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Structural Analysis I

by: Robb McCullough

Structural Analysis I AAE 35200

Robb McCullough
GPA 3.65


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This 36 page Class Notes was uploaded by Robb McCullough on Saturday September 19, 2015. The Class Notes belongs to AAE 35200 at Purdue University taught by Staff in Fall. Since its upload, it has received 109 views. For similar materials see /class/207860/aae-35200-purdue-university in Aerospace Engineering at Purdue University.

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Date Created: 09/19/15
Glossary AAE352 Girish Modgil Background The following glossary was created using Dr CT Sun s book as a study guide Important definitions not covered in the Dr Sun s book were included from Dr Merrill s lecture notes The glossary is intended as a quick review of all important concepts covered in AAE352 and will be updated when I take AAE553 in the Fall and in preparation for my Qualifying Exams Table of Contents Chapter 1 Basic Aircraft Structural Elements 9 Chapter 2 Introduction to elasticity 4 Chapter 3 Torsion 12 Chapter 4 Bending 9 Chapter 5 Flexural Shear Flow in Thin Walled Sections 97 Chapter 6 Failure Criteria for Isotropic Materials 11 Chapter 1 Basic Aircraft Structural Elements 1 Axial Member Carries tensile and compressive loads The term EA is called the axial stiffness of the member 2 Shear Panel Carries inplane shear loads For a shear panel of uniform thickness t under uniform shear stress the shear force in the x direction is given by For a curved panel of width a and height b under a state of constant shear stress the 2 components of shear force are given by 3 Bending Member A beam carries bending moments and can also act as an axial member Given a deflection w we get 5 2 M E1x quot For a simple cantilever beam of length L height h and width b with a force V applied to its end Mmh2 m 1 17h2 3V 1 m 2bh Bending stress plays a more dominant role than transverse shear stress Beam with a rectangular cs is not optimal lbeam have bending stress concentrated at the flanges On the contrary the lbeam web carries mostly all the transverse shear stress Torsion Member Torque arises from shear stresses acting in the plane of the crosssection For a hollow cylinder subjected to torque T the torque induced shear stress is linearly distributed in the radial direction T GJB The term GJ is the torsional stiffness For hollow cylinders J is equal to the polar moment of inertia of the crosssection Jnb4 a4aZMr 3 Important Material Properties Chapter 2 Introduction to elasticity 1 Displacement and Strains 11 Displacements For a displacement of point Pto P the transformation is given by x39xu y39yv z zw For an axial member 1D body the average strain is given by 12 Strains If the strain is not uniform consider a small segment Ax and the displacement given by the function ux given by ux u0 x xo For 1D body and displacement in the xdirection only coordinates are Pxyz and P x Axyz and the corresponding displacements are uxyz and u x Axyz The new distance is given by xAxu39 xu Axu39 u AxAu So the strains in the 3 axes directions normal strains are given by au 6v 6w an syy SE 6x 0y 62 Shear strain components are derived by considering the xy plane and movement through small angles 614 6v 0 6 ny 1 2 6y ax Y 361 W 192 y Bu 6w Yu Bz Bx Thus a general state of deformation can be defined by 3 normal strain components and 3 shear strain components 13 Rigid Body motion The body undergoes displacement without inducing strains in the body For example the equations below are for rigid body rotation ie no translational strains are induced Bu 4 cry gt 0 6x v x 2 Stress 21 Stress In the most general sense P 0 A Sign convention 1quot is the shear stress acting on the plane perpendicular to the xaxis along the ydirection For an infinitesimal element in 3D with volume AxAyAz we use the equations of equilibrium Consider the forces in xdirection 2w 0 A0quotAyAz 0nAyAz 12r AtnAxAy cquAy 1 At AxAz 1yxAxAz 0 This will simplify to 6x By 62 E 30 6x 6y 6z 61 617le y 60 0 6x 6y az Using the equilibrium equations for the moments we get the relation as shown below for all 3 directions Simplify the stress element to 2D and consider a unit normal vector to the inclined surface as n and the traction vector t pointing in an arbitrary direction t with x and y component tx and ty respectively For this stress state we end up with the linear system as follows on tyx nx tx 117 CW quoty tJ This can be extended to a 3D state of stress and the equation above implies that the stress matrix can be viewed as a transformation matrix that transforms the unit normal vector into the stress vector On any arbitrary surface with unit normal vector n the normal component of t on n is ol t n 22 Principal Stress If we are interested in finding surfaces for which t is parallel to n we find the eigenvalues for the stress tensor shown in 2D For a 3D tensor we get a cubic equation in owith 3 real roots 01 02 and 03 called principal stresses with corresponding mutually orthogonal eigenvectors hm n2 and n3 called principal directions V thout loss of generality we assume 012 0201 01 is the maximum principal stress and 03 is called the minimum principal stress Additionally assumption of n as a unit vector yields the following relationship between normal components and the min and max principal stresses 2 2 2 n ny nz 1 020 ch3 23 Shear Stress The traction vector t can be decomposed into a normal vector and a tangential vector 1 t onn 2 2 2 2 2 2 2 1 o tx ty tz on Assuming the coordinate system is parallel to the principal directions we get the relations simplified Further simplification is obtained by considering the problem in 2D tx 01quot ty 02ny tz 03quot For nx nz 0 and nyO we get 1 nx 1 2 ml 1Iol oal 2 24 Stress Coordinate Transformations Consider the 2D stress state as shown below We need to determine the stresses in a coordinate system defined by x y rotated by arbitrary angle theta counterclockwise Transfurmmg a state m by m a 001 s 0 we 1 1 inn 3 Constitutive Models faressfrumthe gtltry mane m an armramy mutated xuy mane s gwen e mater re atmn shuvvn be uvv 2D transfurmatmn in e mas u can t Iin78 a 1 r 19 29 1w 2m can A cuns ututwe made 5 a re atmnsmp between stress and straw These stressstram depend un mremen E se me matena s caHEd Amsutrupm 31 straws ndueed b Nurma stress 1 an oxx EX v a E quotou v1 an on EX quotx sz fox 71 a Y Ex XX 71 Y1 for n is similarto Poisson s ratio but describes the interaction between normal and induced shear strains The first subscript is the loading direction and the the second set of subscripts indicates the plane of induced shear strain 32 Strains Induced by Shear Stress nxygc 8 TI 1 ny 9 my 8 T W ny l7 71ny E T a ny xy my sz try Gquot 3ng Y1 T G W Y g V1 ny 1 33 Compliance and Stiffness Matrices These matrices define the relation between stress and strain The matrix a is the compliance matrix and c is the stiffness matrix There are 21 independent constants for the full matrix 8agt0 0C8 8 039 SW OW 8 a 0 Yy quot 3 YE 1 ny 1v 34 Orthotropic Materials Contains nine independent elastic constants Top left corner 3X3 matrix contains entries in addition to the diagonal elements ia L i an Ex 22 Ey aaa E v a JV v v a12E a21 E a13 E 1a31 Ex 123 Efv asz E a a La L 44 G 55 Ga 66 ny 35 Isotropic Materials Only two independent elastic constants exist The coefficients for the compliance matrix simplify to 1 a11a22a33 E a12a21a13a31a23a32 E a44a55a66 G Corresponding coefficients for the stiffness matrix are as follows cll c22 c33 AZG 912 c21c13 C31czs csz quot c44c55666G XL quotl vxl zv L 21v 36 Plane Stress and Plane Strain Many types of loading may yield strain and stress fields that are independent of the z direction Sheet of paper skin of airplane fuselage are simple examples of plane stress Concrete dam hollow cylinder pressurized internally and constrained at the ends steel rod with crack in the middle are all examples of plane strain For plane strain and plane stress conditions respectively we have the following For the isotropic case there is a simplification in the stiffness matrix that yields the following for plane strain E on ml VExx V8W E ny WW8 1 VEW vE 039 1v1 2v 39szy nyy en eW v on 0 1 a vm wow 8 1239 v vomr 1 voW 1 my 17W Additionally for plane stress we get the following 1 1 a EH0 0yy XV ezz 0n ow 1 M 61 E 039 xx lv2 an vsw E U WEV8XXEYJJ Ezy nyy 37 Octahedral StressDeviatoric Stress J2 and Von Mises Stress 2 2 2 2 2 2 9c moxx ow oxx ozz ozz ow 613 613 615 3 o 1 vm act Chapter 3 Torsion 1 Uniqueness Compatibility and SaintVenant s Principle a Uniqueness 39 Stress satisfies equilibrium equations 39 Strains are compatible 39 BC s tractions and displacements are satisfied 2 2 1 b Compatibility 39 Relations between strains that must be imposed to ensure simple valuednessquot of displacements 39 StrainCompaitibility Stress Equilibrium c Solution Uniqueness 39 An assigned stress field may not yield strains that are compatible 39 An assigned displacement field may not yield stresses in equilibrium d SaintVenant s Principle The stresses or strains at a point sufficiently far from the locations of two sets of applied loads do not differ significantly is these loads have the same force and moment Torsion Introduction In deriving the deformation and stress fields in a circular solid or hollow shaft the following assumptions were made in undergraduate coursework gt Plane sections of the shaft remain plane and circular after deformation produced by application of torque gt Diameters in plane sections remain straight after deformation These assumptions lead to the result that shear strain is a linear function of the radial distance from the point of interest to the center and that plane sections of the shaft rotate as rigid bodies without inplane deformations or outofplane displacements warping This is too simplistic an approach 2 Derivation for noncircular shafts Consider a straight shaft of noncircular crosssection subjected to equal and opposite torques T at the end The origin of the coordinate system is called the Center of Twistquot about which the crosssection rotates due to application of the torque T By its definition there are no inplane displacements at the COT For the case where the end support is ignored the COT is assumed to lie at the shear center Let or denote the total angle of inplane rotation at a 2 located relative to the end 20 The rate of twist at z is angle per unit length eg Z St Venant assumed that the plane sections warp but their projections on the xy plane rotate as a rigid body Consider an arbitrary point P on the section that moves to P through 13 rotation or CCW If its original position is defined by angle 3 and radius is r we have the following urasinB Gzy vrucosl3 Bzx w0qJxy Y y quotz 5x 52 6x Y 6w6v6w e W by 52 6y Using the equilibrium equations we get 0 6x 6y Prandtl introduced a stress function which satisfies the equilibrium equation above Combining all the equations together we get the compatibility equation as 67 126Yxz29 6x 6y 2 2 QQ40 6x 6y So the torsion problem reduces to finding the stress function phi and requiring that the stresses derived from this function satisfy the boundary conditions On the lateral surface of the bar no loads are applied so the traction vector must vanish on the lateral surface So t 0 n 0 0 In nx tx 0 0 1 ny ty In 1 yz 0 0 tz n sinn n cosn quot ds y ds tz nx n d 6y 6x y ds We assume that the traction free boundary condition on the lateral surface satisfies t 0d 0 ds const To derive the torque we consider an elemental area dx by dy 6i ii T x dxd 2 dxd I 0x y 6y y IA y The derivation clearly indicates that the solution of the torsion problem lies in finding the stress function that vanishes along the lateral boundary of the bar Once phi is found the location of COT is also defined For bars of arbitrary crosssection warping occurs Here we have 5 w Ixze 5x G y 5WTyzex By G The warping displacement can be obtained by integrating this equation The following equations can be used thereafter to get the torsional rigidity GJ i GB 4 J V2 y dxdy Thus once the Prandtl stress function is solved we can derive the torsional rigidity 391 Bars with circular crosssections If the origin of the bar is chosen to coincide with the center of crosssection the boundary contour is given by 2 2 x q C 2 y 2 1 a a Substitution into the compatibility equation gives C lazGB 2 This indicates that the center of the circular crosssection is the COT To derive the relation for torque we have r2 I T 2 dxdy 20 7 1dxdy 2Ca A 091 p I 11 4 2 The shear stresses are 6 Eu 2clz Gey By a 6 x tyz a 2C 2 6936 x On the lateral surface of the cylinder we have tJr ty t nnx 1zyzny x nx ny Zgttz O39lz r r If we consider the surface exposed by cutting along the radial direction x y x n2 0nx ny r r tz 17 Gar w 0 Tr 1 J 4 Bars with 39 crosssections For a square crosssection the shear stress is not perpendicular to the radial direction and its magnitude is not proportional to the radial distance Warping is present Consider a bar with rectangular crosssection subjected to pure torque bgtgtt and Lgtgtgtb Basically a thin plate in zdirection to satisfy SaintVenant principle On the top and bottom face we have a traction free boundary yt217yz0 34gt 1 0 6x 7 Since t is very small we can assume that shear stress quotcyz is 0 through the thickness and that l is independent of x 2 Q 209 6y 4 G9y2 ClyC2 q 0yt2 z Gey 4 The torque is obtained similar to the derivation for the general crosssection and the warping is nonzero 2 T ZGB 39y2 dde gbtaG A 6 wyn6yT 0y 2G6y0y 6y 6x G 39w 6xy Note that wO at the COT and along the centerline of the sheet The results shown here can be used for sections composed of multiple rectangular thinwalled CS members The torsional rigidities are additive The torsion constant derives here is to be used only if b gtgt t If the dimensions are comparable J should be evaluated using the elasticity solution obtained by solving a rectangular CS using the Prandtl Stress function 5 Closed single Cell thin Walled Sections Wall thickness is small compared to length of wall contour t is not constant but a function of s lnner contour is S1 and outer contour is So Boundary conditions on So and S1 yields stress functions as follows as lC0S0 C13S1 Set up a local coordinate system sn along the centerline of the wall section for the saxis and n being perpendicular to s pointing out The equilibrium conditions give t an n as Let I be expressed in terms of coordinates s and n Using only the linear terms and the boundary conditions that require fto be Co on So and C1 on S1 at thickness values t2 and t2 respectively we get the following MSW oS 41101 Sa o 1 C0 t t 41537 0 TE 1 C1 1 1 o 3Co C1a 1 Co C1 1 1C1Co 2 We now define shear flow q and note that the shear flow is constant along wall section regardless of wall thickness The shear flow forms a closed contour so force resultants are equal to zero The shear flow produces a resultant torque A is the enclosed area by the centerline of the wall section q 1t Cl C0 dT pqu T pqu 392qu2qZ A The resultant force R oriented parallel to the line connecting two end points P and Q of the shear flow is as follows Rqd FqbFyqh ReT2Zq Angle of twist if q is not constant 1 dS 2GA t Angle of twist if q is constant 6 i L ds GJ 2GA T ZqZ 422 Nixz 39J 6 MultiCell Thin Walled Sections V ng sections are composed of airfoil skin supported by vertical webs to form multi cell bodies Stiffeners are used to carry bending loads they do not add to the torsional rigidity of the wing box and are often neglected in the consideration of torsional stiffness For a 2 cell section there are 3 boundary conditions to be satisfied This gives us the following set of relations for the multicell body SoC0a S1C1 S2C2 q1 C1 TCoaqz C2 TCo q12 C1 02 T2A1q12Azqz 1 ds 1 ds 9 31T q 92 T Ng ZAlGum t 2142sz12 t For each cell the shear flow and twist angle is taken positive if it rotates CCW For example if q12q1q2 is applied to cell 1 then for cell 2 we should use q12q2q1 7 Warping in Open Thin Walled Sections Except for circular crosssections shafts of noncircular sections exhibit warping ie out of plane displacements occur under pure torques Warping denoted by wO along the centerline of the wall y0 When warping occurs only across the thickness of the wall it is called secondam warping For general thin walled sections the centerline may also warp with magnitude much greater than secondary warping This is called primary warping For a rectangular CS 20 wp 6w xgt 0gt1 0 sz 6y Y yz We derive warping equations again by setting up a local coordinate system consisting of s along the contour and n being perpendicularto the contour Us and un being displacements in the s and n directions k is the area enclosed by the contour s and the two lines connecting the COT or alternatively it is the area swept by the generator line from sO to ss The positive direction of s must be set up so that it is CCW wrt the zaxis to make it consistent with the positive direction of 6 We have 6w Bu aw aux Ysz 0 6s 62 6s 62 u p26gt p6 dZpds ws w0 2Ze amp 6w Bun Ynz an 62 8 Warping in Closed Thin Walled Sections Set up an snz coordinate system with the origin at the COT and with positive s forming CCW about the COT G is shear modulus t is the wall thickness we get the relation for warp at any point relative to that sO qSGt 6 wau aw q as 62 ws w0 f ds 226 0 9 Effect of end constraints 21 The SaintVenant torsion formulation is derived based on the assumption that warping is freely developed and uniform along the shaft In general this assumption is violated because torsion members are connected to other components in an assembly Suppression of warping using end constraints can increase the torsional rigidity of a shaft Consider k to be a length parameter and z the distance from the free end Then we have the following J 1 e zk Jeff Chapter 4 Bending 39 General Definition a Method of Superposition From Beer and Johnston When a beam is subjected to several concentrated or distributed loads it is convenient to compute separately the slope and deflection caused by each of the given loads The slope and deflection caused by the combined load is found by adding or superimposing the contribution from the individual loads 39 Unidirectional Bending The equations defining a state of unidirectional bending can be summarized as shown below Note that uxz is a linear function in z implying that plane crosssections remain plane after deformation but may not be perpendicular to the centroidal axis Afurther simplification that the transverse shear strain 0 suggests that the plane crosssection remains perpendicular to the centroidal axis after deformation and that the amount of rotation is equal to the slope of the deflection note the sign convention for angle wy is CW which is opposite to the slope 22 uuoxzwyx wwox du0 ley IW0 2 z onEsn dwo y E duo N EA ondA dx A dzw My EIy ix g zoadA dM 3 V V EIy 1dr de d4w P1 quot3 Myquot 0 Myz 6 omMyz Bidirectional Bending The equations for bidirectional bending are as shown below The xaxis is chosen to coincide with the centroidal axis and the external load is decomposed into y and z loads py and pz respectively For no torsion to occur the line loads must pass through the center of twist shear center The positive direction of wy is the right hand rule about the positive y axis and wz is about the negative zaxis 23 u uoxleJyxyIlJzx v vox w W005 my vxz 0 dwo dvo Wy 1p dx dx dzw d v dzv dzw My EIy dx EIyz dx 2 EIZ 1x EIyz dx2 Ed2wo IszzIMy Ed v0 IszyIyMz 2 2 2 2 dx 1 Iyz dx 1y 1yz dM 3 3 V yEIdwEI dv0 VdMZEIdvoE dw0 1 dx y dx3 yz de 1 dx 1 dx3 yz de d z d4w d v dV d v d4w pl dx EIy dx4 EIyz dxf py EyEIz dx4 EIyz dx4 IzMy 1yzzMz IyMz IyzzMy 6 0 1y 1y IyI 1 z E tam i IyMz 4sz y IZMy IszI The equations for stress can be simplified further for special cases M2 My on y z a Axis of symmetry 12 1y M on yz b Furtherisz O 1y my 2 2 2 c If Iyz is nonzero and Mz 0 1y I J7 1y I yz Transverse shear stress in narrow rectangular crosssections The exact distribution of sz on the crosssection of a beam is easy to analyze for the narrow rectangular crosssection If hgtgtb the plane stress assumptions adopted for simple beam theory are valid That is sz is assumed to be a function of y Consider a beam with resultant transverse shear force of V2 24 h2 Vzbftndz w at 0 i y 0 4 6x dz 1 6x 62 2 2 1V1 gt1nKc 1Z2 c maxz0 az Iy 21y c 2 Transverse shear stress in general symmetric sections For a general symmetric section all we know is that sz is symmetrical with the zaxis It is more convenient to consider the transverse shear flow qz defined as shown below qztt Tn XZ qz ftudy dM qz f yidAQ A1 g t dx 1y 1y Q 1PM A z Thin Walled Sections In aircraft structures stringers are used to provide bending stiffness and thin webs carry shear flows To maximize bending stiffness we place stringers at the greatest distance from the neutral axis The thin web is assumed to be ineffective in bending Therefore the shear flow between two adjacent stringers can be assumed to be constant Shear Deformation in Thin Walled Sections In general for short beams use Timoshenko theory Otherwise long slender beams the simple beam theory will do In developing the simple beam theory the transverse shear strains are neglected this may lead to substantial errors in estimating the deflection of thin walled beam members unless they are long or under pure bending moments In such scenarios we assume the deflection to be composed of 2 parts 1 part from the bending moment and 1 part from the transverse shear deformation Along beam I gtgtgt h with a high shear rigidity GA or small bending rigidity will fit the description of the simple beam theory Consider a cantilever beam subjected to a shear load at its free end 25 BENDIN G M 2 3 2 3 quot d f0 q hx gtwoq hx q hL x q hL Ely dx Ely 6EIy 2E1y 3E1y From dw w0 amp 0 xL 0 dx SHEAR 1z qo 6w Bu Bu xL 0 Y G tG 6x 62 az woq x M tG tG COMBHVED39 wg 3EIy wig GAL Timoshenko Beam Theory a Timoshenko beam theorythat accounts for transverse shear deformation using the displacement expansions for simple beam theory without imposing the zero transverse shear condition b We also assume that the axial force is zero so only the transverse deflection and rotation of the crosssection are kept in the formulation 26 dll y dw0 870 Z dx Ya E1lly dwo VI g 1udA GAEW Ely dx 12 dw Ely dxzy GAEO1pyO 2 GA d fwd pz0 dx 1 COMBINING E1 d4w0 p Ely dzpz y dx4 2395 4x2 c For a hinged end wOO and My is prescribed For a clamped end wOpsiyO but dwOdx is nonzero For a free end V2 and My are prescribed Shear Lag In web stringer situations distance is required to translate a load applied to one stringer to the other stringers This is called shear lag The effect is that the state of stress in the stringers and web will not be accurate unless considered suf ciently far from points of applied loads or cutouts For example considering three stringers connected by two webs If there is a cutout in the middle a rule ofthumb is that the calculated stress state will be inaccurate within cl 3X the height ofthe cutout distance between top and bottom stringers where dis distance along the stringer from the edge of the cutout Chapter 5 Flexural Shear Flow in Thin Walled Sections 1 Shear Flow in Open Thin Walled Sections Transverse shear stresses require the determination of Tu and Txy This is not the most efficient coordinate system to use It is more advantageous to apply sn system with s along the contour and n perpendicular to the saxis We apply the argument that 1x vanishes on the boundary and the only remaining component is sz 27 511 Symmetric Thin Walled Section I M Either K 0 orVy0 Myz ampthy V I dx 39 y A0 dA Ax 0 quot 0 q xx A q V Q where Q Aszc 1y The shear flow calculated above is flexural shear flow induced solely by bending stress The horizontal position of V2 is never specified because bending moment My does not depend on the horizontal position of V2 but it does not imply that Vz can be applied arbitrarily if torsional shear stresses are to be avoided 512 Unsymmetric Thin Walled Section on kyMz 1szy y kzMy 1le z I o II c I k k 11 12yz k IyIz 12 q39 kyVy kszz QZ kZVZ kylVy Qy Q 39ydA Qy sz A A The equations shown above are for unsymmetric thinwalled sections under bidirectional bending 512 Multiple Shear Flow Junctions ln multicell thin walled sections there are junctions where 3 or more shear flows meet In these instances we may consider equilibrium of shear flows ie inflow outflow For sections consisting of stringers and thin sheets this assumption is not valid F or Junction of 3 walls q 11 qz For stringer web junction M W a g y Vz I dx x y z c I 7 73 q1 q2 Shear flow contour s indicates the assumed direction of the shear flow The contour must begin from a free edge top or bottom free edge is typically chosen For an lsection 5 contours can 28 io 3900 be selected to calculate the shear flow in the flanges separately While calculating the shear flow in the web the areas of both top flanges must be included Shear Center in Open Sections The flexural shear flow resulting from the shear force has a definite resultant force location This is called the shear center or the center of twist If a torque is applied at the shear center the beam will twist without bending Also if a shear force is applied at the shear center the beam with bend without twisting To find the shear center of an open section follow the steps listed below Find 1 Centroid yvzc 2 IIZIyz 3 q39 kyVy kszz QZ kZVz ky39Vy Qy 4R 39 e Eqix momentarm If the shear force V2 is applied through the shear center this shear flow is the complete response of the structure If it is not applied through the shear center then an additional torque Vzd is to be included in the analysis The shear stresses resulting from this additional torque must be added to the flexural shear stress Since open thin walled sections are weak in torsion it is desirable to apply the shear force through the shear center Simple rule for determining a shear center If a section is symmetric about an axis then the shear center lies on that axis Closed Thin Walled Sections Closed sections take both shear forces and torques For closed sections there are no free edges For closed sections we split the problem into two distinct parts That is the actual flow is the superposition of qs and an unknown constant shear flow qo The flexural shear flow is the flow produced by the shear force in the open section obtained by cutting the wall longitudinally at an arbitrary point mm Vzd 22 momemt produced by q about the x axis 29 In The shear flow qs should produce the resultant force Vz about any axis parallel to the xaxis In fact it is the shear flow qs that produces the shear force since the resultant force of the closed constant shear flow qo vanishes The equation above is used to determine qo 531 Shear Center If the applied shear force Vz passes through the shear center ie dysc then the resulting shear flow is pure flexural shear which produces no twist angle The following equation helps us to determine qo in terms of ysc 531 Statically Determinate Shear Flow At any crosssection of a thinwalled beam the shear flow must result in the same resultant forces and moments as the applied ones 2154 25Vy2MVeyVyez Here ey and ez are the distances of Vy and V2 from the axis about which the moments are taken For some sections shear flow can be determined from these equations alone This type of shear flow is statically determinate Closed Multicell Sections Shear flow in a single cell beam can be analyzed by making a fictitious cut so that it can be treated like an open section with an existing constant shear flow qo The shear flow q is obtained from the relations for an open section subjected to applied shear forces and qo is obtained from the requirement that the moment produced by the total shear flow q qo must be equal to the moment produced by the applied shear forces For multicell thin wall crosssections we make a cut in the wall in each cell to make the entire section open For each cell the constant shear flow qi must be added to the shear flow q calculated forthe open section The n equations are provided by the n1 compatibility equations for the angle of twist and the 1 equation relating the moment of applied shear forces to the total resultant moment of all the shearflows in the cells Rules of thumb 1 In making the cuts no part of the section should be completely cut off 30 2 Each wall can be covered only by a single contour 3 Begin each contour from the cut location where q 0 Chapter 6 Failure Criteria for Isotropic Materials Modes of failure of a structure can be 1 material failure or 2 structural failure Material failure is considered here Astate of plane stress is assumed and only isotropic materials are considered 61 Strength Criteria for Brittle Materials In general brittle materials exhibit linear stressstrain curves and have small strains to failure Their uniaxial strengths can be determined by simple tensioncompression tests 2 criteria have been proposed a Maximum Principal Stress Criterion Failure would occur if any of the principal stresses reaches ultimate strength The line forming the square box is called the failure envelope Any stress state within the envelope would not produce failure 01200 for 01 gt0 01 sow for 01 lt 0 02 20 for 02 gt0 02 save for 02 lt 0 b CoulombMohr Criterion Many experiments suggest that presence of 52 could reduce the ultimate value of 51 This criterion accounts for the interaction The CoulombMohr criterion for a state of plane stress s30 is as follows 31 1stQuadrant 1 0 02 0 3rdQuadrant Il 0UC 02 0UC 0 0 2nd Quadrant 2 1 1 0 one o 039 4th Quadrant 1 2 1 0W OUC For pure shear 01 02 U 1 M 2 Weld Criteria for Ductile Materials a Maximum Shear Stress CriterionTresca Weld Criterion in Plane Stress 01 02 2 2cy loll 2 2 7 y lozl 2 2 y For simple tension o1 02 2 0y 3901 2 0y lozl 20y b Maximum Distortion Energy Criterionvon Mises Weld Criterion For an isotropic material deformation can be separated into dilatation volume change and distortion shape change It is easier to produce yielding in a solid under plane stress than under plane strain The equations describing the criteria are as shown below Total strain energy density for plane stress is 32 W onsn Hays Hwy 2G Maximum Distortion Energy 2 0 J2 on 0W2 0 2 0 2 6txy2or J2 012 022 1102 Von Mises for Plane Stress 1 J2 E0 Von Mises for Plane Strain 1 J2 1 v v201 022 1 2v 20102 3 Fracture Mechanics XV a 12 Holezab 00 b Griffith proposed that new crack surfaces are formed during crack extension Creation of crack surfaces requires a supply of energy from the system The strain release rate is denoted by G and given by the relation shown below The crack growth criterion based on strain energy balance concept defines Go as the fracture toughness of the material 33 dW d U dU P62 al tda tda Axial Element 2 U PL 2EA Beam Element L 2 U fM dx 0 2E1 Torsion Member 2 U T L 2GJ Thin Walled Closed Section L zds ZG t U 4 Stress Intensity Factor Symmetric LoadingMode Fracture If the loading and geometry of the cracked structure are symmetric with respect to the cracked surface then the stress field is given by o K cosg1 singsin xx 4an 2 2 2 0 K cosg1singsin W Van 2 2 2 K e e 3e 17 SlIl COS COS W 42 2 2 2 In nitePanel 2K1 cox31a n139 1l 39 39 Fracture 34 If the loading is antisymmetric with respect to the cracked surface then the stress field is given by K1 6 6 3B Un sm 2cos cos 1an 2 2 2 K 11 36 sm cos cos W 4231 2 2 2 1 K cosg 1 singsin xy 4an 2 2 2 In nite Panel KI 10 7 a 039 Relation between Kand G 2 G 1 PlaneStress G1 K PlaneStress K2 1 12 1 E 22 GHKHU Plane Strain Plane Strain Mixed Mode Fracture 35 36


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