 2 m 7 1 b 1 H2 b2 where m is a constant the boundary condition for 547 is obviously satisfied The value of m can be determined by substituting b in 545 whidi yields m 7 7 anZGo C 7 112 b2 Thus the stress function is given by nzbzGo x2 y2 7 71 d 1124472 112 b2 The applied torque can be determined using 548 7 260 2 2 2 2 2 2 7 abaGtx Mfzmmx gbgb x gm y m b gamma b e where 2 7 771723 2 7 77an3 7 7 x dAilyiT y dAilx T J39J39dAiAiimb are the moments of inertia and area of the cross section The angle of twist per unit length is given by 112 b2 2 f a asbsG and the torsional rigidity is given by M2 7 m3b3G 7 6A4 D or 1124472 47r21p where Z Z 111 2 P X y 4 TM Tan Drexel University 5 12 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs is the polar moment of inertia of the elliptic cross section The shearing stress components can be obtained by substituting d in 544 2y 2x QE WMZ Uyz mMz h Since from b and h we have x 112 7 03992 yb2 0u it follows that the resultant shearing stress I along any radius line must be parallel to the line tangential to the boundary at the point where the radius intersects the boundary as shown in Fig 57 Eli dxi From h we see that the maximum values of the shearing stress components are 2x max 7m3b 2M2 mbz Zymm Mz 2 In I mks 03992 z O39I W W and if 11 gt b the absolute maximum shearing stress is M i mbz Ir max occurring at 0147 where the minor axis intersects the surface of the cross section The warping function 1xy can be determined from 543 b27112 112 b2 x Z y D Finally the components of displacement can be obtained by substituting f and j into 535 2 b2 2 2 bzi 2 u7otyz7WMz voocz 1 3173 ZM7 wazM2 k In I G 7111 b G In I G The contour lines of the warped cross section can be obtained by setting the warping function given in j to constants ie b2 7 112 x K or x K H2 172 y y These are hyperbolas having the principal axes of the ellipse as asymptotes The warped surface is convex in the first and third quadrants and concave in the second and fourth quadrants It can be shown easily that if 11 b then the solution of the elliptic bars reduces to that of circular bars discussed in Section 53 56 The Membrane Analogy Method for Torsion Problems The membrane analogy method for solving torsion problems was first introduced by L Prandtl in 1906 Prandtl considered the equilibrium of a homogeneous membrane such as the elliptic one shown in Fig58 supported along its edges and subjected to a uniformly distributed internal pressure p If the membrane deflection zxy is small then we can assume that the membrane force S remain unchanged throughout the membrane and the internal pressure p is directed along the z axis Consider TM Tan Drexel University 5 13 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs an infinitesimal element of membrane nbsd of size Axgtlt Ay as shown in Fig 58 The 2component of the membrane force along edge ad is given by F5 Sm Aysin 6m m SmAy m 42 096 m 2xy Deformed Sm membrane R i 6 a i M P m S P l P a 55 Side View y i 393 Edge of the I membrane I S I S 1 c do 5 4y 5 39 Ax I u b x v quot 5 39 S Top Fig58 An elliptic membrane subjected to a uniform pressure p where for an infinitesimal deflection sin 6 is approximated as 626xm ie the slope in the x direction of the deformed membrane along edge ad of the element Similar expressions can be obtained for membrane forces acting along the other three edges Since the sum of all these forces acting on the four edges must be balanced with the internal pressure p we have 62 62 62 62 13Z SAy 7 SAy j SAx 7 SAx prAy 0 2 596 m 596 b 0y 1 5V d Dividing this equation by SAxAy and letting Ax gt 0 and Ay gt 0 yield the following Poisson s equation 2 Z V22a a 7 549 6x 6y 5 Furthermore since the edges of the membrane are constrained we have E 0 550 do where do is an infinitesimal arc length of the membrane edge Comparing 549 and 550 with 545 and 547 the governing equation and boundary condition for torsion problems in terms of the Prandtl39s stress function we conclude that if the outline of the membrane is identical to that of the cross section of the bar subjected to torsion then the solutions of these two problems are identical provided that the quantity pS in the membrane solution is replaced by 2G0 in that of torsion and vise versa TM Tan Drexel University 5 14 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bendin o Prismatic Burs zx Y Contour lines of membrane Lines of shearing stress dx B l a Membrane problem b Torsion problem Fig59 Membrane analogy for torsion problems Figure 59 a shows the first quadrant of a membrane with contour lines depicting its deformation due to internal pressure p The corresponding lines drawn on the cross section of the torsion subjected to torsion as shown in Fig59 b are called the lines ofshetzring stress Consider a point eg point B on the membrane through whid r a particular contour line is drawn Since along a contour line the deflection of the membrane is constant we have in as where c is the arc length of the contour line The corresponding equation for a bar in torsion is 0 as where 5 is the stress function Using 544 and 546 we can express the above equation as nine W1 7 a Zn 03an 0 551 6c 6x do 6y do 9 9 Since ayzny 03an represents the total component of shearing stress in the direction normal to the line of shearing stress see the insert of Fig 59 we conclude from 551 that the shearing stress resultant r must be tangent to the lines of shearing stress The magnitude of the resultant can be obtained by summing up the components of 039 and aw in that tangent direction ie r awn 7 0tu gm Jrgny 73 552 Thus the magnitude of r at any point is equal to the gradient of the stress function 5 in the direction normal to the line of shearing stress at that point On the corresponding membrane dzdn is the maximum slope at that point Thus the shearing stress resultant on the bar in torsion can be obtained by multiplying this maximum slope by the constant 2G 06 ie TM Tan Drexel University 5 15 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars EJJE 553 an pS an From 548 we also conclude that the resultant torques at the ends of the bar is equal to twice the volume enclosed by the membrane with pS being replaced by 260 ie M2 2 j j 7lxrly 2GTIIzaxay 554 Consider a free body of the membrane obtained by cutting the membrane along a contour line eg the shaded area in Fig 59 Along the boundary of the free body there is a uniformly distributed membrane force S whose zcomponent per unit arc length of the contour line is 8azrln Summing up all the forces acting on the free body we have dz F A S rl 0 Z 2 r f dn 6 where A is the area of the xy projection of the membrane free body Substituting 553 into the above equation yields 17quot 7 A 7 rlc 7 0 P Zsz frds2GaA 555 from whidi the average shearing stress along a line of shearing stress can be determined ie 7 frdc 7 ZszA 1 m fat C where C fat is the length of the line of shearing stress Example 52 Torsion ofA Rectangular Prismatic Bar w Fig510 Torsion of a rectangular prismatic bar Consider a prismatic bar with a rectangular cross section of size 2agtlt2l7 as shown in Fig 510 subjected to end torques By using the method of membrane analogy the problem becomes that of finding the deflection surface of a uniformly loaded membrane having the same rectangular boundary and satisfying the governing equation and boundary condition of 549 and 550 By assuming the deflection of the membrane to be in the following series form TM Tan Drexel University 5 16 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 2w coswgt a where b1bZ are constants and Yny are unknown functions in y the condition of symmetry with respect to the y axis and the boundary condition zx in 0 are satisfied automatically Substituting a in 549 and then expanding the righthand side into a Fourier series for in x E 11 ie 7373 m i E Si 871 n 1 2 cos nx b where 8 nil211 we obtain the following differential equations for Y y H 4 Li Y main 4 lt71 2 The solution of this secondorder differential equation is given by 2 n 16 7 UTl Y A sinh B cosh n n n ns sbns Constant An vanishes as the result of Y being symmetric with respect to the y axis Constant B can be determined by using the condition that Y 0 at y ib This yields Z 1171 n 1361 71 17 cosh ny n It bnS cosh nb Thus the membrane deformation is given by Z16P112 in 71 1 17 cosh nyJCOS nx C 735 My 113 cosh nb Replacing pS by 2G0 in c yields the stress function for the corresponding torsion problem 326 Z w 712 h Z 12 lcosm d and the nonvanishing stress components can be obtained by substituting d in 544 ie cos nx n71 77716Gm i 71 sinh ny X2 6y It2 My 112 cosh nb n71 7716Gm Z 422 licosh nyJSm nx 71 92 6x It2 My cosh nb If h gt 11 the maximum shearing stress occurs at i a 0 where 039CZ 0 16Gm 1 1 r 039 ia0 17 e W 1 gt z cosh5 ltgt Using the relation 2 1 Zi 3 5 8 TM Tan Drexel University 5 17 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs we can rewrite e in the following form 16Gm 1 f IIZ Hays cosh nb rmax 2Gm7 The infinite series on the righthand side whidi converges very rapidly if l7gt 11 becomes negligible compared to the first term for a very narrow rectangular cross section where I gtgt 11 In such a case the maximum shearing stress becomes rmax 26m g For a square cross section where 11 l7 f becomes rmax m 1351Gou h In general we can express the maximum shearing stress as rmax k2Gm i where k is a numerical factor whose value depends on the aspect ratio of the cross section btz The applied torque can be obtained by substituting d in 548 This yields 1 3 19211 w 1 G 2 2b 17 t h b M 3 a a s MESS an A 1 The infinite series term on the righthand side again converges very rapidly In general the torque can be expressed as M2 leo2iz32b k By eliminating G6 from i and k we can also express the maximum shearing stress as Ml r l k2lt2agtzlt2bgt 0 where k1 and k2 are numerical factors whose values depend on the aspect ratio 1711 Values of the factors k k1 and k2 for several different btz ratios are given in Table 51 For a bar with a very narrow rectangular cross section 1711 00 we have k gt1 k1 kZ gt13 26m 211 2b 1 3 M2 3GDt211 212 Table 51 Constants for Torsion of a Rectangular Prismatic Bar btz k k1 kZ btz k k1 k2 10 0675 01406 0208 3 0985 0263 0267 12 0759 0166 0219 4 0997 0281 0282 15 0848 0196 0231 5 0999 0291 0291 20 0930 0229 0246 10 1000 0312 0312 25 0968 0249 0258 00 1000 0333 0333 TM Tan Drexel University 5 18 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs gtI r Za Fig511 Torsion of a narrow rectangular prismatic bar Solutions for bars with narrow rectangular cross sections subjected to torsion also can be obtained in the following manner Consider the deformation of a membrane whose outline is the same as the narrow cross section ie I gtgt a of the bar as shown in Fig 511 The shape of the slightly deflected membrane can be approximated as being cylindrical throughout the length in the y direction if the small regions near yib is ignored see the yz projection in Fig 511 The xz projection of the deflected membrane see Fig 511 can be approximated by that of a string subjected to a uniformly distributed pressure p The maximum deflection of the string is given by 5 112228 and the maximum slope occurring at x in is given by E r m dn max 11 S The volume enclosed by the parabolashaped cylindrical membrane and the xy plane is given by 2 4113b p V 39 2 2b 3 ax gt 3 S n Replacing pS in m and n with 2G0 and using the membrane analog we have 1W 26m M2 6a2a32b 0 which agree with those given in f and j with the series term being neglected for 17 gtgt 11 57 Torsion of Prismatic Bars with Rolled Profile Sections Solution for a narrow rectangular bar subjected to torsion can be easily extended to those for rolled profile sections Let t 211 and l 217 so that the size of the narrow rectangular cross section is now tgtltl where lgtgt t then the maximum shearing stress and the applied torque can be expressed as rmax Gtor M2 Do 556 where 3 D G l 557 is the torsional rigidity Consider a slotted tube whose cross section is shown in Fig 512 It is easy to see that whether the narrow rectangle is straight or curved has very little effect on the volume enclosed under the corresponding membranes Thus the same equations as those in 556 can be used for the slotted tube shown in Fig 512 TM Tan Drexel University 5 19 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs t t a r a t 12 4 I 1 T b Fig 512 Cross section of a slotted tube Fig 513 Rolled profile sections For other types of rolled profile sections such as angles diannels and lbeams shown in Fig 513 we may consider each of them as being a combination of a number of narrow rectangular cross sections The total torsional rigidity is then the sum of those of individual ones Take the angle shown in Fig 513 a as an example the total torsional rigidity is given by 3 3 DDh H31 Gltll rtzlzl 2 3 Thus we have 3MZ 0 3 3 Gltll1 thzlZ The shearing stress in each leg of the angle can be estimated by using the formula for narrow rectangular cross sections as well ie I thor33I zt i12 t1l1t2lZ Similar expressions can be derived for the diannel and lbeam sections shown in Fig 513 ie 0 33M2 3 and r i12 Git1l12t2l2l t1l12t2lZ It should be noted that in deriving the approximated solutions for rolled profile sections we have neglected the detailed geometry of the fillet at the reentrant corners where a considerable stress concentration may develop if the radius of the fillet is small Values of these highly concentrated stresses can be estimated by using the membrane analogy method4 58 Torsion of Prismatic Hollow Bars So far we have discussed torsion problems for bars with solid cross sections only ie those that are bounded by a single curve At the boundary of sudi cross sections the stress functions according to 547 must remain constant and are usually set to zero However for bars with hollow cross sections ie those that have two or more boundaries the values of stress function along the boundaries can no longer be chosen arbitrarily We shall consider first a simple case in which the inner boundary of the hollow cross section coincides with one of the lines of shearing stress of the corresponding solid bar such as the elliptic cross section shown in Fig 514 showing only the first quadrant The outer boundary of the hollow cross section whidi coincides with that of the corresponding solid cross section is given by 4 SP Timoshenko and N Goodier Theory of Elasticity 3rd ed McGrawHill 1970 pp 322324 TM Tan Drexel University 5 20 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Membrane D T It p p a 212 r boundary x Z Lines of shearing stress Inner l boundary k1 V1 Fig 514 Torsion of a hollow bar with elliptic cross section x2 y 7 710 558 The inner boundary of the hollow cross section being geometrically similar to the outer boundary as it coincides with one of the lines of shearing stress can be expressed as x2 y2 W W 71 7 0 559 in which k lt1 is a scaling constant Since the resultant of shearing stress is tangent to hence does not cross the line of shearing stress the hollow bar therefore can be considered as being obtained by removing an inner cylinder from the solid bar without altering the stress distribution in the remaining portion of the bar This can also be verified by using the following membrane analogy Since in the hollow region the stress is zero the membrane must have a zero slope This can be easily adiieved by placing a rigid horizontal plate 6 in Fig 514 over the hollow region It is noted that the pressure distributed over the membrane a in a solid bar is statically equivalent to that distributed over the plate 6 in a hollow bar Consequently the membrane force S along the boundaries of free bodies represented by a and 6 respectively must be the same Thus the equilibrium conditions for the remaining portion of the membranes ie the shaded area in Fig 514 are identical for both cases and the stress function derived in Example 51 for a solid elliptical bar is therefore applicable to the hollow bar ie 7 nzbzGo LQLZA 560 112 b2 11 172 From Fig 514 it can be seen that the volume enclosed by the membrane of the hollow elliptic bar is less than that of the solid elliptic bar by the amount equal to the volume of revolution of area CDD or VHW 17 k4vs Based on the membrane analogy and e of Example 51 we concluded that for a hollow elliptic bar 3 3 The angle of twist per unit length thus is given by TM Tan Drexel University 5 21 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs MZ 1124472 a 17 k4 asbsG 561 The stress function and the corresponding maximum shearing stress are given respectively by M x2 y2 7 z 1 562 mbilik liE Z b2 2M 1 r quot 563 max mbz 17 k4 Substituting the equation of the inner boundary 559 in 562 we have M2 Vim boundary W If the inner boundary of the hollow bar does not coincide with any of the stress lines of the corresponding solid bar then the selfequilibrating condition of the rigid horizontal plate discussed previously is no longer satisfied In sudi a case additional load must be applied on the plate to maintain its horizontal position hence would introduce additional conditions that the membrane must satisfy In terms of the stress function for torsion problems these additional conditions are necessary to ensure that the displacements be singlevalued From 535 and 536b we have 0w 0w o xzG 7oty o zG ooc 6x 9 0y Expressing the resultant shearing stress in terms of its components ie x dy 1039 039 do 92 do and integrating it along the inner boundary of the hollow cross section we have dx dy 6w 0w 1dc an 5 cry Ejdc Jrady 7 Ga ydxi xdy If the displacement w is singlevalued then the first integral in the above equation must vanish as the integration is taken along a closed curve The second integral is equal to twice the area enclosed by the curve Thus we have Medeai This is the additional condition that needs to be satisfied when determining the constant value of stress function 5 along the inner boundary 59 Torsion of ThinWalled Tubes Solutions for torsion of bars made of thinwalled tubes can be easily derived by using the membrane analogy Figure 515 shows the cross section of a thinwalled tube and the corresponding stress function whidi is geometrically similar to the membrane Since the wall is thin we can neglect the slight variation in the slope of the membrane across the thickness of the wall and replace it by a straight line as shown in the insert of Fig 515 Thus the shearing stress can be approximated by TM Tan Drexel University 5 22 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 564 where h is the difference of the values of 5 between the inner and outer boundaries and t is the thickness of the wall It is easy to see that if the wall thickness is uniform then the shearing stress remains constant throughout On the other hand for a tube of nonuniform thickness the maximum shearing stress would occur at the location of the smallest wall thickness In either case It h remains constant throughout and is referred to as the shear flow in the wall of the cross section The corresponding torque can be calculated using membrane analogy ie it is twice the volume enclosed by the stress function whidi can be approximated as M I m 2Ah 2At r where A is the area enclosed by the dashed line in Fig 515 or the mean value of the areas enclosed by the outer and inner boundaries of the cross section of the tube Thus given a torque M2 the shearing stress can be computed as M Z 565 1 2A To determine the angle of twist per unit length we integrate the shearing stress given in 565 along the centerline ie the dashed line in Fig 515 of the wall and use 555 We f 2610 566 2A t from whidi we obtain M do z 567 0 4AZG t If the tube has a uniform thickness t is constant 567 becomes M Z1 0 T 568 4A Gt where l is the length of the centerline along the tube wall The torsional rigidity is given by 2 D w 569 or Z TM Tan Drexel University 5 23 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs For a circular tube with a mean radius V 1471 2 and 12m we have D47rzr4Gt Comparing this with 557 the torsional rigidity for a slotted tube we see that the ratio of the two torsional rigidities is D Z Dunslouedmbe 570 slotted tube For thinwalled tubes 7 gtgtt Thus having a longitudinal slot in the tube would drastically reduce the torsional rigidity Profile of 5 and membrane Fig 516 Torsion of a typical aircraft fuselage For thinwalled tubes with multiple holes sudi as an aircraft fuselage with a typical cross section as shown in Fig 516 the shearing stress in ead portion of the wall can be obtained by using the membrane analogy ie h h2 h1 7 hZ TF l 72 13 a ti Integrating the shearing stress along the walls enclosing the two compartInents denoted as A1 and AZ respectively in the figure then using 555 we have 1111 1313 2G04A1 b 1le 71313 ZGoa lZ c where ll lz and 13 are the arc lengths of ACE ADB and AEB respectively The total torque can be determined by using the membrane analogy M2 2 dA 2A1h1 A2h2 2A1t111 12512 d Shearing stresses 1391 and the angle of twist per unit length or can be obtained by solving a b c and d simultaneously They are M 71 NZ t3lZAl tzlsA1Azl M 12 NZ t3llAZ t113A1 A2l TM Tan Drexel University 5 24 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs M2 73 N tilei tzliAzl M2 0 ZGN tllZl3 tZl3ll t3lllZ where N 2t1t312A12t2t311A t1t213A1AZZ 510 Pure Bending of Prismatic Bars Fig 517 Pure bending of a prismatic bar by and moments A prismatic bar of length L is fixed at xy plane and subjected to a bending moment M at z L as shown in Fig 517 If the z axis coincides with the centerline of the bar and the yz plane where moment M is acting on is one of the principal planes then from the elementary theory of bending we know that the bar is in a state of pure bending and that the components of stress are given by 039ZZ am 7W cryZ an o xy 0 571 where R is the radius of curvature of the bar after bending It can be seen that the equilibrium is satisfied if the body force is absent The compatibility is also satisfied automatically as the stress components are either zero or linear functions of 2 For the boundary conditions it can be easily shown that the traction free condition on the lateral surface of the bar is satisfied At 2 L however we must use the following statically equivalent conditions F A andl 0 F9 A azydA 0 F2 A crudl 0 572a My A 0 22di 0 M2 L 7 o znyr 0 Wth 0 57213 Ey2 E1 M jAauydAjA R dick 572c where Ix is the moment of inertia of the cross section about the x axis This is the wellknown moment 7 curvature equation derived in the elementary theory of bending Substituting 572c in 571 yields whidi is again identical to the flexure formula in elementary theory of bending To determine the components of displacement we substitute 571 in 52 This yields TM Tan Drexel University 5 25 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 1 573 v v R ax R By R 62 0 0 f qj 574 02 By 6x 02 By ax Integrating the last equation in 573 we have w7w0xy 575 where m0 is a function of x and y Substituting 575 in the first two equations in 574 yields 614 72770100 607 0w 2 awe Eax ax E ayRay or after carrying out the integration with respect to z 0 0w 0w 2 6x u0xy 07 72 0y voxy 576 where 140 and 00 are functions of x and y as well Substituting further 576 in the first two equations of 573 yields 2 2 6x 6x R 0y By R from which the following conclusion can be drawn by comparing terms on both sides 2 Z a 206 200 ivl 577 6x 6y 6x By R Integrating the second equation in 577 gives 2 x uoxyev7yf1y voxyev2y Rfzx 578 Substituting 576 and 578 into the last equation of 574 yields azwimyqzqkwhwrvq axay R dy dx Since this equation is true for any value of 2 we have Z a woxy0 d ydfzxiv 0 579 may dy dx R From the first equations in 577 and 579 we conclude that w0 can only be a linear function of x and y ie w0xy mxnyp 580 From the second equation in 579 we conclude dfiy df2x7v a R dy dx TM Tan Drexel University 5 26 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs whidq upon integration yields 2 x f1ywj fzxv m7 581 Combining 575 576 578 580 and 581 gives 9 7 u 7v 7 7 mz R 00 5 07izzivxziy2 oocinz7 2R w mx 11 R V P in which m n p or 8 and 7 are constants to be determined by the displacement boundary conditions Since the bar is fixed at z 0 we can assume that at 0 0 0 uvw0 and 0 Oz Oz 0y These conditions whidi prevent the point 000 from having any movement or rotation yield 0 8 7 m n p 0 Thus the components of displacement are given by uiivjz y v7zzivxziy2 w 582 We now examine the validity of the basic assumptions used in deriving the elementary theory of bending by comparing the solution of elementary theory with that of theory of elasticity shown here 1 By substituting y 0 in the last equation of 582 we have w 0 This implies that the xz plane remains undeformed and is indeed the neutral plane 2 Displacement component w as shown in 582 is a linear function of y indicating that plane cross sections will indeed remain plane 3 Equation 574 requires that the shearing strain vanish throughout the entire bar Thus cross sections perpendicular to the neutral plane will indeed remain perpendicular after bending The deflection curve along the z axis the centerline of the bar can be obtained by setting x y 0 in 582 ie 22 07 uw0 583 2R whidi is identical to those given in the elementary theory of bending Consider the cross section of a rectangular prismatic bar at a distance 2 c from the fixed end The 211x2l7 rectangular cross section shown by solid lines in Fig 518 would deform to the shape represented by the dashed lines The two vertical edges at x in would deform linearly ie my i ux 11 R while the top and bottom edges at y would deform to parabolic curves ie TM Tan Drexel University 5 27 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bendin o Prismatic Burs k r 1 lt w y l l l k b I 1L Fig 518 Deformation of the cross section of Fig 519 Optical image Showing contour lines a rectangular bar due to bending on top surface of a bar in bending vltyibgt7 czvxzebz For small deformations the curvatures of the two parabolic lines can be approximated by the second derivatives of the curves ie dzy d2 1 Z Z Z v ibi c vx 7b dxz dxz 2R R The top or bottom face of the bar forms an anticlas c surface with curvatures being convex down in the lengthwise direction but convex upward in the widthwise direction Taking the top face where y b as an example the contour lines of the deformed surface can be obtained by setting the displacement v in 582 constant This yields the following hyperbolas 22 ivxz constant whose asymptotes are given by zzivxz0 584 Figure 519 shows the hyperbolic contour lines of sudi a surface obtained using an optical tedinique5 By measuring the angle or as shown in the figure we can determine the Poisson s ratio 1 of the material using 584 ie v cot2xz cot2 or 511 Bending of a Prismatic Bar by an End Force A prismatic bar of length L is fixed at the xy plane and subjected to a concentrated force Py at zL as shown in Fig 520 The coordinate system has been setup in such a way so that the z axis coincides with the centerline of the bar and the x and y axes coincide with the principal centroidal axes of the cross section In addition the force Py is on a plane parallel to the yz plane and is applied at sudi a distance6 from the centroid that twisting of the bar does not occur Based on the elementary theory of bending we assume that the normal stress on all cross sections is the same as that of pure bending derived in Section 510 ie 5 SP Timoshenko amp N Goodier Theory ofElusticity 301 ed McGrawHill Inc 1970 p 288 6 The significance of this distance and the method of determining it will be discussed in Section 512 TM Tan Drexel University 5 28 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars 585 The elementary theory also assumes that all the remaining stress components except an and 72V should vanish We shall now examine these assumptions to see if they satisfy the governing equations of elasticity and the boundary conditions The equilibrium equations 51 require that in the absence of body force the assumed stress components satisfy 6039 6039 P 602 0 yz 0 6 yz W 586 62 6x 6y I It is noted that the first two equations in 586 imply that the shearing stresses are independent of 2 hence the distribution must remain unchanged throughout the length of the bar Fig521 Unit normal on the boundary of an arbitrary cross section Next we examine the boundary conditions 55 for an arbitrary cross section as shown in Fig 521 showing only the first quadrant It is easy to see that the assumed stress components satisfy 55a and 55b identically From 55c we have 03211 ayzny 0 587 where from Fig 521 do 9 do Substituting these components of unit normal into 587 yileds TM Tan Drexel University 5 29 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs W M e 0 588 039 7039 7 xzalt yzdc Finally the compatibility equations 53 in the absence of body force require that 2 7 Z 7 P9 Va 0 Vazii 589 y 1x1v Thus the bending problem reduces to that of finding for GE and 039 functions of x and y that satisfy 2 f the equilibrium equation 586 the boundary condition 588 and the compatibility equations 589 This can be further simplified by using the following method of stress function Let xy be a stress function for problems of bending not to be confused with the stress function for the problems of torsion discussed previously in this diapter from whidi the components of stress can be derived as follows c cpyyz 0X2 2 0y 9 6x 21 fx 590 It is easy to show that these stress components satisfy 586 the equilibrium equations Function fx which has no effect on the conditions of equilibrium will be determined later from the boundary condition Substituting 590 in the compatibility equations of 589 yields 3infl j 362 62 V Zu 0y of 6x 6x2 of 51va dxz From whidi we conclude Z Z Px VZ 6 6 V Lid x 6x 0y 1v Ix dx 3 591 where 8 is a constant To find the meaning of 8 we consider the infinitesimal rotation of an area element of the cross section with respect to the z axis given by see Chapter 3 1 60 014 ml 26x The rate of change of this rotation in the z direction is given by a w iin 111mg 3p jaewn 652 az azzax 0y 76x262 By 262 6x 76x 726 6x 0y Substituting 590 in the above equation and comparing the resulting equation with 591 we have Px wifi V Lopg 39 2c 1v1x Since wzyzx 0 ZG we conclude that ZG is the rate of diange of rotation of an area element along the centerline of the bar where x 0 Assuming that the end force is applied in sudi a way so that the bar is not being twisted then 8 0 and 591 becomes V Pyx d x 1v Ix dx W 592 TM Tan Drexel University 5 30 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Finally in terms of the stress function the boundary condition 588 becomes dxdy Pyy27fxdx 593 6c 6x dc ayrlc 21 Pic If the function is known then the values of stress function 5 along the boundary of the cross section can be determined and the solution can be obtained by using 592 and 593 In the following examples the function will be chosen in such way as to make the righthand side of 593 vanish The value of stress function 5 therefore remains constant along the boundary Furthermore for a sirnpleconnected cross section we can make the constant equal to zero then the problem of bending of a prismatic bar reduces to that of finding the solution of 592 with boundary condition 5 0 Once the stress function is determined the stress components can be computed using 590 and the displacement components can be obtained by substituting the stress components in 52 and carrying out the integrations ln terms of stress function the displacement components are given by P L 14 awo21tvgtzvr xyiw p 594 ax E 0y E1 P P 2 v 7 6Eny 23 73L 227 Vx27 yz 77 fEZerch 7 595 P wzzi2Lzwoxy 596 where or 8 and 7 are arbitrary constants to be determined using the displacement boundary conditions and m0 is a function of x and y that can be determined using the following equations Z 2 VP 6 740 721V 6 yy 59721 6x E may EIX azwo 21v 62 7 Pyr vl yr 597 of E may Ix E1 62 1 62 62 d w0 v Z 7r x 597C may E x dx Example 53 Bending ofa Circular Prismatic Bar by an End Force Consider a circular prismatic bar whose boundary is given by x2 y2 R2 a in whidi R is the radius of the cross section The righthand side of 593 becomes zero if we dioose P Ami 3022798 b Substituting b in 592 yields TM Tan Drexel University 5 31 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 712v 1v Ix v2 C Thus the problem reduces to that of finding the solution of c subjected to the boundary condition of 5 0 This boundary condition is satisfied if we dioose mx2y27R2x d where m a constant factor can be determined by substituting d in C Le 7 12v Py 81v1x Thus the stress function becomes 12v Pyx Z Z Z x 7R f 81v Ix y and the stress components can be obtained by using 590 a 77 12v Pyxy 2 41v I g P c 02 32Vy27y271 21ij h 9 81v1x 32v Along y0wehave P sz0 az32V yR271 21362 y 81v1x 32v The maximum shearing stress occurs at the center of the cross section where x 0 ie a 7 32v PyRZ7 32v P 91 max 81v I 21vA in whidi A IrRZ is the crosssectional area of the bar At each end of the horizontal diameter ie at x irR the shearing stress is a 7 12v 17sz 712vPy 9quot Xi 41v Ix 1v A Taking v 03 the value of Poisson s ratio for most engineering materials we have P P c y c y cl MW 71387 cl 937 71237 Comparing with the elementary solution P o 2 ii 9 3 A obtained by assuming that the shearing stress is uniformly distributed along the horizontal diameter the maximum error is seen to be approximately 4 percent TM Tan Drexel University 5 32 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars To find the displacement components we need to determine first the function w0xy by substituting the stress function f in 597 This yields P weby y y3x2yc x77yg j 4EIX where 5 77 and g are arbitrary constants to be 39 using the J39 J boundary 4quot Substituting j and the stress function f in 594 595 and 596 and assume constants or 8 7 5 77 and g all vanish valid if the origin of the coordinate systems undergoes no translation or rotation ie no rigidbody motion we have VP uyLiz k P v izs3L2273VLizx27yzRzz l P w4 ly 22274inx27y2 m We now examine again the basic assumptions used in deriving the elementary theory of bending against the elasticity solution presented here 1 By substituting y0 in m we have w 0 Thus the xz plane remains undeformed and is indeed the neutral plane 2 Equation m reveals that a cross section will deform into a cubic surface in y Thus the assumption that cross section remains plane is no longer valid 3 The shearing stress component 7 whidi can be easily obtained from h does not vanish along y 0 Thus the assumption that cross sections remain prependicular to the neutral plane is no longer valid either Substituting x y 0 in 1 gives the deflection curve of the centerline of the bar P 290 y 23 F3LZZ FMRZZ n 90 6EIX 2 The maximum deflection of the centerline occurs at z L ie f l P9L31m5 o 0x 0 90ij 3E1 2 L The coefficient on the righthand side of o is recognized as the solution from the elementary theory of bending The second term in the bracket is the correction factor whidi is less than 1 for bars with aspect ratio 2RL less than 012 if v 03 Example 54 Bending 7le Rectangular Prismatic Bar by an End Force The boundary of a rectangular cross section as shown in Fig 522 is given by xziaZXEZibz0 a By dioosin g TM Tan Drexel University 5 33 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs W W x z Membrane A lt1 3 m WV Pressure Fig 522 Bending of a rectangularbar by an and force 7 Pybz fx7 2 b x the righthand side of 593 vanishes along the boundary ie Z 2 Pyy 7 Pyb 0 21 along y ib 21 alongxrtz d x0 do Thus the boundary condition becomes 0 Substituting b in 592 yields V Pyx C v2 1v Ix Using the membrane analogy similar to that of a torsion problem we determine the deflection of a membrane subjected to a pressure load proportional to V Px y d 1v Ix This pressure which remains constant in the y direction but varies linearly in the x direction will cause the membrane to deform in a shape whose intersection with the xz plane is as shown in Fig 522 Consequently 5 must be an odd function of x and an even function of y These conditions can be satisfied by taking 7 in the following Fourier series form mmm e 11 2b ZZAZWM Sin m1n1 Substituting e in c and applying the standard procedure for determining the coefficients of a Fourier series we can obtain TM Tan Drexel University 5 34 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 71 1 Z sin m x cos 2n T wily 2 f 2 71 Z 2 712 mn m n 4b2 V Py8 3mm 4 1v Ix 7r mam To find the shearing stress components we first substitute b in 590 this yields P g 02yb2y2 0y 9 6x 21 which can be resolved into the following two systems Oxfalzto39 z 0920 03952 E where P oquot 0 oquot 9 bzi 2 h x2 92 21X y represent a parabolic stress distribution given by the elementary theory of bending and 6 6 a 7 5 0 lt1 6y 6x represent the corrections to the elementary solution due to the stress function Along the x axis 6 0y0 as the result of 5 being an even function of y the corrections to the elementary solution therefore affect only the vertical component of the shearing stress contributed by 052 Substituting f in the second equation of i and letting y 0 we have Tnch 71Ver COS V pg 8H2 no no agile 1vI n3 1 1 112 1 X m 2n71m22n712m Since Ix 411bS3 AbZB where A 411k is the crosssectional area j can be written as m x 71mn72 COS 3P 1 16 112 Oils 9 3 222 2 k 9 2A 1v II b mam 2 2 F 21171 m 4421171 Z 4b The coefficient on the righthand side of k is recognized as the average shearing stress along x 0 from the elementary theory of bending Thus given the ratio nb of a rectangular cross section and Poisson s ratio of the material we can evaluate the term in the brackets of k and make a correction to the elementary solution The infinite series in the equation converges very rapidly particularly for nb gt 1 Table 52 lists for a number of nb values the ratios of exact solutions to the elementary solutions for the stress component cry at x y 0 center of the cross section and at y 0 x in middle of the vertical sides of the rectangle The Poisson s ratio is set to 025 It is seen that the elementary solution yields very accurate estimates for nb 2 2 about 10 error for a square cross section where lZb 1 but become less accurate as the value of nb decreases TM Tan Drexel University 5 35 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Table 52 Comparison of exact and elementary solutions for shear stress in a bar subject to an end force 51 b 5 4 3 2 1 1 2 12 1 2 1 5 x 0 y 0 0997 0996 0993 0983 0940 0856 0817 0805 0801 x id y 0 1005 1008 1015 1033 1126 1396 1691 1988 2285 512 Shear Center In Section 511 we have assumed that the z axis coincides with the centerline of the bar and the x and y axes coincide with the principal centroidal axes of the cross section In addition the force Py is on a plane parallel to the 32 plane and is applied at such a distance from the centroid that twisting of the bar does not occur This distance can be easily obtained once the stress components as given by 590 are known ie we evaluate the moment about the centroid produced by the shearing stress components 0x2 and oyz as Mz 7ny oxzybxdy 598 Since the stress resultant must be statically equivalent to the applied force Py we conclude that the distance from the application point of the force to the centroid of the cross section must be Mz P 1 d 599 For a positive value of Mz dx must be taken in the positive direction of x If a force denoted by Px is applied on a plane parallel to the xz plane then we can follow the same procedure and determine the distance for the application point denoted by dy so that the bar will not be twisted by the applied force The intersection of the two lines drawn parallel to the x and y axes and at distances dy and dx respectively from the centroid is called the shear center Any force that is applied at the shear center and perpendicular to the axis of the bar will not cause the bar to twist Problems 51 A prismatic bar with a 2n x 2h square cross section and of length L is supported at the four corners of the top surface by four rigid bars and is being stretched by its own weight as shown a Specify the displacement boundary conditions b Use the boundary conditions specified in a to determine the constants in 522 523 and 524 hence obtain the expressions for the components of displacement c Compare the results you obtained in b with those in 526 9 2 y x l l dz 39139 391 G x lt L gt Problem 51 Problem 52 TM Tan Drexel University 5 36 November 13 2007 THEORY OF ELASTI CI TY Solution for Chugter 3 31 The deformation of a 2D body is given by x1 02111 01112 xZ 01111 021Z a Compute the components of displacement 141139 1 2 in terms of the Lagrangian variables a and Eulerian variables x respectively Displacement in terms of the Lagrangian variables 141 x1 711 02121 01112 7111 708111 01112 Problem 31 142 x2 7 112 01121 02112 7 112 01121 7 08122 Displacement in terms of the Eulerian variables We first express a in terms of x 111 6667x1 7 3333xz 112 73333x1 6667xZ Then we obtain the displacements in terms of x 141 x1 7111 x1 7 6667x1 7 3333x2 75667x1 3333xZ 142 x2 7112 x2 77 3333x1 6667x2 3333x1 75667x2 b Compute the Lagrangian strain tensor E17 and Eulerian strain tensors 217 Let C1ll 0 8 0 1 then E1761c1c1c c 741475 0 02 1 an 01 7 08 002 70475 a 7 5667 3333 727278 22222 LtD 1 th iDD7DTD e l 6le 3333 75667 en e in l l l 22222 727278 c Given line elements E R and E in the initial configuration as shown in the figure ie EN 59 RH i2 and LEAD 45 determine the deformed lengths of these line elements if they are of unit length originally From Eq316 Vls2 7ulsoZ 2E17d111d117 Since dso 1 we have ds J12E17d111d117 For E dial 1 d z 0 ds 12Endtzldtzl J005 02236 For E V1121 0 d z 1 ds1 12E22d Zd 1005 02236 For E V1121 0707 d z 0707 ds J1 2E11d111d111 1616616 Emdtzzdal Ezzdazdaz 1009 03 d Determine the angle LBAC in the deformed configuration Use Eq350 cos lz 612 cos39108 3687 2E12 2002 08 mm M 1 260475 32 Given the following displacement field in a 2D space 141 05112 142 05111 XZ a Compute E17 and 217 00 05 C B For E we first compute C then 6117 05 00 1 0125 05 E i C C C T C Hz HHn u5 0125 0 1A x1 For 217 we first express the displacements in terms of x Problem 32 7 1 2 7 2 1 141 7 7 x1 x2 142 7 x17 xZ 3 3 3 3 TM Tan Drexel University 1 November 1 2007 THEORY OF ELASTICITY 33 Solution for Chapter 3 The Eulerian strain tensor is given by 1 2 Then we com ute D l P l ax 1 2 J 3 3 1 11 16 0611 0889 l D D D T D 6 in H H H 18 16 11 0889 0611 b Sketch the deformed shape of an element OABC 1515 which is a unit square in the initial configuration as 4 shown in the flgure 00 10 05 10 I 1010 c Determine the angle between two line elements in Iquot It the initial undeformed configuration if they are Iquot 139 parallel to the X1 and X2 axes respectively in the 139 x 10 05 deformed configuration I Xxx Use equation 355 page 311 Iquot 10 00 26 sin 2 sin 90 cos 12 3 1 2e11 1 2e22 2612 08 we have 12 90 sin l 08 14313 1 1 2121111 2e22 The deformation of a body is defined by the following displacement field 111 k3alZ a 112 k211 113 113 k4a 111 where a are the Lagrangian variables and k is a positive constant Determine the deformed length Since of a line element originally of unit length and passes through the point 111 in the direction 1J 1J 1J First compute the Lagrangian strain tensor Since a 6a1 1 0 k 0 4012 1 a J 1 0 8113 T T E l 17 2 6a 6a 6a 6a we have 12ka1k2136af k6k2a1 k8k2a3 k6kzal ska2 k2116a k4k2aZ k 8k2a3 k 4k2a2 16ka3 k21 64a gt At 1 1 1 the Lagrangian strain tensor is given by 12k37kZ k6k2 k8k2 1517111 k6k2 8k17k2 k4k2 k8k2 k4k2 16k65kZ For a line element of unit length oriented in the direction 1J31J31J3 we have doll daz 5103 1J3 ds2 ds 2Eijdaida and recognizing that dso 1 we get 5152 1 42k 155k23 or ds 2 1 1 l42k 155kZ Substituting E17 and dai in the equation TM Tan Drexel University 2 November 1 2007 THEORY OF ELASTI CI TY 34 Solution for Chugter 3 Consider the following displacement field in terms of the Lagrangian variables a 141 711117 cos 7 112 sin 5 142 alsin 7nz17cos 143 0 where 5 is a constant a Compute the components of the Cauchy39s strain tensor 8 if the deformation is infinitesimal Describe the characteristics of such deformations What approximate range of values must the constant 5 be so that the infinitesimal deformation assumption could be justi ed The components of Cauchy s strain tensor are given by 811 cos 71 22 cos 71 12 l0 6x1 6H1 6x2 622 2 an 6 Since there is no shear deformation and the normal strain in both x1 and x2 directions are negative cos 1 and equal valued the deformation is a uniform compression The assumption of infinitesimal deformation is true only if 5 m 0 so that Icos 7 1 ltlt 1 if the deformation is finite 17 3 Compute the components of the Lagrangian strain tensor E Describe the characteristics of sud a deformation Hintz Sketch the deformed configurations of a unit square whose four corners initially occupy the positions 0 0 1 0 11 and 01 respectively for the cases of 5 0 7r2 7 and 37r2 All the components of Lagrangian strain tensor can be easily shown to be zero Thus there is no deformation occurred to the body and the only displacement is that of rigidbody motion This can be illustrated by substituting 0 7r2 7 and 37r2 into the displacement equations The deformation configuration is given by x1 111 141 711117cos 7112sin 111 cos 7tzzsin xZ nzu2 111 sin 751217cos 111 sin 112 cos Thus the body will rotate counterclockwise about the x3 axis by an angle of 7r2 Please try it yourself c Discuss the significance of the results obtained in a and b The following are the measurements taken from a strain rosette see figure attad ed to point A on a deformable body that undergoes an infinitesimal deformation 5 000200 g 000135 87 000095 Determine the principal strains and their directions at pointA Let 817 be the strain tensor with respect to the X1 7 X2 coordinate system as shown then the strain in any other direction can be obtained by transforming 17 to a coordinate system coinciding with that direction Thus 7 7 7 Z a Z a a 9 8a 7 SULHU 727111118 330 7 11 cos 30 822 s1n 30 72812 s1n30 cos30 11 cosZ 75 22 sin2 75 2812 sin 30 cos 75 7 f 7 g 511L775 mi nigmn 375 gm cosz120 522 51112120 2 1Zsin120 c05120 97 gliladzo ml nlgm 3120 Solving these three equations yields an 0001846 22 0001104 and 12 0001568 The principal strains are 81 0003086 and 2 70000136 and the principal directions are 3834quot and 128340 from the X1 axis TM Tan Drexel University 3 November 1 2007 THEORY OF ELASTI CI TY 41 Solution for Chugter 4 A thin plate made of a homogeneous isotropic material is subjected to a proportional biaxial loading as shown Two strain gages are being placed at the center of the plate to measure the strain components 211 and 222 As the load reaches 522 03911 2039 lt 0220 039 35gtlt103 psi the readings from the two gages 211 all 20 are 211 35 gtlt10 3 and 222 10gtlt10 3 respectively Determine the Young s modulus E and Poisson s ratio V of this material 03922 039 SOLUTION Use equation 418 2D version Problem 41 en iv 1700007 35000v 00035 E E E 522 0 27 vi 135000 7 70000v 0001 E E E Solving the above two equation yields E 175 gtlt106psi v 025 A parallelogramshaped thin plate as shown in the figure is made of a material with elastic constants E 30gtlt106 psi and v03 The plate is subject to a uniform shearing stress of 14140 psi as shown If the deformation is infinitesimal determine a the dqange in length of line E and b the principal strains and their directions 20in 14140psi A lt B 14140 psi 14140 psi Problem 42 SOLUTION First determine the components of stress tensor Since this is the same as Problem 26 the stress components therefore are 03911 728280 03922 0 03912 714140 and the principal stresses are 0391 5860 675 from X1 axis and 039Z 734140 225 from Xl axis a To determine the change in length of line E we first use the Hooke s law to find the strain components Substituting the components 418 yields e11 700009427 222 00002828 712 2e12 70001225 Line element E originally is 20 in long and parallel to the X1 axis therefore the change in length is AL BULB 70001885 in b For the principal strains we can either solve the eigenvalue problem using 6 stress into U or use the principal stresses that we have already obtained This yields 21 00005367 675 from X1 axis and 039Z 70001197 225 from X1 axis Show that for isotropic materials the principal directions of stresses coincide with those of principal strains SOLUTION To find the principal stresses and the corresponding principal directions we solve the following eigenvalue problem a 71 1 0 a TM Tan Drexel University 1 November 1 2007 THEORY OF ELASTI CI TY Solution for Chugter 4 where A the eigenvalue is the principal stress and 11 the eigenvector defines the principal direction Substituting the Hooke s law 7 15ekk 2116 in a yields 1611121115 1 o a Performing a tensor contraction on the Hooke s law by letting i gives 0391 151811 2116 31 21081 11 e c 1 3 2 lt gt where 11 7 is the first invariant of the stress tensor Substituting c in b rearranging terms and then divided the equation by 211 we have e17iA7 711 5 110 7 211 31211 7 7 217M712 70 D where Ki A7 711 e 211 31211 Equation d is recognized as the eigenvalue equation for solving the principal strains and their corresponding directions Since in deriving d from a we have never made any coordinate transformation therefore the principal directions in d must be the same as those in a It is noted that the relations between the principal stresses and strains are given by e Show that if a material possesses two planes of symmetry that are mutually orthogonal to each other then the third orthogonal plane must be a plane of symmetry SOLUTION Assume that the material is symmetric about the X1 7X2 and X17X3 planes The direction cosine matrices corresponding to symmetries with respect to X1 7 X2 and X1 7X3 plane 1 0 0 1 0 0 are given by 0 1 0 see Section 42 and 0 71 0 respectively It s easy to show that 0 0 71 0 0 1 1 0 0 1 0 0 1 0 0 71 0 0 0 71 0 0 1 0 0 71 0 whid is equivalent to 0 1 0 the direction cosine 0 0 1 0 0 71 0 0 71 0 0 1 matrix corresponding to symmetry with respect to X1 7 X2 plane Show that the number of independent elastic constants reduces to nine for an orthotropic material and to five for a transversely isotropic material SOLUTION Follow the same procedure as outlined in Section 42 start with a monoclinic material and use a new coordinate system with X X1 X2 7X7Z X73 X73 to get the matrix for an orthotropic material then do another transformation with respect to a plane of symmetry making a 45 angle with X1 7 X3 and X2 7 X3 planes to get the matrix for a transversely isotropic material TM Tan Drexel University 2 November 1 2007 THEORY OF ELASTI CI TY Solution for Chugter 4 46 Given the following state of stress cx vxlzix 7251x1x2 0 039 72cm x cx2vx27xz 0 17 1 z 1 2 1 0 0 c 1x1Z xg a Do the components satisfy the equilibrium equations 6011 6012 6013 6x1 6x2 6x3 6012 6022 6023 6x1 6x2 6x3 6013 6023 6033 6x1 6x2 6x3 25wc172cvx1 0 7251xZ 251 Z 0 0 b Are they a possible solution of a problem in elasticity Why Substitute the stress components in 428 the compatibility equations in terms of stress and see if with 171 0 they satisfy TM Tan Drexel University 3 November 1 2007 THEORY OF ELASTICITY 3 Analysis ofStmin Chapter 3 Analysis ufStmin In classical medaanics a solid body is often assumed to be rigid that is the distances between particles comprising the body are assumed to remain constant under the action of forces ln reality however there is no rigid body in nature All bodies are deformable to a greater or lesser degree and the distance between the particles of a solid body always undergoes some daange under the action of forces The question of whether or not a body may be treated as being rigid depends on the range of validity of the rigidbody approximation In this daapter we shall study the deformation of a solid body in terms of strain tensors The deformation will be treated as a mapping of the body from its initial undeformed state to the final deformed state We shall assume that the configuration of the solid body is described by a continuous mathematical model whose geometrical points are identified with the positions of the material particles of the body When such a continuous body changes its configuration under some physical action we assume that the change is also continuous ie neighboring points are daanged into neighboring points Any introduction of new boundary surfaces suda as those caused by tearing of a membrane or fracture of a test specimen must be regarded as an extraordinary circumstance requiring special attention and explanation 31 Lagrangian and Eulerian Variables Fig 31 Displacemmt of a deformable body Let X1 be a Cartesian coordinate system with respect to whida the motion of a deformable body is described Let the coordinates of a material point P on the body be 15 or a at time t t0 and a or x at time t as shown in Fig 31 then the displacements of P denoted by 141 are given by 141 x in 31 If the history of displacement is known then x can be expressed as a function of t and its initial coordinates a ie x1 x102 mam t 32 If this expression describes a continuous and onetoone relation between the initial and deformed positions then an inverse relation can always be found in which the initial position is expressed in terms of the deformed positions ie 2 aim 2 x1 x2 x3 33 TM Tan Drexel University 3 1 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStrain Thus the displacements given by 31 can be expressed in terms of either a or x The variables a and x1 are called the Lagrangian and Eulerian variables respectively In the Lagrangian representation all quantities are expressed in terms of the coordinates of initial position a and time whereas in the Eulerian representation they are expressed in terms of the coordinates of deformed position x and time Consider a quantity lt t expressed in terms of the Lagrangian variables and time The material derivative of the quantity is given by d 34 at at If it is expressed in terms of the Eulerian variables and time ie agt then its material derivative is given by VILLQ Vl 35 at at 6x 6t 6t 6x1 where V1 is the rate of deformation or velocity If only the initial and final positions of the deformation are of interest then the time t can be treated as a parameter rather than a variable Consequently 32 and 33 can be written as x x1011 22 as W 36 212x1mpx3 w 37 32 Deformation and Strain Tensors When a body is subjected to certain loading material points in the body will undergo motions The motions that cause the body to move as a whole without changing the relative distance between any two points in the body are called rigiarboay motions A rigidbody motion does not produce any stress in the body and is of little interest in the study of continuum medianics Deformations on the other hand diange the relative positions of material points in the body Therefore the easiest way to distinguish between a deformation and a rigidbody motion is to consider the change in distance between two material points lnitial X3 Fig 32 Deformation of a line segmmt in a deformable body Figure 32 shows in a deformable body two neighboring material points whose positions in the initial undeformed state are defined by vectors a and a ala1 respectively The distance between TM Tan Drexel University 3 2 October 17 2010 THEORY OF ELASTI CI TY 3 Analysis of Strain these two points is defined by the vector an of length rls0 After the body deforms to its final state those two points move to positions defined by vectors x and x1 ax1 respectively and the distance between them is defined by the vector ax of length as The squares of the initial and final distances respectively between these two neighboring points can be expressed as as aakaak dafaazzaa 38 rls2 axkaxk lel2 ax le32 39 If the relation between the initial and final 390 39 is 39 and i t i ie x1 x1a1 and 111 111 1 then as can be expressed in terms of the Eulerian variables in the following manner as aakaak axl ax axlax 310 6x1 6x7 7 6x1 6x7 7 while Vls2 can be expressed in terms of the Lagrangian variables in the following manner rlsZ axkaxk nial aiaa aa a 311 6a 6a 7 6a 6a 7 The difference of the two distance squares in terms of the Lagrangian and the Eulerian variables respectively are given by aszias aalaa imam a ia main 312 6a 6a 7 611 617 7 7 aszias axlaxlia axlax 17axlax 313 ix ax 7 7 6x ax 7 By defining the following two secondorder tensors E1 1 ai cgl 314 7 2 an an 7 e 1 5 315 77 2 77 6x ax then 312 and 313 can be written as asz idsg 2573mm 316 as2 idsg 22dxdx 317 The tensor E1 is called the Lagrangian strain tensor or Green s strain tensor and the tensor e17 is called the Eulerian strain tensor or Almansi s strain tensor Recall 31 the displacements are given by 141 xlia1 31 Taking partial differentiations with respect to a and x1 respectively of the above equation gives TM Tan Drexel University 3 3 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin 6151 gj ai 6117 6117 7 6x7 7 6x7 Substituting these expressions in 314 and 315 we have E 318 172 1 6141 a 614k 614k 6a 6111 6111 6117 e 17 319 It can be seen from 318 and 319 that both Lagrangian and Eulerian strain tensors include nonlinear terms of displacement gradients If the components of displacement 141 are sud that their first derivatives are small then the products of derivatives become negligible and those higher order terms in 318 and 319 can be discarded Furthermore when displacements are small it is immaterial whether the derivatives of the displacements are computed at the position of a point before or after deformation ie 6 Van H 6 6x1 Consequently for infinitesimal deformations the distinction between the Lagrangian and the Eulerian strain tensors disappears and both of them reduce to the socalled Cnuchy39s in nitesimal strain tensor 7 320 817 7 2 6x 6x The geometrical interpretations of the Caud y39s Lagrangian and Eulerian strain tensors are discussed in detail in the following sections 33 Geometric Interpretation of Cauchy s Infinitesimal Strain Components Consider a line element of length Vls0 originally parallel to the X1 axis ie an am1 am V1513 dso 0 0 321 Substituting the values of da in 316 yields VlsZ 7 dsg 2E11d111d111 2E11ds 2E11ds ds7ds 0 VlerVls0 For an infinitesimal deformation E11 m 811 ds m Vls0 but ds 7 Vls0 Ads 5 0 The equation above becomes Ads ds 7 Vls0 u Ends0 m suds0 322 Thus the Cauchy s strain tensor component 811 represents the dqange of length per unit length of a line element initially parallel to the X1 axis It can be seen from Fig 33 a that 322 becomes TM Tan Drexel University 3 4 October 17 2010 THEORY OF ELASTICITY 3 Analysis of Strain X2 A ds A 1 39 gt 4 gt X V 0 dx A 1 oisO 61611 z abc1 1 91129x1 a Normal strain b Shearing strain Fig33 Geometric interpretation of Cauchy s infinitesimal strain components u1da1 u1 1211dx1 n1 5611 N 8x1 511 Z N 61611 dx1 8x1 Similar expressions for line elements parallel to the X 2 and X 3 axes can be obtained ie 8112 8113 522 a I 533 8 x2 x3 Next we consider a deformation in the X1 X2 plane as shown in Fig 33 b where a rectangle of size dx1 xdx2 deforms into a parallelogram For an infinitesimal deformation we assume that the angles LAOA39 and LBOB39 can be approximated by 8112 8x1 and 5u18x2 respectively From Fig 33 b we see that the sum of 8111 8x2 and 8112 8x1 represents the total change to the angle LAOB Thus 8 1 1 12 21 2 8x2 8x1 2712 represents onehalf of the total change to the angle LAOB where 712 is the socalled engineering shearing strain It is noted that 8112 8x1 and 8111 8x2 are positive when the deformation is as shown in Fig 33 b ie dx1 rotates counterclockwise and dx2 rotates clockwise Similarly components 923 and 931 represent the changes to the angles in X2 X3 and X3 X1 planes respectively ie g g 1 1 g g 1 1 23 32 2 8x3 8x2 2723 31 13 2 8x1 8x3 2731 34 General Infinitesimal Deformations Consider the deformation of a body where two neighboring points PO and P in the initial state are displaced by it and iii respectively to point PO39 and P39 in the deformed state as shown in Fig 34 The distance between these two points is denoted by doll in the initial configuration and by dxi in the deformed configuration For an infinitesimal deformation 8 851 z 8 8x1 Liz may be expanded into a Taylor series around point P0 as Tll Tan Drexel University 3 5 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin X3 Fig34 Deformation of two neighboring points in a body 1 6214 0 Out 141 14 dx ax 7 Zaxzaxk dxzdxk 323 Neglecting terms of the second and higherorder derivatives in the series we have u 14 dx 324 6x7 7 which can be further expressed as 014 014 141 140 Jrl 7 dx 1 7 7 dx 1410 1 aix 325 2 6x7 6x1 7 2 6x7 6x1 7 7 7 7 where 014 811 7 81 326 7 2 ax 0x 7 is recognized as the Cauchy39s infinitesimal strain tensor and 014 5011 7 14 327 7 2 ax 6x 7 is called the in nitesimal rotation tensor Thus for an 39 deiunuatiuu the quot of point P in the neighborhood of P0 is the sum of the displacement at point P0 denoted by 140 and the relative motion of P with respect to P0 denoted by 817 w xz The physical meaning of the strain tensor 817 has been discussed in Section 33 The physical meaning of the rotation tensor ail can be illustrated by the example shown in Fig 35 in which the rectangle OABC is displaced to OA B39C39 in the XliXZ plane For an infinitesimal deformation the angle LAOl7 and ACOC can be approximated by BuZaxl and Bulsz respectively It is noted that Oiizax1 as shown in the figure is positive whereas Bulaxz is negative If lath69 Bibax1 then 812 21 0 and 012 represents the angle whidi the rectangle rotates as a rigid body about the X3 axis Similarly we can show that 023 and 031 represent the rigidbody rotations about X1 and X2 axes respectively TM Tan Drexel University 3 6 October 17 2010 THEORY OF ELASTICITY 3 Analysis of Strain X2 1 B C C B au1ax2 I i IA O A gtX1 duZdx1 Fig35 Geometric interpretation of infinitesimal rotation tensor In a more general case let us consider the vector 52 defined as z curluz qu 328 2 2 Written in indexical notations 328 becomes 1 8L1 1 1 812 51139 1 812 51139 1 1 Q 2 5 k g k k 2 5 a 5 2 5 3 329 1 2 quotax 2 Zk2dxj an 2axj dxkl 2 A Jquot Jquot 2 Vquot Jquot where 17k is the alternating tensor and gijksjk O as sjk being a symmetric tensor Multiplying both sides of 329 with aim and using the relation gilmsijk lj mk lk mj we have 1 1 1 539 939 2 5 gijkwjk 255395 lk mjwjk Ewlm wml mm 330 zlm z ilm l mk 2 J or by changing the subscripts 331 gkink wji wz j In view of 329 and 331 21 is commonly referred to as the dual vector of wjk or wjk referred to as the dual tensor of 21 Substituting 331 in 325 and letting 61 O we have dui u 1210 sk kdxj sikj kdxj 52 x d 332 Initial Fig 36 Displacement as a result of rotation Thus in the absence of strain the displacement field is the cross product of vectors 2 and 613 The geometric interpretation of 332 can be illustrated by the example shown in Fig 36 in which a line element 6155 is rotating about an axis 0039 by an infinitesimal angle of 52 Since TM Tan Drexel University 3 7 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin quot2 X aquot anyme a Hannaquot quotmiquot hence Q can be considered as the angle of rotation per unit length of a line element 35 RigidBody Motions As pointed out earlier a rigidbody motion does not dqange the distance between any two material points hence produces no strains We shall show that in the case of infinitesimal deformations a rigidbody motion is represented by a set of linear functions whose coefficients must satisfy certain constraint conditions Since a rigidbody motion produces no strain we have 614 614 614 11 1 0 522 Z 0 533 3 0 333a 6x1 6x2 6x3 6 43 0 2 0 1 6 43 0 2 0 1 732823 0 7312831 0 7122812 0 333b 6x2 3 6x3 1 6x1 6x2 Differentiating the three equations in 333b with respect to x1 x2 and x3 respectively yields 62143 1 62142 62141 1 02143 62142 1 62141 0 334 6x26x1 6x36x1 6x36xZ 6x16xZ 6x16x3 6x26x3 Solving the above equations gives 2 Z Z 6 141 7 6 142 7 6 143 0 335 6x26x3 7 63636361 7 6x16x2 From 333a and 335 we conclude that 141 142 and 143 must be of the following forms 1 f2ltxzfax3 2 83x381x1 3 h1x1h2xz 336 where f2x2 is a function of xZ only etc Substituting the expressions for 142 and 143 from 336 in the first equation of 333b we have dhzx2dgsx30 337 dxz dx3 which implies dhzxz 7 d83x3 338 dxz dx3 C1 a where cl is a constant Similarly we can show that df3x3idh1x1cz d 19 17dfzxzC3 3381 dx3 le1 le1 dxz Integrating 338 and substituting the results back into 336 we have 141 14j czx3 icsxz 339a 142 143 3x1 iclxs 339b 143 14 clxziczx1 339c TM Tan Drexel University 3 8 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin Furthermore by substituting 339 into 327 we have 51 5023 52 5031 53 9 12 340 Thus 14 143 and 143 are the rigidbody translations in the X1 X2 and X3 directions respectively and c1 2 and c3 are the components of the infinitesimal rotation tensor 36 Geometrical Interpretation of Lagrangian and Eulerian Strain Tensors 3 a Normal strain b Shearing strain Fig37 Geometric interpretation of Lagrangian strain components When the deformation is finite Cauchy39s infinitesimal strain tensor is no longer valid and the Lagrangian or Eulerian strain tensor must be employed to describe the deformation We shall now examine the physical meanings of these finite strain tensors Consider first a line element vial of length Vls0 parallel to the X1 axis in the initial configuration as shown in Fig 37 a In the deformed configuration this line element is displaced to le1 with length ds Recall 316 the definition for the Lagrangian strain tensor Since V1111 V150 and V1112 d 3 0 we have dsz 7 as 21311315112 2Euds 341 Define E1 as the extension per unit original length of the line element ie E1 ds 7 V150 V150 01 ds 1E1dso 342 Substituting 342 in 341 and eliminating also we have E1 1211 71 343 Thus E11 is related to the dqange of length of a line element parallel to the X1 axis Applying binomial expansion to the first term on the righthand side of 343 yields 31 1E11 7Ef1271 E117 EflZ 344 For an infinitesimal deformation I E11 ltlt1 344 is then reduced to E1 u E11 The error involved in this approximation can be quantified as being TM Tan Drexel University 3 9 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin Error m Bil2 i E11 2 which is less than 1 if E11 lt 002 Similar relations for line elements parallel to the X2 and X3 directions respectively in the initial configuration can be obtained ie EZ 1222 71 1322 if quotE22quot ltlt 1 345 E3 12E33 71 E33 if quotE33quot ltlt 1 346 A deformation can also exhibit a distortion in the configuration or a change of angle between two line elements Consider line elements V1111 of length V150 and VIE of length V150 that are parallel to X1 and X2 axes respectively in the initial configuration as shown in Fig 37 b The components of these two line elements perpendicular to each other initially are given by V1111 dso V1112 d 3 0 347a d z a150 d l d 3 O 347b After deformation these two line elements are displaced to dx of length ds and d5 of length d5 1 respectively Taking a dot product of vectors dx and d5 gives dxldil dsdgcos 612 g ml damd n 348 an 611 where 612 is the angle between dx and VIZ Substituting 347 in 348 and using the definition of Lagrangian strain tensor in the form of 314 we have dsdgcos lz dso o 2E12dsod 0 349 0 a z By substituting 342 and the similar relation d5 1 EZ 150 into 349 we obtain 2E c056 12 sino 350 12 1E11E2 12 where 012 Zi 612 is the total dqange in the angle between the two line elements from the initial state to the deformed state By further substituting 343 and 345 in 350 we have 21512 sino 12 11211 112E22 Thus for nite deformations not only E12 but also E11 and E22 contribute to the dqange of angle For 351 infinitesimal deformations E17 ltlt1 and a12ltlt1 354 is reduced to E12 m 05122 The components of Eulerian strain tensor can be interpreted in a similar manner Consider in the deformed configuration a line element dxl of length ds that is parallel to the X1 axis as shown in Fig38 a In the initial configuration this line element denoted by d511 is of length dso Substituting le1 ds and dxz dx3 0 in 317 for the Eulerian strain tensor we have Vls2 ids Zenulxlylx1 ZenylsZ 352 TM Tan Drexel University 3 10 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin X3 X3 21 Normal strain b Shearing strain Fig38 Geometric interpretation of Eulerian strain components Define 81 as the extension per unit deformed length of a line element ie dsidso 81 T dso ds17 21 353 Substituting 353 in 352 and rearranging terms we have e1 1 7 m 354 Applying the binomial expansion to the righthand side yields 21 171ie11 ef127n e11 81212quot39 In the case of infinitesimal deformation Ielllltlt1 we have 21 m 211 Next consider two line elements dx and VIZ parallel to X1 and X2 axes respectively in the deformed configuration with components being given by le1 ds dxz dx3 0 dfzd V151 df30 Denoting these two line elements as V111 and VIE respectively and the angle between them as 12 in the initial configuration as shown in Fig 38 b we have da dsOdEO cos 12 i 6 dxmdy dxld 2 2e12ng 6x 6x 6x1 6x2 Using 354 and the similar relation 22 171 17 2222 we have 2512 quot172211 quot17 2222 where u is the total diange in the angle between the two line elements from the initial state to the sin u cos 12 7 7cos 12 355 deformed state For an infinitesimal deformation where lewlltlt1 and 12ltlt1 355 is reduced to 512 m 18122 TM Tan Drexel University 3 11 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin 37 Further Discussion on the Theory of InfinitesiJnal Strains and Rotations Recall 318 the definition of the Lagrangian strain tensor 014 E1 l7 3 18 7 2 6a 6111 6111 6117 whidi can be written in the following form El 51 1 3 56 7 7 2 6111 6117 where 014 51 l 7 51 357 7 2 an an 7 is recognized as the linear part of E1 The term Oak6a in 356 can be decomposed into a symmetric part and an antisymmetric part as l 1 gkliakl thF lk 358 6a 2 6111 611k 2 6111 ak where 014 014 5 l 17 7 75 359 2 611k 61 is called the Lagrangian rotation tensor Substituting 358 and a similar expression for nk6a in 356 yields 11 E17 17E k1 wlkgk7 60 17Eghgq ghw7k 8 a aikw7k 360 It can be seen from 360 that if the deformation is infinitesimal sudi that 7ltlt1 ltlt1 and 6 6a1 u 6 6x1 then the components of Lagrangian strain tensor reduce to those of Cauchy s infinitesimal strain tensor However in many engineering applications such as postbuckling analysis of thin shells even though the values of 517 may be small those of the rotation tensor 717 may not In such cases we may simplify 360 by neglecting the higherorder terms involving 517 and replacing 517 and 717 with Caudiy s infinitesimal strain tensor 17 and infinitesimal rotation tensor 5017 defined in 326 and 327 respectively Thus 1 E m 817 lszk 361 38 Homogeneous Deformation A homogeneous deformation is given by x 517117 l7 362 TM Tan Drexel University 3 12 October 17 2010 THEORY OF ELASTICITY 3 Analysis of Strain where Ci and bl are constants Due to the linear relationship between a and xi it is easy to see that a straightline element will deform into another straightline element Furthermore since 6x1 651 Z J are constants the Lagrangian strain tensor E17 and Eulerian strain tensor ei will also be constant everywhere We shall discuss three special cases of homogeneous deformation a Simple extension A simple extensional deformation in the X1 direction for a rectangular prism is shown in terms of its X1 X2 projection in Fig 39 where a rectangle OABC is deformed into another rectangle OA39B39C 39 The deformation can be described by the following equations x1 611611 91 x2 2 61an 92 x3 610613 193 363 where b1 b2 and b3 representing rigidbody motions in the X1 X2 and X3 directions respectively can be neglected without the loss of generality The displacements are given by 1 2 d1 1511 2 2 do 1512 3 2 do 1513 364 The components of the Lagrangian strain tensor can be computed accordingly using 318 2 d1 1 O O 2 2 510 1 E17 0 2 20 365 O 0 510 1 2 and the corresponding extensions per unit original length in the X1 X2 and X3 directions are E112E11 1d1 1 152 12E22 1d0 1 133 121333 1d0 1 X2 X2 A A C B C C39 BB39 C39 39 39 m l B I I I I L o A A39 gt X1 0 A X1 Fig39 Simple extension Fig310 Simple shear 17 Simple shear A simple shear deformation parallel to the X1 X2 plane is depicted in Fig 310 where a rectangle OABC is deformed into a parallelogram OAB39C39 The deformation is given by Tll Tan Drexel University 3 13 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStrain 710 0 0 570 405571 0 0 0 Thus for finite deformations the existence of nonzero 57 or 57 may not be interpreted as the presence of strain 39 Principal Strains and Strain Invariants The principal strains and the corresponding principal directions can be determined following the same procedure as that for the principal stresses outlined in Chapter 2 ie by solving the eigenvalue problem 5 715 1 0 373 For nontrivial solutions of 373 to exist it must be that 511 339 512 513 395 7 254 an 522 7 a an 0 374 531 532 533 A or 3 2 a 7191 I 1710 375 where 19 g gm 522 533 376 1 511 512 522 523 533 531 I g g 7 517517 377 2 521 522 532 533 513 511 1 511 512 513 7 7 13 qukgzmngizgmgm 521 522 523 378 531 532 533 are the first second and third invariants respectively of the strain tensor Solving 375 yields the three principal strains 81 2 and 3 The corresponding principal directions can be determined using 373 and the orthogonal condition 111711 1 If we let the coordinate axes to coincide with the principal directions then the shearing strain components 812 23 and 13 vanish and 511 51 522 52 533 93 379 Therefore in the absence of rigidbody motions the principal directions remain coinciding with the coordinate axes during deformation Sudi a deformation is called pure strain or dilatation Consequently any displacement field in a solid body can be represented by the combination of a pure strain and a rigid body motion TM Tan Drexel University 3 15 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStrain 310 Compatibility Equations for Cauchy s Infinitesimal Strain Tensor When the displacements of a deformable body are known the strain components can be calculated by dii 39 39 O the r of J39 1 using 318 319 or 320 with little difficulty However if the strain components are given instead the question of how to determine the displacements naturally arises From a mathematical point of view since there are six strain displacement equations that are to be integrated to obtain the three unknown functions a in general we cannot expect to have singlevalued solutions unless the strain components satisfy certain conditions From a physical point of view since the strain components only determine the relative positions of points in the body and since any rigidbody motion corresponds to zero strain we expect that the solution a can be determined only up to an arbitrary rigidbody motion and a singlevalued displacement field can be obtained only if the strain components satisfy certain constraint conditions The constraint conditions that the strain components must satisfy in order to have singlevalued solutions for displacements are called the compatibility conditions The compatibility conditions for the Lagrangian and Eulerian strain tensors whidi include nonlinear terms are difficult to derive We shall therefore consider the Caudiy39s infinitesimal strain tensor only Recall 320 whidi defines the Cauchy39s strain tensor 1 g Ea M 320 The compatibility equations can be derived by quot 39 39 the J39 1 t r from 320 Differentiating the above equation twice gives 1 51714 5 1414 J Hum lnterdianging subscripts gives 1 5km EQ m z1a 1 gym 3amp4th z71k elm hyk 14W From these we can show that 8W1 8W T 81W T 87W 0 380 This is the equation of compatibility first obtained by St Venant in 1860 Since eadi term in 380 is a fourthorder tensor this expression actually represents 81 equations Of these 81 equations however only six are essential and the rest are either identities or repetitions on account of the symmetry of 17 Written in the unabridged notations these six equations are 6 gm 62523 7 62531 7 62512 0 381a 6x26x3 6x16x1 6x16x2 63636361 2 Z Z Z 6 22 6 31 6 12 6 23 63636361 6x26x2 6x26x3 6x16x2 0 381b TM Tan Drexel University 3 16 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin Z Z Z 7 7 0 381c 6x16x2 6x36x3 63636361 6x 6x Z 0 381d Z Z Z 2 6 31 7 6 33 76 8amp1 0 3816 6x36x1 6x1 6x3 2 6 12 7 62811 7 62822 6x16xz ax 6x12 0 381f The compatibility equations thus obtained are the necessary conditions for the displacements to be singlevalued It can be shown that they are also the sufficient conditions Detailed proof is given in Appendix 3A PROBLEMS The deformation of a 2D body is given by x1 02111 01112 xZ 01121 021Z a Compute the components of displacement 141139 12 in terms of the Lagrangian variables a and Eulerian variables x respectively b Compute the Lagrangian strain tensor E17 and Eulerian strain tensors 27 c Given line elements E E and E in the initial configuration as shown in the figure ie EH 1 RH i2 and LEAD 45 determine the deformed lengths of these line elements if they are of unit length originally d Determine the angle LBAC in the deformed configuration X2 C B 1 O 1 A X1 Problem 31 Problem 32 32 Given the following displacement field in a 2D space 141 05112 142 05111 a Compute E17 and 27 b Sketch the deformed shape of an element OABC whid is a unit square in the initial configuration as shown in the figure c Determine the angle between two line elements in the initial undeformed configuration if they are parallel to the X1 and X2 axes respectively in the deformed configuration 33 The deformation of a body is defined by the following displacement field 141 k 3212112 142 k22 113 143 k 4a tzl TM Tan Drexel University 3 17 October 17 2010 THEORY OF ELASTI CI TY 36 3 Analysis ofStmin where a are the Lagrangian variables and k is a positive constant Determine the deformed length of a line element originally of unit length and passes through the point 111 in the direction 71 1J 1J31J3 Consider the following displacement field in terms of the Lagrangian variables a 141 711117 cos 7 112 sin 5 142 111 sin 7az17cos 143 0 where 5 is a constant a Compute the components of the Cauchy39s strain tensor 8 if the deformation is infinitesimal Describe the characteristics of such deformations What approximate range of values must the constant 5 be so that the infinitesimal deformation assumption could be justified b Compute the components of the Lagrangian strain tensor E17 if the deformation is finite Describe the characteristics of sudi a deformation Hintz Sketch the deformed configurations of a unit square whose four corners initially occupy the positions 0 0 1 0 1 1 and 01 respectively for the cases of 5 0 7r2 7 and 37r2 c Discuss the significance of the results obtained in a and b The following are the measurements taken from a strain rosette see figure attadied to point A on a deformable body that undergoes an infinitesimal deformation 5 000200 g 000135 8y 000095 Determine the principal strains and their directions at pointA Problem 35 Derive 318 and 319 from 314 and 315 TM Tan Drexel University October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin Appendix 3A Further Discussion on the Compatibility Equations Fig3A1 Displacement path of point P in a deformable body We have shown in Section 310 that the compatibility equations as expressed in 380 are the necessary 4quot for the J39 J t to be Lug ale d To show that they are also the sufficient conditions we consider the displacements of a point P on a deformed body as shown in Fig 3A1 Without loss of generality we assume that the rigidbody motions at the origin vanish The total displacement of point P is given by m7 P 7 0141 u 7 jodu can 3A1 7 where C is a continuous curve connecting the origin O and the point P displacement field is singlevalued then the path of integration ice the curve C can be dqosen arbitrarily and the resulting displacement of point P must remain the same It is noted that if the Decomposing the righthand side of 3A1 into a symmetric part and an antisymmetric part we have 14513 Ic dxz C7 g dx J glzdxz Jr 071de 3A2 The second term of 3A2 when integrating by parts yields Law PX 5ka BM where the constant term pr is the displacement of point P due to rotation as described in Section 34 The com term in 3A3 can be further decomposed in the following manner 1 6214 6214 a 3 6x 6x 7 6x 6x 7 k 1 k 2 2 62 2 1 6141 1 Wk 1 7 1 Wk 3A4 2 axZax axZax 2 axpx 6x76x1 681k 7 687k 6x7 6x1 Substitution of 3A3 and 3A4 in 3A2 gives 513 wipxip J c 81a 7 x7 811m 7 87k11xk wipxip ICQ1kdxk 3A5 where TM Tan Drexel University 3 19 October 17 2010 THEORY OF ELASTICITY 3 Analysis ofStmin Q 2 inlet 7 61 Since 145p must be singlevalued ie its value must be independent of the integration path C consequently Qlkdxk in 3A5 must be an exact differential The necessary and sufficient condition for Qlkdxk being an exact differential is QM QM 3A7 Substituting 3A6 in 3A7 gives 51am 67mg1k7 57k1 3 8116m 57km gmmc 571681m4 gm17 3 81m7k gmmk 3 5mm gylmm 1m7k Spay 0 Since pointP was dqosen arbitrarily the expression in the parenthesis must vanish ie 51mm 57km 1m7k smug 0 Therefore this proves that the compatibility equations are also the sufficient conditions for the displacements to be singlevalued It should be noted that these equations of compatibility are necessary and sufficient conditions for quotsiInpleconnected bodies only A body is said to be simpleconnected if any closed contour drawn in the region can be shrunk continuously to a point without leaving the region otherwise the region is said to be multipleconnected For example a hollow sphere is simpleconnected while an openended hollow cylinder is a multipleconnected body For a multipleconnected body 380 is necessary but no longer sufficient Additional conditions must be imposed to ensure that the displacement field is single valued TM Tan Drexel University 3 20 October 17 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Chapter 5 Extension Torsion and Bending ofPrismatic Bars In this diapter we shall study the behavior of prismatic bars subjected to various types of loading conditions that cause the bars to extend bend or twist The primary solution method we shall employ is the socalled semifinverse method in which certain assumptions are made regarding the components of stress strain or displacement yet leaving enough freedom in those quantities so that the governing equations and boundary conditions can be satisfied It should be noted that in many cases it is often difficult if not impossible to specify the displacement andor traction boundary conditions exactly In sudi cases statically equivalent conditions based on the SaintVenant s Principle must be used instead In deriving the governing equations of elasticity we used the tensor or the abridged notation to take the advantage of its compactness When solving specific problems however no particular benefit can be gained by using sudi a notation Therefore in this diapter we shall use the variables x y and z to replace x1 x2 and x3 and denote the components of displacement in the X1 X2 and X3 directions by 14 v and w respectively Thus the governing equations derived in the previous chapters will be rewritten in the following forms a Equilibrium Equations From 419 we have 6039 67mxybx 0 51a 6z 6039 6039 6039 quoty W 9quot 12 0 51b 6x 6y 62 y 6039 Law W Lam 122 0 Sic 6x 6y 62 b Hooke s Law Substituting 420 in 421b and using the relations in Table 41 we obtain the following stressdisplacement relations in terms of Young s modulus E shear modulus G and Poisson s ratio 1 em 039m 71039W 022 52a eW gy v039wiv039ZZ 0m 52b eZZ 039uiv039m039W 52c 039 52d 7U 251x QJrQ 21TVo m Uzx 526 Oz 6x E G 7 7 25 MU 9 520 x9 6x 6y E W G c Compatibility Equations From 430 we have 1 6211 7 v 1 1 6x 2 1 7 v VZ039 v2 2 53a a v TM Tan Drexel University 5 1 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Z 2 via La 121 V Viv26 531 W 1v By 171 6y 2 Z V2022 v2w2 f 53c Z Z via y LQ a V 53d 1v ayaz Oyaz Z Z VZUZXLa 11 2 a V 53e 1v 626x 626x 621 62 via W 1 e V 530 1v may 5 My where 11 is the first invariant of the stress tensor and 11 is a potential function from which the components of body force can be derived iie 6y 6y 6y I 7 b 7 1727 54 quot 6x y ax y 62 d Traction Boundary Conditions From 422 we have t awn axyny runZ 55a ty awn aWny ayznz 55b tZ o xznx ayzny 72211 55c where 71 11 my 712 is the unit vector normal to the surface of the boundary 51 Extension of a Prismatic Bar by an Axial Force Fig 51 Extension of a prismatic bar by an axial force Consider a prismatic bar of length L with one end fixed at the xy plane and the other end subjected to an axial force P as shown in Fig 51 Force P is directed along the z axis Which coincides with the centerline of the bar The elementary ie Mechanics of Materials theory assumes that on cross sections perpendicular to the zaxis shearing stress vanishes and normal stress distributes uniformly ie 03922 PA a 7W cryZ an o xy 0 56 TM Tan Drexel University 5 2 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs where A is the crosssectional area It is easy to show that in the absence of body force the state of stress given in 56 satisfies the equilibrium equations 51 and the compatibility equations 53 The tractionfree boundary condition ie t1 0 on the lateral surface of the bar is also satisfied as can be shown by substituting the components of the unit normal to the surface 111 15119 0 and 56 into 55 On the z L surface 11 0 01 the exact traction boundary condition should be 0 to 0 t039 t1 717117 03912 or tx a y W z 22 In order for the stress components given in 56 to be the solution of this problem the normal stress 039ZZ must be uniformly distributed over the entire cross section at the free end This condition however is rarely satisfied exactly The concentrated force in Fig 51 for instance would have to be represented by an infinitely large stress at the point of application and zero stress everywhere else on the cross section For a slender bar whose length is much larger than the typical dimension of the cross section a set of statically equivalent condition based on the SaintVenant s principle see Chapter 4 is often used to replace the exact condition For the particular example shown in Fig 51 the statically equivalent conditions at 2 L are the resultant forces in the x and y directions and the resultant moments about the three coordinate axes must vanish and the resultant force in the z direction must be equal to the applied concentrated force P These conditions can be expressed in the following equations 2F L crudl 0 ZFy L 0 9sz 0 2132 L crudi P 57a 2M L auydA 0 ZMy 7 A 0 2sz 0 2M2 L iyam x039yz7lA 0 571 Since the state of stress given in 56 satisfies 57 it is therefore the solution of the problem It should be noted however that the first two equations in 57b are satisfied only if P is applied at the centroid of the cross section J The col r J39 o 39 J t r can be determined by substituting 56 into 52 ie the Hooke s law and then integrating the resulting equations provided that the displacement boundary conditions at z 0 are properly specified For instance by constraining the displacements and the rotations with respect to the three coordinate axes at 000 we obtain va wg 14 AE AE AE 58 Further discussion on the determination of quot 1 r will be given in Section 52 The total amount of extension of the bar is characterized by the zdisplacement at the free end From 58 we have wz L PLAE which agrees with the solution of elementary theory 52 A Prismatic Bar Stretched by Its Own Weight A slender prismatic bar of length L supported at its top surface is being stretdied by its own weight as shown in Fig 52 We set up a coordinate system so that the origin is located at the centroid of the bottom surface and the z axis is directed along the centerline of the bar The components of body force are given by bxl7 0 bz7pg 59 where pg is the weight density From 59 and 54 we obtain the potential function 1 as being wng 510 TM Tan Drexel University 5 3 October 31 2010 THEORY OF ELASTICITY 5 Extension Torsion and Bending of Prismatic Bnrs l H V x Fig 52 Stretching of a prismatic bar by its own weight It can be shown that the equilibrium 51 compatibility 53 and tractionfree boundary conditions on all but the top surface are satisfied if the following state of stress is assumed Uzz ngl Gxx ny Gyz sz ny O To find the displacement components we first substitute 511 in 52 which yields dn do Vogz dw L2Z I 512a dx dy E dz E a lzol al5w0 iv zo 512b dy dz dz dx dx dy Integrating the second equation of 512a with respect to z gives 2 P82 w w x 513 E 0 y lt gt where wo x y is an unknown function to be determined Substituting 513 into the first two equations of 512b and carrying out the integrations with respect to z we have 2 awe xly o no xy n z awoax39y no xy 514 x where no xy and no xy are also unknown functions resulting from the integration To determine functions no oo and zoo we substitute 514 into the first equation of 512a This gives 2 2 dx dx dy dy E Comparing the coefficients of 515 term by term we have 5140 xry a00 xy 0 516a dx dy dzwoxy 62w0xy 516b dxz of E Next we substitute 514 in the last equation of 512b This gives 2 26 mo z a a 0 dxdy dy dx Tll Tan Drexel University 5 4 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars The coefficients of eadi individual terms in the above equation must vanish ie 0140xyavoxy0 517 5y 59 62w x y 0 518 axay Equations 516a and 517 imply that 140 is a linear function of y 00 is a linear function of x and the coefficients of the two linear terms must be equal in value but opposite in sign ie 140 ayl7 519 00 fax 5 520 in whidi a b and c are arbitrary constants From 516b and 518 we conclude that m0 is a quadratic function of x and y and that the coefficients of the xy term must vanish ie w0xzy2dxeyf 521 where d e andf are unknown constants Substituting 519 520 and 521 in 514 and 513 yields u7xzidz2yb 522 v7yziezioxc 523 wzzvx2y2dxeyf 524 Constants in 522 7 524 can be determined using the displacement boundary conditions at 2 L the top surface of the bar Since the exact condition ie 14 v w 0 everywhere on the cross section is impossible to satisfy results can only be obtained for certain approximate boundary conditions specified based on the SaintVenant s principle Consider for instance the following set of displacement boundary conditions at 0 0 L ie at the centroid of the cross section on the top surface 14 v w 0 to prevent translations in x y and 2 directions 525a 60 6w a 0 or a 0 to prevent rotation of cross section about the x aXis 525b z 6 0 or a 0 to prevent rotation of cross section about the y aXis 525c z 60 614 a 0 or a 0 to prevent rotation of cross section about the z aXis 525d x It is easy to show that these conditions would yield 11 b c d e 0 and f ingZ2E Thus the components of displacement are given by u7xz v7yz wZZLZlx2yz 526 TM Tan Drexel University 5 5 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 00L lt L 000 max Fig 53 Deformation of a bar stretched by its own weight Figure 53 shows the cross section y 0 of the bar before dashed lines and after solid lines deformation has occurred Points along the z axis would displace vertically only as can be seen by substituting x y 0 in 526 ie 7 pg 2 2 g g 7 0 wig 7 E z L 527 and the maximum vertical displacement wmax ingZ2E occurs at 0 0 0 Other points on the cross section however would displace both vertically and horizontally It can be seen from 526 that 14 and v are linear functions of x and y respectively Therefore longitudinal fibers that were parallel to the z axis before deformation would deform into straight lines inclined to the z axis Furthermore points on the cross section 2 c where 0 g c S L would displace to ZCw6CZLZlx2yz whidi represents a surface of paraboloid to which all longitudinal fibers would be perpendicular after deformation Therefore there is no shearing strain associated with this deformation Due to the assumed uniformly distributed normal stress given in 511 and the displacement boundary condition given in 525 the top surface of the bar would deform concavely as shown in Fig 53 This of course would not have happened if the entire top surface were completely constrained In the latter case even though the stress distribution on the top surface would be very different from being uniform the resultant force must still be equal to the total weight of the bar in view of the overall equilibrium Thus based on the SaintVenant s principle the difference between the two solutions based on different displacement boundary conditions would be appreciable only in a small region near the top surface 53 Torsion of a Prismatic Bar of Circular Cross Section Consider a prismatic bar of circular cross section and of length L with one end fixed on the xy plane and the other end subjected to a torque MZ as shown in Fig 54a Under the action of the torque the bar would be twisted and a longitudinal line on the lateral surface eg the dashed line in Fig 54a would deform into a helical curve the solid line in Fig 54a The elementary theory of torsion of circular bars assumes that w 0 ie plane cross sections would remain plane after deformation and that any TM Tan Drexel University 5 6 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs cross section at a distance 2 from the xed end would simply rotate by an angle 6 Consider for instance the cross section shown in Fig 54b showing only the first quadrant A point originally located at P would move by an angle 6 to P after deformation The inplane displacement components are given by a A generator deforms b Displacement and stress at into a helical curve point P on a cross section Fig54 Torsion of a circular prismatic bar 14rcos667rcos xcos6717ysin6 528a vrsins 67rsin xsin6ycos671 528b where x cos and y sin If 6 is infinitesimal then cos6 m 1 and sin6 m 6 528 becomes 147y6 vx6 529 If we further assume that 6 varies linearly in z ie 6 20 where or is known as the angle of twist per unit length of the bar then 529 can be written as 147yzot vxzo 530 The corresponding stress components can be readily derived from 52 0x2 iGozy cry Gooc 531a am039W crmcrxy 0 531b in which the assumption that w 0 has been employed It is easy to show that these stress components satisfy equilibrium 51 and compatibility 53 To deck the traction boundary conditions we first examine the lateral surface of the bar For a circular cross section as shown in Fig 54b the unit normal has the following components 71 n n2 0 532 Substituting 531 and 532 in 55 we have t awn axyny 03211Z 0 533a ty 03911 799119 79211 0 533b TM Tan Drexel University 5 7 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs circular bars The same pattern of deformations has also been observed in prismatic bars of varying cross sections Based on these observations SaintVenant in 1855 hypothesized the following displacement components for prismatic bars of general cross sections subjected to end torques 14 foxyz v oocz w byway 535 The corresponding stress components can be derived by substituting 535 in 52 ie am 7W 03922 7xy 0 536a 039CZ GaKa Zi y 039 2 Go a Zx 536b 6x 9 0y Substituting 536 in 51 yields the following Laplace s equation for 1xy 62 62 V2 2 5 6x 0y Thus warping functions x are harmonic functions Furthermore by substituting 536 into 55 0 537 it can be shown that the tractionfree boundary condition on the lateral surface of the bar is satisfied if the warping function satisfies the following equation on the boundary 7yjnxxny 0 538a or ll Z ynx 7 amy 538b in whid 71 and my are the x and y components of the unit normal At the two ends of the bar the statically equivalent traction conditions based on the SaintVenant s principle will be used It is easy to show that the stress components given in 536 would yield FZ Mx My 0 The resultant force in the x direction is given by2 F Laud4 Leggy Gangpg j Hizw j dv GaJLxEi ynx xny st 0 in which 537 538b and the Theorem of Gauss see Chapter 1 have been used Similar result can be obtained for the resultant force in the y direction ie ry j 039 dA j Gaa WxdA 0 A 91 A W Finally the resultant moment about the z axis must be equal to the applied torque MZ Thus M2 IJ039xzyo39yzxdAGog x2y2xiydfl 53 2 LS Sokolnikoff Mathematical Theory of Elasticity 2 d Ed McGrawHill 1956 p 112 TM Tan Drexel University 5 9 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs or M2 Do 5391 where 7 2 2 62 6 DiGHEx y xaiyajdl 540 is the torsional rigidity of the bar From 539b we conclude that the angle of twist per unit length 0 must be proportional to the applied torque M2 The torsion problem is thus reduced to that of finding a harmonic function 1x y satisfying the boundary condition 538 associated with the particular cross section It is noted that since any arbitrary constant satisfies both 537 and 538 the warping function therefore can be determined only to within an arbitrary constant This constant however will have no effect on the stresses as can be seen from 537 55 Prandtl39s Stress Function for Torsion Problems Consider the function 1 2 Z Ga7 x y 541 where 7 is the conjugate harmonic function of the warping function 1xy so that they satisfy the CauchyRiemann equations ELLE 121 542 6x 6y 0y 6x Differentiating 541 with respect to x and y respectively and using 542 we have sza Zixisz alx 543a 6x 6x 0y 51 a 52 J G 7 G 7 543b la 4 lt gt Comparing 543 with 536b we have 039 7 544 x2 0y yz ax The function xy introduced by L Prandtl in 1903 is called the stress function3 for torsion problems It is easy to show that 544 satisfies the equilibrium equations Eliminating the warping function I from the two equations in 543 yields the following Poisson s equation for the stress function xy 62 62 v2 7 772sz 545 V5 M By 3 A rigorous derivation of the stress functions for torsion can be found in LS Sokolnikoff s Mathematical Theory of Elasticity 2 01 ed McGrawHill 1956 pp 114119 TM Tan Drexel University 5 10 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Fig56 A unit normal on the boundary of an arbitrary cross section As for the traction boundary condition on the lateral surface of the bar let us consider an arbitrary cross section as shown in Fig 56 showing only the first quadrant The components of unit normal to the surface can be expressed as d dx y n n x dc 9 dc 546 where dc is an infinitesimal arc element counterclockwise being positive along the boundary of the cross section Substituting 546 in the boundary condition 538a and using 543 we have dijiw 547 By dc 6x dc dc Equation 547 implies that the stress function xy must remain constant along the boundary of the cross section For bars with simpleconnected cross sections eg solid bars this constant can be set to zero ie 0 as the value of the constant has no effect on the stresses Thus eadi torsion problem is reduced to that of finding a stress function xy satisfying the governing Poisson s equation 545 and the boundary condition 5 0 For bars with multipleconnected cross sections eg hollow bars the value of 5 can still be set to zero along its outer boundary Along the inner boundaries however the values of 5 can no longer be chosen arbitrarily We shall discuss this in more detail in Section 58 The torque applied at the two ends of the bar can be obtained by taking the moment of the stress components on a cross section about the z axis ie M2jjiaxzyayzxiA7 yxm2jj m 548 in which 0 along the boundary has been observed Therefore the torque is numerically equal to twice the volume enclosed by the surface representing function xy and the xy plane It is noted that each term in the integrand of 548 contributes to one half of the total torque Thus one half of the torque is due to the stress component 039CZ and the other half due to cry Example 51 Torsion ofan Elliptic Prismatic Bar Consider a prismatic bar of elliptic cross section as shown in Fig 57 The boundary of the cross section is described by the equation TM Tan Drexel University 5 11 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs w Fig 57 The cross section of an elliptic prismatic bar 2 Z y 2710 a NIX N Q in whidi 11 and b are the major and minor axes respectively of the ellipse By assuming x2 y 2 m 7 1 b 1 H2 b2 where m is a constant the boundary condition for 547 is obviously satisfied The value of m can be determined by substituting b in 545 whidi yields m 7 7 11217260 C 7 112 172 Thus the stress function is given by 11217260 x2 y2 7 71 d 1124472 112 b2 The applied torque can be determined using 548 7 260 2 2 2 2 2 2 7 abaGtx Mfzmmx gbgb x gm y m b gamma b e where 2 7 771723 2 7 77an3 7 7 x dAilyiT y dAilx T J39J39dAiAiimb are the moments of inertia and area of the cross section The angle of twist per unit length is given by 112 b2 2 f a asbsG and the torsional rigidity is given by M2 7 m3b3G 7 6A4 D or 1124472 47r21p where Z Z 111 2 P X y 4 TM Tan Drexel University 5 12 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs is the polar moment of inertia of the elliptic cross section The shearing stress components can be obtained by substituting d in 544 2y 2x QE WMZ Uyz mMz h Since from b and h we have x 112 7 03992 yb2 0u it follows that the resultant shearing stress I along any radius line must be parallel to the line tangential to the boundary at the point where the radius intersects the boundary as shown in Fig 57 Eli dxi From h we see that the maximum values of the shearing stress components are 2x max 7m3b 2M2 mbz Zymm Mz 2 In I mks 03992 z lemax a max and if 11 gt b the absolute maximum shearing stress is 2M Z i mbz Ir max occurring at 0147 where the minor axis intersects the surface of the cross section The warping function 1xy can be determined from 543 b27112 112 b2 x Z y D Finally the components of displacement can be obtained by substituting f and j into 535 2 b2 2 2 bzi 2 u7otyz7WMz voocz 1 3173 ZM7 wazM2 k In I G 7111 b G In I G The contour lines of the warped cross section can be obtained by setting the warping function given in j to constants ie b2 7 112 x K or x K H2 172 y y These are hyperbolas having the principal axes of the ellipse as asymptotes The warped surface is convex in the first and third quadrants and concave in the second and fourth quadrants It can be shown easily that if 11 b then the solution of the elliptic bars reduces to that of circular bars discussed in Section 53 56 The Membrane Analogy Method for Torsion Problems The membrane analogy method for solving torsion problems was first introduced by L Prandtl in 1906 Prandtl considered the equilibrium of a homogeneous membrane such as the elliptic one shown in Fig58 supported along its edges and subjected to a uniformly distributed internal pressure p If the membrane deflection zxy is small then we can assume that the membrane force S remain unchanged throughout the membrane and the internal pressure p is directed along the z axis Consider TM Tan Drexel University 5 13 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs an infinitesimal element of membrane nbsd of size Axgtlt Ay as shown in Fig 58 The 2component of the membrane force along edge ad is given by F5 Sm Aysin 6m m SmAy m 42 096 m 2xy Deformed Sm membrane R a 6 a i M P m S P l P a 55 Side View y i 393 Edge of the I membrane I S I S 1 c do 5 4y 5 39 Ax I u b x v quot 5 39 S Top Fig58 An elliptic membrane subjected to a uniform pressure p where for an infinitesimal deflection sin 6 is approximated as 626xm ie the slope in the x direction of the deformed membrane along edge ad of the element Similar expressions can be obtained for membrane forces acting along the other three edges Since the sum of all these forces acting on the four edges must be balanced with the internal pressure p we have 62 62 62 62 13Z SAy 7 SAy j SAx 7 SAx prAy 0 2 596 m 596 b 0y 1 5V d Dividing this equation by SAxAy and letting Ax gt 0 and Ay gt 0 yield the following Poisson s equation 2 Z V22a a 7 549 6x 6y 5 Furthermore since the edges of the membrane are constrained we have E 0 550 do where do is an infinitesimal arc length of the membrane edge Comparing 549 and 550 with 545 and 547 the governing equation and boundary condition for torsion problems in terms of the Prandtl39s stress function we conclude that if the outline of the membrane is identical to that of the cross section of the bar subjected to torsion then the solutions of these two problems are identical provided that the quantity pS in the membrane solution is replaced by 2G0 in that of torsion and vise versa TM Tan Drexel University 5 14 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bendin o Prismatic Burs zx Y Contour lines of membrane Lines of shearing stress dx B l a Membrane problem b Torsion problem Fig59 Membrane analogy for torsion problems Figure 59 a shows the first quadrant of a membrane with contour lines depicting its deformation due to internal pressure p The corresponding lines drawn on the cross section of the torsion subjected to torsion as shown in Fig59 b are called the lines ofshetzring stress Consider a point eg point B on the membrane through whid r a particular contour line is drawn Since along a contour line the deflection of the membrane is constant we have in as where c is the arc length of the contour line The corresponding equation for a bar in torsion is 0 as where 5 is the stress function Using 544 and 546 we can express the above equation as nine W1 7 a Zn 03an 0 551 6c 6x do 6y do 9 9 Since ayzny 03an represents the total component of shearing stress in the direction normal to the line of shearing stress see the insert of Fig 59 we conclude from 551 that the shearing stress resultant r must be tangent to the lines of shearing stress The magnitude of the resultant can be obtained by summing up the components of 039 and aw in that tangent direction ie r awn 7 0tu gm Jrgny 73 552 Thus the magnitude of r at any point is equal to the gradient of the stress function 5 in the direction normal to the line of shearing stress at that point On the corresponding membrane dzdn is the maximum slope at that point Thus the shearing stress resultant on the bar in torsion can be obtained by multiplying this maximum slope by the constant 2G 06 ie TM Tan Drexel University 5 15 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars EJJE 553 an pS an From 548 we also conclude that the resultant torques at the ends of the bar is equal to twice the volume enclosed by the membrane with pS being replaced by 260 ie M2 2 j j 7lxrly 2GTIIzaxay 554 Consider a free body of the membrane obtained by cutting the membrane along a contour line eg the shaded area in Fig 59 Along the boundary of the free body there is a uniformly distributed membrane force S whose zcomponent per unit arc length of the contour line is 8azrln Summing up all the forces acting on the free body we have dz F A S rl 0 Z 2 r f dn 6 where A is the area of the xy projection of the membrane free body Substituting 553 into the above equation yields 17quot 7 A 7 rlc 7 0 P Zsz frds2GaA 555 from whidi the average shearing stress along a line of shearing stress can be determined ie 7 frdc 7 ZszA 1 m fat C where C fat is the length of the line of shearing stress Example 52 Torsion ofA Rectangular Prismatic Bar w Fig510 Torsion of a rectangular prismatic bar Consider a prismatic bar with a rectangular cross section of size 2agtlt2l7 as shown in Fig 510 subjected to end torques By using the method of membrane analogy the problem becomes that of finding the deflection surface of a uniformly loaded membrane having the same rectangular boundary and satisfying the governing equation and boundary condition of 549 and 550 By assuming the deflection of the membrane to be in the following series form TM Tan Drexel University 5 16 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 2w coswgt a where b1bZ are constants and Yny are unknown functions in y the condition of symmetry with respect to the y axis and the boundary condition zx in 0 are satisfied automatically Substituting a in 549 and then expanding the righthand side into a Fourier series for in x E 11 ie 7373 m i E Si 871 n 1 2 cos nx b where 8 nil211 we obtain the following differential equations for Y y H 4 Li Y main 4 lt71 2 The solution of this secondorder differential equation is given by 2 n 16 7 UTl Y A sinh B cosh n n n ns sbns Constant An vanishes as the result of Y being symmetric with respect to the y axis Constant B can be determined by using the condition that Y 0 at y ib This yields Z 1171 n 1361 71 17 cosh ny n It bnS cosh nb Thus the membrane deformation is given by Z16P112 in 71 1 17 cosh nyJCOS nx C 735 My 113 cosh nb Replacing pS by 2G0 in c yields the stress function for the corresponding torsion problem 326 Z w 712 h Z 12 lcosm d and the nonvanishing stress components can be obtained by substituting d in 544 ie cos nx n71 77716Gm i 71 sinh ny X2 6y It2 My 112 cosh nb n71 7716Gm Z 422 licosh nyJSm nx 71 92 6x It2 My cosh nb If h gt 11 the maximum shearing stress occurs at i a 0 where 039CZ 0 16Gm 1 1 r 039 ia0 17 e W 1 gt z cosh5 ltgt Using the relation 2 1 Zi 3 5 8 TM Tan Drexel University 5 17 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs we can rewrite e in the following form 16Gm 1 f IIZ Hays cosh nb rmax 2Gm7 The infinite series on the righthand side whidi converges very rapidly if l7gt 11 becomes negligible compared to the first term for a very narrow rectangular cross section where I gtgt 11 In such a case the maximum shearing stress becomes rmax 26m g For a square cross section where 11 l7 f becomes rmax m 1351Gou h In general we can express the maximum shearing stress as rmax k2Gm i where k is a numerical factor whose value depends on the aspect ratio of the cross section btz The applied torque can be obtained by substituting d in 548 This yields 1 3 192 a 1 M G 2 2b 17 t h b 3 a a s 5 an 6 j 1 The infinite series term on the righthand side again converges very rapidly In general the torque can be expressed as M2 leo2iz32b k By eliminating G6 from i and k we can also express the maximum shearing stress as Ml r l k2lt2agtzlt2bgt 0 where k1 and k2 are numerical factors whose values depend on the aspect ratio 1711 Values of the factors k k1 and k2 for several different btz ratios are given in Table 51 For a bar with a very narrow rectangular cross section 1711 00 we have k gt1 k1 kZ gt13 26m 211 2b 1 3 M2 3GDt211 212 Table 51 Constants for Torsion of a Rectangular Prismatic Bar btz k k1 kZ btz k k1 k2 10 0675 01406 0208 3 0985 0263 0267 12 0759 0166 0219 4 0997 0281 0282 15 0848 0196 0231 5 0999 0291 0291 20 0930 0229 0246 10 1000 0312 0312 25 0968 0249 0258 00 1000 0333 0333 TM Tan Drexel University 5 18 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs gtI r Za Fig511 Torsion of a narrow rectangular prismatic bar Solutions for bars with narrow rectangular cross sections subjected to torsion also can be obtained in the following manner Consider the deformation of a membrane whose outline is the same as the narrow cross section ie I gtgt a of the bar as shown in Fig 511 The shape of the slightly deflected membrane can be approximated as being cylindrical throughout the length in the y direction if the small regions near yib is ignored see the yz projection in Fig 511 The xz projection of the deflected membrane see Fig 511 can be approximated by that of a string subjected to a uniformly distributed pressure p The maximum deflection of the string is given by 5 112228 and the maximum slope occurring at x in is given by E r m dn max 11 S The volume enclosed by the parabolashaped cylindrical membrane and the xy plane is given by 2 4113b p V 39 2 2b 3 ax gt 3 S n Replacing pS in m and n with 2G0 and using the membrane analog we have 1W 26m M2 6a21232b 0 which agree with those given in f and j with the series term being neglected for 17 gtgt 11 57 Torsion of Prismatic Bars with Rolled Profile Sections Solution for a narrow rectangular bar subjected to torsion can be easily extended to those for rolled profile sections Let t 211 and l 217 so that the size of the narrow rectangular cross section is now tgtltl where lgtgt t then the maximum shearing stress and the applied torque can be expressed as rmax Gtor M2 Do 556 where 3 D G l 557 is the torsional rigidity Consider a slotted tube whose cross section is shown in Fig 512 It is easy to see that whether the narrow rectangle is straight or curved has very little effect on the volume enclosed under the corresponding membranes Thus the same equations as those in 556 can be used for the slotted tube shown in Fig 512 TM Tan Drexel University 5 19 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs t t a r a t 12 4 I 1 T b Fig 512 Cross section of a slotted tube Fig 513 Rolled profile sections For other types of rolled profile sections such as angles diannels and lbeams shown in Fig 513 we may consider each of them as being a combination of a number of narrow rectangular cross sections The total torsional rigidity is then the sum of those of individual ones Take the angle shown in Fig 513 a as an example the total torsional rigidity is given by 3 3 DDh H31 Gltll rtzlzl 2 3 Thus we have 3MZ 0 3 3 Gltll1 thzlZ The shearing stress in each leg of the angle can be estimated by using the formula for narrow rectangular cross sections as well ie I thor33I zt i12 t1l1t2lZ Similar expressions can be derived for the diannel and lbeam sections shown in Fig 513 ie 0 33M2 3 and r i12 Git1l12t2l2l t1l12t2lZ It should be noted that in deriving the approximated solutions for rolled profile sections we have neglected the detailed geometry of the fillet at the reentrant corners where a considerable stress concentration may develop if the radius of the fillet is small Values of these highly concentrated stresses can be estimated by using the membrane analogy method4 58 Torsion of Prismatic Hollow Bars So far we have discussed torsion problems for bars with solid cross sections only ie those that are bounded by a single curve At the boundary of sudi cross sections the stress functions according to 547 must remain constant and are usually set to zero However for bars with hollow cross sections ie those that have two or more boundaries the values of stress function along the boundaries can no longer be chosen arbitrarily We shall consider first a simple case in which the inner boundary of the hollow cross section coincides with one of the lines of shearing stress of the corresponding solid bar such as the elliptic cross section shown in Fig 514 showing only the first quadrant The outer boundary of the hollow cross section whidi coincides with that of the corresponding solid cross section is given by 4 SP Timoshenko and N Goodier Theory of Elasticity 3rd ed McGrawHill 1970 pp 322324 TM Tan Drexel University 5 20 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Membrane D T It p p a 212 r boundary x Z Lines of shearing stress Inner l boundary k1 V1 Fig 514 Torsion of a hollow bar with elliptic cross section x2 y 7 710 558 The inner boundary of the hollow cross section being geometrically similar to the outer boundary as it coincides with one of the lines of shearing stress can be expressed as x2 y2 W W 71 7 0 559 in which k lt1 is a scaling constant Since the resultant of shearing stress is tangent to hence does not cross the line of shearing stress the hollow bar therefore can be considered as being obtained by removing an inner cylinder from the solid bar without altering the stress distribution in the remaining portion of the bar This can also be verified by using the following membrane analogy Since in the hollow region the stress is zero the membrane must have a zero slope This can be easily adiieved by placing a rigid horizontal plate 6 in Fig 514 over the hollow region It is noted that the pressure distributed over the membrane a in a solid bar is statically equivalent to that distributed over the plate 6 in a hollow bar Consequently the membrane force S along the boundaries of free bodies represented by a and 6 respectively must be the same Thus the equilibrium conditions for the remaining portion of the membranes ie the shaded area in Fig 514 are identical for both cases and the stress function derived in Example 51 for a solid elliptical bar is therefore applicable to the hollow bar ie 7 nzbzGo LQLZA 560 112 b2 11 172 From Fig 514 it can be seen that the volume enclosed by the membrane of the hollow elliptic bar is less than that of the solid elliptic bar by the amount equal to the volume of revolution of area CDD or VHW 17 k4vs Based on the membrane analogy and e of Example 51 we concluded that for a hollow elliptic bar 3 3 The angle of twist per unit length thus is given by TM Tan Drexel University 5 21 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs MZ 1124472 a 17 k4 asbsG 561 The stress function and the corresponding maximum shearing stress are given respectively by M x2 y2 7 z 1 562 mbilik liE Z b2 2M 1 r quot 563 max mbz 17 k4 Substituting the equation of the inner boundary 559 in 562 we have M2 Vim boundary W If the inner boundary of the hollow bar does not coincide with any of the stress lines of the corresponding solid bar then the selfequilibrating condition of the rigid horizontal plate discussed previously is no longer satisfied In sudi a case additional load must be applied on the plate to maintain its horizontal position hence would introduce additional conditions that the membrane must satisfy In terms of the stress function for torsion problems these additional conditions are necessary to ensure that the displacements be singlevalued From 535 and 536b we have 0w 0w o xzG 7oty o zG ooc 6x 9 0y Expressing the resultant shearing stress in terms of its components ie x dy 1039 039 do 92 do and integrating it along the inner boundary of the hollow cross section we have dx dy 6w 0w 1dc an 5 cry Ejdc Jrady 7 Ga ydxi xdy If the displacement w is singlevalued then the first integral in the above equation must vanish as the integration is taken along a closed curve The second integral is equal to twice the area enclosed by the curve Thus we have Medeai This is the additional condition that needs to be satisfied when determining the constant value of stress function 5 along the inner boundary 59 Torsion of ThinWalled Tubes Solutions for torsion of bars made of thinwalled tubes can be easily derived by using the membrane analogy Figure 515 shows the cross section of a thinwalled tube and the corresponding stress function whidi is geometrically similar to the membrane Since the wall is thin we can neglect the slight variation in the slope of the membrane across the thickness of the wall and replace it by a straight line as shown in the insert of Fig 515 Thus the shearing stress can be approximated by TM Tan Drexel University 5 22 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 564 where h is the difference of the values of 5 between the inner and outer boundaries and t is the thickness of the wall It is easy to see that if the wall thickness is uniform then the shearing stress remains constant throughout On the other hand for a tube of nonuniform thickness the maximum shearing stress would occur at the location of the smallest wall thickness In either case It h remains constant throughout and is referred to as the shear flow in the wall of the cross section The corresponding torque can be calculated using membrane analogy ie it is twice the volume enclosed by the stress function whidi can be approximated as M I m 2Ah 2At r where A is the area enclosed by the dashed line in Fig 515 or the mean value of the areas enclosed by the outer and inner boundaries of the cross section of the tube Thus given a torque M2 the shearing stress can be computed as M Z 565 1 2A To determine the angle of twist per unit length we integrate the shearing stress given in 565 along the centerline ie the dashed line in Fig 515 of the wall and use 555 We f 2610 566 2A t from whidi we obtain M do z 567 0 4AZG t If the tube has a uniform thickness t is constant 567 becomes M Z1 0 T 568 4A Gt where l is the length of the centerline along the tube wall The torsional rigidity is given by 2 D w 569 or Z TM Tan Drexel University 5 23 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs For a circular tube with a mean radius V 1471 2 and 12m we have D47rzr4Gt Comparing this with 557 the torsional rigidity for a slotted tube we see that the ratio of the two torsional rigidities is D Z Dunslouedmbe 570 slotted tube For thinwalled tubes 7 gtgtt Thus having a longitudinal slot in the tube would drastically reduce the torsional rigidity Profile of 5 and membrane Fig 516 Torsion of a typical aircraft fuselage For thinwalled tubes with multiple holes sudi as an aircraft fuselage with a typical cross section as shown in Fig 516 the shearing stress in ead portion of the wall can be obtained by using the membrane analogy ie h h2 h1 7 hZ TF l 72 13 a ti Integrating the shearing stress along the walls enclosing the two compartInents denoted as A1 and AZ respectively in the figure then using 555 we have 1111 1313 2G04A1 b 1le 71313 ZGoa lZ c where ll lz and 13 are the arc lengths of ACE ADB and AEB respectively The total torque can be determined by using the membrane analogy M2 2 dA 2A1h1 A2h2 2A1t111 12512 d Shearing stresses 1391 and the angle of twist per unit length or can be obtained by solving a b c and d simultaneously They are M 71 NZ t3lZAl tzlsA1Azl M 12 NZ t3llAZ t113A1 A2l TM Tan Drexel University 5 24 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs M2 73 N tilei tzliAzl M2 0 ZGN tllZl3 tZl3ll t3lllZ where N 2t1t312A12t2t311A t1t213A1AZZ 510 Pure Bending of Prismatic Bars Fig 517 Pure bending of a prismatic bar by and moments A prismatic bar of length L is fixed at xy plane and subjected to a bending moment M at z L as shown in Fig 517 If the z axis coincides with the centerline of the bar and the yz plane where moment M is acting on is one of the principal planes then from the elementary theory of bending we know that the bar is in a state of pure bending and that the components of stress are given by 039ZZ am 7W cryZ an o xy 0 571 where R is the radius of curvature of the bar after bending It can be seen that the equilibrium is satisfied if the body force is absent The compatibility is also satisfied automatically as the stress components are either zero or linear functions of 2 For the boundary conditions it can be easily shown that the traction free condition on the lateral surface of the bar is satisfied At 2 L however we must use the following statically equivalent conditions F A andl 0 F9 A azydA 0 F2 A crudl 0 572a My A 0 22di 0 M2 L 7 o znyr 0 Wth 0 57213 Ey2 E1 M jAauydAjA R dick 572c where Ix is the moment of inertia of the cross section about the x axis This is the wellknown moment 7 curvature equation derived in the elementary theory of bending Substituting 572c in 571 yields whidi is again identical to the flexure formula in elementary theory of bending To determine the components of displacement we substitute 571 in 52 This yields TM Tan Drexel University 5 25 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 1 573 v v R ax R By R 62 0 0 f qj 574 02 By 6x 02 By ax Integrating the last equation in 573 we have w7w0xy 575 where m0 is a function of x and y Substituting 575 in the first two equations in 574 yields 614 72770100 607 0w 2 awe Eax ax E ayRay or after carrying out the integration with respect to z 0 0w 0w 2 6x u0xy 07 72 0y voxy 576 where 140 and 00 are functions of x and y as well Substituting further 576 in the first two equations of 573 yields 2 2 6x 6x R 0y By R from which the following conclusion can be drawn by comparing terms on both sides 2 Z a 206 200 ivl 577 6x 6y 6x By R Integrating the second equation in 577 gives 2 x uoxyev7yf1y voxyev2y Rfzx 578 Substituting 576 and 578 into the last equation of 574 yields azwimyqzqkwhwrvq Bxay dy dx R Since this equation is true for any value of 2 we have Z a woxy0 d ydfzxiv 0 579 may dy dx R From the first equations in 577 and 579 we conclude that w0 can only be a linear function of x and y ie w0xy mxnyp 580 From the second equation in 579 we conclude dfiy df2x7v a R dy dx TM Tan Drexel University 5 26 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs whidq upon integration yields 2 x f1ywj fzxv m7 581 Combining 575 576 578 580 and 581 gives 9 7 u 7v 7 7 mz R 00 5 07izzivxziy2 oocinz7 2R w mx 11 R V P in which m n p or 8 and 7 are constants to be determined by the displacement boundary conditions Since the bar is fixed at z 0 we can assume that at 0 0 0 uvw0 and 0 Oz Oz 0y These conditions whidi prevent the point 000 from having any movement or rotation yield 0 8 7 m n p 0 Thus the components of displacement are given by uiivjz y v7zzivxziy2 w 582 We now examine the validity of the basic assumptions used in deriving the elementary theory of bending by comparing the solution of elementary theory with that of theory of elasticity shown here 1 By substituting y 0 in the last equation of 582 we have w 0 This implies that the xz plane remains undeformed and is indeed the neutral plane 2 Displacement component w as shown in 582 is a linear function of y indicating that plane cross sections will indeed remain plane 3 Equation 574 requires that the shearing strain vanish throughout the entire bar Thus cross sections perpendicular to the neutral plane will indeed remain perpendicular after bending The deflection curve along the z axis the centerline of the bar can be obtained by setting x y 0 in 582 ie 22 07 uw0 583 2R whidi is identical to those given in the elementary theory of bending Consider the cross section of a rectangular prismatic bar at a distance 2 c from the fixed end The 211x2l7 rectangular cross section shown by solid lines in Fig 518 would deform to the shape represented by the dashed lines The two vertical edges at x in would deform linearly ie my i ux 11 R while the top and bottom edges at y would deform to parabolic curves ie TM Tan Drexel University 5 27 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bendin o Prismatic Burs k r 1 lt w y l l l k b I rL Fig 518 Deformation of the cross section of Fig 519 Optical image Showing contour lines a rectangular bar due to bending on top surface of a bar in bending vltyibgt7 czvxzebz For small deformations the curvatures of the two parabolic lines can be approximated by the second derivatives of the curves ie dzy d2 1 Z Z Z v ibi c vx 7b dxz dxz 2R R The top or bottom face of the bar forms an anticlas c surface with curvatures being convex down in the lengthwise direction but convex upward in the widthwise direction Taking the top face where y b as an example the contour lines of the deformed surface can be obtained by setting the displacement v in 582 constant This yields the following hyperbolas 22 ivxz constant whose asymptotes are given by zzivxz0 584 Figure 519 shows the hyperbolic contour lines of sudi a surface obtained using an optical tedinique5 By measuring the angle or as shown in the figure we can determine the Poisson s ratio 1 of the material using 584 ie v cot2xz cot2 or 511 Bending of a Prismatic Bar by an End Force A prismatic bar of length L is fixed at the xy plane and subjected to a concentrated force Py at zL as shown in Fig 520 The coordinate system has been setup in such a way so that the z axis coincides with the centerline of the bar and the x and y axes coincide with the principal centroidal axes of the cross section In addition the force Py is on a plane parallel to the yz plane and is applied at sudi a distance6 from the centroid that twisting of the bar does not occur Based on the elementary theory of bending we assume that the normal stress on all cross sections is the same as that of pure bending derived in Section 510 ie 5 SP Timoshenko amp N Goodier Theory ofElusticity 301 ed McGrawHill Inc 1970 p 288 6 The significance of this distance and the method of determining it will be discussed in Section 512 TM Tan Drexel University 5 28 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars 585 The elementary theory also assumes that all the remaining stress components except an and 72V should vanish We shall now examine these assumptions to see if they satisfy the governing equations of elasticity and the boundary conditions The equilibrium equations 51 require that in the absence of body force the assumed stress components satisfy 6039 6039 P 602 0 yz 0 6 yz W 586 62 6x 6y I It is noted that the first two equations in 586 imply that the shearing stresses are independent of 2 hence the distribution must remain unchanged throughout the length of the bar Fig521 Unit normal on the boundary of an arbitrary cross section Next we examine the boundary conditions 55 for an arbitrary cross section as shown in Fig 521 showing only the first quadrant It is easy to see that the assumed stress components satisfy 55a and 55b identically From 55c we have 03211 ayzny 0 587 where from Fig 521 do 9 do Substituting these components of unit normal into 587 yileds TM Tan Drexel University 5 29 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs W M e 0 588 039 7039 7 xzalt yzdc Finally the compatibility equations 53 in the absence of body force require that 2 7 Z 7 P9 Va 0 Vazii 589 y 1x1v Thus the bending problem reduces to that of finding for GE and 039 functions of x and y that satisfy 2 f the equilibrium equation 586 the boundary condition 588 and the compatibility equations 589 This can be further simplified by using the following method of stress function Let xy be a stress function for problems of bending not to be confused with the stress function for the problems of torsion discussed previously in this diapter from whidi the components of stress can be derived as follows c cpyyz 0X2 2 0y 9 6x 21 fx 590 It is easy to show that these stress components satisfy 586 the equilibrium equations Function fx which has no effect on the conditions of equilibrium will be determined later from the boundary condition Substituting 590 in the compatibility equations of 589 yields 3infl j 362 62 V Zu 0y of 6x 6x2 of 51va dxz From whidi we conclude Z Z Px VZ 6 6 V Lid x 6x 0y 1v Ix dx 3 591 where 8 is a constant To find the meaning of 8 we consider the infinitesimal rotation of an area element of the cross section with respect to the z axis given by see Chapter 3 1 60 014 ml 26x The rate of change of this rotation in the z direction is given by a w iin 111mg 3p jaewn 652 az azzax 0y 76x262 By 262 6x 76x 726 6x 0y Substituting 590 in the above equation and comparing the resulting equation with 591 we have Px wifi V Lopg 39 2c 1v1x Since wzyzx 0 ZG we conclude that ZG is the rate of diange of rotation of an area element along the centerline of the bar where x 0 Assuming that the end force is applied in sudi a way so that the bar is not being twisted then 8 0 and 591 becomes V Pyx d x 1v Ix dx W 592 TM Tan Drexel University 5 30 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Finally in terms of the stress function the boundary condition 588 becomes dxdy Pyy27fxdx 593 6c 6x dc ayrlc 21 Pic If the function is known then the values of stress function 5 along the boundary of the cross section can be determined and the solution can be obtained by using 592 and 593 In the following examples the function will be chosen in such way as to make the righthand side of 593 vanish The value of stress function 5 therefore remains constant along the boundary Furthermore for a sirnpleconnected cross section we can make the constant equal to zero then the problem of bending of a prismatic bar reduces to that of finding the solution of 592 with boundary condition 5 0 Once the stress function is determined the stress components can be computed using 590 and the displacement components can be obtained by substituting the stress components in 52 and carrying out the integrations ln terms of stress function the displacement components are given by P L 14 awo21tvgtzvr xyiw p 594 ax E 0y E1 P P 2 v 7 6Eny 23 73L 227 Vx27 yz 77 fEZerch 7 595 P wzzi2Lzwoxy 596 where or 8 and 7 are arbitrary constants to be determined using the displacement boundary conditions and m0 is a function of x and y that can be determined using the following equations Z 2 VP 6 740 721V 6 yy 59721 6x E may EIX azwo 21v 62 7 Pyr vl yr 597 of E may Ix E1 62 1 62 62 d w0 v Z 7r x 597C may E x dx Example 53 Bending ofa Circular Prismatic Bar by an End Force Consider a circular prismatic bar whose boundary is given by x2 y2 R2 a in whidi R is the radius of the cross section The righthand side of 593 becomes zero if we dioose P Ami 3022798 b Substituting b in 592 yields TM Tan Drexel University 5 31 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 712v 1v Ix VZ C Thus the problem reduces to that of finding the solution of c subjected to the boundary condition of 5 0 This boundary condition is satisfied if we dioose mx2y27R2x d where m a constant factor can be determined by substituting d in C Le 7 12v Py 81v 1x Thus the stress function becomes 12v Pyx Z Z Z x 7R f 81v Ix y and the stress components can be obtained by using 590 a 77 12v Pyxy 2 41v I g P c 02 32Vy27y271 21ij h 9 81v1x 32v Along y0wehave P sz0 az32V yR271 21362 y 81VIX 32v The maximum shearing stress occurs at the center of the cross section where x 0 ie a 7 32v PyRZ7 32v P 91 max 81v I 21VA in whidi A IrRZ is the crosssectional area of the bar At each end of the horizontal diameter ie at x irR the shearing stress is a 7 12v 17sz 712vPy 9quot Xi 41v Ix 1v A Taking v 03 the value of Poisson s ratio for most engineering materials we have P P c y c y cl MW 71387 cl 937 71237 Comparing with the elementary solution P o 2 ii 9 3 A obtained by assuming that the shearing stress is uniformly distributed along the horizontal diameter the maximum error is seen to be approximately 4 percent TM Tan Drexel University 5 32 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars To find the displacement components we need to determine first the function w0xy by substituting the stress function f in 597 This yields Py 4E1Xy3xzy xw j w0xyT 1 1 11 Substituting j and the stress function f in 594 595 and 596 and assume constants or 8 7 5 77 and g all vanish valid if the origin of the coordinate systems undergoes no translation or rotation ie where 5 77 and g are arbitrary constants to be using the J39 boundary no rigidbody motion we have VP uyLiz k P v izs3L2273VLizx27yzRzz l P w4 ly 22274inx27y2 m We now examine again the basic assumptions used in deriving the elementary theory of bending against the elasticity solution presented here 1 By substituting y 0 in m we have w 0 Thus the xz plane remains undeformed and is indeed the neutral plane 2 Equation m reveals that a cross section will deform into a cubic surface in y Thus the assumption that cross section remains plane is no longer valid 9quot The shearing stress component 7 whidi can be easily obtained from h does not vanish along y 0 Thus the assumption that cross sections remain prependicular to the neutral plane is no longer valid either Substituting x y 0 in 1 gives the deflection curve of the centerline of the bar P 290 y 23 F3LZZ FMRZZ n 90 6EIX 2 The maximum deflection of the centerline occurs at z L ie f l P9L31m5 o 0x 0 90ij 3E1 2 L The coefficient on the righthand side of o is recognized as the solution from the elementary theory of bending The second term in the bracket is the correction factor whidi is less than 1 for bars with aspect ratio 2RL less than 012 if v 03 Example 54 Bending 7le Rectangular Prismatic Bar by an End Force The boundary of a rectangular cross section as shown in Fig 522 is given by xziaZXEZibz0 a By dioosin g TM Tan Drexel University 5 33 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs W W x z Membrane A lt1 3 m WV Pressure Fig 522 Bending of a rectangularbar by an and force 7 Pybz fx7 2 b x the righthand side of 593 vanishes along the boundary ie Z 2 Pyy 7 Pyb 0 21 along y ib 21 alongxrtz d x0 do Thus the boundary condition becomes 0 Substituting b in 592 yields V Pyx C v2 1v Ix Using the membrane analogy similar to that of a torsion problem we determine the deflection of a membrane subjected to a pressure load proportional to V Px y d 1v Ix This pressure which remains constant in the y direction but varies linearly in the x direction will cause the membrane to deform in a shape whose intersection with the xz plane is as shown in Fig 522 Consequently 5 must be an odd function of x and an even function of y These conditions can be satisfied by taking 7 in the following Fourier series form mmm e 11 2b ZZAZWM Sin m1n1 Substituting e in c and applying the standard procedure for determining the coefficients of a Fourier series we can obtain TM Tan Drexel University 5 34 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 71 1 Z sin m x cos 2n T wily 2 f 2 71 Z 2 712 mn m n 4b2 V Py8 3mm 4 1v Ix 7r mam To find the shearing stress components we first substitute b in 590 this yields P 2 02yb2y2 0y 9 6x 21 which can be resolved into the following two systems Oxfalzto39 z 0920 03952 E where P oquot 0 oquot 9 bzi 2 h x2 92 21X y represent a parabolic stress distribution given by the elementary theory of bending and 6 6 a 7 5 0 lt1 6y 6x represent the corrections to the elementary solution due to the stress function Along the x axis 6 0y0 as the result of 5 being an even function of y the corrections to the elementary solution therefore affect only the vertical component of the shearing stress contributed by 052 Substituting f in the second equation of i and letting y 0 we have Tnch 71Ver COS V pg 8H2 no no agile 1vI n3 1 1 112 1 X m 2n71m22n712m Since Ix 411bS3 AbZB where A 411k is the crosssectional area j can be written as m x 71mn72 COS 3P 1 16 112 Oils 9 3 222 2 k 9 2A 1v II b mam 2 2 F 21171 m 4421171 Z 4b The coefficient on the righthand side of k is recognized as the average shearing stress along x 0 from the elementary theory of bending Thus given the ratio nb of a rectangular cross section and Poisson s ratio of the material we can evaluate the term in the brackets of k and make a correction to the elementary solution The infinite series in the equation converges very rapidly particularly for nb gt 1 Table 52 lists for a number of nb values the ratios of exact solutions to the elementary solutions for the stress component cry at x y 0 center of the cross section and at y 0 x in middle of the vertical sides of the rectangle The Poisson s ratio is set to 025 It is seen that the elementary solution yields very accurate estimates for nb 2 2 about 10 error for a square cross section where lZb 1 but become less accurate as the value of nb decreases TM Tan Drexel University 5 35 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Table 52 Comparison of exact and elementary solutions for shear stress in a bar subject to an end force 51 b 5 4 3 2 1 1 2 12 1 2 1 5 x 0 y 0 0997 0996 0993 0983 0940 0856 0817 0805 0801 x id y 0 1005 1008 1015 1033 1126 1396 1691 1988 2285 512 Shear Center In Section 511 we have assumed that the z axis coincides with the centerline of the bar and the x and y axes coincide with the principal centroidal axes of the cross section In addition the force Py is on a plane parallel to the yz plane and is applied at such a distance from the centroid that twisting of the bar does not occur This distance can be easily obtained once the stress components as given by 590 are known ie we evaluate the moment about the centroid produced by the shearing stress components 0x2 and oyz as Mz 7ny oxzybxdy 598 Since the stress resultant must be statically equivalent to the applied force Py we conclude that the distance from the application point of the force to the centroid of the cross section must be Mz P 1 d 599 For a positive value of Mz dx must be taken in the positive direction of x If a force denoted by Px is applied on a plane parallel to the xz plane then we can follow the same procedure and determine the distance for the application point denoted by dy so that the bar will not be twisted by the applied force The intersection of the two lines drawn parallel to the x and y axes and at distances dy and dx respectively from the centroid is called the shear center Any force that is applied at the shear center and perpendicular to the axis of the bar will not cause the bar to twist Problems 51 A prismatic bar with a 2n x 2h square cross section and of length L is supported at the four corners of the top surface by four rigid bars and is being stretched by its own weight as shown a Specify the displacement boundary conditions b Use the boundary conditions specified in a to determine the constants in 522 523 and 524 hence obtain the expressions for the components of displacement c Compare the results you obtained in b with those in 526 9 2 y x l l dz 39139 391 G x lt L gt Problem 51 Problem 52 TM Tan Drexel University 5 36 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 52 A prismatic bar of length L is rotating about the y axis with a constant angular velocity Q a Consider this as a onedimensional 1D problem first Draw a freebody diagram for an infinitesimal bar element of length dz as shown Show that in the absence of gravitational effect the stress in the bar due to the centrifugal force is 039ZZ pQZQZ 7 22 where p is the mass density of the bar b Show that the following 3D state of stress 039ZZ pQZQZ 7 22 amUWayzazxaxy0 obtained from the 1D analysis in part a is a solution to this problem in threedimension if and only if v 0 Hint Check the compatibility equations 53 Prove equation 548 54 Prove equation b of Example 54 on page 516 55 Consider a prismatic bar with an equallateral triangular cross section as shown in the figure subjected to torsion show that 1 2 2 1 3 2 2 2 7G 7 7 3 7 4 y 24x xy is the solution of the problem Determine the distribution of shearing stress component 039yz and its maximum value along the x axis 2 b3 Perfectly plastic l1 n gt Problem 55 Problem 56 56 The membrane analogy method also can be employed to analyze torsion problems in which the material deforms beyond the elastic limit For instance consider a circular bar made of the material having an elasticperfectly plastic stressstrain relation as shown in the figure If the shearing yield stress is 19 show by using the membrane analogy method that the ultimate torque the bar can sustain is MZ Z RsryB 57 Show that the stress distribution in a bar subjected to torsion remains the same regardless of the location of the origin that the axis of rotation passed through so long as the axis is parallel to that of the bar TM Tan Drexel University 5 37 October 31 2010 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Chapter 5 Extension Torsion and Bending ofPrismatic Bars In this diapter we shall study the behavior of prismatic bars subjected to various types of loading conditions that cause the bars to extend bend or twist The primary solution method we shall employ is the socalled semifinverse method in which certain assumptions are made regarding the components of stress strain or displacement yet leaving enough freedom in those quantities so that the governing equations and boundary conditions can be satisfied It should be noted that in many cases it is often difficult if not impossible to specify the displacement andor traction boundary conditions exactly In sudi cases statically equivalent conditions based on the SaintVenant s Principle must be used instead In deriving the governing equations of elasticity we used the tensor or the abridged notation to take the advantage of its compactness When solving specific problems however no particular benefit can be gained by using sudi a notation Therefore in this diapter we shall use the variables x y and z to replace x1 x2 and x3 and denote the components of displacement in the X1 X2 and X3 directions by 14 v and w respectively Thus the governing equations derived in the previous chapters will be rewritten in the following forms a Equilibrium Equations From 419 we have 6039 Law e Lam bx 0 M 6y 62 6039 6039 6039 quoty W 92 b 0 51b 6x 6y 62 y 6039 Law W Lam H72 0 Sic 6x 6y 62 b Hooke s Law Substituting 420 in 421b and using the relations in Table 41 we obtain the following stressdisplacement relations in terms of Young s modulus E shear modulus G and Poisson s ratio 1 039m 7 1039W 039ZZ 52a 039W 7103922 03 52b 03922 7 1039m 039W 52C ZONE 52d 6y 62 E 91 G 2 ZONE quotzx 52e 62 6x E m C Q 21V A 520 6x 6y E 9 G c Compatibility Equations From 430 we have 1 621 v 62w V2 1 V2 2 53 0 1v 6x2 171 W 6962 a TM Tan Drexel University 5 1 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Z 2 via La 121 V Viv26 531 W 1v By 171 6y 2 Z V2022 v2w2 f 53c Z Z via y LQ a V 53d 1v ayaz Oyaz Z Z VZUZXLa 11 2 a V 53e 1v 626x 626x 621 62 via W 1 e V 530 1v may 5 My where 11 is the first invariant of the stress tensor and 11 is a potential function from which the components of body force can be derived iie 6y 6y 6y I 7 b 7 1727 54 quot 6x y ax y 62 d Traction Boundary Conditions From 422 we have t awn axyny runZ 55a ty awn aWny ayznz 55b tZ o xznx ayzny 72211 55c where 71 11 my 712 is the unit vector normal to the surface of the boundary 51 Extension of a Prismatic Bar by an Axial Force Fig 51 Extension of a prismatic bar by an axial force Consider a prismatic bar of length L with one end fixed at the xy plane and the other end subjected to an axial force P as shown in Fig 51 Force P is directed along the z axis Which coincides with the centerline of the bar The elementary ie Mechanics of Materials theory assumes that on cross sections perpendicular to the zaxis shearing stress vanishes and normal stress distributes uniformly ie 03922 PA a 7W cryZ an o xy 0 56 TM Tan Drexel University 5 2 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs where A is the crosssectional area It is easy to show that in the absence of body force the state of stress given in 56 satisfies the equilibrium equations 51 and the compatibility equations 53 The tractionfree boundary condition ie t1 0 on the lateral surface of the bar is also satisfied as can be shown by substituting the components of the unit normal to the surface 111 15119 0 and 56 into 55 On the z L surface 11 0 01 the exact traction boundary condition should be 0 to 0 t039 y yz z 22 t1 717117 03912 or tx a In order for the stress components given in 56 to be the solution of this problem the normal stress 039ZZ must be uniformly distributed over the entire cross section at the free end This condition however is rarely satisfied exactly The concentrated force in Fig 51 for instance would have to be represented by an infinitely large stress at the point of application and zero stress everywhere else on the cross section For a slender bar whose length is much larger than the typical dimension of the cross section a set of statically equivalent condition based on the SaintVenant s principle see Chapter 4 is often used to replace the exact condition For the particular example shown in Fig 51 the statically equivalent conditions at 2 L are the resultant forces in the x and y directions and the resultant moments about the three coordinate axes must vanish and the resultant force in the z direction must be equal to the applied concentrated force P These conditions can be expressed in the following equations 2F L crudl 0 ZFy L 0 9sz 0 2132 L crudi P 57a 2M L auydA 0 ZMy 7 A 0 2sz 0 2M2 L iyam x039yz7lA 0 571 Since the state of stress given in 56 satisfies 57 it is therefore the solution of the problem It should be noted however that the first two equations in 57b are satisfied only if P is applied at the centroid of the cross section The LUL r J39 o J39 J ie the Hooke s law and the boundary conditions at z 0 are properly specified For instance by constraining the displacements and the rotations with respect to the three coordinate axes at 000 we obtain t r can be determined by substituting 56 into 52 n integrating the resulting equations provided that the displacement va wg 14 AE AE AE 58 Further discussion on the determination of quot 1 r will be given in Section 52 The total amount of extension of the bar is characterized by the zdisplacement at the free end From 58 we have wz L PLAE which agrees with the solution of elementary theory 52 A Prismatic Bar Stretched by Its Own Weight A slender prismatic bar of length L supported at its top surface is being stretdied by its own weight as shown in Fig 52 We set up a coordinate system so that the origin is located at the centroid of the bottom surface and the z axis is directed along the centerline of the bar The components of body force are given by bxl7 0 bz7pg 59 where pg is the weight density From 59 and 54 we obtain the potential function 1 as being wng 510 TM Tan Drexel University 5 3 November 13 2007 THEORY OF ELASTICITY 5 Extension Torsion and Bending of Prismatic Bnrs l H V x Fig 52 Stretching of a prismatic bar by its own weight It can be shown that the equilibrium 51 compatibility 53 and tractionfree boundary conditions on all but the top surface are satisfied if the following state of stress is assumed Uzz ngl Gxx ny Gyz sz ny O To find the displacement components we first substitute 511 in 52 which yields dn do Vogz dw L2Z I 512a dx dy E dz E a lzol al5w0 iv zo 512b dy dz dz dx dx dy Integrating the second equation of 512a with respect to z gives 2 P82 w w x 513 E 0 y lt gt where wo x y is an unknown function to be determined Substituting 513 into the first two equations of 512b and carrying out the integrations with respect to z we have 2 awe xly o no xy n z awoax39y no xy 514 x where no xy and no xy are also unknown functions resulting from the integration To determine functions no oo and zoo we substitute 514 into the first equation of 512a This gives 2 2 dx dx dy dy E Comparing the coefficients of 515 term by term we have 5140 xry a00 xy 0 516a dx dy dzwoxy 62w0xy 516b dxz of E Next we substitute 514 in the last equation of 512b This gives 2 26 mo z a a 0 dxdy dy dx Tll Tan Drexel University 5 4 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars The coefficients of eadi individual terms in the above equation must vanish ie 0140xyavoxy0 517 By 6x 2 6 w x y 0 518 ix Equations 516a and 517 imply that 140 is a linear function of y 00 is a linear function of x and the coefficients of the two linear terms must be equal in value but opposite in sign ie 140 ayl7 519 00 fax 5 520 in whidi a b and c are arbitrary constants From 516b and 518 we conclude that m0 is a quadratic function of x and y and that the coefficients of the xy term must vanish ie wox2y2dxeyf 521 where d e andf are unknown constants Substituting 519 520 and 521 in 514 and 513 yields u7xzidz2yb 522 v7yziezinyc 523 wzzvx2y2dxeyf 524 Constants in 522 7 524 can be determined using the displacement boundary conditions at 2 L the top surface of the bar Since the exact condition ie 14 v w 0 everywhere on the cross section is impossible to satisfy results can only be obtained for certain approximate boundary conditions specified based on the SaintVenant s principle Consider for instance the following set of displacement boundary conditions at 0 0 L ie at the centroid of the cross section on the top surface 14 v w 0 to prevent translations in x y and 2 directions 525a 6 0 or 0 to prevent rotation of cross section about the x axis 525b z 60 6w a 0 or a 0 to prevent rotation of cross section about the y ax1s 525c z 60 614 a 0 or a 0 to prevent rotation of cross section about the z ax1s 525d x It is easy to show that these conditions would yield 11 b c d e 0 and f ingZ2E Thus the components of displacement are given by u7xz v7yz wZZLZlx2yz 526 TM Tan Drexel University 5 5 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 00L lt L 000 max Fig 53 Deformation of a bar stretched by its own weight Figure 53 shows the cross section y 0 of the bar before dashed lines and after solid lines deformation has occurred Points along the z axis would displace vertically only as can be seen by substituting x y 0 in 526 ie u 0 0 v 0 0 aux0 wi 527 y 0 y 0 90 2E and the maximum vertical displacement wmax ingZ2E occurs at 0 0 0 Other points on the cross section however would displace both vertically and horizontally It can be seen from 526 that 14 and v are linear functions of x and y respectively Therefore longitudinal fibers that were parallel to the z axis before deformation would deform into straight lines inclined to the z axis Furthermore points on the cross section 2 c where 0 g c S L would displace to ZCw6CZLZlx2yz whidi represents a surface of paraboloid to which all longitudinal fibers would be perpendicular after deformation Therefore there is no shearing strain associated with this deformation Due to the assumed uniformly distributed normal stress given in 511 and the displacement boundary condition given in 525 the top surface of the bar would deform concavely as shown in Fig 53 This of course would not have happened if the entire top surface were completely constrained In the latter case even though the stress distribution on the top surface would be very different from being uniform the resultant force must still be equal to the total weight of the bar in view of the overall equilibrium Thus based on the SaintVenant s principle the difference between the two solutions based on different displacement boundary conditions would be appreciable only in a small region near the top surface 53 Torsion of a Prismatic Bar of Circular Cross Section Consider a prismatic bar of circular cross section and of length L with one end fixed on the xy plane and the other end subjected to a torque MZ as shown in Fig 54a Under the action of the torque the bar would be twisted and a longitudinal line on the lateral surface eg the dashed line in Fig 54a would deform into a helical curve the solid line in Fig 54a The elementary theory of torsion of circular bars assumes that w 0 ie plane cross sections would remain plane after deformation and that any TM Tan Drexel University 5 6 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs cross section at a distance 2 from the xed end would simply rotate by an angle 6 Consider for instance the cross section shown in Fig 54b showing only the first quadrant A point originally located at P would move by an angle 6 to P after deformation The inplane displacement components are given by a A generator deforms b Displacement and stress at into a helical curve point P on a cross section Fig54 Torsion of a circular prismatic bar 14rcos667rcos xcos6717ysin6 528a vrsins 67rsin xsin6ycos671 528b where x cos and y sin If 6 is infinitesimal then cos6 m 1 and sin6 m 6 528 becomes 147y6 vx6 529 If we further assume that 6 varies linearly in z ie 6 20 where or is known as the angle of twist per unit length of the bar then 529 can be written as 147yzot vxzo 530 The corresponding stress components can be readily derived from 52 0x2 iGozy cry Gooc 531a am039W crmcrxy 0 531b in which the assumption that w 0 has been employed It is easy to show that these stress components satisfy equilibrium 51 and compatibility 53 To deck the traction boundary conditions we first examine the lateral surface of the bar For a circular cross section as shown in Fig 54b the unit normal has the following components 71 n n2 0 532 Substituting 531 and 532 in 55 we have t awn axyny 03211Z 0 533a ty 03911 799119 79211 0 533b TM Tan Drexel University 5 7 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs circular bars The same pattern of deformations has also been observed in prismatic bars of varying cross sections Based on these observations SaintVenant in 1855 hypothesized the following displacement components for prismatic bars of general cross sections subjected to end torques 14 foxyz v oocz w byway 535 The corresponding stress components can be derived by substituting 535 in 52 ie am 7W 03922 7xy 0 536a 039CZ GaKa Zi y 039 2 Go a Zx 536b 6x 9 0y Substituting 536 in 51 yields the following Laplace s equation for 1xy 62 62 V2 2 5 6x 0y Thus warping functions x are harmonic functions Furthermore by substituting 536 into 55 0 537 it can be shown that the tractionfree boundary condition on the lateral surface of the bar is satisfied if the warping function satisfies the following equation on the boundary 7yjnxxny 0 538a or ll Z ynx 7 amy 538b in whid 71 and my are the x and y components of the unit normal At the two ends of the bar the statically equivalent traction conditions based on the SaintVenant s principle will be used It is easy to show that the stress components given in 536 would yield FZ Mx My 0 The resultant force in the x direction is given by2 F Laud4 Leggy Gangpg j Hizw j dv GaJLxEi ynx xny st 0 in which 537 538b and the Theorem of Gauss see Chapter 1 have been used Similar result can be obtained for the resultant force in the y direction ie ry j 039 dA j Gaa WxdA 0 A 91 A W Finally the resultant moment about the z axis must be equal to the applied torque MZ Thus M2 IJ039xzyo39yzxdAGog x2y2xiydfl 53 2 LS Sokolnikoff Mathematical Theory of Elasticity 2 d Ed McGrawHill 1956 p 112 TM Tan Drexel University 5 9 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs or M2 Do 5395 where 7 2 2 62 6 DiGHEx y xaiyajdl 540 is the torsional rigidity of the bar From 539b we conclude that the angle of twist per unit length 0 must be proportional to the applied torque M2 The torsion problem is thus reduced to that of finding a harmonic function 1x y satisfying the boundary condition 538 associated with the particular cross section It is noted that since any arbitrary constant satisfies both 537 and 538 the warping function therefore can be determined only to within an arbitrary constant This constant however will have no effect on the stresses as can be seen from 537 55 Prandtl39s Stress Function for Torsion Problems Consider the function Ga77xzyz 541 where 7 is the conjugate harmonic function of the warping function shy so that they satisfy the CauchyRiemann equations 6 756ZI 6 757675 542 6x 6y 0y 6x Differentiating 541 with respect to x and y respectively and using 542 we have sza Zixisz alx 543a 6x 6x 0y aw a Ziy Goia Ziy 5435 0y 0y 6x Comparing 543 with 536b we have a a 7 544 6y 9 x The function xy introduced by L Prandtl in 1903 is called the stress function3 for torsion problems It is easy to show that 544 satisfies the equilibrium equations Eliminating the warping function I from the two equations in 543 yields the following Poisson s equation for the stress function xy 62 62 Vz mf6yfizca 545 3 A rigorous derivation of the stress functions for torsion can be found in LS Sokolnikoff s Mathematical Theory of Elasticity 2 01 ed McGrawHill 1956 pp 114119 TM Tan Drexel University 5 10 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Fig56 A unit normal on the boundary of an arbitrary cross section As for the traction boundary condition on the lateral surface of the bar let us consider an arbitrary cross section as shown in Fig 56 showing only the first quadrant The components of unit normal to the surface can be expressed as d dx y n n x dc 9 dc 546 where dc is an infinitesimal arc element counterclockwise being positive along the boundary of the cross section Substituting 546 in the boundary condition 538a and using 543 we have dijiw 547 By dc 6x dc dc Equation 547 implies that the stress function xy must remain constant along the boundary of the cross section For bars with simpleconnected cross sections eg solid bars this constant can be set to zero ie 0 as the value of the constant has no effect on the stresses Thus eadi torsion problem is reduced to that of finding a stress function xy satisfying the governing Poisson s equation 545 and the boundary condition 5 0 For bars with multipleconnected cross sections eg hollow bars the value of 5 can still be set to zero along its outer boundary Along the inner boundaries however the values of 5 can no longer be chosen arbitrarily We shall discuss this in more detail in Section 58 The torque applied at the two ends of the bar can be obtained by taking the moment of the stress components on a cross section about the z axis ie M2jjiaxzyayzxiA7 yxm2jj m 548 in which 0 along the boundary has been observed Therefore the torque is numerically equal to twice the volume enclosed by the surface representing function xy and the xy plane It is noted that each term in the integrand of 548 contributes to one half of the total torque Thus one half of the torque is due to the stress component 039CZ and the other half due to cry Example 51 Torsion ofan Elliptic Prismatic Bar Consider a prismatic bar of elliptic cross section as shown in Fig 57 The boundary of the cross section is described by the equation TM Tan Drexel University 5 11 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs w Fig 57 The cross section of an elliptic prismatic bar 2 Z y 2710 a NIX N Q in whidi 11 and b are the major and minor axes respectively of the ellipse By assuming x2 y 2 m 7 1 b 1 H2 b2 where m is a constant the boundary condition for 547 is obviously satisfied The value of m can be determined by substituting b in 545 whidi yields m 7 7 anZGo C 7 112 b2 Thus the stress function is given by nzbzGo x2 y2 7 71 d 1124472 112 b2 The applied torque can be determined using 548 7 260 2 2 2 2 2 2 7 abaGtx Mfzmmx gbgb x gm y m b gamma b e where 2 7 771723 2 7 77an3 7 7 x dAilyiT y dAilx T J39J39dAiAiimb are the moments of inertia and area of the cross section The angle of twist per unit length is given by 112 b2 2 f a asbsG and the torsional rigidity is given by M2 7 m3b3G 7 6A4 D or 1124472 47r21p where Z Z 111 2 P X y 4 TM Tan Drexel University 5 12 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs is the polar moment of inertia of the elliptic cross section The shearing stress components can be obtained by substituting d in 544 2y 2x QE WMZ Uyz mMz h Since from b and h we have x 112 7 03992 yb2 0u it follows that the resultant shearing stress I along any radius line must be parallel to the line tangential to the boundary at the point where the radius intersects the boundary as shown in Fig 57 Eli dxi From h we see that the maximum values of the shearing stress components are 2x max 7m3b 2M2 mbz Zymm Mz 2 In I mks 03992 z O39I W W and if 11 gt b the absolute maximum shearing stress is M i mbz Ir max occurring at 0147 where the minor axis intersects the surface of the cross section The warping function 1xy can be determined from 543 b27112 112 b2 x Z y D Finally the components of displacement can be obtained by substituting f and j into 535 2 b2 2 2 bzi 2 u7otyz7WMz voocz 1 3173 ZM7 wazM2 k In I G 7111 b G In I G The contour lines of the warped cross section can be obtained by setting the warping function given in j to constants ie b2 7 112 x K or x K H2 172 y y These are hyperbolas having the principal axes of the ellipse as asymptotes The warped surface is convex in the first and third quadrants and concave in the second and fourth quadrants It can be shown easily that if 11 b then the solution of the elliptic bars reduces to that of circular bars discussed in Section 53 56 The Membrane Analogy Method for Torsion Problems The membrane analogy method for solving torsion problems was first introduced by L Prandtl in 1906 Prandtl considered the equilibrium of a homogeneous membrane such as the elliptic one shown in Fig58 supported along its edges and subjected to a uniformly distributed internal pressure p If the membrane deflection zxy is small then we can assume that the membrane force S remain unchanged throughout the membrane and the internal pressure p is directed along the z axis Consider TM Tan Drexel University 5 13 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs an infinitesimal element of membrane nbsd of size Axgtlt Ay as shown in Fig 58 The 2component of the membrane force along edge ad is given by F5 Sm Aysin 6m m SmAy m 42 096 m 2xy Deformed Sm membrane R i 6 a i M P m S P l P a 55 Side View y i 393 Edge of the I membrane I S I S 1 c do 5 4y 5 39 Ax I u b x v quot 5 39 S Top Fig58 An elliptic membrane subjected to a uniform pressure p where for an infinitesimal deflection sin 6 is approximated as 626xm ie the slope in the x direction of the deformed membrane along edge ad of the element Similar expressions can be obtained for membrane forces acting along the other three edges Since the sum of all these forces acting on the four edges must be balanced with the internal pressure p we have 62 62 62 62 13Z SAy 7 SAy j SAx 7 SAx prAy 0 2 596 m 596 b 0y 1 5V d Dividing this equation by SAxAy and letting Ax gt 0 and Ay gt 0 yield the following Poisson s equation 2 Z V22a a 7 549 6x 6y 5 Furthermore since the edges of the membrane are constrained we have E 0 550 do where do is an infinitesimal arc length of the membrane edge Comparing 549 and 550 with 545 and 547 the governing equation and boundary condition for torsion problems in terms of the Prandtl39s stress function we conclude that if the outline of the membrane is identical to that of the cross section of the bar subjected to torsion then the solutions of these two problems are identical provided that the quantity pS in the membrane solution is replaced by 2G0 in that of torsion and vise versa TM Tan Drexel University 5 14 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bendin o Prismatic Burs zx Y Contour lines of membrane Lines of shearing stress dx B l a Membrane problem b Torsion problem Fig59 Membrane analogy for torsion problems Figure 59 a shows the first quadrant of a membrane with contour lines depicting its deformation due to internal pressure p The corresponding lines drawn on the cross section of the torsion subjected to torsion as shown in Fig59 b are called the lines ofshetzring stress Consider a point eg point B on the membrane through whid r a particular contour line is drawn Since along a contour line the deflection of the membrane is constant we have in as where c is the arc length of the contour line The corresponding equation for a bar in torsion is 0 as where 5 is the stress function Using 544 and 546 we can express the above equation as nine W1 7 a Zn 03an 0 551 6c 6x do 6y do 9 9 Since ayzny 03an represents the total component of shearing stress in the direction normal to the line of shearing stress see the insert of Fig 59 we conclude from 551 that the shearing stress resultant r must be tangent to the lines of shearing stress The magnitude of the resultant can be obtained by summing up the components of 039 and aw in that tangent direction ie r awn 7 0tu gm Jrgny 73 552 Thus the magnitude of r at any point is equal to the gradient of the stress function 5 in the direction normal to the line of shearing stress at that point On the corresponding membrane dzdn is the maximum slope at that point Thus the shearing stress resultant on the bar in torsion can be obtained by multiplying this maximum slope by the constant 2G 06 ie TM Tan Drexel University 5 15 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars EJJE 553 an pS an From 548 we also conclude that the resultant torques at the ends of the bar is equal to twice the volume enclosed by the membrane with pS being replaced by 260 ie M2 2 j j 7lxrly 2GTIIzaxay 554 Consider a free body of the membrane obtained by cutting the membrane along a contour line eg the shaded area in Fig 59 Along the boundary of the free body there is a uniformly distributed membrane force S whose zcomponent per unit arc length of the contour line is 8azrln Summing up all the forces acting on the free body we have dz F A S rl 0 Z 2 r f dn 6 where A is the area of the xy projection of the membrane free body Substituting 553 into the above equation yields 17quot 7 A 7 rlc 7 0 P Zsz frds2GaA 555 from whidi the average shearing stress along a line of shearing stress can be determined ie 7 frdc 7 ZszA 1 m fat C where C fat is the length of the line of shearing stress Example 52 Torsion ofA Rectangular Prismatic Bar w Fig510 Torsion of a rectangular prismatic bar Consider a prismatic bar with a rectangular cross section of size 2agtlt2l7 as shown in Fig 510 subjected to end torques By using the method of membrane analogy the problem becomes that of finding the deflection surface of a uniformly loaded membrane having the same rectangular boundary and satisfying the governing equation and boundary condition of 549 and 550 By assuming the deflection of the membrane to be in the following series form TM Tan Drexel University 5 16 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 2w coswgt a where b1bZ are constants and Yny are unknown functions in y the condition of symmetry with respect to the y axis and the boundary condition zx in 0 are satisfied automatically Substituting a in 549 and then expanding the righthand side into a Fourier series for in x E 11 ie 7373 m i E Si 871 n 1 2 cos nx b where 8 nil211 we obtain the following differential equations for Y y H 4 Li Y main 4 lt71 2 The solution of this secondorder differential equation is given by 2 n 16 7 UTl Y A sinh B cosh n n n ns sbns Constant An vanishes as the result of Y being symmetric with respect to the y axis Constant B can be determined by using the condition that Y 0 at y ib This yields Z 1171 n 1361 71 17 cosh ny n It bnS cosh nb Thus the membrane deformation is given by Z16P112 in 71 1 17 cosh nyJCOS nx C 735 My 113 cosh nb Replacing pS by 2G0 in c yields the stress function for the corresponding torsion problem 326 Z w 712 h Z 12 lcosm d and the nonvanishing stress components can be obtained by substituting d in 544 ie cos nx n71 77716Gm i 71 sinh ny X2 6y It2 My 112 cosh nb n71 7716Gm Z 422 licosh nyJSm nx 71 92 6x It2 My cosh nb If h gt 11 the maximum shearing stress occurs at i a 0 where 039CZ 0 16Gm 1 1 r 039 ia0 17 e W 1 gt z cosh5 ltgt Using the relation 2 1 Zi 3 5 8 TM Tan Drexel University 5 17 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs we can rewrite e in the following form 16Gm 1 f IIZ Hays cosh nb rmax 2Gm7 The infinite series on the righthand side whidi converges very rapidly if l7gt 11 becomes negligible compared to the first term for a very narrow rectangular cross section where I gtgt 11 In such a case the maximum shearing stress becomes rmax 26m g For a square cross section where 11 l7 f becomes rmax m 1351Gou h In general we can express the maximum shearing stress as rmax k2Gm i where k is a numerical factor whose value depends on the aspect ratio of the cross section btz The applied torque can be obtained by substituting d in 548 This yields 1 3 19211 w 1 G 2 2b 17 t h b M 3 a a s MESS an A 1 The infinite series term on the righthand side again converges very rapidly In general the torque can be expressed as M2 leo2iz32b k By eliminating G6 from i and k we can also express the maximum shearing stress as Ml r l k2lt2agtzlt2bgt 0 where k1 and k2 are numerical factors whose values depend on the aspect ratio 1711 Values of the factors k k1 and k2 for several different btz ratios are given in Table 51 For a bar with a very narrow rectangular cross section 1711 00 we have k gt1 k1 kZ gt13 26m 211 2b 1 3 M2 3GDt211 212 Table 51 Constants for Torsion of a Rectangular Prismatic Bar btz k k1 kZ btz k k1 k2 10 0675 01406 0208 3 0985 0263 0267 12 0759 0166 0219 4 0997 0281 0282 15 0848 0196 0231 5 0999 0291 0291 20 0930 0229 0246 10 1000 0312 0312 25 0968 0249 0258 00 1000 0333 0333 TM Tan Drexel University 5 18 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs gtI r Za Fig511 Torsion of a narrow rectangular prismatic bar Solutions for bars with narrow rectangular cross sections subjected to torsion also can be obtained in the following manner Consider the deformation of a membrane whose outline is the same as the narrow cross section ie I gtgt a of the bar as shown in Fig 511 The shape of the slightly deflected membrane can be approximated as being cylindrical throughout the length in the y direction if the small regions near yib is ignored see the yz projection in Fig 511 The xz projection of the deflected membrane see Fig 511 can be approximated by that of a string subjected to a uniformly distributed pressure p The maximum deflection of the string is given by 5 112228 and the maximum slope occurring at x in is given by E r m dn max 11 S The volume enclosed by the parabolashaped cylindrical membrane and the xy plane is given by 2 4113b p V 39 2 2b 3 ax gt 3 S n Replacing pS in m and n with 2G0 and using the membrane analog we have 1W 26m M2 6a2a32b 0 which agree with those given in f and j with the series term being neglected for 17 gtgt 11 57 Torsion of Prismatic Bars with Rolled Profile Sections Solution for a narrow rectangular bar subjected to torsion can be easily extended to those for rolled profile sections Let t 211 and l 217 so that the size of the narrow rectangular cross section is now tgtltl where lgtgt t then the maximum shearing stress and the applied torque can be expressed as rmax Gtor M2 Do 556 where 3 D G l 557 is the torsional rigidity Consider a slotted tube whose cross section is shown in Fig 512 It is easy to see that whether the narrow rectangle is straight or curved has very little effect on the volume enclosed under the corresponding membranes Thus the same equations as those in 556 can be used for the slotted tube shown in Fig 512 TM Tan Drexel University 5 19 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs t t a r a t 12 4 I 1 T b Fig 512 Cross section of a slotted tube Fig 513 Rolled profile sections For other types of rolled profile sections such as angles diannels and lbeams shown in Fig 513 we may consider each of them as being a combination of a number of narrow rectangular cross sections The total torsional rigidity is then the sum of those of individual ones Take the angle shown in Fig 513 a as an example the total torsional rigidity is given by 3 3 DDh H31 Gltll rtzlzl 2 3 Thus we have 3MZ 0 3 3 Gltll1 thzlZ The shearing stress in each leg of the angle can be estimated by using the formula for narrow rectangular cross sections as well ie I thor33I zt i12 t1l1t2lZ Similar expressions can be derived for the diannel and lbeam sections shown in Fig 513 ie 0 33M2 3 and r i12 Git1l12t2l2l t1l12t2lZ It should be noted that in deriving the approximated solutions for rolled profile sections we have neglected the detailed geometry of the fillet at the reentrant corners where a considerable stress concentration may develop if the radius of the fillet is small Values of these highly concentrated stresses can be estimated by using the membrane analogy method4 58 Torsion of Prismatic Hollow Bars So far we have discussed torsion problems for bars with solid cross sections only ie those that are bounded by a single curve At the boundary of sudi cross sections the stress functions according to 547 must remain constant and are usually set to zero However for bars with hollow cross sections ie those that have two or more boundaries the values of stress function along the boundaries can no longer be chosen arbitrarily We shall consider first a simple case in which the inner boundary of the hollow cross section coincides with one of the lines of shearing stress of the corresponding solid bar such as the elliptic cross section shown in Fig 514 showing only the first quadrant The outer boundary of the hollow cross section whidi coincides with that of the corresponding solid cross section is given by 4 SP Timoshenko and N Goodier Theory of Elasticity 3rd ed McGrawHill 1970 pp 322324 TM Tan Drexel University 5 20 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs Membrane D T It p p a 212 r boundary x Z Lines of shearing stress Inner l boundary k1 V1 Fig 514 Torsion of a hollow bar with elliptic cross section x2 y 7 710 558 The inner boundary of the hollow cross section being geometrically similar to the outer boundary as it coincides with one of the lines of shearing stress can be expressed as x2 y2 W W 71 7 0 559 in which k lt1 is a scaling constant Since the resultant of shearing stress is tangent to hence does not cross the line of shearing stress the hollow bar therefore can be considered as being obtained by removing an inner cylinder from the solid bar without altering the stress distribution in the remaining portion of the bar This can also be verified by using the following membrane analogy Since in the hollow region the stress is zero the membrane must have a zero slope This can be easily adiieved by placing a rigid horizontal plate 6 in Fig 514 over the hollow region It is noted that the pressure distributed over the membrane a in a solid bar is statically equivalent to that distributed over the plate 6 in a hollow bar Consequently the membrane force S along the boundaries of free bodies represented by a and 6 respectively must be the same Thus the equilibrium conditions for the remaining portion of the membranes ie the shaded area in Fig 514 are identical for both cases and the stress function derived in Example 51 for a solid elliptical bar is therefore applicable to the hollow bar ie 7 nzbzGo LQLZA 560 112 b2 11 172 From Fig 514 it can be seen that the volume enclosed by the membrane of the hollow elliptic bar is less than that of the solid elliptic bar by the amount equal to the volume of revolution of area CDD or VHW 17 k4vs Based on the membrane analogy and e of Example 51 we concluded that for a hollow elliptic bar 3 3 The angle of twist per unit length thus is given by TM Tan Drexel University 5 21 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs MZ 1124472 a 17 k4 asbsG 561 The stress function and the corresponding maximum shearing stress are given respectively by M x2 y2 7 z 1 562 mbilik liE Z b2 2M 1 r quot 563 max mbz 17 k4 Substituting the equation of the inner boundary 559 in 562 we have M2 Vim boundary W If the inner boundary of the hollow bar does not coincide with any of the stress lines of the corresponding solid bar then the selfequilibrating condition of the rigid horizontal plate discussed previously is no longer satisfied In sudi a case additional load must be applied on the plate to maintain its horizontal position hence would introduce additional conditions that the membrane must satisfy In terms of the stress function for torsion problems these additional conditions are necessary to ensure that the displacements be singlevalued From 535 and 536b we have 0w 0w o xzG 7oty o zG ooc 6x 9 0y Expressing the resultant shearing stress in terms of its components ie x dy 1039 039 do 92 do and integrating it along the inner boundary of the hollow cross section we have dx dy 6w 0w 1dc an 5 cry Ejdc Jrady 7 Ga ydxi xdy If the displacement w is singlevalued then the first integral in the above equation must vanish as the integration is taken along a closed curve The second integral is equal to twice the area enclosed by the curve Thus we have Medeai This is the additional condition that needs to be satisfied when determining the constant value of stress function 5 along the inner boundary 59 Torsion of ThinWalled Tubes Solutions for torsion of bars made of thinwalled tubes can be easily derived by using the membrane analogy Figure 515 shows the cross section of a thinwalled tube and the corresponding stress function whidi is geometrically similar to the membrane Since the wall is thin we can neglect the slight variation in the slope of the membrane across the thickness of the wall and replace it by a straight line as shown in the insert of Fig 515 Thus the shearing stress can be approximated by TM Tan Drexel University 5 22 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 564 where h is the difference of the values of 5 between the inner and outer boundaries and t is the thickness of the wall It is easy to see that if the wall thickness is uniform then the shearing stress remains constant throughout On the other hand for a tube of nonuniform thickness the maximum shearing stress would occur at the location of the smallest wall thickness In either case It h remains constant throughout and is referred to as the shear flow in the wall of the cross section The corresponding torque can be calculated using membrane analogy ie it is twice the volume enclosed by the stress function whidi can be approximated as M I m 2Ah 2At r where A is the area enclosed by the dashed line in Fig 515 or the mean value of the areas enclosed by the outer and inner boundaries of the cross section of the tube Thus given a torque M2 the shearing stress can be computed as M Z 565 1 2A To determine the angle of twist per unit length we integrate the shearing stress given in 565 along the centerline ie the dashed line in Fig 515 of the wall and use 555 We f 2610 566 2A t from whidi we obtain M do z 567 0 4AZG t If the tube has a uniform thickness t is constant 567 becomes M Z1 0 T 568 4A Gt where l is the length of the centerline along the tube wall The torsional rigidity is given by 2 D w 569 or Z TM Tan Drexel University 5 23 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs For a circular tube with a mean radius V 1471 2 and 12m we have D47rzr4Gt Comparing this with 557 the torsional rigidity for a slotted tube we see that the ratio of the two torsional rigidities is D Z Dunslouedmbe 570 slotted tube For thinwalled tubes 7 gtgtt Thus having a longitudinal slot in the tube would drastically reduce the torsional rigidity Profile of 5 and membrane Fig 516 Torsion of a typical aircraft fuselage For thinwalled tubes with multiple holes sudi as an aircraft fuselage with a typical cross section as shown in Fig 516 the shearing stress in ead portion of the wall can be obtained by using the membrane analogy ie h h2 h1 7 hZ TF l 72 13 a ti Integrating the shearing stress along the walls enclosing the two compartInents denoted as A1 and AZ respectively in the figure then using 555 we have 1111 1313 2G04A1 b 1le 71313 ZGoa lZ c where ll lz and 13 are the arc lengths of ACE ADB and AEB respectively The total torque can be determined by using the membrane analogy M2 2 dA 2A1h1 A2h2 2A1t111 12512 d Shearing stresses 1391 and the angle of twist per unit length or can be obtained by solving a b c and d simultaneously They are M 71 NZ t3lZAl tzlsA1Azl M 12 NZ t3llAZ t113A1 A2l TM Tan Drexel University 5 24 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs M2 73 N tilei tzliAzl M2 0 ZGN tllZl3 tZl3ll t3lllZ where N 2t1t312A12t2t311A t1t213A1AZZ 510 Pure Bending of Prismatic Bars Fig 517 Pure bending of a prismatic bar by and moments A prismatic bar of length L is fixed at xy plane and subjected to a bending moment M at z L as shown in Fig 517 If the z axis coincides with the centerline of the bar and the yz plane where moment M is acting on is one of the principal planes then from the elementary theory of bending we know that the bar is in a state of pure bending and that the components of stress are given by 039ZZ am 7W cryZ an o xy 0 571 where R is the radius of curvature of the bar after bending It can be seen that the equilibrium is satisfied if the body force is absent The compatibility is also satisfied automatically as the stress components are either zero or linear functions of 2 For the boundary conditions it can be easily shown that the traction free condition on the lateral surface of the bar is satisfied At 2 L however we must use the following statically equivalent conditions F A andl 0 F9 A azydA 0 F2 A crudl 0 572a My A 0 22di 0 M2 L 7 o znyr 0 Wth 0 57213 Ey2 E1 M jAauydAjA R dick 572c where Ix is the moment of inertia of the cross section about the x axis This is the wellknown moment 7 curvature equation derived in the elementary theory of bending Substituting 572c in 571 yields whidi is again identical to the flexure formula in elementary theory of bending To determine the components of displacement we substitute 571 in 52 This yields TM Tan Drexel University 5 25 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 1 573 v v R ax R By R 62 0 0 f qj 574 02 By 6x 02 By ax Integrating the last equation in 573 we have w7w0xy 575 where m0 is a function of x and y Substituting 575 in the first two equations in 574 yields 614 72770100 607 0w 2 awe Eax ax E ayRay or after carrying out the integration with respect to z 0 0w 0w 2 6x u0xy 07 72 0y voxy 576 where 140 and 00 are functions of x and y as well Substituting further 576 in the first two equations of 573 yields 2 2 6x 6x R 0y By R from which the following conclusion can be drawn by comparing terms on both sides 2 Z a 206 200 ivl 577 6x 6y 6x By R Integrating the second equation in 577 gives 2 x uoxyev7yf1y voxyev2y Rfzx 578 Substituting 576 and 578 into the last equation of 574 yields azwimyqzqkwhwrvq axay R dy dx Since this equation is true for any value of 2 we have Z a woxy0 d ydfzxiv 0 579 may dy dx R From the first equations in 577 and 579 we conclude that w0 can only be a linear function of x and y ie w0xy mxnyp 580 From the second equation in 579 we conclude dfiy df2x7v a R dy dx TM Tan Drexel University 5 26 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs whidq upon integration yields 2 x f1ywj fzxv m7 581 Combining 575 576 578 580 and 581 gives 9 7 u 7v 7 7 mz R 00 5 07izzivxziy2 oocinz7 2R w mx 11 R V P in which m n p or 8 and 7 are constants to be determined by the displacement boundary conditions Since the bar is fixed at z 0 we can assume that at 0 0 0 uvw0 and 0 Oz Oz 0y These conditions whidi prevent the point 000 from having any movement or rotation yield 0 8 7 m n p 0 Thus the components of displacement are given by uiivjz y v7zzivxziy2 w 582 We now examine the validity of the basic assumptions used in deriving the elementary theory of bending by comparing the solution of elementary theory with that of theory of elasticity shown here 1 By substituting y 0 in the last equation of 582 we have w 0 This implies that the xz plane remains undeformed and is indeed the neutral plane 2 Displacement component w as shown in 582 is a linear function of y indicating that plane cross sections will indeed remain plane 3 Equation 574 requires that the shearing strain vanish throughout the entire bar Thus cross sections perpendicular to the neutral plane will indeed remain perpendicular after bending The deflection curve along the z axis the centerline of the bar can be obtained by setting x y 0 in 582 ie 22 07 uw0 583 2R whidi is identical to those given in the elementary theory of bending Consider the cross section of a rectangular prismatic bar at a distance 2 c from the fixed end The 211x2l7 rectangular cross section shown by solid lines in Fig 518 would deform to the shape represented by the dashed lines The two vertical edges at x in would deform linearly ie my i ux 11 R while the top and bottom edges at y would deform to parabolic curves ie TM Tan Drexel University 5 27 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bendin o Prismatic Burs k r 1 lt w y l l l k b I 1L Fig 518 Deformation of the cross section of Fig 519 Optical image Showing contour lines a rectangular bar due to bending on top surface of a bar in bending vltyibgt7 czvxzebz For small deformations the curvatures of the two parabolic lines can be approximated by the second derivatives of the curves ie dzy d2 1 Z Z Z v ibi c vx 7b dxz dxz 2R R The top or bottom face of the bar forms an anticlas c surface with curvatures being convex down in the lengthwise direction but convex upward in the widthwise direction Taking the top face where y b as an example the contour lines of the deformed surface can be obtained by setting the displacement v in 582 constant This yields the following hyperbolas 22 ivxz constant whose asymptotes are given by zzivxz0 584 Figure 519 shows the hyperbolic contour lines of sudi a surface obtained using an optical tedinique5 By measuring the angle or as shown in the figure we can determine the Poisson s ratio 1 of the material using 584 ie v cot2xz cot2 or 511 Bending of a Prismatic Bar by an End Force A prismatic bar of length L is fixed at the xy plane and subjected to a concentrated force Py at zL as shown in Fig 520 The coordinate system has been setup in such a way so that the z axis coincides with the centerline of the bar and the x and y axes coincide with the principal centroidal axes of the cross section In addition the force Py is on a plane parallel to the yz plane and is applied at sudi a distance6 from the centroid that twisting of the bar does not occur Based on the elementary theory of bending we assume that the normal stress on all cross sections is the same as that of pure bending derived in Section 510 ie 5 SP Timoshenko amp N Goodier Theory ofElusticity 301 ed McGrawHill Inc 1970 p 288 6 The significance of this distance and the method of determining it will be discussed in Section 512 TM Tan Drexel University 5 28 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars 585 The elementary theory also assumes that all the remaining stress components except an and 72V should vanish We shall now examine these assumptions to see if they satisfy the governing equations of elasticity and the boundary conditions The equilibrium equations 51 require that in the absence of body force the assumed stress components satisfy 6039 6039 P 602 0 yz 0 6 yz W 586 62 6x 6y I It is noted that the first two equations in 586 imply that the shearing stresses are independent of 2 hence the distribution must remain unchanged throughout the length of the bar Fig521 Unit normal on the boundary of an arbitrary cross section Next we examine the boundary conditions 55 for an arbitrary cross section as shown in Fig 521 showing only the first quadrant It is easy to see that the assumed stress components satisfy 55a and 55b identically From 55c we have 03211 ayzny 0 587 where from Fig 521 do 9 do Substituting these components of unit normal into 587 yileds TM Tan Drexel University 5 29 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs W M e 0 588 039 7039 7 xzalt yzdc Finally the compatibility equations 53 in the absence of body force require that 2 7 Z 7 P9 Va 0 Vazii 589 y 1x1v Thus the bending problem reduces to that of finding for GE and 039 functions of x and y that satisfy 2 f the equilibrium equation 586 the boundary condition 588 and the compatibility equations 589 This can be further simplified by using the following method of stress function Let xy be a stress function for problems of bending not to be confused with the stress function for the problems of torsion discussed previously in this diapter from whidi the components of stress can be derived as follows c cpyyz 0X2 2 0y 9 6x 21 fx 590 It is easy to show that these stress components satisfy 586 the equilibrium equations Function fx which has no effect on the conditions of equilibrium will be determined later from the boundary condition Substituting 590 in the compatibility equations of 589 yields 3infl j 362 62 V Zu 0y of 6x 6x2 of 51va dxz From whidi we conclude Z Z Px VZ 6 6 V Lid x 6x 0y 1v Ix dx 3 591 where 8 is a constant To find the meaning of 8 we consider the infinitesimal rotation of an area element of the cross section with respect to the z axis given by see Chapter 3 1 60 014 ml 26x The rate of change of this rotation in the z direction is given by a w iin 111mg 3p jaewn 652 az azzax 0y 76x262 By 262 6x 76x 726 6x 0y Substituting 590 in the above equation and comparing the resulting equation with 591 we have Px wifi V Lopg 39 2c 1v1x Since wzyzx 0 ZG we conclude that ZG is the rate of diange of rotation of an area element along the centerline of the bar where x 0 Assuming that the end force is applied in sudi a way so that the bar is not being twisted then 8 0 and 591 becomes V Pyx d x 1v Ix dx W 592 TM Tan Drexel University 5 30 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Finally in terms of the stress function the boundary condition 588 becomes dxdy Pyy27fxdx 593 6c 6x dc ayrlc 21 Pic If the function is known then the values of stress function 5 along the boundary of the cross section can be determined and the solution can be obtained by using 592 and 593 In the following examples the function will be chosen in such way as to make the righthand side of 593 vanish The value of stress function 5 therefore remains constant along the boundary Furthermore for a sirnpleconnected cross section we can make the constant equal to zero then the problem of bending of a prismatic bar reduces to that of finding the solution of 592 with boundary condition 5 0 Once the stress function is determined the stress components can be computed using 590 and the displacement components can be obtained by substituting the stress components in 52 and carrying out the integrations ln terms of stress function the displacement components are given by P L 14 awo21tvgtzvr xyiw p 594 ax E 0y E1 P P 2 v 7 6Eny 23 73L 227 Vx27 yz 77 fEZerch 7 595 P wzzi2Lzwoxy 596 where or 8 and 7 are arbitrary constants to be determined using the displacement boundary conditions and m0 is a function of x and y that can be determined using the following equations Z 2 VP 6 740 721V 6 yy 59721 6x E may EIX azwo 21v 62 7 Pyr vl yr 597 of E may Ix E1 62 1 62 62 d w0 v Z 7r x 597C may E x dx Example 53 Bending ofa Circular Prismatic Bar by an End Force Consider a circular prismatic bar whose boundary is given by x2 y2 R2 a in whidi R is the radius of the cross section The righthand side of 593 becomes zero if we dioose P Ami 3022798 b Substituting b in 592 yields TM Tan Drexel University 5 31 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 712v 1v Ix v2 C Thus the problem reduces to that of finding the solution of c subjected to the boundary condition of 5 0 This boundary condition is satisfied if we dioose mx2y27R2x d where m a constant factor can be determined by substituting d in C Le 7 12v Py 81v1x Thus the stress function becomes 12v Pyx Z Z Z x 7R f 81v Ix y and the stress components can be obtained by using 590 a 77 12v Pyxy 2 41v I g P c 02 32Vy27y271 21ij h 9 81v1x 32v Along y0wehave P sz0 az32V yR271 21362 y 81v1x 32v The maximum shearing stress occurs at the center of the cross section where x 0 ie a 7 32v PyRZ7 32v P 91 max 81v I 21vA in whidi A IrRZ is the crosssectional area of the bar At each end of the horizontal diameter ie at x irR the shearing stress is a 7 12v 17sz 712vPy 9quot Xi 41v Ix 1v A Taking v 03 the value of Poisson s ratio for most engineering materials we have P P c y c y cl MW 71387 cl 937 71237 Comparing with the elementary solution P o 2 ii 9 3 A obtained by assuming that the shearing stress is uniformly distributed along the horizontal diameter the maximum error is seen to be approximately 4 percent TM Tan Drexel University 5 32 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars To find the displacement components we need to determine first the function w0xy by substituting the stress function f in 597 This yields P weby y y3x2yc x77yg j 4EIX where 5 77 and g are arbitrary constants to be 39 using the J39 J boundary 4quot Substituting j and the stress function f in 594 595 and 596 and assume constants or 8 7 5 77 and g all vanish valid if the origin of the coordinate systems undergoes no translation or rotation ie no rigidbody motion we have VP uyLiz k P v izs3L2273VLizx27yzRzz l P w4 ly 22274inx27y2 m We now examine again the basic assumptions used in deriving the elementary theory of bending against the elasticity solution presented here 1 By substituting y0 in m we have w 0 Thus the xz plane remains undeformed and is indeed the neutral plane 2 Equation m reveals that a cross section will deform into a cubic surface in y Thus the assumption that cross section remains plane is no longer valid 3 The shearing stress component 7 whidi can be easily obtained from h does not vanish along y 0 Thus the assumption that cross sections remain prependicular to the neutral plane is no longer valid either Substituting x y 0 in 1 gives the deflection curve of the centerline of the bar P 290 y 23 F3LZZ FMRZZ n 90 6EIX 2 The maximum deflection of the centerline occurs at z L ie f l P9L31m5 o 0x 0 90ij 3E1 2 L The coefficient on the righthand side of o is recognized as the solution from the elementary theory of bending The second term in the bracket is the correction factor whidi is less than 1 for bars with aspect ratio 2RL less than 012 if v 03 Example 54 Bending 7le Rectangular Prismatic Bar by an End Force The boundary of a rectangular cross section as shown in Fig 522 is given by xziaZXEZibz0 a By dioosin g TM Tan Drexel University 5 33 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs W W x z Membrane A lt1 3 m WV Pressure Fig 522 Bending of a rectangularbar by an and force 7 Pybz fx7 2 b x the righthand side of 593 vanishes along the boundary ie Z 2 Pyy 7 Pyb 0 21 along y ib 21 alongxrtz d x0 do Thus the boundary condition becomes 0 Substituting b in 592 yields V Pyx C v2 1v Ix Using the membrane analogy similar to that of a torsion problem we determine the deflection of a membrane subjected to a pressure load proportional to V Px y d 1v Ix This pressure which remains constant in the y direction but varies linearly in the x direction will cause the membrane to deform in a shape whose intersection with the xz plane is as shown in Fig 522 Consequently 5 must be an odd function of x and an even function of y These conditions can be satisfied by taking 7 in the following Fourier series form mmm e 11 2b ZZAZWM Sin m1n1 Substituting e in c and applying the standard procedure for determining the coefficients of a Fourier series we can obtain TM Tan Drexel University 5 34 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 71 1 Z sin m x cos 2n T wily 2 f 2 71 Z 2 712 mn m n 4b2 V Py8 3mm 4 1v Ix 7r mam To find the shearing stress components we first substitute b in 590 this yields P g 02yb2y2 0y 9 6x 21 which can be resolved into the following two systems Oxfalzto39 z 0920 03952 E where P oquot 0 oquot 9 bzi 2 h x2 92 21X y represent a parabolic stress distribution given by the elementary theory of bending and 6 6 a 7 5 0 lt1 6y 6x represent the corrections to the elementary solution due to the stress function Along the x axis 6 0y0 as the result of 5 being an even function of y the corrections to the elementary solution therefore affect only the vertical component of the shearing stress contributed by 052 Substituting f in the second equation of i and letting y 0 we have Tnch 71Ver COS V pg 8H2 no no agile 1vI n3 1 1 112 1 X m 2n71m22n712m Since Ix 411bS3 AbZB where A 411k is the crosssectional area j can be written as m x 71mn72 COS 3P 1 16 112 Oils 9 3 222 2 k 9 2A 1v II b mam 2 2 F 21171 m 4421171 Z 4b The coefficient on the righthand side of k is recognized as the average shearing stress along x 0 from the elementary theory of bending Thus given the ratio nb of a rectangular cross section and Poisson s ratio of the material we can evaluate the term in the brackets of k and make a correction to the elementary solution The infinite series in the equation converges very rapidly particularly for nb gt 1 Table 52 lists for a number of nb values the ratios of exact solutions to the elementary solutions for the stress component cry at x y 0 center of the cross section and at y 0 x in middle of the vertical sides of the rectangle The Poisson s ratio is set to 025 It is seen that the elementary solution yields very accurate estimates for nb 2 2 about 10 error for a square cross section where lZb 1 but become less accurate as the value of nb decreases TM Tan Drexel University 5 35 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Bars Table 52 Comparison of exact and elementary solutions for shear stress in a bar subject to an end force 51 b 5 4 3 2 1 1 2 12 1 2 1 5 x 0 y 0 0997 0996 0993 0983 0940 0856 0817 0805 0801 x id y 0 1005 1008 1015 1033 1126 1396 1691 1988 2285 512 Shear Center In Section 511 we have assumed that the z axis coincides with the centerline of the bar and the x and y axes coincide with the principal centroidal axes of the cross section In addition the force Py is on a plane parallel to the 32 plane and is applied at such a distance from the centroid that twisting of the bar does not occur This distance can be easily obtained once the stress components as given by 590 are known ie we evaluate the moment about the centroid produced by the shearing stress components 0x2 and oyz as Mz 7ny oxzybxdy 598 Since the stress resultant must be statically equivalent to the applied force Py we conclude that the distance from the application point of the force to the centroid of the cross section must be Mz P 1 d 599 For a positive value of Mz dx must be taken in the positive direction of x If a force denoted by Px is applied on a plane parallel to the xz plane then we can follow the same procedure and determine the distance for the application point denoted by dy so that the bar will not be twisted by the applied force The intersection of the two lines drawn parallel to the x and y axes and at distances dy and dx respectively from the centroid is called the shear center Any force that is applied at the shear center and perpendicular to the axis of the bar will not cause the bar to twist Problems 51 A prismatic bar with a 2n x 2h square cross section and of length L is supported at the four corners of the top surface by four rigid bars and is being stretched by its own weight as shown a Specify the displacement boundary conditions b Use the boundary conditions specified in a to determine the constants in 522 523 and 524 hence obtain the expressions for the components of displacement c Compare the results you obtained in b with those in 526 9 2 y x l l dz 39139 0 G x lt L gt Problem 51 Problem 52 TM Tan Drexel University 5 36 November 13 2007 THEORY OF ELASTI CI TY 5 Extension Torsion and Bending of Prismatic Burs 52 A prismatic bar of length L is rotating about the y axis with a constant angular velocity Q a Consider this as a onedimensional 1D problem first Draw a freebody diagram for an infinitesimal bar element of length dz as shown Show that in the absence of gravitational effect the stress in the bar due to the centrifugal force is 039ZZ pQZQZ 7 22 where p is the mass density of the bar b Show that the following 3D state of stress 039ZZ pQZQZ 7 22 amUWayzazxaxy0 obtained from the 1D analysis in part a is a solution to this problem in threedimension if and only if v 0 Hint Check the compatibility equations 53 Prove equation 548 54 Prove equation b of Example 54 on page 516 55 Consider a prismatic bar with an equallateral triangular cross section as shown in the figure subjected to torsion show that 1 2 2 1 3 2 2 2 7G 7 7 3 7 4 y 24x xy is the solution of the problem Determine the distribution of shearing stress component 039yz and its maximum value along the x axis 2 b3 Perfectly plastic l1 n gt Problem 55 Problem 56 56 The membrane analogy method also can be employed to analyze torsion problems in which the material deforms beyond the elastic limit For instance consider a circular bar made of the material having an elasticperfectly plastic stressstrain relation as shown in the figure If the shearing yield stress is 19 show by using the membrane analogy method that the ultimate torque the bar can sustain is MZ Z RsryB 57 Show that the stress distribution in a bar subjected to torsion remains the same regardless of the location of the origin that the axis of rotation passed through so long as the axis is parallel to that of the bar TM Tan Drexel University 5 37 November 13 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors Chapter Cartesian Tensurs 11 Vectors Fig 11 A vector in twodimensional Cartesian coordinate systems Consider a vector 7 in a twodimensional 2D Cartesian coordinate system XliXZ as shown in Fig 11 Unit vectors El and E2 in Fig 11 are the base vectors in the X1 and X2 directions respectively The vector 7 can be expressed in terms of its components V1 and V2 in these two directions as 7V1E1 VZEZ 11 or simply 17V1V2 12 With respect to a new coordinate system X17 X39z having base vectors E and E obtained by rotating the X1 7 X2 coordinates counterclockwise at angle 6 Fig 11 the components are V1 and V2 and the vector 7 or 7 can be expressed in terms of these new components as V Via V5 13 7 VJVD 14 From Fig 11 it is easy to see that the relations between the components of 7 or 7 in the unprimed original and primed new systems are given by v V1 cosX1 x VZ cosXZ x V1 cos 6 VZ sin 6 15a V V1 cosX1 x VZ cosXZ X Z 7V1 sin VZ c056 1513 where cosX1X is the cosine of the angle between the X1 and X axes etc Defining a11 cosX1 X cos 6 16 H12 cosX1Xgcos7r267sin 9 17 H21 cosXZX cos7r276sin6 18 H22 cosXZX c056 19 TM Tan Drexel University 1 1 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors then 15 can be written as V anv1 amVZ 110a VZ ale1 aZZVZ 11013 Quantities an an a21 and an defined in 16 19 are called the direction cosines Equations 11 110 can be easily extended to vectors in three dimensions 3D ie 7V1E1 VZEZV3E3 V1VZV3 111 in X coordinates 1739 Vie VZ39E V3 Eg Vl39VZ V3 112 in X39 coordinates and the relations between the primed and unprimed components are Vl allV1 asz a31V3 113a VZ ale1 aZZVZ pigzv3 1131 V3 a13V1 a23VZ a33V3 113c 12 The Summation Convention Equation 113 can be written in the following condensed form 3 3 3 Vi Zanvw Vi 25sz Vi Z 13 1 114 11 11 11 which may be further consolidated into 3 V gem 1123 115 It is noted that 115 represents three equations one for each value of the subscript j and within ead equation the summation extends over the range of the repeated subscript i This equation can be further simplified by introducing a summation convention whereby the symbol 2 is redundant as follows Ifa repeated alphabetic subscript appears in one monomial an automatic summation over the range of that subscript is required By using this convention we can rewrite 115 in the following form V a V 7 17 1 ij 123 116 It should be noted that the ranges of both subscripts i and j must be specified in the equation 11 w A repeated subscript a subscript that appears twice in a monomial such as i in the above equation is called a dummy index whereas a nonrepeated subscript a subscript that appears only once in a monomial sudl as j in the above equation is called a free index Since a dummy index merely indicates summation over its range it is immaterial which symbol or letter is used so long as the ranges of the symbols are the same Hence 116 can be written as v739aqvk jk123 117 The symbol or letter used for the free index is also arbitrary so long as the same is used in every monomial Therefore 117 or 116 can be written as TM Tan Drexel University 1 2 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors W alek kl 123 118 The rules for an equation written in the indexical or subscript form such as those of 116 7 118 can be summarized as follows I An index can only appear either once a free index or twice a dummy index in a monomial An index that appears more than twice in a monomial is not allowed I A repeated dummy index in a monomial automatically requires a summation over its range I An index that appears only once in one monomial hence a free index must also appear just once in each and every other monomial in the equation 13 Interpretation of the Free Indices In a 3D space a vector 7 has three components V1 V2 and V3 and can be written in any of the following forms 7 Vial VZEZ V3E3 VszVs 119 139 123 The free index i ranging from 1 to 3 indicates that V1 has three components Examples of vector include velocities forces etc There are also many physical quantities that have less or more than three components in a 3D space For instance the direction cosines defined previously can be written in the following array form a11 a12 a13 cosX1X cosX1X z cosX1X 3 a1 a21 a22 a23 cosXZX cosXZX39Z cosXZX393 120 32 a33 cosX3X cosX3X39z cosX3X 11 11 31 Thus a whidi has two free indices i and j both ranging from 1 to 3 represents a quantity with nine components including every permutation of i 1 2 3 and 1 2 3 In the form of 120 the first index represents the row number and the second index represents the column number in the array Similarly the expression R lmn 123 lmn implies that Rm has 27 components in a 3D space including every permutation of the three free indices l m and n It is noted that in a 2D space V a and R 1 17 m with each index ranging from 1 to 2 represent quantities with two four and eight components respectively The free indices need not appear with only one quantity in a monomial For instance the expression AlBZ has two free indices associated with two quantities A and B respectively Accordingly AlBZ has nine components and can be written in the following array form A131 A132 A133 A137 A231 AZB2 A233 121 A331 A332 A333 TM Tan Drexel University 1 3 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors Consider next the quantity Ctzk There are two free indices 139 and k and one dummy index Thus C17 D77 has nine components including all permutations of 139 1 2 3 and k 1 2 3 ead component is the sum of three terms found by summing over the dummy indexj Written in an array form they are 3 3 3 2C17D71 2C17D72 2C17D73 C17D71 CMD72 CUD73 71 71 71 C713 C271371 3271372 3271373 gem371 ngD 2127ng C37D71 C37D72 C37D73 393 393 393 122 2C31D11 2C31D12 2C31D13 39 71 71 71 C11D11 C1ZDZ1 C13D31 C11D1Z C1ZDZZ C13D3Z C11D13 C1ZD23 C13D33 CZ1D11 CZZDZ1 CZBD31 CZ1D12 CZZDZZ CZBD3Z CZ1D13 CZZDZ3 CZBD33 C31D11 C3ZDZ1 C33D31 C31D1Z C3ZDZZ C33D3Z C31D13 C3ZDZ3 C33D33 14 The Kronecker Delta Consider a 3D vector 7 whose components are denoted as V 139 1 2 3 in an X coordinate system and as W in an X coordinate system Since the length or the magnitude of a vector is constant regardless of which coordinate system is used we have 2 2 V V V12V22V32 V V Vl Z VZ Z V3 Z 123 Substituting 113 in 123 and rearranging terms we have V1Z V2Z V3Z 11121 11122 anVf p1 1 112 11sz 11321 1amp2 11323V3Z 2 11 121 12 122 13 123 Vivz 2 11 131 12 132 13 133 V1V3 124 2 21 131 22332 23333 V2V3 Comparing terms on both sides of 124 we have 121 1zz 123 1 3amp1 32 32 1 31 H 2H 3 1 11 121 12 22 13 23 0 11331 1z 32 113 33 0 21331 zz 3z z3 33 0 These six equations can be written in the following indexical form 11171117 1 51271127 1 51371137 1 k 123 125a 111an7 0 11171137 0 51271137 0 k 123 125b which can be further condensed to alknjk 1 for 139 139jk 123 126a 117117 0 foriij 139jk 123 126b If we introduce a quantity 5 called Kronecker delta defined as 1 if 1 739 51 127 7 0 1f 1 it 1 4 September 23 2007 TM Tan Drexel University THEORY OF ELASTI CI TY 1 Cartesian Tensors then 126 can be written as wag 517 ijk 123 128 In a matrix form the Kronecker delta 51 is actually a 3x3 identity matrix ie 511 512 513 1 0 0 517 521 522 523 0 1 0 129 531 532 533 0 0 1 Recall 116 V 21711 ij123 116 which relates the components of a vector 7 from the X1 coordinate system to the X coordinate system Multiplying both sides of the equation by HM and using the relation in 128 we have qu tzkpsz1 5616 lel 5kZVZ 5k3V3 k 123 130 For k 1 we can show that 511Vl 6le2 513V3 V1 0 0 V1 Similarly for k 2 and k 3 we have 521V V2 and 531V V3 Thus 130 becomes 639th Vk ik 123 131 Equation 131 illustrates the use of the Kronecker delta 617 that is when one of the two indices of 517 is a dummy index 517 can be eliminated from the expression by changing that dummy index to the other index in the entire monomial Substituting 131 in 130 yields HMV Vk ik 123 132 or by changing the indices V7 12 V ij123 133 1 1 This is the inverse expression of 116 whidl relates the components of vector 7 from the X coordinate system to the X1 coordinate system By dqanging again the indices in 133 to V7 asz jk 123 134 k and substituting it in 116 we have V amvk 139 13k 123 135 Since both V and Vk represent components of 7 in the same coordinate system ie Xl system we conclude that V Vk if and only if j k Therefore 71 ifjk kiln 17 0 ifjik quotJ TM Tan Drexel University 1 5 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors aa6 ijk123 136 15 Coordinate Transformation Fig 12 A position vector in two Cartesian coordinate systems Consider two Cartesian coordinate systems X with base vectors 5 and X1 with base vectors 51 in a 3D space as shown in Fig 12 The position vector 2 or 5C that defines a point P can be expressed in terms of its components and the base vectors as 2 X xga 137 The relations between the 2 sets of components are given by1 x 5199 138 X a X 139 Taking the dot product of both sides of 137 with E we have XE E X9 6 xga x 140 in whid1 the orthonormal property of Cartesian coordinate systems ie E E 51 has been employed Comparing 140 with 138 we see that 1117 E e 141 Switd ing the indices and rearranging terms yield 1171 1 e 142 Thus once the base vectors of two Cartesian coordinate systems are known the direction cosines 1117 or 1171 can be obtained using 141 or 142 and the transformation of any quantities from one system to another can be established Since the dot product of any two vectors is equal to the product of their 1 Hereafter unless otherwise speci ed the ranges of all the indices are from 1 to 3 in 3D and from 1 to 2 in 2D TM Tan Drexel University 1 6 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors lengths times the cosine of the angle between them and since a base vector has unit length the fact that 141 or 142 results in direction cosines becomes obvious The necessary and sufficient conditions to insure that a transformation is reversible and has a onetoone correspondence in certain region V of the variables X are a The relation given in 138 and 139 are singlevalued continuous and possess continuous first partial derivatives in the region V and b The Jacobian determinant does not vanish at any point of the region V where 11 21 31 J 71 12 22 32 143 13 23 33 Coordinate transformations with the properties a and b described above are called admissible transformations If the Jacobian is positive everywhere then a righthand set of coordinates is transformed into another righthand set and the transformation is said to be proper If the Jacobian is negative everywhere a righthand set of coordinates is transformed into a lefthand set and the transformation is said to be improper 16 Scalars Vectors and Tensors Let X and X be two sets of Cartesian coordinate systems related by the transformation laws X a X 144 X aX 145 where a the direction cosine is defined as a cosX X 146 A physical quantity is called a scalar a vector or a tensor depending on how the components of the quantity are defined in the X and X coordinate systems and how they are transformed from one system to the other A physical quantity is called a scalar or a tensor of order zero if it has one component say 5 in the X system and one component in the X system and if 5 and are numerically equal at the corresponding points Thus if is a scalar then 11509 Xl 147 In terms of indexical notation a scalar has no free index since it has only one component and its value remains constant regardless of whidi reference coordinate system is used Examples of a scalar include the length of a vector temperature distribution energy etc A physical quantity is called a vector or a tensor of order one if it has three components say 5 i 123 in the X system and three components 5 i 1 2 3 in the X system and if the components follow the transformation laws 5 7 148 5 25 149 TM Tan Drexel University 1 7 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors It is easy to see that a tensor of order one has one and only one free index Examples of a firstorder tensor include displacements velocities forces etc A physical quantity is called a tensor oforder two if it has nine components say t ij 1 2 3 in 1 the X1 system and nine components tl z ij 12 3 in the X1 system and if the components follow the transformation laws t 11 11 t 150 r 17 mmmn t 11 11 t 151 int n mn Examples of a secondorder tensor whidi has two free indices include stresses and strains Similarly we can define an n order tensor as a quantity having 3 components in each of the X1 and X systems denoted as twyupu and tarp respectively and the components follow the transformation laws t 152 I 7 P1P2quot39P m w HauntingW 153 7 I trimquotPu Wham12 mutinyr where p and q i 1 2n are subscripts eadi ranging from 1 to 3 in a 3D space We shall show later in Chapter 4 that material properties such as Young s modulus and Poisson s ratio are fourthorder tensors 17 Vector Operations Using Tensor Notation Consider two vectors A and B in a Cartesian coordinate system The dot product of the two vectors is given by 213 AlBl AZBZA3B3 AlB 154 It is noted that there is no free index in the above equation hence the dot product of two vectors yields a scalar or a tensor of zero order Next consider the cross product of the two vectors 5 A X B 155 The components of C are given by C1 AZB3 7 A332 CZ A3B1 7 A133 C3 AlBZ 714231 156 In the indexical notations 156 can be written as C 8 17k AZBk 157 where 17k ijk 123 called alternating tensor is defined as 17k 1 if the numerals taken by any two of the subscripts 139 j and k are unequal and in a cyclic order of 123 ie 123 231 312 1 SW 71 if the numerals taken by any two of the subscripts 139 j and k are unequal and in a non cyclic order of 123 ie gm 213 132 71 TM Tan Drexel University 1 8 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors 17k 0 if the numerals taken by any two of the subscripts 139 j and k are equal ie all components except 123 231 312 321 213 and 8132 are equal to zero Take 139 1 in 157 as an example we have C1 511114131 gliZAlBZ 51134133 51214231 512214sz 512314233 51314331 51324332 51334333 O0OOOAZB3 O7ASBZO A233 71332 Components C2 and C3 can be easily obtained by letting i 2 and 139 3 respectively in 157 It is noted that the alternating tensor SM is a thirdorder tensor ie 7 7 7 I 17k all my nkglmn 17k all fm knglmn 158 One can also show that the Kronecker delta 61 is a secondorder tensor Finally the triple scalar product A X can be expressed as A X B E QM173 57ABC 571437 C 159 where according to the definition of the alternating tensor 17k 75M gm 18 Paltial Derivatives of Tensors Recall 138 X le 138 Differentiating both sides of 138 with respect to X yields OX a gamma 160 6X 7 6X 7 7 in which the following orthonormal property of Cartesian coordinates have been employed 1 if 139 m 5m I 6X7 0 if i m Similarly we can show that 6X7 1 61 a 6X 77 Now consider a scalar function X1 which remains constant in any coordinate system ie Xl X1 Differentiating the above equation with respect to X and using 161 we have 6 7 6 6 6X17 6 162 ax 7 ax 6X ax 7 77 ax By using the following indexical notation to represent the partial derivative of a quantity TM Tan Drexel University 1 9 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors we can express 162 in the following form V5 M1 163 Based on 163 whid resembles the transformation law for a firstorder tensor we conclude that d is a firstorder tensor with three components in a 3D space One may recall that in vector analysis the gradient of a scalar function r is indeed a vector ie v E EZ E3 3 164 Next consider the derivatives of a vector V The transformation law for vectors is given by Vi lzvl Differentiating the above equation with respect to X gives 6V av av aX av 1 t l lm aX 7 aX 7 ax aX 7 ax Wm amid11 165 ie the gradient of a firstorder tensor is a secondorder tensor Thus we conclude that taking the partial differentiation of a tensor yields a new tensor one order higher than that of the original one Other useful vector operations in terms of index notation include avl avZ av3 Divergence of a vector V V a scalar 166 6X1 6XZ 6X3 le t39 vve aVkev t 167 ur o avec or X 7817165781716 k a vec or 7 Laplace VZ VV 2 a scalar 168 19 Tensor Contraction and Quotient Rule The operation of equating two subscripts of a tensor and summing accordingly is known as contraction In general contraction of a tensor yields another tensor of order two less than that of the original one For instance given a secondorder tensor t we have 1 t amamtm A contraction operation by equating subscripts i and gives t a a t 5 t 7m in mn mn mn itmm where tW is a tensor of zero order or a scalar TM Tan Drexel University 1 10 September 23 2007 THEORY OF ELASTICITY 1 Cartesian Tensors Another useful rule in the tensor analysis is the socalled Quotient Rule The following example illustrates how the quotient rule can be used Let V1 be an arbitrary firstorder tensor and B jk be another arbitrary secondorder tensor If Bjk AijkVi 169 is true regardless of the reference coordinate system then Aijk is a thirdorder tensor The proof is as Z follows Since Bjk is a secondorder tensor and V is a firstorder tensor we have by using the tensor transformation laws and 169 BJ39k amjankan amjankApman amjankApmnaWVq39 170 Meanwhile since the relation given in 169 is true in the system also we have Bj k AZTJkVi39 171 Substituting 171 in 170 and rearranging terms we have Agljk apqamjankApmn V 6139 Z 0 Since Vq39 is an arbitrary vector it must be that Agljkza a a A 172 pg 111 nk pmn Equation 172 is precisely the transformation law for a thirdorder tensor Therefore AM is a third 11 order tensor It is a straightforward process to extend quotient rule to more general cases 110 The Theorem of Gauss Integral Theorem gtegte as 39 39 X3 Fig 13 A prism in region Vbounded by surface S Consider a tensor tijkm defined in a region V bounded by a surface S as shown in Fig 13 The Theorem of Gauss states that Tll Tan Drexel University 1 11 September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors jvtdv Isnmtkd8 173 where 11 is the unit normal vector along the exterior of S To prove this theorem we first let m 1 and carry out the integration in the X1 direction for a prism as shown in the figure at 7 17k 7 7 jvt1dV V 6X1 XmdXZdXS StdXZdX3 jstdxzdxg StkdX2dX3 174 Since dXZdX3 is the projection area of the prism on the X2 7 X3 plane we have dX dX 7 VIS cos E1 dSn in the7 X1 direction 2 3 7 d8 cos71MEl dSMn in the X1 direction Thus 174 becomes jvtdv j 51n ds j 51nds Istkn1ds Similar expressions can be obtained for m 2 and m 3 Together they form the Gauss Theorem 173 The Gauss Theorem can be used to convert a volume integral to a surface integral or vice versa When applied to firstorder tensors vectors the Theorem of Gauss becomes Divergence Theorem ie VtdV Sands 175 or in vector form vatdvjSt ds 176 PROBLEMS 11 Given the components of firstorder tensors A B and C as 4110 142715 1436 B1 3 B2 4 33 1 C116 C212 C33 Evaluate for i jk l 2 3 31 A1933 b AlBlAz C A B C d Cszcz e AlBlchk 117 12 Given the components of firstorder tensors A and C and a secondorder tensor 3 as 412 423 134 B11 0 B12 2 B13 72 321 3 322 1 323 1 B31 6 332 73 B33 71 C1 CZ3 C38 Evaluate for ijk 123 a A133 b 371C c ABMCW d 36 e A1C6Bm f Aqumak g AiB m 6m 13 Given two Cartesian coordinate systems defined by the following two sets of base vectors respectively 2165 715 0 EZO 01 E34715 715 0 TM Tan Drexel University 1 12 September 23 2007 THEORY OF ELASTICITY 11 11 0 1 N 1 Curtes inn Tensors 0 51 0 0 Eg0 a Determine the transformation matrix 1117 a a a a b Is the transformation proper or improper c If E is a vector whose components in the E system are 50 30 0 find the components of this vector with respect to the E system Write the following expressions in indexical notations a AxBCD b v2v2 c VZ 171B Show that a secondorder tensor can be expressed as the sum of a symmetric secondorder tensor and an antisymmetric secondorder tensor ie t1 tf Hf where tfzs tgf and tsz 76 Show that A173 2 0 if A1 is a symmetric secondorder tensor and B1 is an antisymmetric second order tensor Show that 8k14k is an antisymmetric tensor where 8k is the alternating tensor and 14k is a firstorder tensor ie a vector Prove the following identities in whid r 517 and 8 W are the Kronecker delta and the alternating tensor respectively a 17 17 b glzkAZAk 0 c 873 0 if B is symmetric ie B Bh d glfkglmn aym m ayn km Hint show that each side of the above equation 1if jm kn unless jk 71 if j71 km unless jk Oif jk or m71 e swam 6 Hint Use the identity in d Use the identit in 18 d to show that Ax x 216 42136 V If 7 3x1E1 2x 210x E3 compute a b sz c VW Check the divergence theorem by using the vector T xlzszl 3x1x3EZ for a cubic domain Venclosed by x1 0 x2 0 x3 0 x1 1 x2 1 and x3 1 Express the following integral theorems in terms of tensor notations in which V S and L represent volume surface and line respectively a J39VVWZVJ39S FWZS b va dvj 1ads c jpwv pms d J39S Vx liSjgdi e IVVZWZVIS V 7ZS TM Tan Drexel University September 23 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors APPENDIX 1A The transformations of first and secondorder tensors can be conveniently expressed in the matrix form Recall 120 the direction cosines can be expressed as 11 12 13 l 21 22 23 1A1 31 32 33 Now consider the transformation of the firstorder tensors as given in 150 and 151 Let 6121 52 63 1A2 6 ft fl EST 1A3 then 148 and 149 can be written as 6 21 1A4 6 ME 1A5 respectively For the transformation of secondorder tensors let t t21 t22 t23 1A6 t39 t 21 t 22 H23 1A7 then 150 and 151 can be written as V W tll l 1A8 t a M W 1A9 respectively TM Tan Drexel University 1 14 September 23 2007 THEORY OF ELASTICITY 2 Analysis of Stress Chapter 2 Analysis of Stress In particle mechanics we study two types of interaction between particles ie those by collision and those by action at a distance and the manner in which one particle is being influenced by all the others must be specified exactly In continuum mechanics in addition to the aforementioned interactions we also have to consider the interaction between one part of the body and another However since even the smallest volume contains a very large number of particles it would be futile to approach the interaction problem using the particle concept The new concept used in the study of continuum mechanics is stress which is a unique way that continuum mechanics uses to describe the interaction between one part of a material body and another 21 Body Forces Surface Forces and Stress Vectors X3 P m Fig 21 Forces on an infinitesimal surface element passing through a point in a deformable body In continuum mechanics forces in general can be grouped into two categories one for the body forces denoted by 1 and the other for the surface forces denoted by f Body forces are associated with the mass of the body They are usually distributed throughout the entire body and are quantified by forces per unit volume Examples of body forces include the gravitational force the magnetic force and the inertia force Surface forces which are quantified by forces per unit area result from physical contact between two bodies or more subtly they may represent the force that an imaginary surface within a body exerts on the adjacent surface Consider an infinitesimal surface element AS on an imaginary plane S passing through a certain point in a deformable body as shown in Fig 21 Let 171 be the unit outer vector perpendicular to the surface element and AF be the total force on AS exerted from the adjacent surface element with a gt acting on this surface element is given we I negative normal vector 17z The average force per unit area t by 1 AT tF 21 we Tll Tan Drexel University 2 1 September 26 2010 THEORY OF ELASTI CI TY 2 Analysis ofStmss Assuming that as AS tends to zero Em tends to a definite limit dIEdS and the resultant moment due to A13 about any point vanishes then AF VIP 22 where 2a is called the stress vector or sur zce traction representing force per unit area acting on the surface element d8 with unit normal 71 On the surface opposite to AS the unit normal is 71 and the stress vector is denoted by EM In view of equilibrium it is easy to see that 77 72H 22 Stress Tens ors 21 tfo 77 E1 Fig 22 Stress vector on an X1 surface The magnitude and direction of a stress vector as defined in 22 depend on the direction of the surface element it is acting on Since there are infinite numbers of surfaces that can pass through a point the state of stress at that point therefore cannot be uniquely defined by using the stress vector itself Consider an infinitesimal surface element AS perpendicular to the X1 axis as shown in Fig 22 Since the unit normal 71 to this surface element is the same as the base vector El the stress vector on this surface element therefore can be denoted as E It is noted that the direction of F in general may not coincide with that of El Denoting the components of E in the X1 X2 and X3 directions respec tively as fl 011 ff 012 t 03913 23 then the stress vector 6 can be expressed as F90 011E 0 1252 alga 24 Similarly the stress vectors acting on the two surface elements passing through the same point and perpendicular to the X2 and X3 axes respectively can be expressed as 52 021 022E 023 25 53 031 032E 0353 26 TM Tan Drexel University 2 2 September 26 2010 THEORY OF ELASTI CI TY 2 Analysis of Stress 762 7E2 X3 Fig 23 Forces on the surfaces of an infinitesimal tetrahedron Now consider an oblique surface element of unit normal 71 passing through the same point in the body This oblique surface element and those that are perpendicular to the three coordinate axes but each with unit normal pointing in the negative coordinate direction form an infinitesimal tetrahedron as shown in Fig 23 If the body is in a state of equilibrium so must be the infinitesimal tetrahedron Consequently by summingup all forces acting on the surfaces of the tetrahedron we have 213t APQRtH1gtAOQRt E2AOPRt E3AOPQO 27 where APQR etc are the triangular areas formed by vertices O P Q and R etc Since AOQR AOPR and AOPQ are simply the projections of APQR on the surfaces perpendicular to X1 X2 and X3 axes respectively ie AOQR APQR cos e1 APQR 411 etc we have AOQR 7 n AOPR AOPQ 7 71 APQR 1 APQR Z APQR 3 where 111139 123 are the components of 71 Substituting these relations in 27 and observing that tH it etc we have M tg n1tgznz t n3 28a Substituting further 24 25 and 26 in 28 and rearranging terms we have EU 011711 021712 U31n3lg1 012711 Uzznz Usznslgz 013711 Uzsnz Ussnslgs 28b Comparing 28a and 28b yields the following expressions for the components of the stress vector to till 711111 721112 731113 29 t 712111 722112 44732113 210 t 713111 723112 733113 211 or in terms of the indexical notation 77 7 39 392 t1 7 771117 1 7 1 2 3 212 The quantity 03917 in 212 is called the stress tensor Once the values of the components of 03917 at a point are given the stress vector on any surface element passing through this point can be obtained using 212 Thus the state of stress is uniquely defined by 03917 It is noted that in defining the stress tensor the TM Tan Drexel University 2 3 September 26 2010 THEORY OF ELASTI CI TY 2 Analysis of Stress rst subscript denotes the direction of the axis to which the surface element is perpendiculur and the second subscript denotes the direction of the component For example 012 is the component of the stress tensor in the X2 direction acting on the surface perpendicular to the X1 axis Since both tf and 711 are firstorder tensors it is easy to show by using the Quotient Rule that the stress tensor is a secondorder tensor that obeys the transformation laws 0391 umumo m and 0391 umumayln A more detailed discussion on the transformation of stress tensors is given in Appendix 2 at the end of this dqapter 23 Equations of Equilibrium Fig 24 A deformable body subjected to surface tractions and body forces Consider a deformable body of volume Vbounded by surface S and subjected to surface tractions tf and body forces bl as shown in Fig 24 If the body is in a state of equilibrium then the sum of all the forces acting on the body must vanish ie 2F J39Vb1dVIstf gtdS0 213 Substituting 212 in 213 and using the Theorem of Gauss 175 we have 25 J39VbldVJrISo nzdSJ39VbldVJrIVo yde Mam bdV 0 Since this equation is true for any arbitrary volume V ie if the body is in a state of equilibrium so must be any subportion of the body the integrant itself must vanish Thus a b 0 214 In the unabridged notations 214 is given by 0111 0212 0313 b1 0 215a 0121 022 0323 b2 0 215b 0131 0232 0333 173 0 215c Furthermore the resultant moments due to ti and h must vanish as well if the body is in a state of equilibrium Thus summing up all the moments due to t1 and b about the origin we have ZMJ gtltthJS gtltt dSO 216 TM Tan Drexel University 2 4 September 26 2010 THEORY OF ELASTI CI TY 2 Analysis of Stress in whidl X1 as shown in Fig 24 are position vectors In the indexical notation 216 becomes 2M V glszzlzkalVJ39Sglsz7tgHS V 7291dejsgXc nmds 217 V 57X7bkdvjvgkx 039 de 0 7 where the Theorem of Gauss has been employed Carrying out the partial differentiation in the second integral of 217 gives IV 1799 amkymalV IV 17kX7 mamk 17kX7 a ka alV IV 17k 39mo mk 17kX7 a ka alV IV glzkazk glzkxz 0 mka dV Substituting this results back into 217 and rearranging terms we have 2M jvgx am b 07ker 0 Since a ka bk 0 and the volume taken in the above equation is arbitrary we conclude that 570 0 218 07k 7k7 219 Thus the stress tensors are symmetric second order tensors In the unabridged notations 219 becomes 03912 021 023 032 03931 03913 220 Equations 214 and 218 are called the equations of equilibrium 24 Principal Stresses and Stress Invariants Recall 212 a 7 7 t1 7 awn 1 7 1 2 3 212 The direction of a stress vector to varies from one surface to another and in general does not coincide with that of the unit normal 71 to the surface One may therefore ask if there are particular surfaces on whidi the direction of stress vector coincides with that of the unit normal If such surfaces do exist then the stress vector and unit normal on these surfaces would differ only by a scalar ie W a 221 or in the indexical notations if am 222 where it is the scalar representing the magnitude of the stress vector on this particular surface Sudi a surface is called a principal surface its normal a principal direction and the magnitude of the stress vector it a principal stress If a is the unit normal to a principal surface then from 212 and 222 we have TM Tan Drexel University 2 5 September 26 2010 THEORY OF ELASTI CI TY 2 Analysis ofStmss t9 771117 in 223 By using the Kronecker delta 51 see Chapter 1 we can write the above equation as 0 7 2571 0 224 Equation 224 consisting of three homogeneous equations for the four unknowns it and n1 represents a typical eigenproblem in which the eigenvalues it are the principal stresses and the eigenvectors n are the corresponding principal directions For a nontrivial solution of 224 to exist the determinant of its coefficient matrix must vanish ie la 7 25 0 225 Solving 225 yields the three principal stresses denoted as 0391 0392 and 0393 with 0391 2 0392 2 0393 whid can then be substituted in 224 to determine the corresponding principal directions Upon expanding 225 we have 03911 A 03921 03931 03912 03922 7 it 03932 0 226 03913 03923 03933 A 3 2 A 011 03922 03933 1 01102 022033 033011 012021 023032 031013 1 011022033 021032013 031012023 013022031 7 012021033 011023032 0 227 which can be written in the following compact form 237112212271f0 228 where the coefficients I1 03911 03922 03933 03911 229 12 a a an 032 a 0 laa 70101 230 03912 03922 72s 03933 03931 03911 2 77 7 7 03911 03921 03931 1 I3 03912 03922 03932 swamawayyak7 231 6 03913 03923 03933 are called the first second and third stress invariants respectively It is noted that for a given state of stress at a point the principal stresses must remain the same regardless of which coordinate system is being used or mathematically the roots of 228 must remain the same regardless of whid coordinate system is being used Consequently ll 12 and I3 whid are the coefficients of 228 must remain constant with respect to any coordinate transformation hence are called invariants Furthermore knowing that the stress tensor is symmetric one can conclude that 228 must have three real roots eigenvalues and the three corresponding principal directions eigenvectors must be mutually orthogonal If we set up a coordinate system where the X1 X2 and X3 axes coincide with the principal directions then from 223 we have TM Tan Drexel University 2 6 September 26 2010 2 Analysis ofStmss THEORY OF ELASTICITY a 7 0 if 139 i j 17 A if i j or in the matrix form 0391 0 0 0391 0 0392 0 232 0 0 0393 where 0391 0392 and 0393 are the three principal stresses ln sud a case 226 becomes 17 nix17 0392 70393 0 and the stress invariants can be expressed in terms of the principal stresses as I1 0391 039Z 0393 233a lZ 0102 0203 7301 233b 233c 13 0391 72 73 25 Normal Stress and Shearing Stress X2 Fig 25 Normal and shearing stresses on an infinitesimal surface element The normal stress on a surface element with unit normal 71 is the projection of the stress vector in the direction of 71 as shown in Fig 25 Denoting the magnitude of normal stress by a we have if Em 1 tf nl 071117111 234 If the coordinate axes coincide with the principal directions then the stress tensor is given by 232 and the normal stress can be expressed in terms of the principal stresses as a 711112 7211 731132 235 71 To find the extreme values of the normal stress and the corresponding directions we maximize the function in 234 subjected to the following constraint condition m1 nfn n i 236 Using the method of Lagrange multipliers this becomes finding the extreme values of a function lt1gt 0711171117 20111 7 1 237 whid leads to the following three equations in it and n1 ie 2 7 September 26 2010 TM Tan Drexel University THEORY OF ELASTICITY 2 Analysis 0 Stness ML 0n0n1 ma alumen ail 0 7 m 1 03971 1 m 1 1 m 039715 1110115 121115 039 111039 n 121115 7W1 717m 11m mil 7m 11m 201m 351mlnl 0 for m 1 2 3 Since this equation is identical to 224 for determining the principal stresses and principal directions we conclude that the maximum normal stresses are in fact the principal stresses The shearing stress denoted by Ufa is the projection of the stress vector on the surface element see Fig 25 The magnitude of the shearing stress can be obtained by a 4 2 UW wWZ 1202 41214091 238 Again if the coordinate axes are set to coincide with the principal directions then gm tl Ulnl t Uznz t 03713 Substituting the above equation and 235 in 238 and using the constraint condition of 236 we can obtain the following expression for the shearing stress in terms of the principal stresses A Z Z Z Z Z Z Z Z Z of 4711712071 0392 2713 72 73 n3n1 73 71 239 Since on a surface element normal to one of the principal directions the unit normal 11 is 100 010 or 001 it is clear from 239 that on sud a surface Ufa 0 Thus on the principal surfaces the values of shearing stresses are zero To determine the maximum values of Ufa and their directions we maximize the function Z 2 2 2 2 2 2 2 2 2 lm 1712 71 72 n2n3 72 7 73 n3n1 73 01 subjected to the constraint condition of 236 Again by using the method of Lagrange multipliers this becomes that of solving the following extremevalue problem animlaf lz Mn h 1 0 m 1 2 3 The results without getting into the details of mathematical derivation are shown in Table 21 below Table 21 Extreme Values of if and Ufa n1 n2 n3 Ufa meme Us 11 0 0 0 Io39ll 0 11 0 0 Iazl 0 0 11 0 Iasl 0 1422 1422 laziaslZ I0210393I2 1422 0 1422 agial2 agal2 142 2 1422 0 logical2 39011039ZI2 If 0391 2 0392 2 0393 then the absolute maximum shearing stress is 0391 1 0393 2 acting on the surface with unit normal 11 1422 01422 TM Tan Drexel University 2 8 September 26 2010 THEORY OF ELASTICITY 2 Analysis of Stress 26 Mohr39s Circle The results of the previous section can be further illustrated by constructing the Mohr s circles proposed by O Mohr in 1882 By setting up a coordinate system such that the axes coincide with the principal directions we can rewrite 238 in the following form 2 a 2 052 602 012n12022n 0 n 240 11 Solving 240 together with 235 and 236 we obtain 0e 2 age 7 X0 c7 quot12 t cl azxaf a 3 13 n 2 day 01213 03 X01313 01 24110 72 0302 01 n2 0912 057 0147571 02 241c 73 0103 02 Since 01 2 72 2 03 and 1112 2 0 we conclude from 241 that 2 0 C71 72 X0 03 2 0 242 243 244 Fig 26 The Mohr s circle for a threedimensional state of stress We now construct a of cf space and plot in the Cartesian plane the values of of as abscissas and those of of as ordinates for the following three equations of 2 of 72 X0 03 2 0 245 of 2 of 73 X0 71 0 246 of 2 of 71 X0 72 2 0 247 Tll Tan Drexel University 2 9 September 26 2010 THEORY OF ELASTI CI TY 2 Analysis ofStmss This results in three circles C1 C2 and C3 with radii equal to 0392 7 0393 2 0391 7 0393 2 and 0391 7 03922 and centers located at 0203932 0391 03 2 and 0391 az2 respectively as shown in Fig 26 It follows from the inequalities 242 243 and 244 that the admissible values of if and 7 lie in the shaded region bounded by the circles C1 C2 and C3 The maximum shearing stress is clearly represented by the greatest ordinate OZQ of circle CZ Hence A 039 7 039 cl Wquotax 1 3 246 2 To determine the direction of the surface element on which the shearing stress is a maximum we first find the normal stress 039 on the surface whid is the abscissa OOZ in Fig 26 ie 710393 a 247 Substituting 246 and 247 in 241 yields n i1J2 01JE whose components coincide with those given in Table 21 PROBLEMS 21 The state of stress at a point in a deformed body with respect to the coordinate system E1 100 E2 010 E3 001 is given by 1 2 0 0391 2 4 0 psi 0 0 3 a Find the stress vector f on the plane with a unit normal 1 1J3 113 b Find the normal stress and shearing stress on the plane given in a c Find the components of the stress tensor with respect to the a new coordinate system defined by the following base vectors E1JE1J 0 Eg1JE1J 0 Eg001 d Find the components of stress vector P with respect to the new coordinate system on the same plane as given in a e Compare llllland quotI ll f Show that 11 12 and I3 are invariants in E and E system g Find the principal stresses and their directions 22 ln the Medlanics of Materials course you have learned that the stresses due to bending in a beam of circular crosssection are My VRZf 2 WT aw Where Mx and Vx are the bending moment and the shear force respectively R is the radius of ayazaxzayz0 the crosssection and I 7rR44 is the moment of inertia a Do these stresses satisfy the equations of equilibrium Why b Do these stresses satisfy the boundary conditions in general Why TM Tan Drexel University 2 10 September 26 2010 THEORY OF ELASTI CI TY 2 Analysis ofStmss 23 Is the following stress distribution possible for a body in equilibrium if the body forces are absent A B and C are arbitrary constants 7Ax1x2 ABz7x 2 7sz 03917 ABz7x 2 0 0 7 sz 0 0 24 Given the following state of stress 3x123x 7x3 x376x1x2734 x1x2732 03917 x3 7 6393cZ 7 34 3x x1x2732 0 3x1x27x354 a Show that it is in equilibrium if the body forces are absent b Determine the principal stresses at point 12 1 34 25 A rectangular flat panel shown in the figure is subjected to certain surface tractions not shown along its edges which produce the following state of stress 2x1xZ 7 x 0 0391 7x 0 0 psi 0 0 0 E Show that the panel is in equilibrium Find the tractions applied along the edges of the panel that produce this state of stress Find the normal and shearing stress components of the stress vector on the surface A i A at point 210 as shown in the figure 3 X2 2 39A 71 30a Problem 25 26 A 2D parallelogramshaped panel is subject to a uniformly distributed shearing stress along its four edges as shown in the figure Use the method of Mohr s circle to determine the principal stresses and principal directions 14140 psi 45 14140 psi 14140 psi A 45 14140 psi Problem 26 TM Tan Drexel University 2 11 September 26 2010 THEORY OF ELASTICITY 2 Analysis Distress 27 An octahedral plane is the plane that makes the same angles with all three principal directions By setting up a coordinate system sudi that its three axes coincide with the three principal directions show that the octahedral shearing stress the resultant shearing stress on the octahedral plane denoted by 10 can be expressed in terms of the principal stresses in the following form The secondorder tensor 1 151 3 7 is called the stress deviation tensor where 11 039 is the first invariant of stress tensor The principal 03917 03917 values of stress deviation tensor can be obtained using the same procedure as outlined in Section 24 by solving the following eigenvalue problem la 7 25 0 Expanding the above equation and rearranging terrns yield Ag li zi JZA h 0 where J1 12 and 3 are called the first second and third invariants of stress deviation tensor respectively Show that the second invariant of stress deviation tensor 12 can be expressed in terms of stress invariants or octahedral shearing stress as 2 lz 12 1 1370Z 3 2 TM Tan Drexel University September 26 2010 THEORY OF ELASTI CI TY 2 Analysis ofStmss Appendix 2 TwoDimensional 2D State of Stress The stress tensor 03917 being a secondorder tensor obeys the transformation law 0391 11mm 0 2A1 In a 2D case the direction cosines can be expressed as all cosX1 X cos6 2Al2a 012 cosX1XEcos7r267sin6 2A2b 021 cosXZ Xi cos7r27 6 sin 6 2A2c 022 cosXZX Z cos6 2A2d where 6 is the angle measured counterclockwise from X1 to X1 axes Expanding 2A1 and using 2A2a 7 2Al2d we have 0391 1 03911 cosz603922 sin26203912 sin6cos6 2A3a 0E2 03911 sin2 6 03922 cos2 67 203912 sin 6cos6 2A3b 0391 Z 03922 7 03911sin6cos6 0391ZcosZ 67 sin2 6 2A3C By using the following trigonometric identities sin2625in6cos6 cosZ6cosZ67sinZ6 CO5261c526 sin2617c526 we can express 2A3a 7 2A3c as 0391 1 M Mcos 26 0 12 sin 26 2A4a 2 o39gz Wf39n 7 cos 26 7 a12 sin 26 2A4b 0391 Z sm 26 03912 cos 26 2A4c The principal stresses 0391 and 039Z can be obtained by solving the following eigenvalue problem 03911 339 03912 an an 7 A AZ 011 03922 1 011022 0122 0 2A5 which yields 039 039 039 7 039 2 A 0 1ch 11 2 22 i 11 2 22 0 52 2A6 and the corresponding principal directions are given by 1 Z n 039 7 039 039 7 039 21 tan 6 7 22 22 1 2A7 n1 203912 203912 TM Tan Drexel University 2 13 September 26 2010 THEORY OF ELASTI CI TY 2 Analysis ofStmss 22 7t 6371 61 7 tan 2A 8 74 37 an F 7 an p 7 CO V Or tan 261 tan 26 tan 7 26 2A9 p 03911 03922 The maximum shearing stress is given by 039 7 039 Z lrmaxl 0 52 2A10 The Mohr39s circle for a 2D case in an X1 X2 system is given by 245 where the center of the circle is located at 03911 022 01 t 02 2A11 01106 2 2 on the 039 axis and the radius is given by 0391 7 0392 2 or Z R 052 2A12 TM Tan Drexel University 2 14 September 26 2010 THEORY OF ELASTI CI TY Solution for Chugter 2 21 The state of stress at a point in a deformed body with respect to the coordinate system El 1 0 0 E2 01 0 E3 0 0 1 is given by 1 2 0 0391 2 4 0 psi 0 0 3 a Find the stress vector f on the plane with a unit normal 7 1 f3 1 5 1 1 2 0 F 1 3 1 tan 2 4 0 7 6 z 0 0 3 f 3 1 b Find the normal stress and shearing stress on the plane given in a 7 an tlnl tT nJ31 2 1 4psi c Find the components of the stress tensor with respect to the a new coordinate system defined by the following base vectors E1J 1J 0 Eg1JE11 0 Eg001 717 717 0 45 715 0 1117 E1 E7 12 7L2 0 0391 117111170 11F 03911 715 05 0 psi 0 0 1 0 0 3 d Find the components of stress vector P with respect to the new coordinate system on the same plane as given in a 243 9N3 Need to find 111 first 117 alznz 0 then t o 1 7n7 73 5 1J3 3N3 Alternatively you can use t gift to obtain the same result Try it Compare and Both yield IE f Show that 11 12 and I3 are invariants in E and E system Both yield I1 812 15 and I3 0 e V V Find the principal stresses and their directions Knowing 03933 3psi is a principal stress we may simplify the problem to that of a two dimensions Thus 17 2 i 2 4 7 J g 75AAZ 0 yields A 0 5 The principal stresses are 5 3 and 0 psi 175 2 n1 0 a 1 5 For 715solv1ng 2 4 5 n 0 y1elds n ZJg 7 Z 0 TM Tan Drexel University 1 October 17 2007 THEORY OF ELASTI CI TY Solution for Chugter 2 7 2 5 F 701391 02 1170391d 71g ora37 sov1ng 2 470 Hz 7 0 y1e S117 0 22 In the Mechanics of Materials course you have learned that the stresses due to bending in a beam of circular crosssection are M R2 7 Z Uy 090555950 where Mx and Vx are the bending moment and the shear force respectively R is the radius of the crosssection and I 7rR44 is the moment of inertia a Do these stresses satisfy the equations of equilibrium Why Substituting the stress components in the equilibrium equations yield a amp zj zjzgj Note aa w at 6x 0y 6x I 3 I 3 M 6x 0y 6x 31 It is obvious that the equilibrium is satisfied only if the shear force V vanishes ie if the beam is subjected to a loading that produces a pure bending situation Do these stresses satisfy the boundary conditions in general Why It s easy to show that the stress components in general do not satisfy the boundary condition t1 017117 This will be discussed in more details in Chapter 5 b V 23 Is the following stress distribution possible for a body in equilibrium if the body forces are absent A B and C are arbitrary constants 7143cle ABzix 2 7sz 03917 ABzix 2 0 0 7 CxZ 0 0 Substituting the stress components in the equilibrium equations yield 011 6012 6013 7Ax iAx 72Ax 6x1 6x2 6x3 2 Z Z 6012 6022 6023 0 6xi 6x2 6x3 6013 6023 6033 0 6x1 6x2 6x3 Thus to satisfy the equilibrium A must vanish and B and C can be arbitrary constants 24 Given the following state of stress 3x123x 7x3 x376x1x2734 x1x2732 03917 x3 76x1x2734 3x x1x2732 0 3x1x27x354 a Show that it is in equilibrium if the body forces are absent TM Tan Drexel University 2 October 17 2007 THEORY OF ELASTI CI TY Solution for Chugter 2 6011 I 6012 6013 6x176x100 6x1 6x2 6x3 6012 6022 6023 776x2 6x20 0 6x1 6x2 6x3 LU a 23 a 33 1071 0 6x1 6x2 3 b Determine the principal stresses at point 12 1 34 3 73 0 The state of stress at 12 1 34 is given by 07 73 3 0 Solving the 0 eigenvalue problem we get 01 6 0392 3 and 0393 0 25 A rectangular flat panel shown in the figure is subjected to certain surface tractions not shown along its edges which produce the following state of stress 2x1xZ 7x 0 X 2 A 03917 7x 0 0 psi 0 0 0 30 a Show that the panel is in equilibrium gt21 0 12111 aw 0 for 139 123 All three are satisfied A I X1 b Find the tractions applied along the edges of Em the panel that produce this state of stress F 8 in 1rst s1mpllfy the problem to that of a 2D Left edge X10 f171 0 If 7 0 ixg 71 7 0 I lial7n7l a07 7x 0 0 7 xg PSI 7 2 1 16 Right edge X1 8 1 0 t 03917117 8 16 H x2psi 1 7x 0 0 ix 7 4x1 74 0 7 74 I 42 74 0 1 0 P l file 4 39 PSI 74 0 71 0 Problem 25 Top edge X22 f10 1 501an Bottom edge X2 7271 0 71 t1 037117 c Find the normal and shearing stress components of the stress vector on the surface AiA at point 210 as shown in the figure The unit normal to surface A A is given by 1 7 sin 30 c0530 7 05 0866 4 71 705 72866 I Thus39 5 Wig 71 0 H0866 l 05 lPS39 and a tn 18661151 a ltt 703 2232psi TM Tan Drexel University 3 October 17 2007 THEORY OF ELASTI CI TY Solution for Chugter 2 26 A 2D parallelogramshaped panel is subject to a uniformly distributed shearing stress along its four edges as shown in the figure Use the method of Mohr s circle to determine the principal 14140 psi 14140 psi 14140 psi stresses and principal directions p 14140 P51 Problem 26 0 14140 34140 0 P 0 5860 0 0 14140 Refer to Figure in Problem 26 we first locate the two points A and B on the Mohr circle plane that represent the stresses on surfaces A and B respectively ie 0714140 and 014140 Since the angle measured from surface A to surface B is 45 clockwise they must be separated by a 90 angle on the Mohr circle as shown in figure above Therefore the center has to be located at point C 7141400 and the radius is 20000 The principal stresses are located at P1 5860 0 and P2 7341400 The corresponding principal directions are 225 clockwise from surface A and 6750 counterclockwise from surface A respectively Skip Problems 27 and 28 TM Tan Drexel University 4 October 17 2007 THEORY OF ELASTI CI TY 1 Cartesian Tensors Chapter Cartesian Tensurs 11 Vectors Fig 11 A vector in twodimensional Cartesian coordinate systems Consider a vector 7 in a twodimensional 2D Cartesian coordinate system X1 7X2 with base vectors E1 and EZ as shown in Fig 11 Vector 7 can be expressed in terms of its components V1 and V2 in the X1 and X2 directions respectively as VV1E1 VZEZ 11 or simply VV1Vz 12 With respect to a new coordinate system X iX Z whidi has base vectors 61 and E and is obtained by rotating the X1 7X2 coordinates counterclockwise at angle 6 Fig 11 the components of 7 or 739 are V1 and V239 and the vector can be expressed in terms of these new components as V V514 V5 13 V v39vz 14 From Fig 11 it is easy to see that the relations between the components of 7 or 7 in the original unprimed and the new primed systems are given by V V1 cosX1 X VZ cosXZ X V1 cos 6 VZ sin 6 15a VZ V1 cosX1 x VZ cosXZ X Z 7V1 sin VZ c056 151 where cosX1X is the cosine of the angle between the X1 and X axes etc Defining a11 cosX1 X cos 6 16 H12 cosX1Xgcos 26isin6 17 H21 cosXZX cos7r276sin6 18 H22 cosXZX c056 19 TM Tan Drexel University 1 1 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors then 15 can be written as V anv1 amVZ 110a VZ ale1 aZZVZ 11013 Quantities an an a21 and an defined in 16 19 are called the direction cosines Equations 11 110 can be easily extended to vectors in three dimensions 3D ie 7V1E1 VZEZV3E3 V1VZV3 111 in X coordinates 1739 Vie VZ39E V3 Eg Vl39VZ V3 112 in X39 coordinates and the relations between the primed and the unprimed components are Vl allV1 asz a31V3 113a VZ ale1 aZZVZ pigzv3 1131 V3 a13V1 a23VZ a33V3 113c 12 The Summation Convention Equation 113 can be written in the following condensed form 3 3 3 Vi Z tlvw Vi 25sz Vi Z tsvt 114 11 11 11 which may be further consolidated into 3 V Zaqu j 123 115 It is noted that 115 represents three equations one for each value of the subscript j and within ead equation the summation extends over the range of the repeated subscript i This equation can be further simplified by introducing a summation convention whereby the symbol 2 is redundant as follows Ifa repeated alphabetic subscript appears in one monomial an automatic summation over the range of that subscript is required By using this convention we can rewrite 115 in the following form V m 13123 116 It should be noted that the ranges of both subscripts i and j must be specified in the equation 11 w A repeated subscript a subscript that appears twice in a monomial such as i in the above equation is called a dummy index whereas a nonrepeated subscript a subscript that appears only once in a monomial sudl as j in the above equation is called a free index Since a dummy index merely indicates summation over its range it is immaterial which symbol or letter is used so long as the ranges of the symbols are the same Hence 116 can be written as VZ39aqVk jk123 117 The symbol or letter used for the free index is also arbitrary so long as the same is used in every monomial Therefore 117 or 116 can be written as TM Tan Drexel University 1 2 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors W alek kl 123 118 The rules for an equation written in the indexical or subscript form such as those of 116 7 118 can be summarized as follows I An index can only appear either once a free index or twice a dummy index in a monomial An index that appears more than twice in a monomial is not allowed I A repeated dummy index in a monomial automatically requires a summation over its range I An index that appears only once in one monomial hence a free index must also appear just once in each and every other monomial in the equation 13 Interpretation of the Free Indices In a 3D space a vector 7 has three components V1 V2 and V3 and can be written in any of the following forms 7 Vial VZEZ V3E3 VszVs 119 139 123 The free index i ranging from 1 to 3 indicates that V1 has three components Examples of vector include velocities forces etc There are also many physical quantities that have less or more than three components in a 3D space For instance the direction cosines defined previously can be written in the following array form a11 a12 a13 cosX1X cosX1X z cosX1X 3 a1 a21 a22 a23 cosXZX cosXZX39Z cosXZX393 120 32 a33 cosX3X cosX3X39z cosX3X 11 11 31 Thus a whidi has two free indices i and j both ranging from 1 to 3 represents a quantity with nine components including every permutation of i 1 2 3 and 1 2 3 In the form of 120 the first index represents the row number and the second index represents the column number in the array Similarly the expression R lmn 123 lmn implies that Rm has 27 components in a 3D space including every permutation of the three free indices l m and n It is noted that in a 2D space V a and R 1 17 m with each index ranging from 1 to 2 represent quantities with two four and eight components respectively The free indices need not appear with only one quantity in a monomial For instance the expression AlBZ has two free indices associated with two quantities A and B respectively Accordingly AlBZ has nine components and can be written in the following array form A131 A132 A133 A137 A231 AZB2 A233 121 A331 A332 A333 TM Tan Drexel University 1 3 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors Consider next the quantity Ctzk There are two free indices 139 and k and one dummy index Thus C1 D7 has nine components including all permutations of 139 1 2 3 and k 1 2 3 ead component is the sum of three terms obtained by summing over the dummy index Written in an array form they are 3 3 3 ZCUDzl chzD 2C11D13 CUB1 CUBZ CUB3 71 71 71 leD C C27D71 C27D72 C27D73 EcmD ECZZDyz EC27D73 C3JD71 C3JD72 C37D73 393 393 393 122 EICSZDZ1 ECMDZZ EICSZDZ3 C11D11 C12DZ1 C13D31 C11D12 C12DZZ C13D32 C21D11 C22D21 CZBD31 C21D12 CZZDZZ CZBD32 C31D11 C3ZD21 C33D31 C31D12 C32D22 C33D32 C11D13 C12D23 C13D33 C21D13 CZZDZ3 CZBD33 C31D13 C32DZ3 C33D33 14 The Kronecker Delta Consider a 3D vector 7 whose components are denoted as V 139 1 2 3 in an X coordinate system and as W in an X coordinate system Since the length or the magnitude of a vector denoted as IIquot or IIV I is constant regardless of whid coordinate system is used we have W 1717 V12 sz ng 1747 Vl Z VZ Z M2 123 Substituting 113 in 123 and rearranging terms we have V12 VZZ V32 V02 Vz39 2 V3 2 121 122 123V12 a 2 5amp3sz 131 32 33V32 124 2 11 121 12 122 13323V1V2 25 11 131 12 132 13 133 V1V3 2 21 131 22 132 23 133 V2V3 Since 124 is true for any vector 7 it follows by comparing terms on both sides in 124 that 51121 nfztzf3 1 51 11Z 113 1 a 1n 2tz 3 1 11321 12 122 13 123 0 11 131 12 132 13 Z33 0 21331 22 132 23333 0 These six equations can be written in the following indexical form alkalk 1 nanZk 1 askask 1 k 123 125a alank 0 alknsk 0 22km 0 k 123 125b which can be further condensed to alknzk 1 for139 139jk 123 126a alkazk 0 foriij 139jk 123 126b If we introduce a quantity 5 called Kronecker delta defined as TM Tan Drexel University September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors a 1 if ij 127 17 0 if ii j then 126 can be written as 2W2 ijk123 128 In a matrix form the Kronecker delta 5 is actually a 3x3 identity matrix ie 511 512 513 1 0 0 517 521 522 523 0 1 0 129 531 532 533 0 0 1 Recall 116 V 21711 ij123 116 which relates the components of a vector 7 in the X1 coordinate system to those in the X coordinate system Multiplying both sides of the equation by HM and using the relation in 128 we have MV 51711V 56V lel 5kZVZ 5k3V3 k 123 130 For k 1 we can show that 511V1 6le2 613V3 V1 0 0 V1 Similarly for k 2 and k 3 we have 521V V2 and 531V V3 Thus 130 becomes 639th Vk ik 123 131 Equation 131 illustrates the use of the Kronecker delta 67 that is when one of the two indices of 57 is a dummy index 517 can be eliminated from the expression by changing that dummy index to the other index in the entire monomial Substituting 131 in 130 yields HMV Vk ik 123 132 or by dqanging the indices V7 a V ij123 133 71 1 This is the inverse expression of 116 which relates the components of vector 7 from the X coordinate system to the X1 coordinate system By dqanging again the indices in 133 to V aszk jk 123 134 and substituting it in 116 we have V alfalka ijk 123 135 Since both V and Vk represent components of 7 in the same coordinate system ie Xl system we conclude that V Vk if and only if j k Therefore TM Tan Drexel University 1 5 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors 7 1 ifjk quotk7123 a ik 0 ifjik L i 27216 ijk123 136 15 Coordinate Transformation Fig 12 A position vector in two Cartesian coordinate systems Consider two Cartesian coordinate systems X with base vectors 5 and X with base vectors a in a 3D space as shown in Fig 12 The position vector 2 or 52 that defines a point P can be expressed in terms of its components and the base vectors as 2 X21 XE 137 The relations between these two sets of components are given by1 x 59le 138 X 11 X 139 71 1 The dot product of 2 in 137 with respect to E yields gt2 5 x151 a ma E X1617 x 140 in whid the orthonormality of 51 ie E E 5 has been employed Comparing 140 with 138 we 1 7 17 see that 217 E1 E7 141 Switd ing the indices and rearranging terms yield a a E 142 1 Hereafter unless otherwise specified the ranges of all the indices are from 1 to 3 in 3D and from 1 to 2 in 2D TM Tan Drexel University 1 6 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors Thus once the base vectors of two Cartesian coordinate systems are known the direction cosines a or 17 a71 can be obtained using 141 or 142 and the transformation of any quantities from one system to another can be established Since the dot product of any two vectors is equal to the product of their lengths times the cosine of the angle between them and since a base vector has unit length the fact that 141 or 142 results in direction cosines becomes obvious The necessary and sufficient conditions to insure that a transformation is reversible and has a onetoone correspondence in certain region Vof the variables X are a The relation given in 138 and 139 are singlevalued continuous and possess continuous first partial derivatives in the region V and b The Jacobian determinantz J does not vanish at any point of the region V where 11 12 13 J lal an an an 143 31 32 33 Coordinate transformations with the properties a and b described above are called admissible transformations If the value of Jacobian determinant is positive everywhere then a righthand lefthand set of coordinates is transformed into another righthand lefthand set and the transformation is said to be proper If it is negative everywhere then a righthand lefthand set of coordinates is transformed into a lefthand righthand set and the transformation is said to be improper 16 Scalars Vectors and Tensors Let X and X be two sets of Cartesian coordinate systems related by the transformation laws X a17X1 144 x x 145 where am the direction cosines are defined as aw cosX x 146 A physical quantity is called a scalar a vector or a tensor depending on how the components of the quantity are defined in the X1 and X coordinate systems and how they are transformed from one system to the other A physical quantity is called a scalar or a tensor oforaler zero if it has one component say 5 in the X1 system and one component W in the X system and if 5 and W are numerically equal at the corresponding points Thus if is a scalar then We Xi 147 In terms of indexical notation a scalar has no free index since it has only one component and its value remains constant regardless of whid1 reference coordinate system is used Examples of scalar include the length of a vector temperature energy etc 2 It will be shown later that am aXlBX TM Tan Drexel University 1 7 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors A physical quantity is called a vector or a tensor of order one if it has three components say 1139 123 in the X1 system and three components 5139 139 1 2 3 in the X system and if the components follow the transformation laws 5 51 148 5 2 X 149 It is easy to see that a tensor of order one has one and only one free index Examples of firstorder tensor include displacements velocities forces etc A physical quantity is called a tensor oforder two if it has nine components say t1 ij 1 2 3 in the X system and nine components ti ij 12 3 in the X system and if the components follow the transformation laws 11 11 t 150 r 17 mlmmn t t with 151 Examples of secondorder tensor whid has two free indices include stresses and strains Similarly we can define an 11 order tensor as a quantity having 3 components in each of the X1 and X systems denoted as tmyupu and thy respectively and the components follow the transformation laws 152 I 7 trimquotPu ma12h metre11 153 7 t71P2quot39P 7 P1111 P2112 Puin t1 112quot where p and q i 1 2n are subscripts ead ranging from 1 to 3 in a 3D space We shall show later in Chapter 4 that material properties such as Young s modulus and Poisson s ratio are fourthorder tensors 17 Vector Operations Using Tensor Notation Consider two vectors A and B in a Cartesian coordinate system The dot product of the two vectors is given by 213 AlBl AZBZA3B3 AB 154 It is noted that there is no free index in the above equation hence the dot product of two vectors yields a scalar or a tensor of zero order Next consider the cross product of the two vectors a 21 gtlt B 155 The components of C are given by C1 AZB3 iAgBZ C2 143B1 7A1B3 C3 zilleiAzB1 156 In the indexical notations 156 can be written as C 17k 13 157 TM Tan Drexel University 1 8 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors where 17k139jk 123 called alternating tensor is defined as follows 17k 1 if the numerals taken by any two of the subscripts 139 j and k are unequal and in the cyclic order of 123 ie 123 231 gm 1 17k 71 if the numerals taken by any two of the subscripts 139 j and k are unequal and in the reverse cyclic order of 123 ie 321 213 gm 71 17k 0 if the numerals taken by any two of the subscripts 139 j and k are equal ie all components except 123 231 312 321 213 and 8132 are equal to zero Take 139 1 in 157 as an example we have C1 511114131 511214132 51134133 51214231 512214sz 512314233 51314331 51324332 51334333 0OOOOAZB307A3B20 A233 AsBz Similarly components C2 and C3 can be obtained by letting 139 2 and 139 3 respectively in 157 It can be shown that the alternating tensor 17k is a thirdorder tensor ie it has 27 components and follows the transformation laws 81397k ak my nkglmn 17k 1l ym lmgl39mn 158 It can also be shown that the Kronecker delta 51 is a secondorder tensor Finally the triple scalar product gtlt can be expressed as A X B E QM173 57ABC 571437 C 159 where according to the definition of the alternating tensor 17k 7 71k 87h 18 Paltial Derivatives of Tensors Recall 138 X 51 X 138 Differentiating both sides of 138 with respect to Xm yields 6X 1 H1 mlammm 160 6X7 7 6X7 7 7 in whid the orthonormality of Cartesian coordinates have been employed 1 if 139 m 5m I 6X7 0 1f 1 i m Similarly we can show that 6X7 1 61 a 6X 7397 TM Tan Drexel University 1 9 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors Now consider a scalar function X which remains constant in any coordinate system ie X X Differentiating the above equation with respect to X and using 161 we have 6115 0r axl H1 E 162 6X 6X 6X 6X 7 6X By using the following indexical notation to represent the partial derivative of a quantity 4 we can express 162 in the following form V5 ll51 163 Based on 163 whidi resembles the transformation law for a firstorder tensor we conclude that is a firstorder tensor with three components in a 3D space One may recall that in vector analysis the gradient of a scalar function 5 is indeed a vector ie v E1232E3 Mmsl l 1 64 Next consider the derivatives of a vector V The transformation law for vectors is given by V7 lzvl Differentiating the above equation with respect to X gives 6V av av ax av r711 711 r 1 lm 6X 7 6X 7 6X 6X 7 6X I Vm amid11 165 ie the gradient of a firstorder tensor is a secondorder tensor Thus we conclude that taking the partial ali 39erentiation of a tensor yields a new tensor one order higher than that of the original one Other useful vector operations in terms of index notation include av1 avZ av3 Divergence of a vector V47 V a scalar 166 6X1 6XZ 6X3 le t39 vve aVkev t 167 ur o avec or X 7817165781716 k a vec or 7 Laplace VZ VV 2 a scalar 168 TM Tan Drexel University 1 10 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors 19 Tensor Contraction and Quotient Rule The operation of equating two free indices of a tensor hence they become a dummy index and summing accordingly is known as contraction In general contraction of a tensor yields another tensor of order two less than that of the original one For instance given a secondorder tensor t we have 17 5912 at m1 n mn A contraction operation by equating subscripts 139 andj gives t nat t 1 m m m m mftmm where tW is a tensor of zero order or a scalar Another useful rule in the tensor analysis is the socalled Quotient Rule The following example illustrates how the quotient rule can be used Let V be an arbitrary firstorder tensor and sz be another arbitrary secondorder tensor If ng 41ka 169 is true regardless of the reference coordinate system then AW is a thirdorder tensor The proof is as follows Since sz is a secondorder tensor and V is a firstorder tensor we have by using the tensor transformation laws and 169 I 7 7 7 1 7k mankan my nkApman my nkApmn quq 170 Meanwhile since the relation given in 169 is true in the X system also we have Bk Al szl 171 Substituting 171 in 170 and rearranging terms we have A zk 7 amnmnnkAWn M 0 Since Vq is an arbitrary vector it must be that A zk amnmnnklmn 172 Equation 172 is precisely the transformation law for a thirdorder tensor Therefore AW is a third order tensor It is a straightforward process to extend quotient rule to more general cases 110 The Theorem of Gauss Integral Theorem Consider a tensor t defined in a region V bounded by a surface S as shown in Fig 13 The W Theorem of Gauss states that Iva de Sn t as 173 m 17kquot where nm is the unit normal vector along the exterior of S To prove this theorem we first let m 1 and carry out the integration in the X1 direction for a prism as shown in the figure TM Tan Drexel University 1 11 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors 12 Given the components of firstorder tensors A1 and C1 and a secondorder tensor Bl as 412 423 434 Bu 0 3122 Bis 2 B21 73 B22 1 32371 B31 6 332 73 33371 C11 CZ3 C38 Evaluate for ijk 123 a A133 13 371C C AlBkaw d 351 e A1C51Bm f AmCBm5w g A3 Cm 515m 17k 13 Given two Cartesian coordinate systems de ned by the following two sets of base vectors respectively 5141 0 EZO 01 E3 7 0 EH1 0 0 5110 a EHO 7 a Determine the transformation matrix 117 b Is the transformation proper or improper c If E is a vector whose components in the E1 system are 50 30 0 nd the components of this vector with respect to the E system 14 Write the following expressions in indexical notations a a a a Z Z Z a a a lAgtltBCJD b V V c v AB 15 Show that a secondorder tensor can be expressed as the sum of a symmetric secondorder tensor and an antisymmetric secondorder tensor ie s A II 1 1 12 S S A A where t5 tgl and t 7t1 16 Show that A173 2 0 if A1 is a symmetric secondorder tensor and Bl is an antisymmetric second order tensor 17 Show that 60 uk is an antisymmetric tensor where 87 is the alternating tensor and 14k is a firstorder tensor ie a vector 18 Prove the following identities in whid 57 and SW are the Kronecker delta and the alternating tensor respectively a 551 3 b glzkAZAk O c glszzk 0 if sz is symmetric ie sz Bk7 d glfkglmn aym m ayn km Hint show that each side of the above equation 1 if jm k71 unless jk 71 if jn km unless jk Oif jk or m71 e swam 6 Hint Use the identity in d 19 Use the identit in 18 d to show that ax xegagge 110 If x73x1E12x E210x353 compute a VW b sz c VW 111 Check the divergence theorem by using the vector TM Tan Drexel University 1 13 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors 7 Z a a Tixlxzel3xlxseZ for a cubic domain Venclosed by x1 0 x2 0 x3 0 x1 1 x2 1 and x3 1 H a N Express the following integral theorems in terms of tensor notations in whid V S and L represent volume surface and line respectively a J39VVWZVJ39S FWZS b va dvj 1ads c Wadi45mins d jS Vx Sjgdi e IVVZMVIS V 7ZS TM Tan Drexel University 1 14 September 19 2010 THEORY OF ELASTI CI TY 1 Cartesian Tensors APPENDIX 1A The transformations of first and secondorder tensors can be conveniently expressed in the matrix form Recall 120 and 161 Given two Cartesian coordinate systems X1 and X the direction cosines can be expressed as l 21 22 23 1A1 in whidl 6X1 1117 cosX1X W 7 Now consider the transformation of rstorder tensors as given in 148 and 149 g 7 148 6 215 149 Let Q 1 f f 52 and 61 6 61 1A2 53 then 148 and 149 can be written as 6 MT 6 1A3 6 W5 1A4 respectively For the transformation of secondorder tensors as given in 150 and 151 t amnmtm 150 t1 almnmtim 151 Let in tn tn tn t12 t13 t t t21 t2 t2 and t t 21 t zz 23 1A5 t31 ta ta tgl tgz 533 then 150 and 151 can be written as t 21 ma lt1A8gt t allt ll lT 1A9 respectively TM Tan Drexel University 1 15 September 19 2010 MEM66O Theory of Elasticity I Final Examination Solution Fall 2007 1 Extension A circular bar of length L crosssectional area A radius r and mass density p is hung by a thin rigid rod at the center of its top end The rod is then submerged into a fluid of mass density pf until the top of the bar is a distance L under the surface of the fluid as shown in the figure We wish to find the stress and displacement of the bar under the combined action of its own weight and buoyancy of the fluid a Specify the traction boundary conditions on the lateral and bottom surfaces of the bar Assume that L gtgt Zr b Assume a set of stresses that satisfies the traction Cross Section boundary conditions specified in a c Determine the displacements of the bar Solution Set the origin of the coordinate system at the centroid of the cross section on the bottom surface and the 2 axis to be directed along the centerline of the bar The components of body force are given by bx by 0 172 ipg where pg is the weight density Let the weight density of the fluid be pfg then the pressure exerts on the lateral surface is linear in 2 and varies from being ipfgL at the top of the bar to being 72pfgL at the bottom Assume the following stresses all 932quotpr39g a UHFaw Pfg2L39z 0 an ray 0 which can be shown to satisfy the traction boundary condition on the lateral and bottom surfaces Since 6039 60x9 6039x2 x 6x 6y 62 6039 6039 6039 xy i W i 91 b 0 6x 6y 62 y 6039 6039 6039 914 12447 7 0 ax By 62 2 PS PS the equilibrium equations are thus satisfied as well The compatibility conditions are satisfied automatically since all the stress components are at most linear Substituting these stress components in 52 yields EEivpgi17vpfg272172vpfgL a 6w BEp372vpfg272172vpfgL b 6w 6 014 6w 6 014 0 c 6g 62 62 6x 6x 6y Integrating b with respect to 2 gives 1 Ew 393 39 2W3ng 39 2139 WWW W00quot y d MEM66O Theory of Elasticity I Final Examination Solution Fall 2007 where w0xy is an unknown function Substituting d in the first two equations of c and carrying out the integrations with respect to 2 we have Ev7a02voxy 347033 zu0xy e where 140 xy and 00xy are unknown functions Substitutinge in a gives 62w 014 62w 62 7 6x20 26 xo 7W52E 7 vpg717vpfgz72172vpfgL f Comparing terms in f we have 014 00 EU EU 21 2vpfgL g and 62w 62w 6x20 6y vpg1vpg h Integrating g yields 140 7217 2vpfng 00 7217 2vpfgLy gx i Substituting e and i in the last equation of c gives 2 7 2a wo 2 Jer g 0 may dy 6x This implies that the expression in ead r individual bracket must vanish ie lm 0y 6x azwo xy 0 My 7 Equation j implies that is a linear function of y gx is a linear function of x and the 0 139 k coefficients of the two linear terms must be equal in value but opposite in sign ie fy yb gx7nxc 1 in whid 2 l7 and c are arbitrary constants From h and k we conclude that m0 is a quadratic function of x and y and that the coefficients of the xy term must vanish ie l weE pg7l7vpfglxzy2dxeyf m in whid d e andfare arbitrary constants Substituting i l and m in d and e we have E147hpg717vpngxz72l72vpfngtzy7dzb n Ev7bpg717vpfgtz72l72vpfgLy7tzx7ezc o l 1 Eu Evpg 7 17 1pfngZ y2Epg 7 Zlpfgz2 7 21 7 2vpfng dx ey f p where constants a b c d e and f must be determined using the displacement boundary condition at z L Assume the following displacement boundary conditions at 0 0 L uvw0 q MEM660 Theory of Elasticity I N Final Examination Solution Fall 2007 Ow av 0014 f It is easy to show that these conditions yield 1 b c d e 0 and 1 f 73033 2 303ng s u 079 a 39139 3 r up J 5 L K gfir gih til ega ijml 33v 7 v Torsion Figure on the right shows the cross section of a typical narrowbody transport airplane fuselage without frames and stringers The fuselage is divided by a floor into the upper and lower compartments the upper one for passengers and the lower one for luggage Both the skin and the floor are made of 00625 thick aluminum sheets Determine the maximum shear stress and angle of twist per unit length due to a 50000 ftlb torque resulting from certain flight maneuvers The shear modulus of aluminum is assumed to be G 107 psi Solution The areas of the upper and lower compartInents denoted by A1 and AZ respectively are given by 7r65Z 2 A1 T22065 9237 in 2 AZ 625 6637inz The lengths of the cross section of the fuselage above the floor below the floor and the floor sections denoted by 11 lz and 13 respectively are given by 21 7r65220 2442 in 12 7r65 2042111 13 265 1300m Additionally all three segments have the same thickness ie t1 t2 t3 00625 in From the example in Section 59 of Chapter 5 we have N 2t1t312A12t2t3llA t1tZZSllAZZ 476gtlt108 m7 Thus the angle of twist per unit length is given by 12533 E and the stresses in the three segments denoted by 11 12 and 13 respectively are given by MEM66O Theory of Elasticity I Final Examination Solution Fall 2007 71 131271171213 A1 AAgt311 psi M Tz 311A239tilsA1Az 29v0 l35i 13 1 11121451211142 203919rpsi The maximum shear stress is 311 psi occurring along the upper section of the fuselage skin 3 Torsion The cross section of a circular shaft with a keyway is 7 21 cos 6 shown in the figure The boundary of the keyway is represented by 7l7 while the boundary of the shaft by 72ncos6 Let the Prandtl s stress function of torsion for this shaft be Kb277217 211cos6j 7 where K is a constant to be determined Show that this stress function satisfies the traction boundary condition and then determine the value of constant K Compute the shear stress components 039CZ and 7192 and determine the location and value of the maximum shear stress Solution NOTE It is easier to solve the problem in polar coordinates than in Cartesian coordinates a a a 1f f ms f 6x 01 6x 66 6x 07 7 66 a a 1 smgi f 0y 070g 660y ar 7 66 62 62 62 62 v21 1 2 1 6x 67 707 7 66 The proposed stress function r Kb2 772117 211cos6j satisfies the boundary condition as 0 along the boundary Rewrite function 5 in the following form 5 ltl7Z 7 72 7 211l727 1 cos6 2117 c056 then 2 27 17 27 2711727 Z cos6 2acos6 r lf lt7 2 7 471l727 3 c056 Z K211l727 1 cos6 7 2117 c056 Ilt211l727 1 cos6 7 2117 cos 6 Vzw i il ii xgca Taxi 2 Z 707 72662 K7 2 7 471l727 3 cos6 7 2 72711727393 c056 7217 1 cos6211l727 3 cos6 7 2717 1 cos 6 74K Thus MEM66O Theory of Elasticity I Final Examination Solution Fall 2007 5 The stress function becomes 27 1wa r 117 211cos j 2 r39 39 The stress components are sin 67 COS9 r 66 COS 6 Ilt2ptl72r 1 sin 6 7 2117 sin 6 sin 6Ilt7 27 2nbzr392cos6 211cos 6 r lt7 2rsin6 4abzr zsin6cos6 2 Go7rsin6 b 51126 r azcosg 9 6x 6r 7 66 7cos6Ilt7 27 211bzrizcos 6 2ncos6 SI 6 14211172751 sin 6 7 2117 sin 6 r lt2rcos6 7211bzr392cosz6 7 sinz67 211 Ga 7nrcos67wl 739 The maximum shear stress occurs at 76 l70 is rm cryng 75 7Gotv2117l7 Bending Refer to Example 53 on Page 531 of Chapter 5 Lecture Notes A prismatic cantilever bar of elliptical cross section as shown in the figure is subjected to a centroidal vertical force Py at its free end y a Show that by dqoosing the function P b2 x2 fltxgt 17 the righthand side of Eq 593 would vanish hence simplifies the boundary condition to 5 0 b Choose a stress function xy sud that the boundary condition would be satisfied and then determine the constant by using the governing equation 592 Derive the expressions for the stress components 039x2 and 03992 Determine the maximum stresses and their locations DO NOT SOLVE FOR THE DISPLACEMENT ww lt n lt n 0 Solution The boundary of an elliptic cross section is given by x2 2 75771 0 a If we choose P b2 x2 x 3 172 b then the right side of 593 vanishes and 52 becomes MEM660 Theory of Elasticity I Final Examination Solution Fall 2007 PyxszJr v v2 1 112 1w Solution of this equation that satisfies the condition 0 at the boundary is given by 7 m2 1vl7Z Py bzxz T 21vin2 312 i1 It is easy to see that by letting nl7 r this solution reduces to that of circular cross section The Z y2 7 szx c stress components are given by 7 m i139b2 P y 03a 352 Ig 112 2lvb2 P 2 2 17sz o39yzquot g z 2 Ty 2 39 2 quot e lva 311 21 L a 21vb Along y 0 we have 2 2 p 7 2 a 0 a z 11 212vl7Z y bzi Z1 2vl7 2 x2 f y lvin 3l7 lZIC 11 21vl7 Thus 039yZ follows a parabolic distribution in the transverse direction and the maximum shearing stress occurs at x 0 ie aw I 22f1 vlzz 172 g W 21 If bgtgt11 then Zbz a0 wehave P b2 4P 7 y 9 0912112 T 31 3A h which coincides with the solution of elementary beam theory for bars with a narrow rectangular cross section If on the other hand l7 ltlt 11 then P 9323 179 1 and the stress at the ends of the horizontal diameter x in is 41 P9 039 4 92 1 1 A Thus depending on the value of Poisson39s ratio this stress distribution along y0 could be very different from being uniform Take 1 03 as an example we have UAW 154 032 092 The maximum stress is approximately 14 percent larger than that given by the elementary theory THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Eguutions ofElusticit Chapter 4 StressStrain Relations and Governing Equations of Elasticity When a deformable body is subjected to certain loadings deformation will occur and stresses will be developed within the body In general the level of stress developed in the body depends not only on the final state of deformation but also on the history and rate of the applied loading the temperature diange etc and the deformed body usually does not return to its original configuration after the loading is removed However if the deformation is infinitesimal and the deformation process is adiabatic then it is generally acceptable to assume that the stress and strain are independent of the loading rate and history and that the body does return to its original configuration after the loading is removed Sudi a deformation is called elastic and can be studies by using the Theory of Elasticity 41 Generalized Hooke39s Law The stress in an elastic body if depending only on the deformation can be expressed as1 at f1 em 41 For an infinitesimal deformation we can expand this equation into a Taylor39s series in the neighborhood of 217 0 This yields 0 62 0 71 f1 0 6 7lt ekl i eklemn quot39 7 7 66k 2 aekpem For a linearly elastic deformation the higherorder nonlinear terms in the above equation can be neglected Consequently we have ltogt if 03917 ekl A17 C17klekl 42 kl where A17 f170 and CW 050aek are constants It is easy to see from 42 that A17 represents the initial stress ie the state of stress at EU 0 whidi can be neglected without loss of generality in the context of linear elastic deformation Therefore we have 03917 Clyklekl 43a or in the matrix form all C1111 C1122 C1133 C1123 C1131 C1112 C1132 C1113 C1121 ell 022 C2211 C2222 C2233 C2223 C2231 C2212 C2232 C2213 C221 822 033 C3311 C3322 C3333 C3323 C3331 C3312 C3332 C3313 C3321 e33 023 C2311 C2322 C2333 C2323 C2331 C2312 C2332 C2313 C2321 823 031 C3111 C3122 C3133 C3123 C3131 C3112 C3132 C3113 C3121 e31 012 C1211 C1222 C1233 C1223 C1231 C1212 C1232 C1213 C1221 812 032 C3211 C3222 C3233 C3223 C3231 C3212 C3232 C3213 C321 832 013 C1311 C1322 C1333 C1323 C1331 C1312 C1332 C1313 C1321 e 13 021 C2111 C2122 C2133 C2123 C2131 C2112 C2132 C2113 C2121 821 1 We shall use 81 7 to denote the Cauchy s infinitesimal strain tensor in this and subsequent chapters TM Tan Drexel University 4 1 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uations ofElasticit Equation 43a or 43b is called the generalized Hooke39s law in which CW are the elastic constants Since both the stress and strain are secondorder tensors according to the Quotient Rule CW must be a fourthorder tensor with 81 components representing 81 elastic constants as can be seen in 43b However not all 81 constants are independent of one another This can be demonstrated as follows First due to the symmetry of stress and strain tensors ie 03917 03971 and ek elk we have Cm CW lelk Therefore the number of independent elastic constants reduces to 36 Furthermore we shall show later that for an elastic deformation there exists a strain energy density inction Wkly such that 03917 w 44 ew For linearly elastic materials the strain energy density function takes the following form 1 W 171618178161 45 2 Consequently CW CW and the number of independent elastic constants reduces to 21 for generalized anisotropic linear elastic materials By introducing the following abbreviated notations 71 011 0392 022 73 033 74 023 75 031 76 03912 51 511 52 522 53 533 54 2523 55 2531 56 2512 we can avoid carrying the double sums and reduce the number of subscripts of eadi elastic constant to two Thus the generalized Hooke39s law can be expressed in the following matrix form 71 C11 C12 C13 C14 C15 C16 51 0392 C12 C22 23 C24 C25 C26 52 73 C13 C23 C33 C34 C35 C36 53 4 6 74 C14 C24 C34 C44 C45 C46 54 75 C15 C25 C35 C45 C55 C56 55 76 C16 C26 C36 C46 C56 C66 56 or simply 039 CBS 46a It should be noted that in the above equation 54 2523 723 55 2531 731 56 2512 712 are the socalled engineering shearing strain components 42 Elastic Symmetry Most engineering materials possess some types of symmetry in their crystals whidi can be observed in their elastic properties By making the reference coordinates coincide with these planes of symmetry and by showing that the constants are invariant with respect to the coordinate transformation we can further reduce the number of independent elastic constants TM Tan Drexel University 4 2 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uations ofElastieit Thus the matrix of elastic constants for a monoclinic material becomes C11 C12 C13 0 0 C16 C12 C 22 C 23 0 0 C 26 7 C13 C23 C33 0 0 C36 0 0 0 C 44 C 45 0 0 0 0 C 45 C55 0 C16 C26 C36 0 0 C66 and the number of independent elastic constants reduces to 13 l7 Orthotropie materials An orthotropic material has three planes of symmetry that are mutually orthogonal to one another By making the coordinate planes coincide with the planes of symmetry and following a procedure similar to that for the monoclinic materials we can show that the number of independent elastic constants reduces to nine and the matrix of elastic constants becomes C11 C12 C13 0 0 0 c12 c22 c23 0 0 0 C C13 C23 c33 0 0 0 0 0 0 c44 0 0 0 0 0 0 c55 0 0 0 0 0 0 C66 e Hexagonal transversely isotropic materials A hexagonal or transversely isotropic material has one plane of symmetry and one axis of symmetry perpendicular to the plane of symmetry By making the X3 axis the axis of symmetry hence the X1 7 X2 plane the plane of symmetry we can show that the number of independent elastic constants reduces to five and the matrix of elastic constants becomes C11 C12 C13 0 0 0 C12 C11 C13 0 0 0 C C13 C13 C33 0 0 0 0 0 0 C44 0 0 0 0 0 0 C44 0 0 0 0 0 0 Citedz a Isotropic materials For isotropic materials the material properties are independent of the coordinate systems The number of independent elastic constants reduces to two and the matrix of elastic constants becomes C11 C12 C12 0 0 0 C12 C11 C12 0 0 0 C C12 C12 C11 0 0 0 0 0 0 Cuiclzyz 0 0 0 0 0 0 C117C122 0 0 0 0 0 0 eyed2 TM Tan Drexel University 4 4 October 24 2010 THEORY OF ELASTI CI TY 4 StnessStmin Relations and Governing E uutions ofElusticit which can be rewritten in the following form l2u it A 0 0 0 it it2u it 0 0 0 it it it2u 0 0 0 C 49 0 0 0 u 0 0 0 0 0 0 u 0 0 0 0 0 0 u in which H12 y C11C12 l and u are called the Lum constants and the Hooke s law in terms of the Lam constants becomes 03917 it lzekk 2 410 The inverse relation is given by J 2W1 0391 m5170kk 411 43 Engineering Elastic Constants In the previous section we have shown that for an isotropic material the number of independent elastic constants reduces to two ie the Lam constants it and u While these constants are obtained mathematically their physical meanings are difficult to interpret and hence are rarely used directly in solving engineering problems Instead the socalled engineering elastic constants including the bulk modulus K the Young39s modulus E the shear modulus G and the Poisson39s ratio 1 are more commonly used in practice We shall illustrate the physical significance of these engineering elastic constants and derive the relationship among all the constants using the following examples u Uniform pressure 7 Bulk Modulus When a body is subjected to a uniform pressure the state of stress is given by 03911 03922 03933 P 03912 72s 03931 0 where p is the magnitude of the pressure applied on the surface Since the pressure is always acting in the direction perpendicular to the surface the traction thus can be expressed as t1 7pm 711617117 where n is the unit normal to the surface Furthermore since t 717117 we have 017 ip lz Performing a contraction on the above equation and 410 the generalized Hooke s law yields 0 ir i3r a 039 xt llekk 2ue 31 2ueu b TM Tan Drexel University 4 5 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uations ofElastieit A comparison of a and b results in p 7 392 ujgu 7A2Iu1g 3 3 where e e11 e22 e33 If the first invariant of the strain tensor represents the rate of volume dqange The Bulk modulus or Modulus ofCompression Wthh measures the amount of volume change of a material under uniform pressure load or the rigidity of the material in a dilatation deformation is defined as 2 KLa 412 E u l7 Simple tension 7 Young s Modulus and Poisson s Ratio A state of simple tension in the X1 direction is diaracterized by 03911 00 03922 03933 03912 03923 03931 0 where 0390 is the applied uniform tensile stress Substituting this state of stress in 411 yields 2511 201 I 2 039 7 e 3amp2u 3amp2u 0 The Young 39s modulus or the modulus of elasticity whid r measures the rigidity of the material in a unidirectional extension deformation is defined as Ekw 413 511 439 I It is noted from the generalized Hooke39s law 411 that a simple tension in the X1 direction also produces strains in the X2 and X3 directions even though there is no traction applied in those directions From 411 and using 413 we have 03911 7 811 201w 522533 2u3 l2u where the negative sign indicates that contractions will occur in both the X2 and X3 directions when the material is subjected to a simple tension in the X1 direction The Poisson s ratio whidq measures the ratio of lateral contraction due to the axial extension for a material under a simple tension condition is defined as V2e224 414 511 511 2a 1 e Simple shear 7 Shear Modulus When an elastic body is subjected to a simple shear of To in the X17 X2 plane we have 03912 10 03911 03922 03933 03923 03931 0 Substituting this state of stress in 410 yields 012 To 2mm 712 where 712 is the engineering shearing strain The shear modulus whid r measures the rigidity of the material in a pure shear is defined as TM Tan Drexel University 4 6 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uutions ofElustieit G u 415 12 The relationship among the Lam constants it and y and the commonly used engineering elastic constants E v G and K are given in Table 41 3H2 2 it1v1 72v 9K1lt71 gtlt MHZ4 E 241 1721 K2 3 Ev v 72v 313K7E 3Kv 31172v 1v v Consider the following relations from Table 41 7 E and K E 7 21v 3172v Since physically the values of E G and K cannot be negative we have from the above two expressions 1v20 and 172120 ISVSOE For most of engineering materials the values of Poisson s ratio do not deviate mud from 13 For highly incompressible materials rubber for example where the volume remains almost und ranged regardless of the level of stress applied K gt oo then 1 gt 05 and G u E3 Furthermore a negative Poisson39s ratio although mathematically possible has never been observed in isotropic materials For some anisotropic materials however there has been evidence of effective Poisson 395 ratios with values greater than 05 For isotropic materials with 0 E V g 05 the Lam constant 2 must be positive as can be seen from the equation TM Tan Drexel University 4 7 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElusticit Ev 1v172v By using table 41 we can show that the Hooke s law in terms of the engineering elastic constants E G and 1 can be written in the following alternative forms v v 039171 1217 172V 572 2G e 1721 572 416 1 1 v 51E1V07 V51 Ukk E017 m5170kkj 417 In unabridged notation 417 becomes 1 71 71 0 0 0 l 71 71 0 0 0 511 E 1E E 03911 E 1E E 03911 v v v v 522 7E E 7E 0 0 0 03922 7E E 7E 0 0 0 03922 v v 1 v v 1 E 7 0 0 0 7 7 0 0 0 e33 E E E a E E E 1 a 418 823 0 0 0 11 0 0 023 0 0 0 0 0 03923 2G 531 0 0 0 1V 0 03931 0 0 0 0 i 0 03931 E 2G 2 039 039 12 0 0 0 0 0 H V 12 0 0 0 0 0 i 12 E 2G in whida the Poisson s effect on the normal deformation can be clearly seen 44 SumJnary of the Governing Equations in the Theory of Linear Elasticity So far we have derived the following equations for linear elastic materials I Three equilibrium equations 0W 12 0 419 I Six straindisplacement relations for an infinitesimal deformation 1 e 50 M 420 I Six stressstrain relations Hooke s law 03917 it 39lzekk 20617 421a 2 7 1 a 4 21b W17 017 3A2 170 10 When solving an elasticity problem we consider the deformation of an elastic body of volume V enclosed by a surface S subjected to the actions of body force l7 and surface traction t1 We seek solutions of the six stress components 039 six strain components 2 and three displacement components 141 that satisfy 17 419 420 and 421 at every point in Vand the traction boundary condition on S given by t 039 n 422 TM Tan Drexel University 4 8 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElastieit where n is the unit normal to the surface It is noted that b and t cannot be prescribed arbitrarily as they must satisfy the overall equilibrium Instead of the prescribed surface traction an elasticity problem may also involve a prescribed displacement u on a portion or the entire surface ie u u 423 In fact most of the realworld problems involve both prescribed traction and displacement over its surface In sud a case the surface S is divided into two separated regions namely SL and 8 over which the displacements and surface tractions are prescribed respectively They are referred to as the displacement and traction boundary conditions The solution to an elasticity problem is the one in which the components of displacement stress and strain satisfy the governing equations 419 421 and the boundary conditions 422 and 423 Finally the components of strain must satisfy the compatibility equations 517kzekz17 T 51qu 7 gym 0 424 to ensure the uniqueness of the displacement solution Sometimes it is more convenient to combine some of the governing equations and express them in terms of only one set of variables sud as those of displacement or stress This can be accomplished rather easily using the procedures described below a Governing equations in terms ofaisplacements Differentiating 421a with respect to x and using 420 we have 0 1 Him 141 425 Substituting 425 in 419 gives x1uuW tau 12 0 426 or in vector form x1uVl euVZuli0 427 where If e1 um is the first invariant of the strain tensor Equation 426 or 427 called the Navier39s equation is the governing equation of elasticity in terms of the displacements Solving this set of equations subject to the boundary conditions yields the displacement solution of the elasticity problem The strain components can be obtained by dii 39 39 O the quot J J according to 420 and the stress components can be subsequently computed using the Hooke39s law given in 421 It is noted that as the solution of the Navier39s equation is in terms of the components of displacement the compatibility conditions are satisfied automatically 1 Governing equations in terms ofstresses Substituting 421b in 424 and using 419 we obtain wk V m 171 4771 428 0 kk 1v 17 where v it21u is the Poisson39s ratio Equation 428 called the BeltramieMiehell compatibility equation by Mid ell in 1900 and by Beltrami in 1892 Without body force terms is the governing equation TM Tan Drexel University 4 9 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElusticit of elasticity in terms of stresses Solving this set of equations and the equilibrium equations 419 subject to the boundary conditions yields the solution to the elasticity problem If the body force field is conservative then there exists a potential function 1 such that 171a V 6x 1 er 429 Substituting 429 in 428 gives v 171 1 Ulchk makk17 5 Wick 2W 430 If the body forces are constant eg gravitational force or zero 1417 yrkk 0 430 becomes Ulchk liv 0km 0 431 Applying tensor contraction to 431 yields v211 0 432 where 11 01 is the first invariant of the stress tensor Equation 432 implies that 11 is a harmonic mction Furthermore by taking the Laplace of 431 and using 432 we have v40 0 433 Equation 433 is called biharmonic equation whose solutions are called biharmonic functions Thus when body forces are constant or absent the solution of an elasticity problem reduces to that of finding the appropriate biharmonic functions that satisfy the boundary conditions 45 Strain Energy Density Function Consider an elastic body in a state of equilibrium If the body undergoes a further stable incremental deformation denoted by V114 under an adiabatic condition ie there is no heat gained or lost then according to the first law of thermodynamics we have dWe jVdudV 434 where dWe jvbyzuyzw jst u s 435 is the incremental work done by body force l7 and surface traction t1 on the incremental displacements d141 and dU is the incremental increase of the intrinsic energy per unit volume of the body during this process By using the relation t1 717117 and the Theorem of Gauss 435 can be written as dWe Maw b 1udv j V adudv 436 The first integral on the righthand side of 436 vanishes due to the equilibrium conditions The integrant of the second integral can be written as alzdulyz 017d217 dw Harlem 437 TM Tan Drexel University 4 10 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElusticit Combining 434 through 437 gives jVdudV Ivawdequ 438 Since the volume of the integral in 438 can be taken arbitrarily we have du Jade 439 Since the lefthand side of 439 is an exact differential there exists a function Wkly sud that 039 ale de 17 17 e 17 17 or 6W 03917 E 440 The function W called the strain energy density function is the potential energy per unit volume stored in the body The existence of the function Wkly when the deformation in the body takes place isothermally can be proven using the second law of thermodynamics For an elastic body that obeys the generalized Hooke39s law given in 43 the strain energy density function can be easily identified as 1 1 WEC17kleiyekl aneg 441 For isotropic material we have by using 410 W 15172kk 2ue17 gt17 gaujeuekk 7142127 7 217217 3441155 27 21125 442 where I and I are the first and second invariants of the strain tensor respectively Equation 442 can also be written in the following form 7 4 2 2 2 2 2 2 2 W 7 3211 222 233 u 211 222 233 2212 2223 2231 443 Since both it and u are nonnegative quantities we conclude that w 2 0 444 46 Uniqueness of Solution For a given elasticity problem in whid the body force 171 in V surface traction t1 on 8 and displacement 14 on SL are prescribed we seek a solution in whid r the components of stress strain and displacement satisfy the governing equations 419 420 and 421 and boundary conditions 422 and 423 One may ask if sud a solution is unique in the sense that the components of stress and displacement are determined without ambiguity To show that the solution if existing is indeed unique let us assume that there are two possible solutions denoted by 1411 71 21 and 1412 72 22 respectively to the same problem Since both 1717 1717 TM Tan Drexel University 4 11 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing Equations ofElusticit solutions satisfy all the governing equations and the boundary conditions it is easy to see that the difference of these two solutions defined as 7 1 Z 7 1 Z 7 1 Z Auliu 714 Aggie 707 Ae ie iew will satisfy the following conditions A017 0 in V 445a A017 leekk 2yAeW in V 445b A14 0 on SL 445c Aalznz At 0 on S 445d Defining W 7 1A0 A3 i W 7 W 4 46 T 2 17 17 we can show that 2 WW ijaAedV ijaAudv AW Aew 7 Am IVA017AM17 dV 7quot AalzyzAuldV Integration by parts ISAaleulnzdS Theorem of Gauss A017 0 447 jsAtAuds A0n Ag jg AtAuds jg AQAuldS s s 5 Since A14 0 on SL At 0 on S and the volume integral in 447 is arbitrary we have W1A039Ae 0 2 17 17 In view of 445b we conclude that A037 0 and A2 0 The only possible difference between the two solutions is a rigid body motion whid does not produce any strain or stress Finally since the displacements are prescribed on S the rigidbody motion must vanish Consequently the solution if exists must be unique 47 Principle of Superposition Recall the governing equations of linear elasticity 419 421 and the boundary conditions 422 and 423 Since all these equations are linear ie they contain only terms of up to the first degree in the dependent variables and their derivatives therefore the components of stress strain and displacement due to some combined sets of external loads can be obtained by superposing the results due to ead set of load acting separately This is known as the Principle UfSuperposition The proof of this principle is straightforward and can be found in any textbook It should be noted that in deriving 419 421 we have assumed that the deformation is infinitesimal and the distinction between the initial and the deformed configurations of the body is negligible There are however certain situations in whid the deformation may be infinitesimal but the difference between the initial and deformed configurations may not be ignored In such cases the TM Tan Drexel University 4 12 October 24 2010 THEORY OF ELASTICITY 4 StressStrain Relations and Governing Equations of Elasticity justification of the principle of superposition would fail A beam under the simultaneous action of both lateral and axial loads and the stability of thinwalled structures are just two of the examples that are often encountered in solid mechanics where the principle of superposition does not apply 48 SaintVenant39s Principle P P lt 1 N 1 2 01 01 7 01 lt Radiusg 213 1 2 a A selfequilibrated force system b Two statically equivalent force systems Fig42 Two interpretations of SaintVenant39s Principle In 1855 Barre de SaintVenant proclaimed a principle in which he stated that the strains in a body produced by the application to a small region in the body or on the surface of a system of forces statically equivalent to zero force and zero couple are of negligible magnitude at distances which are large compared with the linear dimension of that region Figure 42a illustrates this principle in which a selfequilibrated force system applied in a region characterized by a radius 5 near point A produces negligible strains at point B at distance d from point A where 61 gtgt a An alternative way of interpreting the SaintVenant39s principle is illustrated in the example shown in Fig 42b in which two statically equivalent force systems are applied over a small area on a body The SaintVenant s principle states that the difference in stresses and strains produced by these two systems is significant only in the vicinity of the area where the forces are applied and is negligible at distances large in comparison with the linear dimensions of this area Using this principle we can simplify the solution procedure of a problem by replacing a complicated traction condition with a simple but statically equivalent one When applying the SaintVenant39s principle however one must be careful in interpreting the results It has been shown that in some cases the decay of the selfequilibrating forces may be extremely slow such that their effect is quite significant even at a large distance from the forces In 1945 NJ Hoff presented two interesting examples to illustrate this pointZ The first example is the torsion of bars with different cross sections One end of each bar is fixed while a torque is applied at the free end At the fixed end since the cross section is being restricted from warping normal stresses that are selfbalanced would develop Depending on the cross section of the bar these selfequilibrating normal stresses may propagate very far Figure 43 shows the maximum normal stresses fmax x on a cross section at a distance x from the fixed end normalized with respect to the maximum normal stress at the fixed end fmax O for three bars of different cross sections It can be seen that the normal stresses caused by restricting warping of the cross section at the fixed end is highly localized in the bar with a solid rectangular cross section Case c in Fig 43 while those in bars with thinwalled channel cross sections have appreciable values over the entire length of the bars Cases 51 and b in Fig 43 Consequently reliance on SaintVenant39s principle in the calculation of stresses due to torsion is entirely justified with 2 YC Fung Foundations of Solid Mechanics PrenticeHall Inc 1965 pp 306309 Tll Tan Drexel University 4 13 October 24 2010 THEORY OF ELASTI CI TY 4 StressStrain Relations and Governing E uutions ofElusticit bars of a solid cross section In contrast stresses in the bars of a thinwalled cross section depend largely on the end conditions 10 I xx I a 17 c fwltxgt I x 39 fmax0 I x l x 17 I i O xL 10 Fig 43 Normal stresses due to torsion along bars of various cross section Figure 44 shows the second example by Hoff where a statically determinate truss is subjected to a set of four selfequilibrating concentrated forces at one end The numerical values written on some of the truss elements represent the forces acting in those members It can be seen that the effect of this set of forces decays very slowly over the length of the truss indicating that the SaintVenant39s principle does not work well in this case Fig 44 Atruss subjected to a selfequilibrated force system Problems 41 A thin plate made of a homogeneous isotropic material is subjected to a proportional biaxial loading as shown Two strain gages are being placed at the center of the plate to measure the strain components 211 and 222 As the load readies 039 35 gtlt103 psi the readings from the two gages are 211 35x10 3 and 222 10gtlt10 3 respectively Determine the Young s modulus E and Poisson s ratio 1 of this material TM Tan Drexel University 4 14 October 24 2010 THEORY OF ELASTI CI TY 4 StnessStmin Relations and Govemin E nations 0 Elusticit 7220 20 in 14140 psi A lt B 822 45 7 all 20 Le 011 20 14140 ps1 uo psi 11 A 45 l 14140 psi 2 Problem 41 Problem 42 A parallelogramshaped thin plate as shown in the figure is made of a material with elastic constants E 30gtlt106 psi and v 03 The plate is subject to a uniform shearing stress of 14140 psi as shown If the deformation is infinitesimal determine a the change in length of line E and b the principal strains and their directions Show that for isotropic materials the principal directions of stresses coincide with those of principal strains Show that if a material possesses two planes of symmetry that are mutually orthogonal to eadq other then the third orthogonal plane must be a plane of symmetry Show that the number of independent elastic constants reduces to nine for an orthotropic material and to five for a transversely isotropic material Given the following state of stress cx vx127x 725mg 0 0391 725mg cx vx 7x12 0 0 0 cvxf x a Do the components satisfy the equilibrium equations b Are they a possible solution of a problem in elasticity Why TM Tan Drexel University October 24 2010

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