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# STATICS&MECHCOFMATERIALS2 ENGR0145

Pitt

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##### ENGR 0020: Probability and statistics for Engineers I

###### Emily Binakonsky

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###### Class Notes

##### ENGR 0020: Probability and statistics for Engineers I

###### Emily Binakonsky

verified elite notetaker

###### One Day of Notes

##### ENGR 0020: Probability and statistics for Engineers I

###### Emily Binakonsky

verified elite notetaker

###### One Day of Notes

##### ENGR 0020: Probability and statistics for Engineers I

###### Emily Binakonsky

verified elite notetaker

###### Class Notes

##### ENGR 0020: Probability and statistics for Engineers I

###### Emily Binakonsky

verified elite notetaker

This 242 page Class Notes was uploaded by Shanel Lubowitz on Monday October 26, 2015. The Class Notes belongs to ENGR0145 at University of Pittsburgh taught by Staff in Fall. Since its upload, it has received 25 views. For similar materials see /class/229413/engr0145-university-of-pittsburgh in Engineering and Tech at University of Pittsburgh.

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Date Created: 10/26/15

Flexure of Beams Strains and Stresses ENGROI45 Learning Objeutives What Are Beamsl Flexure of Beams Strains and Stresses PMquot ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Depaltment University of Pittsburgh Student Learning Objectives Students should be able to determine gt the exural strains in pure bending gt the exural stesses in pure bending gt the location of the neutral axis Flexure of Beams Strains and Stresses ENGRO 145 Learning Objectives Are Beam s ll 3 Example Simply Supported Beam s i iii iiiijes ENGRO 145 distributed P concentrated w load load A B pinnedJ roller support support gt 3 support reactions 2 forces at A7 1 force at B gt statically determinate more later Example Cantilevered Beam A B fixed support How many support reactions gt 3 support reactions 2 forces and 1 couple at A gt statically determinate Flexure of Beams Strains and Stresses ENGRO 145 Learning Objectives What Are Beams rts Example Statically Indeterminate Beam 5323 233251215 ENGRO 145 P Learning Objectives What Are Beams rts gt 4 support reactions 2 forces and 1 couple at A7 1 force at B gt statically indeterminate more later Questions About Beams in Flexure What are some questions that an engineer should ask When designing a beam in exure gt What are the normal and shearing stresses at each point in the beam What are the largest stresses and where do they occur Will the beam fail gt What is the de ection at each point along the length of the beam What is the largest de ection and where does it occur Is it too large Flexure of Beams Strains and Stresses ENGRO 145 Learning Objectives What Are Beams Supports Design Issues FBDs Flaggura Fl a39wr al Free Body Diagram of the Entire Beam RAY R5 gt Gives the support reactions if statically determinate more later 2F ZF ZMAO gt RAz7RAy7RB gt Note that RM 0 use RA RAy Flexure of Beams Strains and Stresses ENGRO 145 Learning Objectives What Are B39eamsz orbs Flexure of Beams Free Body Diagram of the Portion Left sum andsmessgs or Right of aa ENGROMS Learning Objectives What Are B39eamsz arts m i w 1 gt Recall that RM 0 and RM RA gt Gives the internal bending moment M and shearing force V at m 2F 0 V ZMoO M Normal and Shearing Stresses in Beams There are internal normal stresses am and shearing stresses 739 acting on each orthogonal cross section of the beam 1 4 These internal normal stresses and shearing stresses are related to the internal bending moment M and shearing force V Flexure of Beams Strains and Stresses ENGRO 145 Learning Objectives What Are Supports Design 135 FBDs Beams Normal and Shearing Stresses in Beams 5323 2332512 ENGR0145 Learning Objectives What Are B39eamsz arts 7 L2 E5 dFmamdA FmamdA0 A dV7TdA gt ViTdA A dM7yamdA gt MiyamdA A Note that the resultant normal force Fm is zero Pure Bending of Beams Flexure ofBeams Strains and Stresses ENGRO 145 M Learning Objectives hat Are Beams orbs M M V Pure bending constant bending moment no shearing force gt Assume plane orthogonal cross sections remain plane gt Symmetry gt longitudinal elements deform into circular arcs Deformation in Pure Bending radius of curvature of neutral surface Ax39 V0111 pressed stretched d neutral surface gt Neutral surface gt Am pA0 gt Am pig A0 Flexure of Beams Strains and Stresses ENGRO 145 Learning Objectives at Are 39eams ll or Flexural Strains in Pure Bending 7 AmiAz i piyA0ipA0 Am p A0 61 p gt y gt 0 gt 61 lt 0 compression gt y lt 0 gt 61 gt 0 extension gt y 0 gt 61 0 neutral surface Flexure of Beams Strains and Stresses ENGRO 145 Learning Objemves Flexural Stresses in Pure Bending s i iii iiiij ENGR0145 Assumlng hnear elastlc deformatlon Learning Objectives t Are Beams rt n Iifra axis Maximum Flexural Stress Ey am if p gt The maximum tensile or compressive exural stress occurs at the point on a cross section farthest from the neutral axis EC p Umax where c is the value of y at the farthest point from the neutral axis gt Note that Flexure of Be s Strains and Stresses ENGRO 145 Learning Objectives 6 Beams rt Locating the Neutral Axis gt Recall that Fm fA am dA 0 amdAiampdAi ydA 0 p A A A gt go fA y 1A is the y coordinate of the centroid of A7 measured from the neutral axis gt Fm 0 gt yo 0 gt The neutral axis passes through the centroid of the cross section Flexure of Beams Strains and Stresses Lear What ENGRO 145 ning Objectives Are Beams rt Relation Between Flexural Stresses and the 532n r fe es Bending Moment ENGROMS Learning Objectives gt Recall that M fA iyam dA What Are Beams rt Ey p 0391 E Miy2dA pA gt I I1 fA yz 1A is the second moment of the area with respect to the neutral axis z axis gt I is a geometric property of the cross section Relation Between Flexural Stresses and the 5323 133251 S Bending Moment ENGROMS Learning Objectives gt Curvature of the neutral surface in terms of the bending moment What Are Beams rt E 1 M 7 M 0 pEI gt Flexural stress in terms of the bending moment Combined Static Loading Continued ENGR0145 Combined Static Loading Continued ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Depaltment University of Pittsburgh Student Learning Objectives Students should be able to gt Given the material properties7 equate stresses and strains gt Interpret data from strain gages gt Determine the stresses in thin walled pressure vessels gt Analyze combined axial7 torsional7 and exural loads Combined Static Loading Continued ENGR0145 Learning Objectives Houke s Law Measuring Stirth Pressure Vessels mam Generalized Hooke s Law Consider uniaxial loading in the z direction y 1dy OX4 OX dz 1sxdx 80611 X gt1szm gt 07 then 61 gt 0 and 81781 lt O gt1szm lt 07 then 6m lt 0 and 81762 gt O Combined Static Loading Continued ENGR0145 Learning Objectives Hooke s Law Measuring strain VegasIs Combined Static Generalized Hooke s Law Loading Continued ENGR0145 y Learning Objectives Hooke s Law Measuring Stirth 1ady OX4 OX dz 1sxdx 80611 X gt According to Hooke s law7 the normal strains are 1 1 61 Eam 6y 61 i am Where E is the elastic modulus Young s modulus and 1 is Poisson s ratio Combined Static 39ng Continued Loa 1 Generalized Hooke s Law ENGR0145 y Learning Objectives Hooke s L W 1 dy dy Ox OX dx dz 1sxdx 15zdz X gt The corresponding shear strains are all zero 39wa sz Yzw 0 39 7 c b 01 St 2 Generalized Hooke s Law Loa iiTng chmaanlid Next7 consider the result of a shear stress my ENGR0145 Learning Objectives y Hooke s L w Measuring Stirth Pressure Vessels mam Tl dx dz 3 ny Z gt According to Hooke s law7 the strains are 1 Yzy5739my7 5y5ygzVyz Yzz0 Where G is the modulus of rigidity shear modulus Generalized Hooke s Law Combined Static Loading Continued ENGR0145 gt Let 6m 8127 and 8mg be the normal strains 8m due A Learning Objectives to am Ty and az respectively Hooks L W Measuring strain 39Emssur 1 i 1 511 E017 12 ani 513 7E0 gt Then from superposition7 the normal strain 61 due to all three at once is Vessels 39 l 5m 811 5m2 5m 7 1 1 1 7 E01 an E01 1 Em e ultay azgti Combined Static Generalized Hooke s Law Loading Continued It follows7 Via similar superposition arguments7 that ENGROI Learning 015139s chives 1 am le 7 way 02gt l 5y Elgy 7 VUz 71l l 61 Ebz 7 101 ay 1 Wm Esz l 39sz 57 22 1 Generalized Hooke s Law in Plane Stress For the special case of plane stress Le when 01 Tm sz 0 1 Em EV V Ty 39wa 5712 1 5y EltUy 7 Vam Viiz Yzm 0 1 61 i wm try gt Note that 171 6n6y E 01ay 1 gt 61 7ng 8y Combined Static Loa ing Continued ENGR0145 Learning Obie ccives Hooke s Law Combined Static Generalized Hooke s Law Loading Continued Alternatively solving for the stresses in terms of the ENGROI strains Learning Ob is at We Hooke s Law E 01 mm 7 m my 62gt E 0y W lt1 7 way m 60 E 01 Wm 7 m m 621 Tmy Glsz Tyz 7 G39sz Generalized Hooke s Law in Plane Strain For the special case of plane strain Le when 52 Yzm Yzy 0 am ml1 7 108 l Vall Twill Glme 1 7 16y 116 2 2m0 Hy 1Vx172m W T Ell 1ux172m 0391 5w l 5y gt Note that E mlt gt 0392 1am l 0y amay 51 l 52 Combined Static Loading Co ntinued ENGR0145 Learning Ob is at We Hooke s Law Measuring Strain Fmssure VegasIs Combined Static Relationship Between E G and V Loading Continued ENGR0145 gt The elastic modulus E modulus of rigidity G and Learning Obj cmes Poisson s ratio 1 are not independent That is if Hooks Law you know the value of any two of them then you Measuring Strain can determine the third Em WSW gt To see this consider the case of uniaxial loading in the z direction y l 1gdy dy o lt 0 dx ix 1mm 1 8m7039m 61177039m yin0 E E Relationship Between E 7 G and V Kl n 1 d sy y dy 39 OX4 ox 5136 ix 1mm 1 1 61E0127 5117E039mi VIM0 gt For a nt coordinate system oriented at a 45 angle to the zy coordinate system7 as shown aloove7 the plane stress transformation equations give Tm 7am 7 0y sin 9 cos 9 7 mycos2 t9 7 sin2 0 7 7701 2 Combined Static Loading Continued ENGR0145 Learning obieccives Hooke s Law Measuring Strain Vessels Relationship Between E 7 G and V ty n i 1eydy v dy OX4 ox x 1 dx 6 ii 6 if 0 m Eo39zi 11 E017 39Yzy gt On the other hand7 the plane strain transformation equations give rym 728m 7 8y sin0cos0 39ymycos2 t9 7 sin2 0 51 7 52 71 1 E Um Combined Static Loading Continued ENGR0145 Learning Obie ctives Combined Static Relationship Between E 7 G and V Loading Continued ENGR0145 Learning Ob is at We 1 11 Tnt iggzv Vnt 7 E Um 21V gt Ym7m gt However7 according to Hooke7s law7 1 Vat 57m gt Thus7 it must follow that E E aim E72G1V V7771 Combined Static Strain Gauges Loading Continued ENGR0145 thin Learning Objectives n d Honke s Law con uctlng Measuring Strain wire Pressure Vessels EX 6 1 gt The strain gauge is glued onto the surface of the sample7 at the point where the strain is to be measured gt Assuming that the glued attachment does not fail which is sometimes a problem7 the strain gauge and the wire embedded within it will deform with the surface of the sample Strain Gauges thin conducting wire gt When there is normal strain an on the surface of the sample7 at the point of attachment and in the direction 71 at which the strain gauge is oriented gt The length of the Wire increases or decreases and its cross sectional area decreases or increases gt The electrical resistance in the Wire increases or decreases Combined Static Loading Continued ENGRO 145 Learning Objectives Hooke s Law Measuring Strain 39 Combined Static Straln Gauges Loading Continued ENGR0145 gt The change in resistance of a strain gauge is calibrated With the normal strain an in the Learning Obj ecmes Hooke s Law direction of gauge orientation Measuring Strain gt This calibration is temperature sensitive Pressure VeSSels Exam 1e 1 gt Many different types of strain gauges With Example 2 different Wire materials backing materials gjgrr gfoads environmental coatings etc have been developed quot quot i gt The selection and use of strain gauges is covered in MElO4l1042 Mechanical Measurements l amp 2 V A single strain gauge just measures the normal strain in one direction strain gauges cannot directly measure shear strains gt An array of three strain gauges called a strain gauge rosette is used to determine the complete state of plane strain eigi 575 5y and My at a point on the surface of a sample Actual Strain Gauge Combined Static Loading Continued ENGR0145 Learning Objectives Hooke s Law Measuring Strain Strain Gauge Rosettes V Three strain gauges with orientations 0m 0177 and 00 With respect to the z axis7 measuring normal strains 6a 65 and 80 gt Want to know 8 6y and My Combined Static Loading Continued ENGR0145 Learning Objectives Homke s Law Measuring Strain Strain Gauge Rosettes gt From the plane stress transformation equations 8a 61 cos2 0a 6y sin2 0a My sin 0a cos 0a Eb 8m cos2 0 6y sin2 05 My sini917 cos 0 so 6m cosz 00 6y sin2 00 V sin 00 cos 00 gt 0a 0b and 0c are known 8 8b and so are measured gt three equations can be solved for the three unknowns 8m 8y and My gt Since the surface is in plane stress 1 lt92 z 5y Combined Static Loa ing Continued ENGR0145 Learning Objectives Hooke s Law Measuring Strain Strain Gauge Rosettes Lfombmed Static ading Continued ENGR0145 Example Learning Objectives 0a 0 7 01 45 00 90 Law Measuring Strain It follows from the plane stress transformation Fmquot 9593 equations that 8a 8m 5b 591 l 5y W21 60 8y Solving for 8 6y and My Ex 8a 8y so 39me 25b 7 Sui 50 Strain Gauge Rosettes Example 0a0 0560quot 90120o It follows from the plane stress transformation equations that 6a 8m 1 3 xg 8b 1 15y T Yzy 1 3 xg 60 1 152 71w Solving for 8m 6y and My Ex 6a 2 1 5y 3a 60 7 35a 2 my tgb Sc Combined Static Loading Continued ENGR0145 Learning Objectives Honke s Law Measuring Strain Ftessu Vessels Combined Static Thin Walled Spherical Pressure Vessels Loading Continued ENGR0145 Learning Objectives HOBke s Law Measuring Strain Pressure Vessels 26a e 1 gt Inner radius r gt Wall thickness 75 gt Thinwalled gt t lt 7 gt Internal pressure p Combined Static Thin Walled Spherical Pressure Vessels Loading Continued ENGR0145 Learning Objectives Hooke s L w Measuring Stirth Pressure Vessels E 1 39 1 gt The axial stress in the vessel wall is an gt It is the normal stress on a surface through the wall thickness7 as shown There is no shear stress on such a surface gt The axial stress in a thinwalled spherical pressure vessel is the same everWherei Thin Walled Spherical Pressure Vessels Cm de Stat Loading Continued ENGR0145 Learning Objectives Houke s Lawr Measur ihg strth Pressure Vessels Kample39 1 gt The area over which an is distributed is 7rr t2 7 W72 27th 7rt2 z 27rrt gt Summing forces in the a direction 2 Fa 27rrtaa 7 wrzp 0 Thin Walled Cylindrical Pressure Vessels Li g ii d ENGR0145 Learning on is at We Hooke s Lawr Measuring Strain Pressure Vessels gt Inner cylinder radius r gt Cylinder wall thickness 25 gt Thinwalled gt t ltltT gt Internal pressure p Thin Walled Cylindrical Pressure Vessels Li ngg o iid ENGR0145 Learning Objectives Honke s Law Measuring Strain Pressure Vessels E 1 1 gt The axial stress in the wall of a cylindrical pressure vessel is the normal stress on a plane with normal in the axial direction gt As in the spherical pressure vessel7 it follows that the axial stress in a thin walled cylindrical pressure vessel is Thin Walled Cylindrical Pressure Vessels V The hoop stress in the wall of a cylindrical pressure vessel is the normal stress on a plane with normal in the transverse direction Combined Static Loading Continued ENGR0145 Learning Objectives Houke s L w Measuring Stirth Pfressure Vessels Kample39 1 Combined Static Thin Walled Cylindrical Pressure Vessels Loading Continued ENGR0145 Learning Objectives Houke s Law Measur ihg Stram Pressure Vessels Kample 1 ZFh 2Ltah e 2er 0 Thin Walled Cylindrical Pressure Vessels Combined Static Loading Co ntinued ENGR0145 Learning Hoske rs Law Measuring straw Eressur Vessels E Example Basketball Consider a basketball7 with an inner radius of 6 in7 a wall thickness of 116 in7 and an inner pressure of 15 psi Determine the axial stress in the wall of the basketball gt The basketball is like a thin walled spherical pressure vessel7 so the axial stress in its wall is 7 1 Ta 7 E 15 psi6 in 21 16 in 7 20 psi Combined Static Loading Continued ENGRoms Learning obieccives Hooke s Law Measuring Strain Emsnre Vessels E 1 Example Pipe With a Welded Seam CDmb ned Static Loading Continued ENGROMS Wm 051mm we 16 l J Hooke s Law Measuring Strain Example 1 Example 2 do 39 dLoads 1E l A 12 in diameter pipe with 1000 psi service pressure has a welded seam running axially along its length Determine the minimum wall thickness for the pipe7 such that the normal stress across the weld does not exceed 10 ksi gt The pipe is like a thin walled cylindrical pressure vessel and the normal stress across the weld is the hoop stress 7 I ah 7 7 Example Pipe With a Welded Seam 39 Th E gt ZL39min L t 07Lmax 1000 ps16 1n 06 in 107 000 ps1 min Combined Static Loading Continued ENGR0145 Learning Objectives Hoske s Law I Example Combined Axial Loading Li ngg O iid Torsion and Internal Pressure ENGROMS Learning Objectives Honke s Law Measuring Strain 10 mm Pressure Vessels 100 mm Consider a thin walled cylindrical pressure vessel subjected to a tensile axial load of 200 kN7 a torsional load of 25 kN In7 and an internal pressure of 2 MPa The inner radius is 100 mm and the wall thickness is 10 mm Example Combined Axial Loading Cm de Stat Loading Continued Torsion and Internal Pressure ENGROMS Learning Objectives 10 mm 100 mm a Determine the principal and maximum shearing stresses at a point at the outer radius 10 Show the stresses of part a and their directions on a sketch Example Combined Axial Loading Torsion and Internal Pressure Tensile axial load gt Using a PA and A 71171 7 7 200 X 103 N a 303152 MPa Combined Static Loading Continued ENGR0145 Learning Obie ccives Houke s Law asufquot Combined Static Example Combined Axial Loading Loading continued Torsion and Internal Pressure ENGR0145 Learning Ob is at We Torsional load Hooke s Law asufquot gt Using Tc TcJ and J an 7T4 i 25 x103 Nm011m c 377226 MP T 011m4701 mm a Example Combined Axial Loading Torsion and Internal Pressure Internal pressure Oh gt Using an pTQt and ah prt 20a 7 2 MPaO1 m 10 MP 20 MP 2001 m a Uh a Ta Combined Static Loading Continued ENGR0145 Learning Ob is at We Houke s Law asufquot Example Combined Axial Loading Li i i id Torsion and Internal Pressure ENGR0145 Learning Objectives Comblned loads Hmkgs Law am a m1 303152 10 403152 MPa 0y ah 20 MPa my in 7377226 MPa Example Combined Axial Loading Cm de Stat Loading Continued Torsion and Internal Pressure ENGROMS Learning Objectives a Determine the principal and maximum shearing stresses at a point at the outer radius 7 2 p939 0171472 i T52 30158 i 39066 4 am 6922 MPa am 0172 7891 MPa lt 0mm 0P3Uz0 1 Tm 5amax 7 amin 3907 MPa 7p Example Combined Axial Loading Torsion and Internal Pressure 10 Show the stresses of part a and their directions on a sketch 017 ltanil 2mg 7375quot 2 am 7 0y 0493750 6922 MPa 3016 MPa 1 375 3907 MPa 6922 MPa 891 MPa Combined Static Loading Continued ENGR0145 Learning Ob is at We Houke s Law Flexure of Beams Beam De ections ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Depaltinent University of Pittsburgh Flexure of Beams Beam De ection ENGRO 145 Learning Objeutives Beam Demandn Theory Student Learning Objectives Students should be able to gt Identify the boundary conditions corresponding to a beam s supports gt Determine the equation of the elastic curve by direct integration of the beam de ection equation V Determine the de ection and slope at any point along a beam Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam De ection Theory Example Flexure of Beams Beam De ection The Beam De ection Equation ENGRO 145 Learning Objectives Beam De ection T eor y Example Recall that the exural strain is related to the radius of curvature p of the neutral surface by 6177 p Curvature Bending Moment Relation 113333 ENGR0145 Given the exural strain and the exure formula Learning Objectives 39 M y Beam De ezmn 51 7 O39z if T eony P I Eis f n39ple39 and assuming linear elastic deformation E 0391E8177y gt Flexure of Beams Beam De ection The Equation of the Elastic Curve ENGRO 145 y Learning Objectives Beam De ection neutral surface T Spry I Example R yx gt is the equation of the elastic curve for the neutral surfaceiit gives the de ection of the beam as a function of z gt Horn differential geometry7 1 i o lzyo lm2 E 1 dydzgt2132 The Beam De ection Equation Assuming dydm ltlt 17 1 dzydmz 1 N dzy Z 1 dydz232 Z N W Since we also have it follows that This is the beam deflection equation Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam De ezb ion T eory Eisei wle39 The Beam De ection Equation 133233 ENGRO 145 Learning Objectives Beam De ection Theory Example gt The beam de ection equation is a second order ordinary differential equation for the elastic curve of the beam gt The boundary conditions are determined by the supports gt Note the sign convention gt z is positive to the right gt y is positive upi Flexure of Beams Beam De ection Example Cantilevered Beam W Uniform Distributed Load ENGROMS Learning Objectives Beam Deflection T eor Examples a Determine the elastic curve in terms of w7 L7 E7 and I b Determine the slope at the left end of the beam 6 Determine the de ection at the left end of the beam Example Cantilevered Beam W Uniform WWW Distributed Load Beam De ection ENGRO 145 Step 1 Find the bending moment distribution M Learning Objectives Beam De ection T eory Examples Y i w m llllllllllllllll Lf39 M a 1 V Example Cantilevered Bearn W Uniform Distributed Load Step 2 Identify the boundary counditions gt What are the boundary conditions For a built in support at z L d l0atzL y0and dm Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam De ection T eor Examples Example Cantilevered Beam W Uniform 332233 Distributed Load ENGRO 145 Step 3 Solve the beam de ection equation7 Ely Mm7 for the equation of the elastic curve7 Learning Objectives B saie39rgrDe ection s Examples WE Since M i wzz dzy 1 2 E w 5 d9 3 E13 iiwm Cl Ely iiwm l Clz 02 gt The constants of integration Cl and Cg are determined by the boundary conditions Example Cantilevered Beam W Uniform FBISXMMBWS Distributed Load earn De ection ENGROI45 Step 3 Solve for cont Learning Objectives Beam De ection Theory Examples dy 1 E17 quotwL3 C1 0 dm mL 6 gt Fiom the boundary condition dydm 0 at z L 1 gt 01 61013 1 1 gt EIy i wm l ng3x Cg Example Cantilevered Beam W Uniform FBISXMMBWS Distributed Load earn De ection ENGROI45 Step 3 Solve for cont Learning Objectives Beam De ection Theor Examples gt Fiom the boundary condition y 0 at z L ElyimL 1 4 7 4 7 24wL 6wL 0270 Example Cantilevered Beam W Uniform 11332332 Distributed Load ENGROMS Learning Objectives a Thus7 the equation of the elastic curve is Ba m De ection e Exafnrples 1 i 1 4 1 3 1 4 yiEI 7247mm 6wLm78wL which can be rewritten as w lt 24E y m4 11ng 7 3L4 Example Cantilevered Beam W Uniform Distributed Load 10 Recalling that 7 1 3 1 3 E 7 610m 6wL the slope ie7 dydm at the left end of the beam is just Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam De ection T eor Examples Flexure of Beams Beam De ection Example Cantilevered Beam W Uniform Distributed Load ENGROMS Learning Objectives 6 Recalling that Beam De ection T eor w Examples 4 3 4 7 7 4L 7 3L y 24ml 3 l a l the de ection ie7 y at the left end of the beam is just wL4 wL4 ylz0 E l l Example The Diving Board77 Problem P Determine in terms of P L7 E7 and I7 a The maximum de ection between A and B b The de ection at C Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam De ection Theor Examples Example The Diving Board77 Problem Step 1 Find the bending moment distribution RA Rs EMAR37PO gtRB3P ZFyRARBPO gtRA72P Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam De ection T eor Examples Example The Diving Board77 Problem Step 1 Find the bending moment cont gt For 0 lt z lt L2 ie7 between A and B 12 Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam Beneathn T eor Examples Flexure of Beams Beam De ection Example The Diving Board77 Problem ENGR0145 Step 1 Fmd the bendmg moment cont gt For L2 lt z lt 3L2 ie7 between B and C T eor Examples 2M0 M2Pz73Pltz7gt M72Pz3Pltz7gt Flexure of Beams Beam De ection Example The Diving Board77 Problem ENGR0145 Step 1 Flnd the bendlng moment cont Learning Objectives Beam De ection T eor Examples M 72Pz7 if0ltzltg 72Pz3Pz7 g ltlt Note For reasons that will become clear later7 do not attempt to simplify the expressions for the bending moment Example The Diving Boaer77 Problem Step 2 Identify the boundary counditions gt What are the boundary conditions For simple supports at A and B who yizL2 0 Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam De ection T eor Examples Example The Diving Board77 Problem 332233 ENGRO 145 Step 2 Identify the boundary counditions cont Learning Objectives Beam De ection P Theor Examples gt What are the continuity conditions at B Assuming that the beam does not break or fold over77 at B7 it follows that y and dydz are continuous at x L2 Example The Diving Board77 Problem Step 3 Solve the beam de ection equation7 Ely Mm7 for the equation of the elastic curve7 Mm When 0 lt x lt L2 gt Since M 72Pz when 0 lt z lt L27 dzy EIW 72Pm L EI Ly7ngZCl 0ltzlti dm 2 Ely 13353 01m 02 Flexure of Beams B am De ection ENGRO 145 Learning Objectives eam De ection Theory Examples Example The Diving Board77 Problem Step 3 Find when 0 lt z lt L2 cont Ely7Pm3ClxCg 0ltmlt gt Using the boundary condition yim0 O7 EIy mO 02 0 gt 020 s ElyiPm3Clx Oltxlt Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam ne ecc ion T eor Examples Example The Diving Boaid77 Problem Step 3 Find when 0 lt z lt L2 cont 0ltxlt Ely 13353 01x 2 gt Using the boundary condition yimL2 O7 1 L 3 L EIyizL2 EP lt5 015 0 1 2 Cli PL 1 1 L E 77133 7PL2 0 i j y 3 90 z ltgglt2 Flexure of Beams Beam De ection ENGRO 145 Learning Objectives Beam De ection T eor Examples Example The Diving Board77 Problem 332233 ENGROI45 Step 3 Find when 0 lt z lt L2 cont Learning Objectives Beam De ection T eor 1 1 L Ely 7313353 EPLZgg 0 lt lt 5 Exam es gt Note7 for later use of the continuity conditions at B7 that Emma2 0 d 1 Ii WPLZ dm mL2 6 Flexure of Beams Beam De ection ENGRO 145 Example The Diving Boaer77 Problem Learning Objectives 1 3 1 2 L Ely 7 ing EFL m7 0 lt lt 3 B e ection s Examples a Determine the maximum de ection between A and B gt The maximum de ection occurs Where dydz 0i dy 2 1 2 L E17 7P iPL 0 7 dz 3 1 12 5 mm m PL3 E 7 MFXW 36 7 PL3 7 PL3 yum 36 E1 108E Example The Diving Board77 Problem Flexure of Beams B am De ection ENGR0145 Step 4 Solve the beam de ectlon equatlon7 Ely Mm7 for the equation of the elastic curve7 Mm When L2 lt z lt 3L2 Learning Objectives Beam De ection Theory Examples gt Slnce M 72Pz 3Pm 7 L2 when L2 lt z lt 3L2 dzy L E13772P9 3Pltz7 gt dyi 2 3 L 2 El iiPz 2Pltz72gt 01 1 3 1 L s E13477ng Pltm7 gt 01e02 Lltlt3L 5357 Flexure of Beams Beam De ection Example The Diving Board77 Problem Step 4 Find Mm when L2 lt lt 3L2 cont ENGROI Learning Objectives 1 1 L 3 Beam De ection Ely 7MB t 5P 95 3 t 0190 02 gt From before7 the continuity conditions at B require that EIy mL2 0 and Emmi2 ePL26 1 gt C1EPL2 020 1 1 L s 1 Ely 7M3 EP 35 7 5 EPLZgg Example The Diving Board77 Problem 1 3 1 L 3 1 2 Ely 7 731335 513 3575 EPL m b Determine the de ection at C 1 3 L3 Ely m3L2 PL gt yo 2E1 Flexure of Beams Beam De ection ENGRO 145 Learning Objemves Beam Be ecmo n T eor Examples Flexure of Beams Beam De ection by Superposition ENGR0145 Learning Objectives Flexure of Beams Beam De ections by Superpyisitim Superposition ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Depaltment University of Pittsburgh Student Learning Objectives Students should be able to gt Use a table of standard de ection curves and the superposition principle to analyze the de ection of complex beams Flexure of Beams Beam De ection by Superposition ENGROMS Learning Objectives Superpyis itimh The Superposition Principle w P yx W 3400 P 32206 Flexure ofBeams Beam De ection by Superposition ENGROMS Learning Objectives Flexure of Beams Example 1 Beam De ection by Superposition ENGROMS Learning Objec ve 600 lbft Superpza39i39i imh Emmgl e A Example 1 4 B 1 0 ft 1500 lb Given that E 12 gtlt 106 psi and I 144 i1147 determine the de ection at A Example 1 A B 10 ft 120 in 1500 lb Case 1 Table A49 Pl L P902 PL3 9 m H 71500110 120 3 1 x 0 5001n 312 x 106 lbin2144 in4 Flexure ofBeams Beam De ection by Superposition ENGROMS Learning objeemves Flexure of Beams Beam De ection Example 1 by Superposition ENGR0145 Learning Objectives Superpp39si imh I e 600 lbft 501bin 10 ft 120 in Case 2 Table A719 L wL4 7 m2 2 4L 6L2 i i y 24EI 3 y FL 8E 50110 120 4 W ux 2 In 4 7750111 812 x 1061b1n 1441n Flexure of Beams Example 1 Beam De ection by Superposition ENGROMS Learnmg Objec ve 600 lbft Superpza39i39itimh Emmgl e A Example 1 4 B 1 0 ft 1500 lb 24A 9A1 yAz 500 in 7750 in gt yA 7250111 250 in i Flexure of Beams Example 2 Beam De ection by Superposition ENGROMS 800 lb ft Learning Objec ve er i Given that E 30 gtlt 106 psi and I 100 in 7 determine the de ection at A xure ofBeams earn De ection by Superposition Example 2 FE ENGROMS 1768 lb A C 15 ft 180 in Case 1 Table A719 P f PH PL3 24 L 95 gt MFL 768 1b 180 in 3 m 7 lt gtlt gt 704977111 330 x 106 110111 1001114 Example 2 8001bft 66671bin B 9ft108in 6ft72in Case 2 Tagla A719 lllllllllllllll 4 403 0 LL3 L SEI FL 6E 66671bin72 in4 sz8 77465 x 102 in 30 x 106 lbin2100 m4 Flexure ofBea 5 Beam De ecti by Superposition ENGROMS Learning Objectives uperpp39si imh E Exa Example 2 8001bft 66671bin B 9ft108in Case 2 Tagla A719 IIIIIIIIIIIIIII 4 ng 77L 0 777 L SEI FL 6E 6ft72in 0 32 630 x 106 lbin2100 M i i 3 66671b1n721n 13824X1073 rad Flexure ofBeams Beam De ection by Superposition ENGROMS Learning Objectives Superp ds39i imh Exampl e 39p g 1 Flexure of Beams Example 2 Beam De ection by Superposition ENGR0145 Learnmg Objectives Superpyj sitimh I Ema es Assuming 9 ltlt 17 one has that Ay z 9 Am yA2 7432 0B2LAB 77465 x 102 in 7 13824 gtlt 103 rad108 in 702239 in yA ya yA2 70722111 0722111 Team Problem FgliirirEi ii i by Superposition ENGROMS L m 2 Determine the de ection at C in terms of P7 L7 E7 and I 240 2401 2402 2401 2432 032L2 QPL3L22 3PL3 3PL2 L Flexure ofBeams Composite Beams ENGRO 145 Flexure of Beams Composite Beams ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Depaltinent University of Pittsburgh Student Learning Objectives Students should be able to gt construct the transformed section of a composite beam gt locate a composite beam s neutral axis gt determine the exural stresses gt analyze reinforced concrete beams Flexure ofBeams Composite Beams ENGR0145 Learning Objectives Composite Beams Transformed 8ch Composite Beams De nition Composite beams are beams Whose cross sections are composed of more than one material gt Material 1 has elastic modulus E1 gt Material 2 has elastic modulus E2 Flexure ofBeams Composite Beams ENGR0145 Learning Ob lie at We Flexural Strains in Composite Beams The exural strains are still given by y 81 if p Where p is the radius of curvature of the neutral surface gt However7 we can no longer assume that the neutral axis passes through the centroid of the composite cross section Flexure ofBeams Composite Beams ENGR0145 Learning Ob is at We 312mm 351 Flexural Stresses in Composite Beams Given that 8m77 p and assuming linear elastic deformation gt The exural stress in material 1 is given by E 0391 Elem 77124 gt The exural stress in material 2 is given by E 0392 Ezgm 77224 Flexure ofBeams Composite Beams ENGR0145 Learning Ob is at We Flexural Stresses in Composite BeamsFlexural Stress 7 P P gt Assuming7 for arguments sake7 that E2 gt E17 the distribution of exural stress would like something like 0391 7 Flexure ofBeams Composite Beams ENGR0145 Learning 12 jeccives Composite Beams Transformed srecg Ex p1e Ed Flexure ofBeams Composite Beams Elemental Forces ENGR0145 01 mg 7 72 fig Learning 39Q i o jsccives p 0 Composite Beams Transformed Sect gt The force on an element 1A of material 1 is E dF1 01 M 7719 M p gt The force on an element 1A of material 2 is E ng asz iiydA p Elemental Forces 1F1 i dA ng 7 M p 0 Let n E2E1 gt E2 nE1 E E sz 7L 1 dA 7amp4 p n dA gt The force on an element 1A of material 2 would be the same if material 2 were replaced with material 1 and 1A were changed to 71 1A gt Contribution to the bending moment is the same if the distance y from the neutral axis is maintained Flexure ofBeams Composite Beams ENGR0145 Learning Objectives Composite Beams Transformed Sect Example Rommel 33011net Elemental Forces dFl p dA dF2idA p Let nE2E1 gt E2nE1 ng y dA E 70171 n dA p gt Important when changing 1A to ndA the widening n gt 1 or narrowing n lt 1 must be parallel to the neutral axis gt The transformed section results from applying this to every 1A in material 2 Flexure ofBeams Composite Beams ENGR0145 Learning Objectives Composite Beams Transformed Sect Example Rginfmgcwonsre e Transformed Section Flexure ofBeams Composite Beams ENGR0145 Learning Objectives Composite Beams Transformed Sect Example Rainfmscwomeqe gt An all material l beam with the transformed section as its cross section has the same neutral axis and distribution of exural strains as the composite beam gt The second moment of the transformed section about its neutral axis is I Transformed Section gt The exural stress in the all material l transformed section is M y Tm if gt Thus the exural strain in the all material l transformed section and the composite beam is gm 51 81 Flexure ofBeams Composite Beams ENGR0145 Learning Ob is at We Composite Beam Flexural Stress FlexureofBeams Co mposit Beams ENGR0145 am My Learning Objectives 5w E 7 gm 7 I Composite Beams 1 Transformed Sect Example gt The exural stress in the material 1 component of the composite beam is Reinfmgcwonsmeeie 71E16m am gt gt The exural stress in the material 2 component of the composite beam is E2 02 E26z for now gt 0 E1 Flexure of Beams Example Composite Beams i i i ENGR0145 Consider a wooden beam reinforced With steel plates as shown Learning Objectives Composite Beams Transformed S39ecgi ple 5 X i in F7 stezel 00d 10 in H wood steel EW 2 x 106 psi ES 30 X 106 psi can 2 ksi can 22 ksi Determine the largest permissible bending moment for the composite beam Example Flexure ofBeams Composite Beams ENGR0145 For an all wood transformed sectlon7 Learning Objectives Es 30 X 106 psi Composite Beams n 7 15 Transformed S39ecg EW 2 x 106 psi Example R inf xcg in 155 in 75 in 10m 10m I T1275m11in3 7 69 in10 in3 2569 m4 Flexure of Beams Example Composite Beams ENGR0145 Learning 390 155 in 75 in Composite Beams harls formr d saggy 55m ina 551m 5X in For the wood7 M MC 0 I gW7Ty gt UmMT gt Mi 0 Question 0 7 Example 155 in 75 in 1 5X21I1 For the wood7 M gall 2 ksi2 569 m4 0 5 in gt This is the moment at which the maximum 1028 kipin exural stress in the wood reaches the woods allowable stress Flexure ofBeams Composite Beams ENGRoms Learning objsccives Composite Beams Tysns forme o srecgi Rainfr r39cscwonsmseie Example 155 in 75 in 55m ina 551m 5X in For the steel7 71M nMC a I 039s77ygtg39max I gtM all 710 Question 0 7 Flexure ofBeams Composite Beams ENGR0145 Learning 390 Composite Beams haqs formso soggy Flexure of Beams Example Composite Beams ENGR0145 Learning Objectives 155 in 75 in an Exam 1e 5 5 in R i f r celi jon e 7 na 55 in 1 5 X 2 1n For the steel7 i i 4 M 0811 22 ks125 9 m 685 kipin no 1555 m gt This is the moment at which the maximum exural stress in the steel reaches the steel s allowable stress Example Flexure ofBea Composite Beams ENGR0145 155 in 75 in Learning Objectives Composite Beams I La fform d 8699 ample 55 in d 9 7 na 55 in S ummary gt The steel hits its allowable stress before the wood a Man 685 kip n Reinforced Concrete Beams concrete steel rebar gt Concrete has gt effectively no strength in tension gt elastic modulus EC in compression gt The steel rebar has gt total cross sectional area As gt elastic modulus ES Flexure ofBeams Composite Beams ENGR0145 Learning Obie ctives Composite Beams narisform d 8ch a mple Rsinifegcei Concrets Transformed Section Flex f Bea Co mposit Beams For an all concrete transformed section7 n EsEC ENGROI L earning a at We Composite ems I r rigfotmf d Sect Example BeinQL Egi Concrete Neutral Axis Flexure of Beams Co mposit Beams ENGR0145 b i Learning 39Q l a jsccives X XZ Composite Beams n a rlg fotm d 8ch c 39 39 Example d i X R 939399517 Coygr First locate the neutral axis nAsd bm nAs gt bzz nAsm 7 nAsd 0 inAs Moms QnAsbd gt m f Second Moment Determine the second moment I IC Is Is Isa d 7 z2nAs x d 7 z2nAs gt I bxg nAsd 7 m2 Flexure ofBeams Composite Beams ENGRO 145 Go mposite Beams foi ma 8ng e r goncretg Flexure of Beams Shear Force and Bending Moment Diagrams ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Department University of Pittsburgh Flexure of Beams Shear Force and Bending Moment ENGRO 145 Learning Q bjectixre Shear as Moment n Student Learning Objectives Students should be able to gt determine the support reactions for statically determinate beams gt construct shear force and bending moment diagrams V determine the maximum shear force in a beam and its location gt determine the maximum bending moment in a beam and its location Flexure of Beams S ENGRO 145 Learning Objectives Shear Forces and Bending Moments Fsli iiep iiei Bending Moment ENGRO 145 P W Learning bj ectivej s X gt For each cross section7 there is a shear force V and a bending moment M Signi cance of Shear Forces and Bending Moments gt Recall the exure formula M 24 am77 I gt The maximum exural stress am occurs on the cross section Where M is maximum gt The maximum shear stress 739 occurs on the cross section where V is maximum more later gt We need to be able to determine Vm and Flexure of Beams S ENGRO 145 Learning objectives Sham as Moment 1 n Sign Convention Fsli iiep iiei Bending Moment E 0145 P NGR Example Using Fiee Body Diagrams 3quot 20kN 15 kNm A rd V 4 4 2m 2m m B Determine the shear force and bending moment distributions Vm and Mm7 for values of z between Oand4m 0ltmlt4m Flexure of Beams Shear Force and Bending Moment ENGRO 145 Learning objective Example Using Fiee Body Diagrams Flexure of Bea S Dia rams Step 1 Free body diagram of the entire beam g gives the reactions at A and B since the beam is ENGR0145 statically determinate Learning bg39emvejs Shsar as Momsn 60 kN I 4 20kN 2m V A Lx 6m l 2m RA RB ZMA 72 m60 kNH6 m20 kN8 mRB 0 RB30kN Example Using Fiee Body Diagrams Flexure of Bea S Step 1 Free body diagram of the entire beam Bend39m Moment Diagram i i i i ENGR0145 gives the reactions at A and B Since the beam is statically determinate Learning abjeccivefs 1mm as Momsnt 60kN l I 4 20kN 2m V A 7x 6m l 2m RA R5 ZFy RA RB 7 60 kN 7 20 kN 0 RA50kN Example Using Fiee Body Diagrams FslexureofBea Bending Step 2 Free body diagram of the portion left or right of the cross section at z 0 lt z lt 4 m ENGR0145 Learning bj emvefs 15 kNmx Shzar a Moment I 39 xZ A gig x V SOkN 2F 50 kNi lt15gtm7V0 gt V 50 715 kN z in meters Example Using Fiee Body Diagrams Fsli iiep iiei Bending Moment Step 2 Free body diagram of the portion left or right of the cross section at z 0 lt z lt 4 m ENGROMS Learning bjectixr 15 kNmx llt gt 39 xZ w M 50 kN 2M0 M 7 50 kNz 15 x 0 M 50 7 75352 kNm Example Using Ree Body Diagrams 15 kNm 4 4m 2m 2m V50715zkN M 503577535 kNm gt Atz0V50kNandM0 gt Atm4m V710kNandM80kNm 3quot 20kN B X f0r0ltzlt4m gt V0atm103m Flexure of Beams Shear Force and Bending Moment ENGRO 145 Learning bg39emvefs Flexure of Beam Shear Force and Example Using Fiee Body Diagrams Bengiiriaggrlzirzgnent ENGR0145 y 20 kN Learning abjectivefs 15 kNm l Shear a Momsnb AWE 6 mpg 4 m 2 m 2 m V 50 715x kN 2 for 0 lt z lt 4 m M 50m 7 75m kNm gt dMdz V gt Maximum and minimum values of M occur Where dMdz V 0 Mum Mim1O3 833 kNm Team Problem 60FN lt 20kN 2m V A Lx 6m 2m SOkN 30kN Determine the shear force and bending moment distributions Vm and Mm7 for values of z between 4mand6m4ltmlt6m Answer V 710 kN7 M 120 710w kNm Flexure of Beams Shear Force and Bending Moment ENGRO 145 Learning bg39emvefs Shmr a Momsnc Die smp Qn Exa les Example Shear Force and Bending Moment ng giepiiiei Bending Moment Dlagrarns ENGR0145 Laaining bj ectbm s 32 Mamsnc n Sham Load7 Shear amp Moment Relations Fsli iiep iiei Bending Moment iagram s ENGRO 145 Learning objective Sham as Momsnt n quot c Wei construction Rules 3 39 39 les AxZ mm x Consider a distributed load w7 concentrated load P7 and concentrated couple C at position x along a beam Construction Rules from Force Equilibrium Wmeem Bending Moment ENGR0145 Learning objectives Shsar 5 Moment D n ZFy VL wavgAx P 7 VL AV 0 gt AV P wavgAz 1 lfPOandwOgt wavg0onAzthen V0 gt Shear force V is constant on beam segments With no concentrated loads and no distributed loads 2 lfP7r 07 then AVHP as AmHO gt There is a jump in V across any concentrated load P7 that has the same magnitude and is in the same direction as Pi Construction Rules frorn Force Equilibrium Fslexm fBeams ENGROMS ZFy VL wavgAz P 7 VL AV 0 Learning objectives Elma 2 M mam gt AV P wavgAz 3 If P 0 and w 7 07 then AV wavgAz a 0 as Ax gt gt The shear force V is continuous has no jumps Where there is no concentrated load 4 If P 07 then AVAx mm dV H 1 Air owan W gt V is linear Where w is constant gt V is quadratic Where w is linear Construction Rules frorn Force Equilibrium dV E WW l gt On any segment Where there is no concentrated load7 gt The Change in the shear force V is equal to the area under the distributed loacl curve wt Flexure of Beams Shear Force and Bending Moment ENGRO 145 Learning objective Shear as Moment n Flexure of Beams Shear Force and Load Shear amp Moment Relations Bending Moment Diagram ENGR0145 Learning bj ectivejs Shear a Moment n 15659 emf gonstruction Rules g AxZ mm x Distributed load w7 concentrated load P7 and concentrated couple C at position x along the beam Construction Rules from Moment Fsli iiep iiei Bend39m Mome t Equlhbrlum g n ENGRO 145 Learning bj activefs ZMoMLAM7ML707VL Am 7 VL AV 7 wmltmgta 0 AM o MM AV 7 wavgAza Note a a 0 and wavg a wm as Am a O Flexure of Beams Construction Rules frorn Moment Shearporceand Bending Moment Equilibrium a ram ENGR0145 Learning objectives Shear as M AM C VLAx AV 7 wavgAma 11fC7 07thenAMHCasAzH0 gt There is a jump in M across any concentrated couple C that has the same magnitude as C and is positive if C is clockwise 21fC390andP07 AM 1 E VL EAV 7 awavg gt as Azgt07agt0and AVAO since 130 Flexure of Beams Construction Rules from Moment Shearporceand Bending Moment Equilibrium ia ram ENGR0145 dM Learning objectives dm V Shear as Moment n gt M is linear where V is constant gt M is quadratic Where V is linear gt On any segment Where there is no concentrated couple7 gt The change in the bending moment M is equal to the area under the shear force curve Vi Example Using the Graphical Method Fsli ziepiiiei Bending Moment ENGRO 145 y 151M MN Ag L I T39LWW Construct the shear force and bending moment diagrams for the beam shown Example Using the Graphical Method Step 1 Free body diagram of the entire beam 60FN lt ZOkN 2m V A L 6m 2m RA RB MAO s RB30kN 2190 RA50kN Flexure of Beams Shear Force and Bending Moment ENGRO 145 Learning bj ectiye fs Shear 2 Moment 1 Example Using the Graphical Method Fsli ziepiiiei Bending Moment Step 2 Construct distributed load diagram ENGR0145 y 20 W Leaning abjectwej s 15 kNm Shsar a5 Moment n A B x 50kN 4m 2m Zm30kN w kN In 4 6 x In 15 Example Using the Graphical Method Step 3 Construct shear force diagram T 20 kN 15 kNm W Wm B 4 5 8 A x x m 715 50kN 4m 2m Zm30kN begin and end at zero 50760 3 Note Vm50 kN atz0 10 gt 0im Flexure of Beams Shear Force and Bending Moment ENGRO 145 Learning bj ectiyefs Shear 2 Moment Example Using the Graphical Method Fsli zgepiiiei Bending Moment ENGRO 145 A1 1lt m 50 kN 833 kNm 2 3 1 10 A2 5 lt4 3 m 710 kN 733 kNm A3 720 kN m A4 760 kNm Example Using the Graphical Method Fsli zgepiiiei Bending Moment Step 4 Construct bending moment diagram ENGRO 145 Y 20 kN 15 kNm A 50kN 4m Zm Zm30kN Team Problem B x 5ft 10ft 5ft Construct the complete shear force and bending moment diagrams for the beam shown Determine the maximum shear force and bending moment either positive or negative and the cross sections on Which they occur Flexure of Beams S ENGRO 145 Learning objectives Shsar as M Team Problem Flexure ofBea Shear Force and Bending Moment ENGROMS Leafnin abject shat egg Moment Example Finding Flexural Stresses Fsli iiep iiei Bending Moment ENGR0145 4m 4000 71b i 1 Learning bgectwe lf 500mm 8m an Air 400mb a Determine the maximum tensile exural stress 10 Determine the maximum compressive exural stress Example Finding Flexural Stresses Fsli iiep iiei Bending Moment M y ENGROMS 7 Step 1 Locate the neutral axis and determine I 35 in na 55 in Tm Learning Q bj39ectixre Shear 25 Momma n I 970 in4 Calculations omitted Example Finding Flexural Stresses Fsli iiep iiei Bending Mornth Step 2 Construct the distributed load diagram 7 ENGROMS Learning G jecti k Shear 2 Momma Flexure of Bea Shear Force and Bending Moment Example Finding Flexural Stresses Step 3 Construct the shear force diagram ENGR0145 Learning 8 gem Shear 25 Moment scrip q n 712000 ftrlb Flexure of Beams Example Finding Flexural Stresses Shearpmeand Beng x ng Moment imam Step 4 Construct the bending moment diagram ENGR0145 Learning bgemmfs Sham a Momma n me 8000 01107 MrnaxH 712000 ft1b Example Finding Flexural Stresses Step 5 Determine the maximum tensile and compressive exural stresses on the critical cross sections My C T I 970 in gt Mmaxm 8000 ftlb 96000 in1b at z 8 ft 7 796 000 in1b755 in 970 m4 5443 p81 96 000 in lb35 in i C 7 970 m4 73464 p81 Flexure of Bea ENGRO 145 Learning b jsq vejs Shsar 2 Moment Example Finding Flexural Stresses gt Mmax 712 000 ftlb 71447000 in1b at 0 UT 7 71447000 111136 1n 5196 psi 970 111 7144 000 lb 755 TC 7w 78165 psi 970 in Flexure of Beams Shear Force and Bending Moment ENGRO 145 Learning VGb j ecbiVeS Example Finding Flexural Stresses Answer a argon 544 ksi at z 8 ft b mum 7817 ksi at 0 Flexure of Beams Shear Force and Bending Moment ENGR0145 Laaining bj ectbm s Shsar as Mcmsnt n Flexure ofBeams Beam De ection by Integration of e oa Distribution ENGRO 13945 Flexure of Beams Beam De ections by L mm WWW Integration of the Load Distribution 39 use No Equation ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Depaltrnent University of Pittsburgh Student Learning Objectives giggggim by Integration of the Load Students should be able to Distribution gt Determine the equation of the elastic curve by ENGROMS direct integration of the load distribution Learning Objectives mlo iEggua lon 4th Order De ection Equation Given the de ection equation7 dzy Ely M and recalling that dMdm V and dVdm w7 it follows by differentiation that dsy Elm3 Vm d4 Eli wm dm4 Flexure of Beams Beam De ection by Integration of e 0a Distribution ENGRO 13945 Learning objectives Us micn Epabion 4th Order De ection Equation F312 2233 by Integration of t e oad Distribution ENGRO 145 Learning objectives i i i i i i D H t E t gt The d1str1buted load 1s posltlve When d1rected 133 qua m E 39I upwards mm 6 V This is a 4th order ordinary differential equation for the equation of the elastic curve7 Mm which is only really useful when there are no concentrated loads or couples and the distributed load can represented by a single expression for the entire beam V There will be four constants of integration7 requiring four boundary conditions Example Flexure ofBeams Beam De ection by Integration of the Loaoi Distribution ENGRO 145 Learning objectives coion Equabion Determine7 in terms of w7 L7 E7 and I7 gt The equation of the elastic curve7 gt The de ection at A gt The support reactions at B Flexure of Beams Example Beam De ection by Integration of t 9 0a Distribution ENGRO 145 Lemming bj ebtjvefs mien EquaMOn E37 Step 1 Identify the four boundary conditions What are they yimL 0 yimL 0 Miw0 0 gt 0 EIy Mx y i 0 Viw0 m 0 Ely Vm y i 0 Example Step 2 Solve EIyOV wm for 3 EIyaW 7 note the sign 4 um ElyH 7m 01 gt Using the boundary condition ymim O7 ElyWM 01 j 01 0 Flexure of Beams Beam De ection by Integration of e 0a Distribution ENGRO 145 L6 him Flexure of Beams Beam De ection Example by 1332 Of Step 2 Solve for cont Distribution ENGROiAS 4 101 E I m if Lg 8 y 4L3 rm mommgumm 5 w i um Ely i 720L3 02 gt Rom the boundary condition y m0 O7 Elyha 02 gt 02 0 Example Step 2 Solve for cont 5 E H y 20L3 6 7 Ely 120L3 03 gt Rom the boundary condition yquotmL O7 3 wL Elyha ier 03 j 03 ng 120 Flexure of Beams Beam De ection by Integration of e 0a Distribution ENGRO 1 45 L lth WWW n mlo nEgguab n 1237 Example Flexure of Beams Beam De ection by Integration of Step 2 Solve for cont t 5 0a Distribution ENGROMS 6 3 um wL EI 77 7 MayHe Whyram y 120L3 120 T 0MEuablon 7 3 um wL m EIy 7840L3 120 C4 gt Horn the boundary condition y mL O7 1011 wL4 wL4 EIy mL 7 Imtmta GE E Flexure of Beams Example Beam De ection by Integration of Distribution 7 wLSm 1011 7840L3 120 7 140 Learning bg39e ctjvsts a The equatlon of the elastlc curve 1s DEHSCMOnEquaMon EIy ENGRO 13945 7 w 77 6 7 7 y 840EIL3 3 ML 6L b For the de ection at A7 Elyim0 wL4 140 Example Flexure of Beams Beam De ection by Integration of c For the support reactions at B7 e 0a Distribution ENGROMS wLZ M mL Ely zL W wL V mL ElyH mL 7 4 Flexure of Beams Design of Beams for ENGR0145 Learnlng Objectives Flexure of Beams Des1gn of Beams for B mmmm m Strength PM ENGR0145 Statics and Mechanics of Materials 2 5 William S Slaughter Mechanical Engineering Depaltment University of Pittsburgh Student Learning Objectives Students should be able to gt select the lightest beam of a given type for a particular application7 such that the allowable exural and shearing stresses are not exceeded Flexure of Beams Design of Beams for ENGROMS Learning Objectives Design for sprengpn g Design Criteria Recall M c M VQ UWT and 77 For a particular beam application ie7 longitudinal dimensions7 method of support7 and external loads and given an allowable exural stress can and shearing stress Tall dictated by the material7 eg7 structural steel7 how does one choose an appropriate ie7 the lightest beam of a particular type Flexure ofBeams Design of Beams for ENGROMS Flexure of Beams Steps in the Design Process Desigmeem for Strength ENGR0145 7 Mo 7 M d 7 VQ Tina T i g an 7 i It Learning Objectives Design for Strength 1 Determine the maximum using the bending gi ples e moment diagram gt Given Jan gt Choose the lightest beam such that S gt Smirk gt Note that the lightest acceptable beam is not necessarily the beam with the smallest section modulus 2 After including the weight of the beam as a uniformly distributed load7 determine the new7 larger max gt Determine the new 3min maxJan gt Check that S gt 5min stilll gt If S lt 5mm pick a new beam and start over Steps in the Design Process M c M VQ 7 7 and 7 Um I S T It 3 Including the weight of the beam7 determine the maximum lVl using the shear force diagram gt Given Tau Check that Tmax lt Tani gt If Tmax gt Tall pick a new beam and start over Note most of the time but not always7 step 1 will give you the right beam Steps 2 85 3 will just be checks that will not require you to change your original choice of beam Exceptions usually are due to relatively low allowable shear stress Flexure of Beams Design of Beams for ENGROMS Example 22kN 40kN 22kN l15ml l l 125m 125m 15m For a beam made of structural steel 0811 152 MPa and Tall 100 MPa7 select the lightest American Standard Beam see Table A 4 for the loading shown Flexure of Beams Design of Beams for ENGROMS Flexure of Beams Example Design Qf Beams for Strength ENDED 145 V kN Example Step 1 determine the minimum allowable section modulus7 neglecting the weight of the beam Mmax 825 kNm Mmax 0511 7 825 x 103 Nm 152 x 106 Nm2 3 543 x 104 m3 x 103 E m gt 5min 543 x 105 mm3 543 x 103 mm3 gt Try an S305gtlt47 beam gt 5330M 596 X 103 mm3 39 73305x47 47 kgm Flexure of Beams Design of Beams for ENGROMS Flexure of Beams Example Design 6 Beam for Strength Step 2 determine the new minimum allowable section ENGROMS modulus7 including the weight of the beam w 47 kgm981ms2 461 Nm Rjeight Rgeight 461 Nm55 m 1268 N Learning Objec ves V N 1268 1743 Nrm 275 55 x In 71268 might 1268 N Ml fjjifht 1743 N m Flexure of Beams Design of Beams for Example ENGR0145 MW 3 MES M n izixght Eli50 g 825 gtlt 103 Nm 1743 NIH 5448914 g 842 x 103 Nm Mmax 711 lt 842 x 103 Nm 7 152 x 106 Nm2 g 554 X1074 m3 g 554 x 103 111mg EX gt Slum 3305X47 596 x 103 mmg still okay Example Diiii rif iii gr Strength Step 3 check the shearing stress7 including the weight ENGROMS of the beam Vm V333 might 42 x 103 N 1268 N g 433 x 103 N Flexure of Beams Example Design of Beams for Strength Special approximation for I type beams only ENGROMS large I and small y Q gives small 1 A web 1 z VA web nearly const Q const I gives nearly const 1 V 739 z 7 max Aweb Example Web Depth Thickness i Flange lTThickness Flange Width For S305x47 beam gt Depth is 3048 mm gt Flange thickness is 138 mm gt Web thickness is 89 mm Aweb 0008903048 7 200138 2467 gtlt 103 m2 Flexure of Beams Design of Beams for ENGROMS Example N max Tmax A Web N 433 x 103 N N 2467 x 103 m2 x 176 MPa Tan 100 MPa gt 8305 gtlt 47 is still okay Answer the lightest acceptable American Standard Beam is an 305x47 Flexure of Beams Design of Beams for ENGROMS Learnlng objecmves Dgssigxi so sgr39e ggh Pr Flexure of Beams Castigliano s ENGRO 145 Learning 39Obje ctives Flexure of Beams Castigliano s Method ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Depaltinent University of Pittsburgh Student Learning Objectives Students should be able to gt Determine beam de ections using Castigliano s method gt Determine beam slopes using Castigliano s method Flexure of Beams Castigliano s ENGRO 145 Learning Objectives Work Done by External Loads First consider simple axial loading axial I load l L P W gt The work done by the axial load in elongating the bar an amount 6 is 6 W Pd6 P6 if linear 0 Flexure of Beams Castigliano s ENGR0145 Learning Objectives Flexure of Beams Castigliano s Strain Energy ENGR0145 gt For elastic deformation7 the work done equals the internal stored energy U7 known as the strain Learning Objectives 77 i Energy Methods energy Theory 1 Exa e 1 W U 5P6 gt Since P 0A 6 6L7 and 6 aE7 1 U 5UAEL 1 TL WA f U2 144 Flexure of Beams Strain Energy Density Castaways Method gt The strain energy density u is the strain energy U ENGROMS per unit volume AL in this case Application to Flexure of Beams Fle i i iiim Method Consider two concentrated loads P1 and P2 applied ENGR0145 simultaneously7 such that the ratio P1 P2 is constant Learning Objectives gt yl and y2 are the de ections under the concentrated loads P1 and P27 respectively gt Each de ection is positive in the same direction as its corresponding load Flexure of Beams Application to Flexure of Beams Castaways Method ENGRO 145 Learning 39ijec vss Methods Theory 6 1 gt The strain energy U equals the total work done by both loads 1 1 U U1 U2 5P1y1 5P2y2 Application to Flexure of Beams Fle i i iiim Method Now7 suppose that the rst load is increased from P1 ENGR0145 to P1 AP17 While P2 is kept xed Learning Obj cmss gt w H 241 A2417 yz H yz A2427 and U a U AU E Tn gggyMgbhm 333 1 yz f yl AM yz Ayz F Application to Flexure of Beams Fle i g ii Finally7 suppose that APl is applied rst7 then P1 and ENGR0145 P2 are applied Simultaneously Learning Obj cmes i E M bh d gt Since linear7 yl Ayl yg and Ayz are the same Egg 8 5 Example 1 gt Utotal GAPiAyi APiyi Piyi 6132342 Application to Flexure of Beams Fle iaillaliZ S Method ENGR0145 gt With P1 and P2 applied rst7 followed by AP17 Learning Objectives 1 1 Energy Methods Utotal 7 513191 513230 AU 335er 1 i 2 gt With APl applied rst7 followed by P1 and P27 1 1 1 Utotal AP1Ay1 APiyi 5131341 5132112 Theorem the nal strain energy is the same7 regardless of the order in which the loads are applied 1 AU EAplAyi APlyl Flexure of Beams Castigliano s Application to Flexure of Beams The change in strain energy due to changing the ENGR0145 concentrated load P1 to P1 AP1 can be given as Learning Objectives Energy Methods 1 Theory AU EAplAyl APlyl W 9 AU 7 1A 3 A131 7 241 2 241 gt Letting APl a 0 gives Ayl a 0 and 8U 873191 Catigliano s Theorem Flexure of Beams Castigliano s If the strain energy U of a linear elastic structure is expressed in terms of external loads eg7 P17 P27 7 M17 M27 then gt The de ection in the direction of applied concentrated load Pl at the point where P is applied is given by ENGR0145 Learning Objectives 1 7 8U yl gt The rotation in the direction of applied concentrated couple Mi at the point Where Mi is applied is given by 8U 07 1 8M1 Catigliano s Theorem Flexure of Beams Flegggggliiz ys Method Recall that the strain energy density due to normal ENGR0145 Stress ls glven by Learning ij ctivss E r Methods 72 U 72 dV g 1 u 7 gt 7 6 2E 2E Vol gt For exure of loeams7 the normal ie7 exural stress is M y I 0 Catigliano s Theorem Flexure of Beams 2 a U 7 2EdV V01 L 2 1 My 7 if dAd 0 2Elt 1 3 area L 2 M 2 70 dA dm area 1 L2 mOde Flexure of Beams Castigliano s ENGRO 145 Learning 39Objze chives Catigliano s Theorem Flexure of Beams 633132 Method ENGR0145 1 L U 7 M2 dz Learning abgames 1 2E 0 Energy Methods Theory Example 1 Exam le 2 gt The de ection at the point where concentrated load P1 is applied is given by gt The rotation ie7 slope at the point where concentrated couple Mi is applied is given by 8U 1 L 8M M d 0 8Mi QB 0F8Mf Flexure of Beams Example 1 Castaways Method ENGR0145 P Leannn Objectives E Methods Y L Use Castigliano s Theorem to determine the de ection and the slope at the free end A of the beam Flexure of Beams Example 1 Castawaan Method Step 1 determine the bending moment M in terms of ENGR0145 P and m P AM A 0 V06 ZMOMPzO gt M7Pm Flexure ofBeams Example 1 Castaways ENGR0145 M 7P Learning 39ijec vss Eneng Methods T h Step 2 apply Castigliano s Theorem to determine yA L 1 8M 7 Mid 1 EIO 8P 3 1 L if 713x 7x dm E lt gtlt gt P L E mzdm 0 7 PL3 3E Example 1 Flexure of Bea Castigliano s Since there is no couple at A7 in order to use ENGR0145 Castigliano s Theorem to determine the slope at A you have to apply a dummy couple MA7 which Will ultimately be set equal to zero Mslf pm V06 Learning Objectives Enexgy Methods Th ioiriv ZMOMPzMAO M7PxiMA Example 1 M7Pm7MA Step 3 apply Castigliano s Theorem to determine 0A 1 L 8M 0 7 Mid A 0 6 MA 35 L 0 713x 7 MA71 dz L PxMAdx I 0 wwwma GPLZ MAL ta I Flexure of Beams Castigliano s ENGRO 145 Learning 39Objze chives Example 1 1 07 A EI GPLZ MAL gt Set the dummy couple equal to zero7 MA 0 2 0A gt Actually7 you can set MA 0 anytime after evaluating 8M8MA L iPz 7 MA71 dz 1 L O Pmdm 7 PL 2EI 0A Flexure of Beams Castigliano s ENGR0145 Learning Objectives Eneng Methods v Flexure of Beams Example 2 Castaways Method ENGR0145 Learning Objectives my V Use Castigliano s Theorem to determine the reaction at A for this statically indeterminate beam Example 2 Consider A RAiii gt Use Castigliano s Theorem to determine the de ection at A due to w and RA gt Then the reaction force RA is such that yA O Flexure of Beams Castigliano s ENGR0145 Learning Objectives v quotechods Example 2 w7 RA7 and m RA ZMOM7RAmwxlt Step 1 determine the bending moment M in terms of 7 0 2 gt MRAzilwz2 2 Flexure of Beams Castigliano s ENGRO 145 Learning 39Objze chives Methods T Example 2 1 M RAm 7 gwxz Step 2 use Castigliano s Theorem to determine the de ection yA due to w and RA 1 L 8M 7 Mid 1 191J aRA 3 L RM 7 gw2gt dm 0 L 2 1 3 I 0 Ftxx igwx gtdm 1 1 7R L377 L4 lt3 A w gt we EM t H m I Flexure of Beams Castigliano s ENGR0145 Learning Objectives Flexure of Beams Example 2 Castawaan Method ENGR0145 1 1 1 7 7R L3 7 7 L4 y E lt3 A 8w gt Step 3 Solve for the reaction force RA that gives 24A 039 Flexure of Beams Example 3 Castaways Method ENGRO 145 L2 L2 Use Castigliano s Theorem to determine the de ection at B Example 3 Flegggsiglgiz fs Method gt Anytime you want to know the de ection at a ENGROMS point that doesn t have a concentrated load7 you Learning Objec vss must add a dummy load gt Add a dummy Enemy Md load P at B 4T4 L2 Flexure of Beams Example 3 Castaways Method ENGR0145 Learning ijec vss L2 L2 gt Rom Ato B Le7 for 0 lt z lt L27 1 1 M EPz ngm M 87M wLm E wLmZ 8P P0 8 2 16 Flexure of Beams Example 3 Castaways Method ENGR0145 Learning ijec vss L2 L2 gt Rom B to C Le7 for L2 lt z lt L7 2 8 i 3wL P0 16 M ltEBwLgtLim7Liz2 MLM 8P Lint lama Flexure of Beams Example 3 Castaways Method ENGROI45 Learning ijec vss Eneng Methods Th dm 1L8M yB2 A A42 3P0 1 W 8M EA M 39 1 L 8M dz i M 7 H E m 8P P0 Example 3 Flexure ofBeams Castigliano s ENGRO 145 LZ 2 dmi sz dm P0 0 16 ung 2 48 wLL3 7482 L2 8M 0 M5 384 Flexure of Beams Example 3 Castaways Method L 8M M7 L2 8p ENGRO 145 Learning ijecmves Flexure of Beams Example 3 Castaways Method ENGR0145 Learning ijec vss P4 L2 L2 1 W 8M 1 L 8M 7 M7 d 7 Mi 1 3 191J 8P P0 mEIL2 8P 7 1 wL4 1 1011 E1 384 E 256 5wL4 i 768E d1 PO Team Problem Flexure of Beams Castigliano s Method ENGR0145 Learning Objectives ENE Methods Use Castigliano s Theorem to determine the reaction at A for this statically indeterminate beam Answer 3M 137 A 2L Combined Static Loading Principal ENGRO 145 Learning 39Obje chives Combined Static Loading Principal Stresses ENGR0145 Statics and Mechanics of Materials 2 William S Slaughter Mechanical Engineering Depaltment University of Pittsburgh Student Learning Objectives Students should be able to gt Determine the normal and shearing stresses on an arbitrary plane in plane stress gt Determine and sketch the principal and maximum shearing stresses V Perform the analogous calculations for normal and shearing strains Combined Static Loading Principal Stresses ENGR0145 Learning Objectives Plane Stress Transformation Equations 0y t n my Em One Ox quot X an am cos2 9 0y sin2 9 27mg sin0 cos 9 k 00s207mysin20 Tm 7am 7 0y sin0 cos 9 7 mycos2 t9 7 sin2 0 7 sin 20 my cos 20 Combined Static Loading Principal ENGR0145 Learning ijsc vss I ransform Eqns St gas Combined Static Symmetry of 0716 Loading principal Stresses ENGR0145 v t Learning 39ijiec vss 0 O 39 t n Transform Eqns Em n Tm On 39 ne on Tn39t39 6 x ty on an am cos2 9 Q sin2 9 2739 sin0 cos 9 Umgigygmgigycos20mysin20 we i180 m0 Combined Static Loading Principal s Symmetry of Tm0 ENGROI45 V t n Learning ijec vss 0 On39 t Transform Eqns Em Em On Pug ragga 039 Tn39t39 x MG a 19 On 0 Tm 7am 7 0y sin0 cos 9 7 mycos2 t9 7 sin2 0 7 7 sin 20 my cos 20 Tm 9 i 180 Tnt0 gt Tnt0 j 900 77 0 Extreme Values of an 0 Gy am cos2 9 Q sin2 9 2739 sin0 cos 9 a a a 7 a i HGOSQ Tmysm20 Question On which planes do the maximum and minimum normal stresses occur Combined Static Loading Principal Stresses ENGR0145 Learning Objectives Tpansf oomi Eqrisi Princi a1 Str Extreme Values of an 0 Laggisgegjgzsg Stresses ENGR0145 039 039 039 7 039 i an l m 2 y c0s20 Tmysm20 gt Maximum and minimum values of an occur at values of 9 for which damd0 0 Learning Objectives d0 7am 7 0y sin 20 2739 cos 20 0 Where maximum and minimum values of an occur when 9 0p Extreme Values of an 0 Lgombmed Static a ing Principal ENGRoms Learning Objectives Transform Eqnsl Principal Stresses i i o Mews Circle gt There are two solutlons for 2017 that dlffer by 180 Gamma gimming gt there are four solutions for 017 at 90 intervals gt Solutions for 91 that differ by 180 correspond to the same value of Uni gt Thus7 any two values of 91 that differ by 90 correspond to the maximum and minimum values of Uni This is suf cient Once you have obtained one value of 91 from your calculator7 say7 then the second can be obtained by simply adding or subtracting 900 V Principal Planes Recall that 7am 7 0y sin 2017 2739 cos 20p 0 and Tm 7 sin 20 my cos 20 gt It follows that Tntl99p 0 planes with maximum or minimum normal stress have zero shearing stress gt Planes without shearing stress are called principal planes gt TWO solutions for 91 give two principal planes gt Plane With normal in the z direction gives a third principal plane since Tm 72y rom t e de nition of plane stress gt There are three principal planes and they are mutually orthogonali Combined Static Loa ing Principal ENGRO 145 Learning objsccivss Transform Eqns Principal Stresses Menis Circle Gumbinnd straining Combined Static Loading Principal Principal Stresses A A ENGR0145 gt The normal stresses on prinmpal planes are called Learning Objectives principal stresses gt These are anmax a mim and a 0i Transmm Eqns Principal Stresses Mews Binder gt Noting that tan 20 Sin 2017 2739 7 cos 2 am 7 0y sin 20p cos 20p 1 it follows that am 7 0y sin20pi 00820pj M where Principal Stresses Recall that Wi cos20iw39mysin20 an gt Thus7 with anl99p Up 19 for principal stress7 it follows that Tm Q 171172 2 i 0121 0nmaX7 0172 anmin7 and 7123 0392 0 are the principal stresses in plane stress gt Note 0P1 012 am Q Combined Static Loading Principal Stresses ENGR0145 Learning Objsccivss I L atnsf39oomi Eqnsi Combined Static Maximum ln Plane Shearing Stress 7 Loading principal Stresses ENGR0145 am 7 y i Tm 7T sin 20 my cos 20 Learning Objectives Transform Eqns gt Maximum and minimum values of Tm occur at values of 9 for which dung10 O dTnt d0 70m 7 aycos20 7 27 mysin20 0 Where maximum and minimum values of Tm occur when 9 0T Maximum ln Plane Shearing Stress 7p gt There are two solutions for 20 that differ by 180 gt there are four solutions for 0 at 90 intervals gt Solutions for 97 that differ by 90 correspond to values of Tm of the same magnitude but opposite sign These are the maximum and minimum values of Tm Their absolute value is the maximum inplane shearing stress Tpi gt Thus any one value of 97 Will suf ce to determine the maximum in plane shearing stress 71 gt Since tan 20 iltan 2017 it follows that the planes with maximum in plane shear form 45 angles with the principal planes corresponding to 011 and 0172 Combined Static Loa ing Principal ENGROMS Learning Objectives Transform Eqnsi Principal Stresses Mohr Circle combined straining Maximum ln Plane Shearing Stress 7p CDmb ned Static Loading Principal Stresses l l ENGR0145 gt It can be shown that the max1mum in plane shearing stress7 7p lelg T is given by Learning Objectives Transform Eqns Principal Stresses Moh x ls Circle Gamblu ed rainirng gt It follows that the maximum in plane shearing stress is onehalf the difference between the two in plane principal stresses gt The normal stress on a plane of maximum in plane shear is generally non zero 01 0y 7 am Upz 7 2 0n least Overall Maximum Shearing Stress7 Tmax gt There are three principal stresses 011 012 an UPS Uz gt One is the maximum normal stress amax gt One is the minimum normal stress 0min gt One is in between gt It can be shown that the overall maximum shearing stress as opposed to the maximum in plane shearing stress7 7 1 is given by V The overall maximum shearing stress Tm acts on the planes that form 45 angles With those for amax and 0mm 39 Tmax 739 if 0172 g 0 3 011 which implies that amax 071 and 0mm 012 on three orthogonal principal planes Combined Static Loading Principal Stresses ENGRO 145 Learning Objectives Transform Eqnsi Principal Stresses Mohr Circle Cbmblned straining

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