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# Mechanics of Deformable Solids ME 222

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This 62 page Class Notes was uploaded by Princess Rolfson on Saturday September 19, 2015. The Class Notes belongs to ME 222 at Michigan State University taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/207537/me-222-michigan-state-university in Mechanical Engineering at Michigan State University.

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Date Created: 09/19/15

IME 222 Mechanics of Deformable Solids Lecture 14 Analysis and Design of Beams for Bending Shear and BendingMoment Diagrams Relations Among Load Shear and Bending Moment Department of Mechanical Engineering A Loos 141 IME 222 Mechanics of Deformable Solids Introduction Department of Mechanical Engineering A Loos Objective Analysis and design of beams Beams structural members supporting loads at various points along the member Transverse loadings of beams are classi ed as concentrated loads or distributed loads Applied loads result in internal forces consisting of a shear force from the shear stress distribution and a bending couple from the normal stress distribution Normal stress is often the critical design criteria a 2 0 2 W10 2 M x 1 m 1 S Requires determination of the location and magnitude of largest bending moment 142 ME 222 Mechanics of Deformable Solids Introduction Classi cation of Beam Supports Stuticully Dch39rmilmto i g L V a Bozlms I LA L gt 1 Simply supported beam 17 Overhanging beam statically v39 lndotvrmimltv it 2 r Booms a Halal L 4 Lil Continuous beam 2 Beam fixed atone end f Fixed beam and simply supported at the other end Department of Mechanical Engineering A Loos 14393 IME 222 Mechanics of Deformable Solids Shear and Bending Moment Diagrams 0 Determination of maximum normal and shearing stresses requires identi cation of maximum internal shear force and bending couple 0 Shear force and bending couple at a point are determined by passing a section through the beam and applying an equilibrium analysis on the beam portions on either side of the section 0 Sign conventions for shear forces Vand V and bending couples M and M 1 Internal forces positive shear and positive bending moment Department of Mechanical Engineering A Loos 14394 IME 222 Mechanics of Deformable Solids Sample Problem 51 SOLUTION Treating the entire beam as a rigid 9 T 0 m 40 m body determme the react10n forces C 0 Section the beam at points near supports and load application points Apply equilibrium analyses on resulting freebodies to determine For the timber beam and loading internal shear forces and bending shown draw the shear and bend couples moment diagrams and determine the maximum normal stress due to bending 25 111lt 3 m m 80 mm Identify the maximum shear and bendingmoment from plots of their distributions Apply the elastic exure formulas to determine the corresponding maximum normal stress Department of Mechanical Engineering A Loos 14395 ME 222 Mechanics of Deformable Solids Sample Problem 51 20kl 40kN D 43 l l 7 I 239 quot3 4 5 3 in Treat1ng the ent1re beam as a r1g1d body determme 39 46 kN 00 H 20 3939 3 IIIquot3 quot1 the react10n forces Ml from ZFy 0 2MB RB 46kN RD 14kN vI 20 W 0 Section the beam and apply equilibrium analyses L M on resulting freebodies szzo 20kN V10 Vlz ZOkN 2M 0 20kN0mM1 0 M1 0 sz0 20kN V20 V22 20kN 2M2 0 20kN25mM2 0 M2 50kNm V3 26kN M3 50kNm V4 2 26kN M4 28kNm V5 14kN M5 28kNm V6 14kN M6 0 M b39 mm V9 Department of Mechanical Engineering A Loos 146 IME 222 Mechanics of Deformable Solids Sample Problem 51 0 Identify the maximum shear and bending moment from plots of their distributions l 3 4 156414 Vm26kN MmMB50kNm 1 2114 j 5 46kN 3 L mgtlt m gt 0 Apply the elastic exure formulas to determine the corresponding maximum normal stress 26 kN 1 2 1 2 S gbh g0080m0250m 83333gtlt106 m3 WB 50x103Nm 0 2 2 m 6 3 S 83333x10 m am 2 600gtlt106 Pa Department of Mechanical Engineering A Loos 14397 IME 222 Mechanics of Deformable Solids Sample Problem 52 SOLUTION 10 kips I 8 Pt I 3 ft 2 ftl 3 ft 0 Replace the 10 klp load W1th an equivalent forcecouple system at D Find the reactions at B by considering the beam as a rigid body 3 kipsft l D 0 Section the beam at points near the support and load application points The structure shown is constructed of a Apply equilibrium analyses on WlOX112 rolledsteel beam 61 Draw resultmg ee39bOdles to determme the Shear and bendingmomem diagrams internal shear forces and bending for the beam and the given loading 9 Couples determine normal stress in sections just to the right and left ofpoint D 0 Apply the elastic exure formulas to determine the maximum normal stress to the left and right of point D Department of Mechanical Engineering A Loos 14398 IME 222 Mechanics of Deformable Solids Sample Problem 52 3 kipsl39t 1 320 kipti 318 PR SOLUTION wquot H 3 0 Replace the 10 kip load With equivalent force couple system at D Find reactions at B 34 kips Section the beam and apply equilibrium analyses on resulting freebodies FromAtoC ZFy0 3x V0 V 3xkips 1 2 le0 3x xM 0 M l5x k1pft FromCtoD szzo 24 V0 V 24kips 2M20 24x 4M0 M96 24xkipft From D to B V 34kips M 226 34xkip ft V l n I 1 Department of Mechanical Engineering A Loos 14399 IME 222 Mechanics of Deformable Solids Sample Problem 52 0 Apply the elastic exure formulas to determine the maximum normal stress to the left and right of point D 10 iP5 34 kjPS From Appendix C for a W10X112 rolled I steel shape S 126 in3 about the XX axis 3 kipsft l 39 20 kip ft 318 kip ft T0 the left of D o m 2M2 20161flp3 m am 160ksi S 126m T0 the right of D M p cm ZMZM am 141ksi S 1261113 8ft 11ft 16ft h 7 961dpft 168kipvft 318 kip ft Department ot39Mechamcal bngmeermg A Loos 143910 IME 222 Mechanics of Deformable Solids Relations Among Load Shear and Bending Moment Department of Mechanical Engineering A Loos Consider a simply supported beam AB carrying a distributed load w per unit length Let C and C be two points of the beam a distance Ax from each other Detach portion of beam CC and draw FBD Relationship between load and shear sz 0 V VAV wa0 AV wa dV w dx Slope of the shear curve is negative Numerical value of the slope at any point is equal to w at that point 14 11 IME 222 Mechanics of Deformable Solids Relations Among Load Shear and Bending Moment ll Integrating the equation between points C and D xD VD VC jw dx x C VD VC area under load curve between C and D Equations are not valid At a point where a concentrated load is applied The shear curve is discontinuous at such a point When concentrated loads are applied between C and D Equations should be applied between successive concentrated loads Department of Mechanical Engineering A Loos 1439 IME 222 Mechanics of Deformable Solids Relations Among Load Shear and Bending Moment 0 Relationship between shear and bending moment ZMCr o MAM M VAxwa o 1 2 AM V Ax 3w Ax Z V dx 0 Slope of the bendingmoment curve is equal to the value of the shear 0 V 0 at points where M is a maximum 0 Integrating equation between points C and D xD MD MC 2 j de xC MD MC area under shear curve between C and D 14 13 Department of Mechanical Engineering A Loos IME 222 Mechanics of Deformable Solids Relations Among Load Shear and Bending Moment 0 Relationship between shear and bending moment Equations are not valid if a couple is applied at a point between C and D Equations are valid when concentrated loads are applied between C and D o The shear curve will always be one degree higher than the loading curve 0 The bendingmoment curve will always be one degree higher than the shear curve and two degrees higher than the loading curve Department of Mechanical Engineering A Loos 1439 IME 222 Mechanics of Deformable Solids Sample Problem 53 SOLUTION 20 kips 12 kips 15 killsll Taking the entire beam as a free body determine the reactions atA and D B C lt 8 ft gti0 ft Draw the shear and bending moment diagrams for the beam and loading shown J 0 Apply the relationship between shear and load to develop the shear diagram gtlt 8ft 6ftgt 0 Apply the relationship between bending moment and shear to develop the bending moment diagram Department of Mechanical Engineering A Loos 1439 IME 222 Mechanics of Deformable Solids Sample Problem 53 Pym SOLUTION A 0 Taking the entire beam as a free body determine the B quot13 reactions atA and D 1l39t gtlt 8ft EMA 0 5 ll gtlt 8 ll gt a 12 kip 15 upsn 0 D24 ft 20 kist6 ft 12kipsl4ft 12 kips28 ft D 26 kips E i Z Fy 0 395 I E 3 I 0 Ay 20 kips 12 kips 26 kips 12 kips ll Rips E Ay 18 laps M 0 Apply the relationship between shear and load to Is my develop the shear diagram 39kips 18 d V w dV 2 w dx 108 12 43 CbC 716 zero slope between concentrated loads H40 linear variation over uniform load segment 14 Department 01 Mechanical Engineering A Loos 143916 IME 222 Mechanics of Deformable Solids Sample Problem 53 IS kips V ki 5 18 20 kips 12 kips 1 I 15 kips 4amp M Lip ft 108 14 Department of Mechanical Engineering A Loos Apply the relationship between bending moment and shear to develop the bending moment diagram zy dx szde bending moment atA and E is zero bending moment variation between A B C and D is linear bending moment variation between D and E is quadratic net change in bending moment is equal to areas under shear distribution segments total of all bending moment changes across the beam should be zero 1417 IME 222 Mechanics of Deformable Solids Sample Problem 55 100 Draw the shear and bending moment diagrams for the beam and loading shown Department of Mechanical Engineering A Loos SOLUTION Taking the entire beam as a free body determine the reactions at C 0 Apply the relationship between shear and load to develop the shear diagram 0 Apply the relationship between bending moment and shear to develop the bending moment diagram 1418 IME 222 Mechanics of Deformable Solids Sample Problem 55 SOLUTION 50 0 Taking the entire beam as a free body determine the reactions at C ZFy0 w0aRC RCw0a Cl Cl AI 0lwch Hw A4 lwch ZC 20E 3 C C 20E 3 Results from integration of the load and shear distributions should be equivalent twig iwww 0 I x 0 Apply the relationship between shear and load to develop the shear diagram i a x x2 a I 21100 VB VA jw0 l jdx W0 36 0 a 261 0 VB w0a area under load curve No change in shear between B and C Compatible With free body analysis Department of Mechanical Engineering A Loos 1419 IME 222 Mechanics of Deformable Solids Lecture 27 Method of Superposition Application of Superposition to Statically Indeterminate Beams Department of Mechanical Engineering A Loos 27391 IME 222 Mechanics of Deformable Solids Method of Superposition Consider a beam subjected to several concentrated or distributed loads Compute separately the slope and de ection caused by each of the given loads The slope and de ection due to the combined loads are obtained by applying the principle of superposition and adding the values of the slope or de ection corresponding to the various loads Department of Mechanical Engineering A Loos 27392 IME 222 Mechanics of Deformable Solids Appendix D Beam De ections and Slopes l Equation of Elastic Curve y P x3 3Lx2 6E1 Maximum De ection Slope at End 3 2 PL PL 2 6 L ymaX 3E1 2E1 Department of Mechanical Engineering A Loos 27393 IME 222 Mechanics of Deformable Solids Appendix D Beam De ections and Slopes B x c v mux I L x 01315 1 L7 L i P EL ymm V M 39 L 439 E h B x A B A 4191 A i ymax Department of Mechanical Engineering A Loos 27394 IME 222 Mechanics of Deformable Solids Example 907 0 Determine the slope and de ection at D for the beam and loading shown knowing that the exural rigidity of the beam is E 100 MNmZ The slope and de ection at any point of the beam can be obtained by superposing the slopes and de ections caused by the concentrated load and the distributed load lSUkN P 130kN 2t kNm u 20 kNm Case 5 Appendix D Case 6 Appendix D Department of Mechanical Engineering A Loos 27395 IME 222 Mechanics of Deformable Solids Sample Problem 97 B For thebeam and loading shown determlne the slope and de ectlon at U2 point B L L2 gt SOLUTION Superpose the deformations due to Loading I and Loading I as shown Loading 1 Loading II Department of Mechanical Engineering A Loos 27396 IME 222 Mechanics of Deformable Solids Sample Problem 97 Loading 1 Loading emf g mpg Loading I lt90 1 g ye 1 Wdi m quot In beam segment CB the bending moment is A C WE zero and the elastic curve is a straight line 2 Hill J L3 W llt 112 gtilt L2 63 6CH 39 48E y I Ker n 1 l quot l WL4 WL3 WWII 7WL4 128E 48E 2 384E x 0 1 Z We Department of Mechanical Engineering A Loos 27397 IME 222 Mechanics of Deformable Solids Sample Problem 97 Loading 1 Loading I Combine the two solutions WL3 M3 6E1 48E 6B QB1 QBH wL4 7wL4 yB V191 VBH JF 384E Department of Mechanical Engineering A Loos 7wL3 B 48E 41wL4 yB 384E IME 222 Mechanics of Deformable Solids Application of Superposition to Statically Indeterminate Beams 0 Method of superposition may be applied to determine the reactions at the supports of statically indeterminate beams Designate one of the reactions as redundant and eliminate or modify the support 0 Treat the redundant reaction as an unknown load which together with the other loads must produce deformations compatible with the original supports The slope or de ection at the point Where the support has been modi ed or eliminated is obtained by computing separately the deformations caused by the given loads and by the redundant reaction and by superposing the results Department of Mechanical Engineering A Loos 27399 IME 222 Mechanics of Deformable Solids Example 908 0 Determine the reactions at the supports for the prismatic beam and loading shown I l 1 l B g L yl 0 Consider the reaction at B as redundant and release the beam from the support The reaction R B is considered as an unknown load and Will be determined from the condition that the de ection of the beam at B must be zero 0 The solution is carried out by considering separately the de ection yBw caused at B by the uniformly distributed load w and by the de ection yBR produced at the same point by the redundant reaction R B Case 2 Appendix D Case 1 Appendix D Department of Mechanical Engineering A Loos 273910 IME 222 Mechanics of Deformable Solids Sample Problem 98 l l l l l l m c For the uniform beam and loading shown B determine the reaction at each support and lt 2L3 lL3 the slope at end A L SOLUTION 0 Release the redundant support at B and nd deformation 0 Apply reaction at B as an unknown load to force zero displacement at B WNW HG Ayn 1L s L 2L3 gt Eli3 L 2143 i L3 gt L 2143 lt 39i43 gt 5 s r B I u U l mu yum 2711 Department of Mechanical Engineering A Loos IME 222 Mechanics of Deformable Solids Sample Problem 98 El 0 Distributed Loading 4 2Lx3 L3x w lx 24E VB U At p01nt Bx3L 2L3 LL3l w 2 4 2 3 3 2 L 2L L L L yBW 24E 3 j 3 j 3 j 4 001132 E1 04 Mb Department of Mechanical Engineering A Loos 2739 IME 222 Mechanics of Deformable Solids Sample Problem 98 0 Redundant Reaction Loading For 61sz and 92L 2 2 RB 2 L L 03 3EIL3 3 y B 2001646RZIL 6101i UBM Department of Mechanical Engineering A Loos 2739 IME 222 Mechanics of Deformable Solids Sample Problem 98 mm 4 For compatibility with original supports yB 0 wL4 RBL3 0 001132 001646 yBw yBR E E RE 2 0688wL T 0 From statics IRA 0271wL T RC 00413wL T Department of Mechanical Engineering A Loos 2739 IME 222 Mechanics of Deformable Solids Lecture 5 MultiaXial loading Shearing strain Department of Mechanical Engineering A Loos 5391 IME 222 Mechanics of Deformable Solids Stress Under General Loadings 0 Consider a body subjected to several r loads P1 P2 P3 etc At some point Q Within the body pass a section through Q using a plane parallel the yz plane The portion of the body to the left of the section is subjected to some of the original loads and to normal and shearing forces distributed over the section Define AFX as the normal forces acting on small area AA surrounding point Q De ne AVX as the shearing forces acting on small area AA surrounding point Q I I The superscript x is used to indicate that the 39 39 forces AFX and AVX act on a surface perpendicular to the x axis Department of Mechanical Engineering A Loos 5392 IME 222 Mechanics of Deformable Solids Stress Under General Loadings Department of Mechanical Engineering A Loos Resolve AVX into two component forces AV and AV in the directions parallel to the y and z axes respectively Dividing each force by the area AA and letting AA approach zero de nes the three stress components AFx 039 lim x AA gt0 AA AVx AVx y z T 11m T 11m xy AA gt0 xz AA gt0 AA The rst subscript in 0x Txy and sz is used to indicate that the stresses are exerted on a surface perpendicular to the x axis The second subscript in Txy and 1x2 identi es the direction of the component 53 IME 222 Mechanics of Deformable Solids Stress Under General Loadings y 0 The stresses are exerted on a surface perpendicular to the positive x axis 0 The normal stress 0x is positive if the corresponding arrow points in the positive x direction The shearing stress components Txy and sz are positive if the corresponding arrows point respectively in the positive y and 2 directions Department of Mechanical Engineering A Loos 5394 IME 222 Mechanics of Deformable Solids Stress Under General Loadings The analysis can be performed by considering the portion of the body located to the right of the vertical plane 9 through Q o The same magnitudes but opposite directions are obtained for the normal and shearing stresses AFX AVyX and AVZX The stresses are now exerted on a surface perpendicular to the negative x axis A normal stress 0x is positive if the corresponding arrow points in the negative x direction Similarly positive signs for Txy and 1x2 will indicate that the corresponding arrows point in the negative y and 2 directions respectively Department of Mechanical Engineering A Loos 55 IME 222 Mechanics of Deformable Solids Stress Under General Loadings Passing a section through Q parallel to the zx plane de nes the stress components 0y 52 and ryx Passing a section through Q parallel to the xy plane de nes the stress components 02 12x and 12y Department of Mechanical Engineering A Loos 5396 IME 222 Mechanics of Deformable Solids Stress Under General Loadings y N Department of Mechanical Engineering A Loos To facilitate the Visualization of the stress condition at Q consider the small cube of side a centered at Q and the stresses exerted on each of the six faces of the cube 0 The nine 9 stress components are 0x rxy and 1x2 on the face perpendicular to the x axis 0y ryx and T Z on the face perpend1cuim t0 the y ax1s 02 In and 12y on the face perpendicular to the z axis Equal and opposite stress components act on the hidden faces 57 IME 222 Mechanics of Deformable Solids Stress Under General Loadings Department of Mechanical Engineering A Loos Important relations among shearing stress components Consider a freebody diagram of the small cube centered at point Q The normal and shearing forces acting on the faces of the cube are determined by multiplying the corresponding stress components by the area AA of each face The combination of forces generated by the stresses must satisfy the conditions for equilibrium 2120 EM 2120 ZMX 0 ZMy 0 2M 0 58 I ME 222 Mechanics of Deformable Solids Stress Under General Loadings 7 AA T Adi w 39 Uxlt ir a A Only 6 components of stress are required to de ne the complete state of stress Department of Mechanical Engineering A Consider the moments about the z axis The only forces With moments about the z axis are the shearing forces 2M2 0 TxyAAa TyxAAa 0 Conclude that the y component shearing stress on a face perpendicular to the x axis is equal to the x component of the shearing stress exerted on a face perpendicular to the y axis Txy Tyx Moments about the x axis give ryz 72y Moments about the y axis give sz 22x Loos 5399 I ME 222 Mechanics of Deformable Solids Poisson s Ratio Department of Mechanical Engineering A Loos Consider a homogeneous slender bar of cross sectional areaA subjected to an axial load P directed along the x axis The normal stress in the bar If the elastic limit of the material is not exceeded then Hooke s law 0x Bag is valid and the normal strain in the bar is Note that the normal stresses 0y and 02 on the faces perpendicular to the y and z axes respectively are zero The corresponding normal strains 8y and 82 are not zero 510 I ME 222 Mechanics of Deformable Solids The elongation produced by the axial tensile force P in the direction of the applied force is U accompanied by a contraction in the transverse directions Poisson s Ratio Homogeneous material mechanical properties are independent of position Isotropic material mechanical properties are independent of direction For a homogeneous and isotropic material 8y8Z 0 8y and 82 are referred to as the lateral or transverse strains De ne Poisson s ratio v Greek letter nu 5y 8z 8x 8x lateral strain axial strain Department of Mechanical Engineering A Loos 53911 I ME 222 Mechanics of Deformable Solids Uniaxial Loading 0 Consider the homogeneous and isotropic bar of cross sectional areaA subjected to an axial load P directed along the X axis P 0x 039 02 0 0 The following relations fully describe the condition of strain 039 VO39 A 8x xand8y82 x E E Department of Mechanical Engineering A Loos 53912 I ME 222 Mechanics of Deformable Solids MultiaxialmILoading Ill 39 Department of Mechanical Engineering A Loos Consider a structural element subjected to loads acting in the directions of the three coordinate axes and producing normal stresses 0x 0y and 02 The condition is referred to as multiaxial loading no shearing stresses are present Consider an element of an isotropic material in the shape of cube The side of the cube is equal to unity Under multiaXial loading the element will deform into a rectangular parallelepiped of sides equal to 18x 18y and 16 513 I ME 222 Mechanics of Deformable Solids Generalized Hooke s Law 0 For an element subjected to multiaXial loading the normal strain components resulting from the Ty stress components may be determined from the principle of superposition This requires 1 strain is linearly related to stress 2 deformations are small 0 Generalized Hooke s law for multiaxial loading of a homogeneous isotropic material Department of Mechanical Engineering A Loos 53914 IME 222 Mechanics of Deformable Solids Sample Problem 25 A circle of diameter d 9 in is scribed on an unstressed aluminum plate of thickness t 34 in Forces acting in the plane of the plate later cause normal stresses 6X 12 ksi and OZ 20 ksi For E 10x106 psi and v 13 determine the change in a the length of diameter AB b the length of diameter CD and c the thickness of the plate Department of Mechanical Engineering A Loos 53915 IME 222 Mechanics of Deformable Solids SOLUTION Apply the generalized Hooke s Law to nd the three components of normal strain 8 XVayvaz V E E E 1 10gtlt106psi 12ksi 0 20ksi 0533 gtlt103inin vo 039 V0 8 xy z yEEE 1067 X 10 31mm V0quot V0quot 039 8 E E E 1600gtlt 10 31mm 0 Evaluate the deformation components 53M 2 excl 0533 gtlt103ininX9in 53M 2 48x10 3in 0131 2 52d 2 1600gtlt10 3ininX9in 0131 2 144x10 3m 5t 2 5y 2 1067 gtlt103ininX075m 5t 2 0800 x10 3in Department of Mechanical Engineering A Loos 53916 IME 222 Mechanics of Deformable Solids Shearing Strain In the general state of stress shearmg stresses rxy ryz sz W111 be present Small deformations shearing stresses have no direct effect on normal stress Fig 246 Shearing stresses deform a cubic element into an oblique parallelepiped The corresponding shear strain is quanti ed in terms of the change in angle between the sides Fig 247 I Txy fiyxy Department of Mechanical Engineering A Loos 517 ME 222 Mechanics of Deformable Solids Shearlng Straln Consider U Cube with slides of unit length Txy and ryx are applied to the faces of the element perpendicular to the x and y axes a 1 The element deforms into a rhomboid of side equal to one Two of the angles formed by the four faces under stress are reduced from 712 to 72 7 yxy while the other two are increased from 722 to 712 yxy The small angle yxy Greek letter gamma expressed in radians de nes the shearing strain Shearing strain is positive when the angle formed by the two faces oriented along the positive x and y direction decreases Fig 247 quot Similar for yz and xz planes Department of Mechanical Engineering A Loos 53918 IME 222 Mechanics of Deformable Solids Hooke s Law for Shearing Stress and Strain U 1 0 A plot of shear stress vs shear strain is similar to the previous plots of normal stress vs normal strain except that the strength values are approximately half 0 For a homogeneous isotropic material Where the values of shearing stress do not exceed the x proportional limit in shear z36y G 7x 0 G is the modulus Ofl igidily or shear modulus units ofpascals 0r psi 13E lt G lt 12E 1 Department of Mechanical Engineering A Loos 53919 IME 222 Mechanics of Deformable Solids Shearing Strains r b Department of Mechanical Engineering A Loos Shear in the yz plane Tyz sz Shearing strains yyz change of angle formed by faces under stress Tyz G yyz Shear in the xz plane TXZ TZX Shearing strains yzx change of angle formed by faces under stress sz G yzx 520 IME 222 Mechanics of Deformable Solids Generalized Hooke s Law 0 General state of stress Homogeneous and isotropic 039C y material 5x E E Stresses are below r0 ortional 1mm p p VO39x 0y 6 y 0 Prmc1p1e of superp0s1t10n E E yox VO39 6 2 2 E E Txy Tyz 7 7 2 xy G yz G E 21 v Department of Mechanical Engineering A Loos 521 IME 222 Mechanics of Deformable Solids Relation Among E V and G Department of Mechanical Engineering A Loos An axially loaded slender bar Will elongate in the axial direction and contract in the transverse directions An initially cubic element oriented as in top gure Will deform into a rectangular parallelepiped The axial load produces a normal strain If the cubic element is oriented as in the bottom gure it Will deform into a rhombus Axial load also results in a shear strain Components of normal and shear strain are related E ilv 522 IME 222 Mechanics of Deformable Solids Example 210 2518 I SOLUTION 0 Determine the average angular deformation 0r shearing strain of the block 2 in 0 Apply Hooke s law for shearing stress and strain to nd the corresponding shearing stress A rectangular block of mater1a1 w1th modulus of gidity G 90 ksi is 0 Use the de nition of shearing stress to bonded to two rigid horizontal plates nd the force P The lower plate is xed while the upper plate is subj ected to a horizontal force P Knowing that the upper plate moves through 004 in under the action of the force determine a the average shearing strain in the material and b the force P exerted on the plate Department of Mechanical Engineering A Loos 53923 ux 3 g 5 T im I amp 8gt fig 253 33 gt a g A NC 5 Q r g L gg Z a x MIM M VWV was a 6i 53 Ma a fix 6 w w 3 3 ff gr v amp 43333 27 if 3 5 it Maia gag v i zii 5 wWMi s g WE Siam i 39 3 a Ww wwm 3 mg F g 11 Er mm kit ng mmmmm 1 cm W m k g v WWW 9 3 J g 3 34 x x gr WWW a MWN r 3333 152 i g 1quot g E V Q f 55 153 b 3w a v mmagzwig j 2amp2sz 1 gt55 2 r 323 f 3 x 3 a og gzg gig MWWWWWMXE fw t 3 m a y W E W g fg ya if W m gig quot gmg 42 g i 13 i i 3 123 W g quota m a mbw kMMwwmwgo f 553 mwd m Egg M 2 ME 3 5 S Cquot L N po 3532 jg i4 59 Ey W95 2 w x as IME 222 Mechanics of Deformable Solids Lecture 26 De ection of Beams Deformation of a Beam Under Transverse Loading Equation of the Elastic Curve Statically Indeterminate Beams Department of Mechanical Engineering A Loos 261 IME 222 Mechanics of Deformable Solids Deformation of a Beam Subjected to Pure Bending 0 Consider a prismatic beam subjected to pure bending The beam is bent into an arc of a circle 0 The curvature Ip of the neutral surface can be expressed as M2 E72 EIH Where 0 radius of curvature M2 bending couple Neutral axis 0 Z moment of inertia of the cross section Department of Mechanical Engineering A Loos 26392 ME 222 Mechanics of Deformable Solids Deformation of a Beam Under Transverse Loading P 0 Relationship between the bending moment and curvature for pure bending remains valid AI 7 f v B for general transverse loadings quot quot 1 W I L 0 l a Cantllever beam AB of length L subjected to a concentrated load P at the free end A l E E B 0 Curvature varies linearly with x l p 0 AtthefreeendA 0 PAZOO 4 m A h B L i 0 a b 0 tt esupport p3 03 PL 39 263 Department of Mechanical Engineering A Loos

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