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# Mechanics of Deformable Bodies 057 019

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This 43 page Class Notes was uploaded by Rahsaan Hoppe on Friday October 23, 2015. The Class Notes belongs to 057 019 at University of Iowa taught by Colby Swan in Fall. Since its upload, it has received 285 views. For similar materials see /class/228005/057-019-university-of-iowa in General Engineering at University of Iowa.

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Date Created: 10/23/15

05701900B Mechanics of Deformable Bodies Instructor CC Swan College of Engineering University of Iowa Period 36 Spring Semester 2009 Textbook Sections 13 3 Topic Buckling of Compression Members Objectives 1 2 3 4 H Review of buckling Critical loads in pinned compression members Critical loads with other types of supports Example problems Role of Buckling 0 When structural or mechanical systems are designed the three factors that must always be considered are 0 Stiffness If a mechanical system deflects or deforms too much under expected design loads then it is not acceptable In this case the system does not have sufficient stiffness 0 Strength if the yield or failure stress in any member in the structure is exceeded the structure does not have sufficient strength 0 Stability Compression loads in long slender structural members can cause buckling behavior which is unstable 0 Pinned Compression Members unloaded st member 1 bUCkhng 2quotd buckling 3rd buckling mode 2 mode mode P z E 47r2E 97r2E 1 57 L2 2 5 L2 3 m L2 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa 0 For a compression member loaded pinned at both ends the critical buckling load is given by PcrElTE2L2 This is the Euler Buckling Load Formula 0 Notes 0 The longer the member is the smaller the critical buckling load 0 The larger the moment of inertia of the cross section the larger the buckling load 0 Long slender members tend to buckle at low load levels The radius ofgyration r of a cross section is defined as r IIA where A is the cross sectional area and is the minimum moment of inertia of the section 2 P o E Here Lr is the slenderness ratio of the compression member A 0 my 2 Critical loads with different types of supports The same concept used to derive the critical buckling load for a pinned compression member can be applied to other support conditions After applying the concept the critical buckling load for different restraint conditions can be expressed by introducing the notion ofthe effective length ofa compression member KL The critical buckling load can then be computed as P EI 392 Er L2 2 a For a compression member pinned at both ends K10 b For a compression fixed at one end and free at the other K2 c For a member fixed at both ends K05 cl For a compression member fixed at one end and pinned at the other K07 where Le KL 05701900B Mechanics of Deformable Bodies College ofEngineering University of Iowa Instructor CC Swan 4 Q lt 7 LP 07L L 05 L 21 L L Pinned and fixed Pinned ends quot39 Fixed ends K1 Fixed and K05 in a free ends Cl K70 4 d b 0 7 E 7 E7rZ 39 A KLrZ When designing axially loaded members the axial stress should be lessthan both the critical buckling stress and the yield stress 05701900B Mechanics ofDeformable Bodies Instructor CC Swan College of Engineering University of Iowa 3 Example Problems Example 1 Determine the maximum allowable load w that can be applied to member BC without causing member AB to buckle Assume that AB is made of steel and is pinned at its ends for xx axis buckling and fixed at its ends for yy axis buckling Use a factor of safety with respect to buckling of FS 3 Est 200 GPa and oy360 MPa 0 l C B i 15 m r 05111 30 mm 5 x LJZO mm J y Al b L 30 mm A x 05701900B Mechanics ofDeformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 2 1342 The 50mm diameter C861OO bronze rod is fixed at A and has a gap of 2mm from the wall at B Determine the increase in temperature DT that will cause the rod to buckle Assume that the contact at B acts as a pin 05701900B Mechanics ofDeformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 3 1347 The ideal column has a weight w forcelength and rests in the horizontal position when it is subjected to the axial load P Determine the maximum moment in the column at mid span El is constant Hint Establish the differential equation for deflection Eq 131 with its origin at the mid span The general solution is v Asinkx Bcoskx wx2 2P wLxZP wEIP2 where k2 PEI 05701900B Mechanics of Deformable Bodies College of Engineering Insiructor CC Swan University of Iowa Example 3 13 47 The ideal column has a weight w forcelength and rests in the horizontal position when it is subjected to the axial load P Determine the maximum moment in the column at mid span El is constant Hint Establish the differential equation for de ection Eq 131 with its origin at the mid span The general solution is V Asinkx Boosx W2 2P wLxZP wEII 2 where k2 PEI 7 i P TEL 39FX VLL L z p M w NW7 wXL V108 Pr 7 L W rx 545 Mm WA wpw alas L lit at r XL xi LO CIXZ39 El ZEI soulch Framed v10 A swig r Bash Mixvi wEI 2P 1 P wlnere 32L PEI U400 Qquot 1 Il ACosl x slul x 1 y w M P 2 05701900B Mechanics of Dcfonnable Bodies College of Engineering Instructor CC Swan University of Iowa B DVM lava ComciiHoMs 1 T0 7 B E 1 C 0 MIL V x 2 o f c 5lt izslm 3 1 EL L an AL A Plz l 1 New evalvaie Mm cl x LL 1 Mm am WW x 1261 4anili5im i 1 CoSlitgt ie4 21 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Period 6 Spring 2009 Semester Section 3 of Textbook Topic Material Properties Objectives 1 Review a Stress b Strain 2 Material Properties that relate stress and strain 3 Examples 1 Review of Stresses and Strains Forces or loads are applied to structures resulting in internal forces and moments On a smaller scale within material of a structure the forces per unit area are represented by shear and normal stresses MN GT 1 Ma a MV cs Stresses in a material give rise to deformations which are represented by dimensionless strains There are essentially two types of strains 1 normal strains and 2 shear strains a Normal strains Normal strain measures the change in length of an infinitesimal fiber of material relative to the original length of that fiber 6 d dso undeformed fiber 5 lt O 3 reduction of length 7 7 5 i d 0 g O 3 no change of length 0 ds deformed fiber 8 gt O 3 increase of length 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa b Shear Strains Shear strain represents the negative change in angle radians during deformation between two infinitesimal fibers that were initially perpendicular 7Z39 shear strain y 2 t9 undeformed deformed 2 Material Properties Relating Stress and Strain Material properties are measured in a variety of mechanical tests Some of the common mechanical tests performed on materials are a the tension test b the compression test and c the torsion test a 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Features of stressstrain behavior 0 Elastic modulus 0 Yield stress 0 Ultimate stress Fracture or Breaking stress Ductility Modulus of toughness Modulus of resilience Strain energy Types of material behavior Elastic behavior 0 Linear behavior Plastic behavior Hardening behavior General Form of Hooke s Law an C11 C12 C13 C14 C15 C16 gxx ayy C21 C22 C23 C24 C25 C26 5 azz C31 C32 C33 C34 C35 C36 522 2 ch C41 C42 C43 C44 C45 C46 7 2Iyz C51 C52 C53 C54 C55 C56 7yz sz C61 C62 C63 C64 C65 C66 79x 8xx S11 S12 S13 S14 S15 S16 Unc yy 521 S22 S23 S24 S25 S26 Uyy gzz 531 S32 S33 S34 S35 S36 Uzz 2 S41 S42 S43 S44 S45 S46 I 7y2 S51 552 S53 S54 S55 S56 2yz S 0M 0M 0M 0M 4 0M 0C0 5 057019 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Hooke s Law for Isotropic Elastic Material 113v 172v 0 0 0 1 17 1 1 0 0 0 a 1721 1721 1721 Eu 0 7 a W V V 1 V 0 0 0 W a i 1721 1721 1721 I 6 22 7 V 1 V 0 0 0 l 0 0 7 z39yz 2 y Ta 0 0 0 0 l 0 72c 2 0 0 0 0 0 l 2 a 1 7 1 7 1 0 0 0 a W 7 1 1 7 1 0 0 0 0W 5 7 1 71 7 1 1 0 0 0 on 7v 7 E 0 0 20 1 0 0 By 7y 0 0 0 0 20 1 0 Ty 7a 0 0 0 0 20 1 1a Special Cases of Hooke s Law a Uniaxial Stress State 039 E8 All other stresses vanish a 822 718DC Here V is known as the Poisson s ratio bi Pure Shear E IV nyy where G Is known as the shear modulus 21 v 3 Examples 057019 Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa Problem 1 The stressstrain diagram for a steel alloy 0Mpa having an initial diameter of 125 mm and an initial gauge 44o length of 50mm is provided in the figure Determine the 400 K approximate modulus of elasticity for the material the 360 load on the specimen that causes yielding and the 320 ultimate load that the specimen will support 23 200 160 120 8O 40 0 J 0 30 c mmmm i 020 min 1 minim mum 057019 Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa Problem 311 The stressstrain diagram for polyethylene which is used km to sheath coaxial cables is determined P from testing a specimen that has a 5 gauge length of 10 in Ifa load P on the specimen develops a strain ofs 0024 determine the approximate length of 3 the specimen measured between the gauge points when the load is removed Assume the specimen 1 recovers elastically p 0 5inin 0 0008 0016 0024 0032 0040 0048 057019 Mechanics of Deformable Bodies Instructor CC Swan College of Engineering University of Iowa Problem 333 A plug has a diameter of 30mm and fits within a rigid sleeve having an inner diameter of 32mm Both the plug and the sleeve are 50mm long Determine the axial stress 0 that must be applied to the plug so that it just makes contact with the sleeve Also how far must the plug be compressed downward in order to do this The plug is made of a material for which E25 MPa and v 045 d ampol gt Jolexxgt 62m 392 Clo Exxquot6i3 5 655 033 3 E 33 CSMPQXw 149 0W0 MPG iii i M p 3 0643 597nm m 25221 0664 2 owg a quoton4a L i 91 gt Al QELO EM325 WMgt AL gt quot7 70 234 119 ComfveSSI39tE 0465 7 057019BBB Mechanics of Deformable Bodies Instructor CC Swan College of Engineering University of Iowa Period 14 Spring Semester 2009 Textbook Section 6364 Topic Bending Kinematics Objectives 1 Discuss Kinematics of Bending Introduce the quotFlexure Formulaquot Solve Example Problems hWN Review of shear and bending moment diagrams 1 Review of shear and bending moment diagrams Transverse loadings in beams give rise to internal transverse shear forces These internal transverse shear forces give rise to bending moments 057 019BBB Mechanics of Deformable Bodies Instructor CC Swan College of Engineering University of Iowa 2 Bending Kinematics of an initially straight member A straight beam has an infinite radius of curvature When a beam bends it curves 0 With curvature there is o a center of curvature o a radius of curvature In quotpositive bending the top fibers of a beam are shortened negative normal strain The bottom fibers are lengthened positive normal strain There exists a quotneutral axis along which fibers are not stretched Crosssections of the beam that were planar before bending remain planar AS A longiludinal axis longitudinal T V axis AA AAquot Undcl ormcd clement Deformed element 21 lb 057019BBB Mechanics ofDeformable Bodies College ofEngineeIing Instructor CC Swan University of Iowa For a homogeneous linearelastic material the neutral axis passes through the geometric centroid of the crosssectional area If quotyquot denotes the distance above the neutral axis ofthe beam the normal strain varies linearly as given by 1 5717yk WhereK 15the curvature P P o For a given positive curvature K the normal strain is extremized at the bottom of the crosssection where y takes the values of c g 77 cl CK For the same positive curvature k the normal strain is minimized at the top of the crosssection where yc Am 0939 symmulry y xg I xllrfucc lnugrmdiual m 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa 3 The Flexure Formula The flexure formula is analogous to the torsion formula Recall the torsion formula related the torque on a section to the distribution of shear stresses on the section riT p rm J J The flexure formula relates the bending moment on a section to the normal bending stresses on the section UiiMy 0 7M0 1 V I i l E Nnmm strain winnunn pro le View cm Bending stress Varimicn in M a y i C v The resultant moment on the cross section is equal tothe moment produced by the linear normal stress distribution about the neutral axis a 71 7w p p MJ7039ydAI7E ydAIEKyZdAIa ZIyZdAIa ZI 0125 Elg 5 K MI Moment Curvature Relation E aniykiiyM E1 0 jam I I 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 16751 The aluminum machine part is subjected to a 20 quotquot1 moment of M75 Nm Determine the bending stress created at points B and Con the cross section Sketch the results on a volume element located at each ofthese points 0 mm 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 2 6758 The control lever is used on a riding lawn mower Determine the maximum bending stress in the lever at section ara if a force of 20 lb is applied to the handle The lever is supported bya pin atA and a wire at B Section arc is square 025 in by 025 in 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa Example 3 6783 The pin is used to connect the three links together Due to wear the load is distributed overthe top and bottom of the pin as shown on the freerbody diagram If the diameter ofthe pin is 04 in determine the maximum bending stress on the crossrsectional area at the center section ara Forthe solution it is first necessaryto determine the load intensities W1 and W2 800 lb quot393 1 400 J 400 lb 057019IBBB Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 3 683 The pin is used to connect the three links together Due to wear the load is distributed over the top and bottom of the pin as shown on the free body diagram If the diameter of the pin is 04 in determine the maximum bending stress on the cross sectional area at the center section a a For the solution it is rst necessary to determine the load intensities WI and w2 800 lb 490 3b 401 lb M 300i ix li 490m WL K V2 Wt 300 lbsin WX Fev a circuiav secitem I 12 I 2 a j 2 ooizm mt 5 3 t 4g sat a gal in 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Period 30 Spring Semester 2009 Textbook Section 123 Topic Deflection of Beams using Singularity Functions Objectives 1 Review of beam differential equations 2 Introduction of singularity functions 3 Demonstration with Examples 1 Review of beam differential equations 4 Given a loading wx applied to a beam one can perform a wx dVEd series offour successive integrations to find the shear dx dx distribution Vx the moment distribution Mx the slope Vx dM E1d3v or rotation distribution 9x and the transverse dx dx3 displacement distribution Vx 619 d2 MxEI EI d dx For complicated loading conditions wx the algebra dv associated with the solving the fourth order differential 906 d x equations can become somewhat complicated and messy One way to reduce the messiness ofthe algebra is to use singularity functions or generalized functions to represent the applied loadings and the reaction loadings on a beam 2 Introduction to Singularity Functions Consider the beam shown below with the loading and boundary conditions To integrate the differential equations would require that the beam be divided into a number of sections on which the loading shear moment etc are all continuous To avoid this singularity functions are used 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa o The fundamental singularity function is the Heavyside step function which can be used to represent a uniformly distributed load 0 xlta WO 1 xgta a A distributed load of magnitude wo beginning at xa and ending at xb can be represented as follows wx w0 lt x a gt0 w0 lt x b gt0 o The derivative ofthe step function is the delta function which can be used to represent point loads P1 d lt x a gt0 d ltx agt391 wxP1 ltx agt391 x a The derivative ofthe delta function is the dipole function which can be used to represent concentrated moment loads M1 d lt x a gt391 dx ltx agt392 wxM1ltx agt392 a The integral ofthe step function ltx agt0 is the ramp function which can be used to represent linearly increasing or decreasing loads 1 0 xSa 1 ltx agt wxmltx agt x a xgta Singularity functions for n 2 0 lt gtn 0 xSa x a x aquot xgta ltx agt 1 Jltx agt dx C n1 05701900B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Singularity functions for nltO Jltx agt dxltx agt39 1 Loading LoadiEEyEllrgction Shear V fwxdx Moment Mthlx 2 w M0ltr agt 2 V M0ltr zzgt l M M0ltt agt 2 P w Pltir agt l V Plt ugt0 M Plt agt1 X i L14 3 w H i m w w0ltx agt0 V w0ltxggtl M 770049 4 slope m DWI i l77 quot7 w mltr ugt39 V 7ltv agt2 M 0709 05701900B Mechanics ofDeformable Bodies Instructor CC Swan College of Engineering University of Iowa Example 1 1239 The beam is subjected to the load shown Determine the displacement at x7m 3 It N 39m 5Lkl39 and the slope at A El is constant a ii A r l 3 In 3 In l I 05701900B Mechanics of Deformable Bodies Instructor CC Swan College of Engineering University of Iowa Example 2 12 42 The beam is subjected to the load shown Determine the equations ofthe slope and the elastic curve El is constant 3 kNIn o L W l I I I 15 kNm O B C 53 111 05701900B Mechanics of Deformable Bodies Instructor CC Swan College of Engineering University of Iowa Example 3 12 27 The beam is subjected to the load shown Determine the elastic curve El is constant wo rfllllmx 3V4 397 Elastic curve 7quot L L a 2 2 05701900B Mechanics of Deformable Bodies College ofEngineen39ng Instructor CC Swan University of Iowa Period 29 Spring Semester 2009 Textbook Sections 1212 Topic Deflection of Beams Objectives Review of beam theory thus far Extension to calculation of rotations and deflections Solving the differential equations Example problems l PWN 1 Review of beam theory thus far W a V 7wx E dIlI Positive distributed load EMquot v V Mx E1 EIKx quot7 l 1 px I Positive internal shear M M Positive inlemal moment Beam sign convention Above px represents the radius of curvature ofthe beam due to flexure and KX represents the local curvature of the beam due to flexu re 2 Extension to rotations and deflections If the centerline axis of a beam is considered its transverse displacement along the length of the beam will be denoted by the variable vx MI Upward displacements or deflections ofthe centerline will be taken as positive and downward deflections negative Bel39m39e deformation deformation 05701900B Mechanics of Deformable Bodies College of EngineeIing Instructor CC Swan University of Iowa 0 The deformed shape of the centerline axis of a beam is usually referred to as the quotelastic curve 0 The local slope dvdx of the elastic curve will be denoted by the angular variable 9x Positive slopes correspond to positive values of dvdx and vice versa Important approximations for linear beam theow The curvature of a beam KX is approximated as follows divdxz Z 1x n S div1x2 ltltl mMir dx 0 The local slope angle 9x is approximated as follows 90 tanquotdvdx dvdx iff ix V 1 v 039 p Elastic curve In u Positive Sign convention Putting all ofthis together the system of differential equationsthat govern load w shear V moment M slope 9 and deflection v are as follows dv 9 1X 1 amp Moo dzv dx E1 dxz 3 V00 5151 dx dx dV dlv EM My 05701900B Mechanics ofDefonnable Bodies College ofEngineerlng Instructor cc Smn University orovm 3 Solving the differential equations Fora given load wlxl applied to a beam with section properties El the bounda condition on the beam must be known Seven common boundawcondilions are shown below39 Pin 7 V M 0 lulcmzil pin 0 hinge A Fl l cntl t m lelonetypicallyfinds the quot quot 39 Mlxlandthen39 39 39 tofindL 39 39 05701900B Mechanics ofDeformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 1 122 A picture is taken of an athlete performing a pole vault and the minimum radius of curvature of the pole is estimated by measurement to be 45m If the pole is 40 mm in diameter and is made of glassreinforced plastic for which E131GPa determine the maximum bending stress in the pole 0572019z00B Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 2 121112 The bar is supported by a roller constraint at B which allows vertical displacement but resists axial load and moment If a the bar is subjected to the loading L L 3 shown determine the slope at A the i quotif i 33 quotl deflection at C and the deflection at B El is constant 1 P f R v M 3 B a x M3 A P J 2 RA 4 P rm manna gal W5 v I A E 2 Q g 39 2 VA 2 V X 20gt 9 902 2 o M a quot 4 X S 52 M x PX a M x 539 ELL it lt X 4 L L mm Elia 429 3 2 4 4 L alX 2 2 lt L ame 55ch L H VSC H G r 39lhwu lwa Cem hl lou 85 8XL 41 salve gr 6 0 QB 8K3L L P CL ng CL C PLZ Ei z 2 L L r 2 7 ow AampxEiE q Zltxlt U 56 COWlzlLwlif m5 51 CV x325 40 Sove 74 CI Z L Dr es Side 61quot 5mm 51960 Z 4 C w vA i El fC 339 8 0 WC 322k a i 1564 W x BILLVi 3 2 PLl if M 2 392 L 6quot E 16 3 quot31 3 7 C 3 2 Ej L 1 3H ma a f 3 e955 2 Z 2 W P 9A 2 9xo 6 39L 3E 4 w L 3 4 EIV CK Px ch 3 am a 3 f3PLZX C3 eix i Use Q V gt 03O 4 39 630 Wx P a g anewz e E1 2 2 On vks m 9qu 39 BI x x 43 F41 x 3CL 3 35 Q i Conknits a f CH X gt Cg 3 LE 3 U 5 2 W quot5 t 451 L 25 3 as L EI U Cx3 541x Lfz ILKfigt P Zltx lt L o ma 2 Am a gt 435T y 0570192008 Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 3 Compute the slope of the elastic curve at A and B and the de ection at C E1 is constant chi 3 amp Wm We W 392quot M M M Z V 30 M jx yg iigm cjmmm L z i 2 31 l Wt m J I V WLX WWME MG viww W MM 2 W 2 EM 1 a 519 Ms M a 5 4 g 3 at v 1 WLX MWW X39 3 QE X Q i Snare Wo lt3 W 22 21 5 M c as E 60 Q viii Z6 3 3 2 w swan 25 3 is k 3 465 m 5quot Mg W i x we p if 2 4 a 3 if r mfg Ms w E g g 3 3 EWW Es 5 iw wwLgt Elt w tquotH l 1 fig 2 2 W t a 2 5 it E1 057019BBB Mechanics ofDeformable Bodies College ofEngineeIing Instructor CC Swan University of Iowa Period 16 Spring Semester 2009 Textbook Section 6667 Topic Composite Beams Objectives Review of bending thus far Composite Beams Reinforced Concrete Beams Example Problems PP Pquot 1 Review of Bending Thus Far 3 Bending about a single principal axis 5 B AVAB If wxdx AMAB jVxdx A A K i E El is the quotflexural rigidibf of the section p E1 Pei pg p I b Flexure Formula for Biaxial Bending 0yz I 1y where y and z are the principal axes for the moments of inertia in the sense that lyzzlzyzo szMyz 2 Composite Beams Beams are often made of different materials in order to efficiently carry a load S i glee plate SCul reinforcing mlm w a 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa 0 How do we compute the effective stiffness of such beams What is the elastic bending stress distribution on the composite crosssection Less sii material d39 i f Normal strain variation 4 pro le View 1 Bending slress variation pro le view Bending tress vanatiun c d The cross section of the beam must be transformed into a single material if the flexure formula which is based on homogenous materials is to be used to compute the bending stress Atransformation factor quotnquot will be used forthis purpose I b 12 Beam transformed to material 139 Beam transformed to material 2 e 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instructor CC Swan University of Iowa The height h of the beam remains constant since the strain distribution shown in fig b must be preserved 7 E1 7 E If the stiffer material 1 is transformed into a less stiff material 2the cross section will look like fig e above and the width ofthe material 1 region must be increased b1 nb If the less stiff material 2 is transformed into a stiffer material 1the cross section will transform as fig f above then the width must be reduced bZ n where E1gt EZ Once a section istransformedthe effective moment of inertia is computed forthe transformed section The effective flexural rigidity E39I is also computed usingthe modulus of the material into which the section was transformed 3 Reinforced Concrete Beams Portland cement concrete is a material with a decent compressive strength but very small tensile strength To make beams with concrete reinforcing steel is embedded in the concrete to carrythe tensile stresses and forces Here we will confine our attention to the elastic behavior of reinforced concrete beams n positive bending material above the NA is in compression and that below is in tension Since it is assumed conservatively that concrete cannot take any tension the bending stress distribution on the section is as shown in Fig b Transformingthe steel area into a concrete of equivalent stiffness we solve for h39 the distance the NA from the top of the beam Afterfinding h39 from the following quadratic equation the solution proceeds in the usual manner for obtaining the stress in the beam b 3h39lnA wih39 0 Once h39 is determinedthe effective moment of inertia for the section is computed The flexural rigidity forthe section is then Emmiquot 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instmctor CC Swan University of Iowa 4 Example Problems Example 16121 A wood beam is reinforced with steel straps at its top and bottom as shown Determine the maximum bending stress developed in the wood and steel ifthe beam is subjected to a bending moment of M5kNm Sketch the stress distribution acting overthe cross section TakeW 1 1GPa E5 200GPa 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instmctor CC Swan University of Iowa Example 2 6126 The composite beam is made of A736 steel A bonded to C83400 red brass B and has the cross section shown Ifthe allowable bending stress for the steel is 0th 1801l1Pa and forthe brass calm b39 601l a determine the maximum moment M that can be applied to the beam 100 mm M 100 min V 057 20192BBB Mechanics of Deformable Bodies College of Engineering Instructor CC Swan University of Iowa Example 2 6126 The composite beam is made of A36 steel A bonded to C83400 red brass B and has the cross section shown If the allowable bending stress for the steel is 7010th 18OMPa and for the brass aallaw br 60Wa determine the maximum moment M that can be applied to the beam A 43er E 206839PQ B r 3700 6 o GPQ e A n LeoA I ll25mm 1 n e 24 ZSYI J w maxi151mm 6 12 52 eoL I 14 90 l7MM bJ lf m 1175 11 3 L L j W rewind 11a 4 50 X SXc39g 4 13gtstxoasoua4 l2 4 39L39L Lolllougbms R oMPq 6039zo fmr 1 Marich39 1559 QHON393B 2 538 JM W855 Cano sw mama lgdo Mun Ii1mm Za llb mgt 92 f7 Mellon st 125 ngt Smau f v are 4W5 QIviabe memew l Mallow SB SAI WZHM 057019BBB Mechanics ofDeformable Bodies College ofEngineering Instmctor CC Swan University of Iowa Example 3 67128 Determine the maximum uniform 0754quot Hummer rod distributed load wothat can be supported by the reinforced concrete beam if the allowable tensile stress forthe steel is CHEW 28ksi and the allowable compressive stress forthe concrete is a 3ksi Assume the cam azzaw concrete cannot su port a tensile stress Take 551291 0 k5quot Em3s 10 ksi

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