1633Moment and Defections in Simply Supported Beams3.1 TheeffecTofaconcenTraTedLoadonShear,MoMenT,anddefLecTionStartGOYA-Stofndawindowshowingasimplysupportedbeam(Figure 3.1.1). Thiswindowshowsabeamwithitsleftandrightendssupportedbyapinanda roller, respectively. A perspective view of the beam is shown in Figure 3.1.2. Because the force F is located at midspan, the reactions from the supports are F/2.Inthelowerright-handcornerofthescreen,youwillfndawindowshowing theshear-forcedistribution(Figure 3.1.3).Theshearforcechangesabruptlyatthe load point (midspan). You can see the reason for the change using Figure 3.1.4if you cut the beam to the left of the load (Figure 3.1.4a), you will obtain Figure 3.1.4b, showing that the shear force is clockwise or positive; if you cut it to the right of the load(Figure 3.1.4c),youwillobtainFigure 3.1.4d,showingthattheshearforceis counterclockwise or negative. Recall that the shear force of a cantilever beam also changed abruptly at the points with concentrated loads (see Section 2.2, Chapter 2).In the lower left-hand corner of the screen, you will fnd a window showing the bendingmomentdistribution(Figure 3.1.5).Themomentiszeroatbothendsand reaches its maximum at the load point (midspan in this case). You can see how the moment distribution is obtained if you cut the beam as shown in Figure 3.1.6a to get the free-body diagram in Figure 3.1.6b, which indicates that a section at a distance x from the left support is subjected to a bending moment of MFx=2(3.1.1)Note that M = 0 at x = 0 and that M increases linearly as x increases. The moment is said to be positive because the top of the beam is compressed. We can arrive at thesameequationevenifweconsiderthepartofthebeamtotherightofthecut (Figure 3.1.6c).Equilibriumofmomentaroundanypointonthesectiongivesthe following equation:(moment at the section, M) + (moment by the external force, F) + (moment by the reaction, F/2) = 0,Au: Please check the numbring of fgures cited or Display maths also.Au: Please check the numbring of fgures cited or Display maths also.6861X_C003.indd 163 2/7/08 4:21:44 PM164Understanding Structures: An Introduction to Structural AnalysisFPinRollerF/2F/2150100500500figure3.1.2 Perspective view of beam.55figure3.1.3 Shear-force window.FF/2 F/2F/2FF/2 F/2F/2(a) Cut to the left of load (c) Cut to the right of load(b) EquilibriumCut Cut(d) EquilibriumF/2 F/2figure3.1.4 Cut the beam to see the shear force.105 50.7figure3.1.1 Window of a simply supported beam.6861X_C003.indd 164 2/7/08 4:21:48 PMMoment and Defections in Simply Supported Beams165or M FLxFL x + ( ) =2 20whereclockwisemomentsaredefnedasbeingpositive.Thisagainresultsin M = Fx/2. If you cut the beam to the right of the load as shown in Figure 3.1.6d, you will obtain the free-body diagram shown in Figure 3.1.6e, which indicates that the 2500 0figure3.1.5 Bending moment.FF/2 F/2F/2(a) Cut to the left of load(b) Equilibrium of left partCutFF/2(c) Equilibrium of right partFF/2 F/2F/2(d) Cut to the right of loadCut(e) EquilibriumM = F . x/2M = F . x/2M = F . (L x)/2F/2xF/2F/2L/2 L/2(L x) x(L x)L2 xfigure3.1.6 Cut the beam to see bending moment.6861X_C003.indd 165 2/7/08 4:21:51 PM166Understanding Structures: An Introduction to Structural Analysisbending moment is linearly proportional to the distance from the right support (M = F.(L x)/2). The bending moment at the loaded point is, therefore,MFL=2(3.1.2)Clickthefree-bodydiagrambuttontoobtainFigure 3.1.7,andconfrmthatthe moment varies depending on the position.Example 3.1.1Evaluatetheforcethatwouldberequiredtobreakachopstickoflength10in.and section 0.2 0.2 in2. The chopstick is supported simply and the force is applied at mid-span. Assume that the tensile strength of the wood for the chopstick is 6000 psi.SolutionAswelearnedinSection2.4Chapter2,thebendingmomentthatcausesatensile stress of 6000 psi is determined as follows. Mbh= = =260 266 000 8 ( . )( , )inpsi 1bf-inSubstitutingtheresultforMintoEquation3.1.1,weobtaintheforcetobreakthe chopstick: PML= ==4 4 8103 20(.lbf-in)inlbfThis is a force that you can apply using your thumb. For comparison, the force required to break the chopstick by pulling is P bh = = = ( . ) ( , ) 0 2 6 000 2402in psi 1bfor 75 times the force required if the chopstick is loaded as a simply supported beam. 1055 50.7150figure3.1.7 Cut the beam in the window.6861X_C003.indd 166 2/7/08 4:21:56 PMMoment and Defections in Simply Supported Beams167Next, we shall evaluate the defection of a simply supported beam. Dividing the bend-ing-moment distribution of Figure 3.1.5 by EI, we obtain the distribution of curvature shown in Figure 3.1.8a, or dvdxFxEIx L222= for 0 /2 Integrating this function, dvdxFxEI= +24A (3.1.3)where qA is a constant of integration representing the slope at the left end. Because the slope should be zero at midspan, we haveA = FLEI216(3.1.4)Integrating this equation with the boundary condition v = 0 at x = 0 leads to vFxEIxFxEIx L x L = + = 32 212 484 3 Afor 0 /2 ( )(3.1.5)Thedefectionintherighthalf(L/2xL)canbeobtainedbyreplacingxwith (L x) as shown in the following equation: vFL xEIL x L L x L = ( )[ ( ) ]484 32 2for /2(3.1.6)The defection at midspan (x = L/2) is vFLEI= 348(3.1.7)AU: Insert in-text figure 3.1.AU: Insert in-text figure 3.1.(c) Deection, vAA(b) Slope, = dv/dx(a) Curvature, = d2v/dx2FL/(4EI)L/2 L/2xfigure3.1.8 Deformation.6861X_C003.indd 167 2/7/08 4:22:02 PM168Understanding Structures: An Introduction to Structural Analysiswherethenegativesignrepresentsthatthebeamdefectsdownward.Notethatthe defection is proportional to the third power of the beam length L. Go back to GOYA-S. Push the Detail of beam button and double the beam length; the defection will be 23 = 8 times.If you substitute a = L/2 and R = F/2 into Equation 3.1.7, you will have vRaEI= 33(3.1.7)which is identical with the defection of the cantilever beam (Section 2.8). The reason can be inferred from Figure 3.1.9each half of the simply supported beam bends like a cantilever beam.Example 3.1.2Evaluate the defection at rupture for the chopstick discussed in Example 3.1.1. Assume that the Youngs modulus is 1000 ksi.SolutionAs discussed in Section 2.5 Chapter 2, the moment of inertia of the section is Ibh= = = 3 44120 2121 3 10.. ( ) in.4Substituting this into Equation 3.1.7 and noting that the force is 3.2 lbf, we obtain the defection vFLEI= = 3 33 4483 2 1048 1 000 10 1 3 100., ... ( ) 5 inThe strain when the chopstick breaks under pure tension is = ==E6 0001 000 100 0063,,.The elongation at the break under pure tension is e L = = = 0 006 10 0 06 . . ( ) inwhich is much smaller than the defection of the simply supported beam.R = F/2FR = F/2a = L/2vfigure3.1.9 Equivalent cantilever beam.6861X_C003.indd 168 2/7/08 4:22:07 PMMoment and Defections in Simply Supported Beams169ExerciseTakeanytwonumbersiandjyouchoose.ConsideratimberbeamofL=1m,b= (5+i)mm,h=(5+j)mm,tensilestrength=60N/mm2,andYoungsmodulus= 10,000N/mm2.Calculatetheforceandthedeformationatrupture(a)ifitisloaded transversely and (b) if it is loaded in tension axially.GobacktoGOYA-SandmovetheloadtotherightasshowninFigure 3.1.10. You will fnd that the reaction (green arrow and digit) at the right support increases, whereas the one at the left decreases as you move the load.Youcanseethereasonforthechangeofthereactionswiththehelpof Figure 3.1.11aif we consider the equilibrium of moments around the right support, we get the following equation with the clockwise moment defned as positive: M RL FbA= = 0which leads to RbL FA =(3.1.8)Similarly,theequilibriumofmomentsaroundtheleftsupportyieldsthefollowing equation:RaL FB = (3.1.9)As you move the load to the right (or increase a), the reaction at the right support RB increases. This is a demonstration of the principle of the lever.In GOYA-S, look at the windows showing the bending moment and the shear force (Figure 3.1.12). The moment and shear diagrams also change as you move the force. If you cut the beam to the left of the applied load as shown in Figure 3.1.11b, you will fnd that V RA=(3.1.10)andM RxA= (3.1.11)If you cut the beam to the right of the applied load as shown in Figure 3.1.11c, you will fnd that V RB=(3.1.12)10370.6figure3.1.10 A simply supported beam with a load.6861X_C003.indd 169 2/7/08 4:22:13 PM170Understanding Structures: An Introduction to Structural Analysisand M R L xB= ( )(3.1.13)The maximum moment takes place at the loaded point, and its magnitude is M RaabLFA max = =(3.1.14)Mmax reaches its highest value if the load is placed at midspan (a = b = L/2).LookatFigure 3.1.10again.Youmaynoticethatthemaximumdefectiontakes place not at the loaded point but near the middle of the beam. The deformation of the FRARARBRBRB(a) Beam(b) Cut to the left of load(c) Cut to the right of loadM = RA . xM = RB . (L x)xa bL(L x)RAxfigure3.1.11 Equilibrium.2100 037figure3.1.12Au: Please provide Figure CaptionAu: Please provide Figure Caption6861X_C003.indd 170 2/7/08 4:22:17 PMMoment and Defections in Simply Supported Beams171beam is illustrated in Figure 3.1.13, where you should note that the maximum defec-tion occurs at the point of zero slope (dv/dx = 0).For interested readers: We can detect the point of zero slope as follows. In the left part of the beam (0 x a), we have the following equation for curvature: dvdxMEIFbxEIL22= =(3.1.15)Integrating this equation leads to dvdxFbxEILA= + 22(3.1.16)where qA is the slope at the left end. Noting that v = 0 at x = 0, we obtain v xFbxEILA= + 36(3.1.17)For the right segment of the beam (a x L), we obtain the following equation: dvdxMEIFa L xEIL22= = ( )(3.1.18)Integrating this equation leads to dvdxFa L xEILB= ( )22(3.1.19)AU: Caption? AU: Caption?vmaxFdv/dx = 0AABB(a) Deection, v(b) Slope, = dv/dx(c) Curvature, = d2v/dx2b aMmax/EIfigure3.1.13 Deformation.6861X_C003.indd 171 2/7/08 4:22:22 PM172Understanding Structures: An Introduction to Structural Analysiswhere q B is the slope at the right end. Integrating this and noting v = 0 at x = L, we obtain v L xFa L xEILB= + ( )( )36(3.1.20)Thedefection(v)shouldbecontinuousattheloadpoint(x=a);i.e.,thedefection computed with Equation 3.1.17 should be equal to that computed with Equation 3.1.20. The slope (dv/dx) should also be continuous at the load point. These boundary condi-tions lead to a set of simultaneous equations about qA and qB: AFb L bEIL= ( )2 26(3.1.21) BFa L aEIL= ( )2 26(3.1.22)If we substitute Equation 3.1.21 into Equation 3.1.16, we get dvdxFbEILL b x = 232 2 2( )(3.1.23)The point of zero slope (dv/dx = 0) is xL b=2 23(3.1.24)Thisisthepointwherethemaximumdefectionoccurs(Figure 3.1.13a).Substitute b = L/10 into Equation 3.1.24, for example, and you will fnd x L 0 57 . , indicating that the maximum defection occurs near midspan. As you increase the load, however, the beamwillbreakattheloadpointbecausethemaximumbendingmomentoccursat that point.ExerciseAssume a simply supported chopstick with L = 10 in., b = h = 0.2 in., tensile strength = 6000psi,andYoungsmodulus=1000ksi(Figure 3.1.14).Takeanynumberiand apply a force at a distance of i in. from the right end of the chopstick. Determine at what force the chopstick will break.This equation is valid only for the case of a > b because we assumed that the point of zero slope is located at the left of the load. You need to use Equation 3.1.19 to analyze the case of a < b.AU: Figure 3.1.14 not cited in text. Tentatively cited here. OK?AU: Figure 3.1.14 not cited in text. Tentatively cited here. OK?i in.L = 10 in.Chopstickfigure3.1.14 Simply supported chopstick.6861X_C003.indd 172 2/7/08 4:22:27 PMMoment and Defections in Simply Supported Beams1733.2 TheeffecTofSeveraLconcenTraTedLoadSonShear,MoMenT,anddefLecTionIn GOYA-S, add a load and move it to suit Figure 3.2.1. We can evaluate the reac-tionsofthisbeamsuperimposingFigures 3.1.1and3.1.10.Wecanalsoevaluate themdirectlyasfollows:Inaccordancewiththeforcesandreactionsshownin Figure 3.2.2,weconsidertheequilibriumofmomentsaroundtheleftsupportto obtain (clockwise moments assumed to be positive) M F L F L RLB= + =1 1 2 20(3.2.1)Equation 3.2.1 can be rearranged to determine the reaction at support B.RF L F LLB =+1 1 2 2(3.2.2)We can similarly evaluate the reaction at the right support, RA. Knowing the reac-tions,wecancalculatetheshearforce,V.NotingthatdM/dx=V,weobtainthe bending moment, M.ExerciseTakeanynumberiyouchoose.Determinetheforcesthatwillproducethebending moments shown in Figures 3.2.3a,b. Check your results using GOYA-S and sketch the deformed shape of the beam.Practice: Construct an interesting bending-moment diagram as you did in Section 2.2, Chapter 2.StartGOYA-CandapplyforcesasshowninFigure 3.2.4(thedeformationis magnifed by a factor of four). The force at the free end is the same as the reaction at the right in Figure 3.2.1. You will note that the equilibrium conditions for the can-tilever beam and the simply supported beam are the sameincluding the bending-moment and shear-force diagrams. The only difference is provided by the boundary conditions.Letdenotethedefectionatthefreeend.RotateFigure 3.2.4by/L clockwise. You will obtain the deformed shape shown in Figure 3.2.1.Figure 3.2.5ashowsasimplysupportedbeamsubjectedtoacoupleattheleft end. You should note that the bending-moment diagram (Figure 3.2.5b) is the same as that for the cantilever beam depicted in Figure 3.2.5c.10810121.3figure3.2.1 A simply supported beam with two loads.6861X_C003.indd 173 2/7/08 4:22:30 PM174Understanding Structures: An Introduction to Structural AnalysisUsing GOYA-S, we can simulate Figure 3.2.5 as shown in Figure 3.2.6. In this case, the load of 60 N is applied at a distance of 5 mm from the left support, and the equivalent couple is M = 60 5 = 300 N-mm; the reaction at the right support is M/L = 300/100 = 3 N.RARBF2L2F1L1Lfigure3.2.2 Reactions.30 40 30100M = 50(i + 10)1000(a)20030 40 30(b)M = 50(i + 10)figure3.2.3 Bending moment.10123.841080figure3.2.4 Bending-moment diagram of cantilever beam.6861X_C003.indd 174 2/7/08 4:22:34 PMMoment and Defections in Simply Supported Beams175Next, we calculate the shear force and the bending moment in the simply sup-ported beam with a uniformly distributed load (Figure 3.2.7a). Because of symme-try, the reactions areR RwLA B= =2(3.2.3)(a) Couple(c) Equivalent cantilever beam(b) Bending momentLMM/L M/LM/LMM/LMfigure3.2.5 Simply supported beam with a couple at the left end.15 1215 1233Force (N) Stress (N/mm2)Strain (10^3)330603605702850.73figure3.2.6 Bending-moment diagram similar to that of a cantilever beam.6861X_C003.indd 175 2/7/08 4:22:39 PM176Understanding Structures: An Introduction to Structural AnalysisAswelearnedinChapter2,wehavedV/dx=w(negativebecausetheloadwis downward). Noting that the shear force at the left support (x = 0) isV RA=(positive because it is clockwise), we have V R wxwLL xA= = 22 ( )(3.2.4)TheresultisshowninFigure 3.2.7b.FromChapter2weknowthatdM/dx=V. Noting that the bending moment at the left support (x = 0) is zero, we write M Rx wxwxL xA= + = 12 22( )(3.2.5)The result is shown in Figure 3.2.7c.Example 3.2.1Calculate the defection of the simply supported beam shown in Figure 3.2.7a.SolutionSubstituting Equation 3.2.5 into the equation d2v/dx2 = M/EI (Chapter 2), we obtain an expression for curvature: dvdxMEIwEIx Lx2222= = ( )(3.2.6)Integrating Equation 3.2.6, we obtain an expression for the slope of the beam at any position x: dvdxwEIx Lx C = +122 33 21( )(3.2.7)whereC1isaconstantofintegration.IntegratingEquation3.2.7,wearriveatthe expression for the defection vwEIx Lx C x C = + +2424 31 2( )(3.2.8)w.L2/8RARB(a) Distributed load (c) Bending moment(b) Shear forceLw +w . L/2w . L/2figure3.2.7 Simply supported beam under uniformly distributed load.6861X_C003.indd 176 2/7/08 4:22:44 PMMoment and Defections in Simply Supported Beams177where C2 is another constant of integration. Knowing the two boundary conditions (the defections at the left and the right supports are zero) helps solve the constants C1 and C2. Thus, we obtain vwEIx Lx L x = +2424 3 3( )(3.2.9)The defection at midspan (x = L/2) is vwLEI= 53844(3.2.10)Wecanalsoderivethisequationbyconsideringthecantileverbeamashavinga length equal to half the length of the simply supported beam (a = L/2), as illustrated in Figure 3.2.8.TheresultsinChapter2indicatethatthedefectionofthecantilever beam of length a subjected to uniformly distributed load w is (see Figure 3.2.8b) vwaEIa = 48(3.2.11)The defection caused by an upward (reaction) force of wa is (see Figure 3.2.8c) vwaEIb =43(3.2.12)The defection caused by both loads is (see Figure 3.2.8a) v v vwaEIa b= + =5244(3.2.13)Substitutinga=L/2intoEquation3.2.13leadstothesamedefectionasEquation 3.2.10 in magnitude.(b) Deection by distributed load (c) Deection by reactionwa = L/2 a = L/2wawawwa(a) Total deformationvavbv = va + vbfigure3.2.8 Defection of cantilever beam.6861X_C003.indd 177 2/7/08 4:22:49 PM178Understanding Structures: An Introduction to Structural AnalysisExample 3.2.2From among the sets of forces shown in Figure 3.2.10(ad), select the correct set that producesthebending-momentdistributionshowninFigure 3.2.9.Selectthecorrect defected shape from among those shown in Figure 3.2.11(ad) for the bending-moment distribution shown in Figure 3.2.9.SolutionApplying dM/dx = V for the moment distribution in Figure 3.2.9, we obtain the shear forceshowninFigure 3.2.12.Becausethechangesintheshearforcearecausedby external forces, the correct force set is in Figure 3.2.10a. To fnd the correct defection pattern, we need to recall that the bending-moment diagram (because of our sign con-vention) indicates whether the top or the bottom of the beam is compressed. Because Figure 3.2.9 indicates that the top is compressed everywhere in the beam, the correct answer must be that shown in Figure 3.2.11a.WecansimulatetheresultusingGOYA-S.First,applytwodownwardforces asshowninFigure 3.2.13.Then,applyanotherforceatthemiddleandincreaseits magnitudeto10N,asshowninFigure 3.2.14,wherethedeformationisamplifed 16 times.Ifyouincreasetheforceto13.8N,youwillfndthatthedefectionatthemid-dle will become zero, as shown in Figure 3.2.15, where the deformation is amplifed 32times.ThisisthedefectedshapeofFigure 3.2.11c.Thebeamdoesnotbend abruptly, as shown schematically in Figure 3.2.11d.Note that the defected shape shown in Figure 3.2.15 is symmetric about midspan. InExample2.5.5,foracantileverbeamwithtwoloadsappliedatthefreeendand midspan (P1 and P2), we saw that the defection at the free end is zero if P1/P2 = 5/16. InFigure 3.2.15,weseethattheratioofthereaction(3.1N)tothedownwardforce (10 N) is 3 110 516 . / / .FL/2 FL/2L L L Lfigure3.2.9 Bending-moment distribution.F2FFFF F3FF F4FF F(a) (b)(c) (d)figure3.2.10 Loads.6861X_C003.indd 178 2/7/08 4:22:52 PMMoment and Defections in Simply Supported Beams179(a) (b)(c)(d)figure3.2.11 Defection.+F/2 +F/2F/2 F/2figure3.2.12 Shear force.25 2525 250 . . 0Force (N)Stress (N/mm2)Strain (10^3)100100250 2501010 100110figure3.2.13 Two downward forces (deformation amplifed 4 times).555550 550Force (N) Stress (N/mm2)Strain (10^3)500125 125510 101000.35figure3.2.14 Upward force of 10 N (deformation amplifed 16 times).6861X_C003.indd 179 2/7/08 4:22:58 PM180Understanding Structures: An Introduction to Structural AnalysisExample 3.2.3CalculatetheshearforceandthebendingmomentatAinthebeamshownin Figure 3.2.16.SolutionWe can represent the total uniform load of 6wL by a concentrated load located at the centroid of the uniform load. The equilibrium of moment around point C gives R wL L L wLB = = 6 4 3 2 / /and the equilibrium of forces in the vertical direction gives R wL R wLC B= = 6 9 2 /Next, we shall cut the beam at A as shown in Figure 3.2.17b and replace the distributed load by a concentrated load of 2wL. From the equilibrium of forces, we get V wLA = /2When you want to calculate the reactions, you may replace all the distributed load with an equivalent single load. When you want to calculate the shear force or the bending moment at a point, however, you need to go through the following procedure:(a) Evaluate the reactions (Figure 3.2.17a).(b) Cut the beam and then replace the distributed load with an equivalent single load (Figure 3.2.17b).(c) Consider equilibrium of forces and moments.10 1013.83.10 03377Force (N)Strain (10^3)Stress (N/mm2)9578 7806.9 + + 6.99.5 2.69.5 2.603.1figure3.2.15 Upward force of 13.8 N (deformation amplifed 32 times).2LwA2L 2Lfigure3.2.16AU: Caption? AU: Caption?6861X_C003.indd 180 2/7/08 4:23:04 PMMoment and Defections in Simply Supported Beams181and from the equilibrium of moments, we get M wLA =2.We can similarly calculate the shear force and the bending moment at other sections. The results are shown in Figure 3.2.17c, d.Example 3.2.4Calculate the bending moment at A in the beam shown in Figure 3.2.18.SolutionBecause the total beam length is 4L, the total vertical load is 4wL. Recognizing sym-metry, we determine each reaction to be 2wL, as shown in Figure 3.2.19a. Next, we cut the beam at A as shown in Figure 3.2.19b and replace the distributed load by a concen-trated load of 2wL. The moment equilibrium gives MA = 0. We can similarly calculate theshearforceandthebendingmomentatotherplaces.Theresultsareshownin Figure 3.2.19c, d.AU: Caption? AU: Caption?(a) Reactions3wL/22wL2wL/2AA(b) Free body+2wL5wL/2(c) Shear force(d) Bending moment3L L2L2wLL2LRB = 3wL/2L6wLC+3wL/2MAQARC = 9wL/2wL2BBfigure3.2.176861X_C003.indd 181 2/7/08 4:23:07 PM182Understanding Structures: An Introduction to Structural AnalysisMinigaMe Using gOYa-sThe green fgures in the GOYA window show the maximum upward and downward defections of the beam. Try producing the largest possible bending moment using four loads while controlling the maximum defections so that they do not exceed 1 mm. Use default values for the size of the beam and Youngs modulus.Design YOUr Own BeaM (Part 6)We want to design a beam that can carry a mini-elephant whose weight is any num-ber you choose plus 10 lbf. Assume that each leg carries the same amount of gravi-tational force. The density of the beam is 0.5 lbf/in.3, the beam width is 1 in., and L L L LwAfigure3.2.18Au: please provide Figure captionAu: please provide Figure captionwL2/2L LA2wLL L L LwA2wL 2wL+wLwL wL+wL(a) Reactions(b) Free body(c) Shear force(d) Bending momentwL2/22wLfigure3.2.19AU: Caption? AU: Caption?6861X_C003.indd 182 2/7/08 4:23:09 PMMoment and Defections in Simply Supported Beams183Youngs modulus is 10 ksi. The elephant is scared of fexible beams. The maximum defection of the beam should not exceed 0.1 in. The beam material cannot resist ten-sile stress exceeding 300 psi. What is the required beam depth, h? Check your results using GOYA-S. (Hint: The beam is symmetric. You may replace it by a cantilever beam as shown in Figure 3.2.8.)3.3 SiMiLariTieSbeTweenbeaMandTruSSreSponSeInthissectionwewillinvestigatethesimilaritiesinthewaysbeamsandtrusses resist transverse loads. We do this to develop an improved perspective of the relation-ships between internal and external forces, a perspective that will help improve our understanding of structural response.In Figure 3.3.1a, b, we compare the internal and external forces in a beam and a truss, both loaded with vertical forces of magnitude F located symmetrically about midspan.Wechooseeachofthe10trusspanelstohavealengthof10mmanda depthof10mm.Becausethepanelsaresquare,thewebmembers,ordiagonals, make an angle of 45 with the horizontal. We choose a beam having the same span and a depth of 15 mm. Because of the symmetry, the magnitude of each reaction of the beam and the truss is determined to be F.Determiningtheshear-forcediagramforthebeamisstraightforward (Figure 3.3.1c).Theshearforceisconstantbetweentheleftreactionandtheforce applied on the left. It is zero between the applied forces. Between the force applied ontherightandtherightreaction,theshearisagainconstant.Thebeamshearis equal in magnitude to the external force F and is shown to be positive on the left and negative on the rightto be consistent with our sign convention. 1 mm 1 mmMax. momentMajor Structural League Ranking (N . mm)> 5,000 > 10,000 > 50,000All Star MVP Hall of Fame> 2,000MajorRookie A AA 3A> 500 > 1,000 > 200 > 100figure3.2.20AU: Caption? AU: Caption?4 in 2 inh4 inTrumpetfigure3.2.21AU: Caption? AU: Caption?6861X_C003.indd 183 2/7/08 4:23:11 PM184Understanding Structures: An Introduction to Structural Analysis3F Compression3F Tension2 F010F10ABCDEGHJ00CompressionTensionFFFF2F2FFF2F2FFKLFF30F15FF3030HJKL(b) Equivalent Truss(d) Axial forces in diagonal members(c) Shear force(f) Axial forces in top chords(g) Axial forces in bottom chords(e) Bending moment(a) A beam2 Ffigure

3.3.1

A beam and an equivalent truss.AU: Caption?AU: Caption?6861X_C003.indd 184 2/7/08 4:23:13 PMMoment and Defections in Simply Supported Beams185The shear force in the truss is resisted by the web members (Figure 3.3.1d). There is a direct relation between the shear in the truss and the force in the web members. Because of the inclination of the web members (45), for a shear force of F, the force in each web member is2F . The sign changes from the left to the right end of the truss because the web members work in tension on the left and in compression on the right.Comparing the diagrams for shear distribution in the beam and force distribu-tion in the web members of the truss, we understand that there is a similarity as well as a proportionality between the internal shear distribution in a beam and the distri-butions of the forces in the web members of a truss.Next, we examine the moment distribution in the beam (Figure 3.3.1e). We have studiedtherelationbetweenshearandchangeinmoment.So,itisnotsurprising to see there is a steady increase in moment in the left portion of the beam, between the reaction and the force F, where the shear is constant. Between the applied forces F, the moment does not change because there is no shear in this region. In the right portion of the beam, where the shear is constant and negative, the moment decreases at a steady rate from the maximum at the point of application of the force F to zero at the point of reaction.In the truss, the bending moment is resisted by the forces in the top and bottom chords. The moment at any section should be equal to the product of the force in one chord and the distance between the two chords. We expect the chord forces in the truss to vary as the moment varies in the beam.When we look at the distribution of forces in the top and bottom chords of the truss (Figure 3.3.1f, g), we notice two surprising features:1.The variation of the forces in the top and bottom chords differ from one another.2.They also differ from the distribution of the moment in the beam (there are abrupt changes, and the force distributions are not symmetrical about midspan).In the following text, we try to understand the reason for these apparent inconsistencies.FirstwelookatasimplecasethechordforcesinpanelHJKL,whichisnot subjected to shear (Figure 3.3.2a). The moment to be resisted is 30F. From this, we deduce that the force in each chord is Chord ForceMomentTruss Height= = =30103FF(3.3.1)The signs for the forces in the top and bottom chords are different. These forces must balance one another.If we consider the internal normal stresses in the beam, we note a similar phe-nomenon. As shown in Figure 3.3.2b, the stress distribution may be assumed to be linear. The internal stresses may be represented by forces at the centroids of the ten-sile and compressive stresses (Figure 3.3.2c). The distance between the two internal forces is 2h/3. Because we assumed h to be 15, we fnd each force to have a magni-tude of 3F to balance the moment of 30F (Figure 3.3.1e).6861X_C003.indd 185 2/7/08 4:23:15 PM186Understanding Structures: An Introduction to Structural AnalysisThere appears to be a similarity as well as a proportionality between the internal normal stresses in beams and chord forces in trusses.WenowexaminetheconditionsinpanelABEDofthetrussnexttotheleft reaction. Our frst deduction is that the vertical component of the force in the web member must be equal to F (Figure 3.3.3). Because the web member makes an angle of 45 with the horizontal, its horizontal component must also be equal to F in ten-sion. The equilibrium of moment around node A requires zero force in the bottom chord;theequilibriumofhorizontalforcesrequiresthatthetopchordmustcarry acompressiveforceF.Thatiswhy,inthispanel,thetopchordsustainsaforce, whereas the bottom chord does not. The moment equilibrium in that section around the top chord givesM = Fx(3.3.2)which agrees with Figure 3.3.1e.We move over to the next panel on the right (Figure 3.3.3b). The shear remains the same. Therefore, the vertical and horizontal components of the web member are equaltoF.ConsideringtheequilibriumatjointB,wedecidethattheforceinthe top chord must be equal to 2F in compression. To maintain horizontal equilibrium Moment equilibrium around the bottom chord also gives M = Fx.AU: Should this be 3.3.4?AU: Should this be 3.3.4?(a) Forces in the truss (c) Equivalent forces (b) Stress in the beamH103F3FHK3F3Fh = 10HKM/ZM/Z23h = 15Kfigure3.3.2 Cut between HJ.xADxFFFFfigure3.3.3 Cut between A and B.6861X_C003.indd 186 2/7/08 4:23:17 PMMoment and Defections in Simply Supported Beams187across any section within the panel, the bottom chord force needs to be F in tension. Now we can consider the moment equilibrium in that section around the top chord: M = + (Contribution of web member) Contribution of ( bbottom chord)( ) = + = F x F Fx 10 10(3.3.3)This result agrees with the linearly distributed bending moment in the beam shown in Figure 3.3.1e.From the foregoing, we deduce that (1) the abrupt changes in chord forces are caused by the condition that forces can change only at the joints of the truss and that (2) the lack of symmetry in the distribution of the chord forces is caused by the orien-tation of the web members. Otherwise, the distribution of the chord forces along the span of the truss represents a good analogue for the changes in the internal normal forces in a beam.Figures 3.3.5 and 3.3.6 compare the results obtained with GOYA-T and GOYA-S, respectively. Note that the deformed shapes are also similar. The compressive and tensile forces in the highlighted region agree with the axial forces in the correspond-ing top and bottom chords.Figure 3.3.7a shows the deformation of panel HJKL, where the top chord short-ens and the bottom chord lengthens, as in the fexural deformation of a beam. The strains of the top and bottom chords are = =PEAFEA3(3.3.4)xBDx 10AE2FFFFF10figure3.3.4 Cut between B and C.Au:pleaseprovide Figure citationAu:pleaseprovide Figure citation30303020201001414203030 30 30 3000 0 0 0 0 0020101014 14.114.114.100101010101010103030figure3.3.5 Results from GOYA-T.6861X_C003.indd 187 2/7/08 4:23:21 PM188Understanding Structures: An Introduction to Structural Analysiswhere P is the axial force, E is the Youngs modulus, and A is the cross-sectional area of each chord. The deformation of each chord is obtained as the product of the strain e and the length of the panel dx = 10. e xFEA= = 30(3.3.5)The fexural rotation dq of the panel in Figure 3.3.7a (the angle between HK and JL) is = =2106 | | e FEA(3.3.6)According to the defnition in Section 2.5 Chapter 2, the curvature f is obtained by dividing dq by the width dx = 10: = =xFEA610(3.3.7)101010101.2figure3.3.6 Results from GOYA-S.(b) Panel ABDEBDAEFFx = 1055(a) Panel HJKLx = 10JKHL553F3F3F3Ffigure3.3.7 Flexural deformation.6861X_C003.indd 188 2/7/08 4:23:29 PMMoment and Defections in Simply Supported Beams189Thebendingmomentaroundthecenterlineofthetruss(thechainedlinein Figure 3.3.7a) is M F F F = + = 3 5 3 5 30(3.3.8)Equations 3.3.7 and 3.3.8 lead to =MEA 50(3.3.9)Ontheotherhand,thebendingmomentatthecenterofpanelABDEisgivenby substituting x = 5 into Equation 3.3.2:M F F = = 5 5 (3.3.10)The curvature f in panel ABDE is obtained in reference to Figure 3.3.7b:= =xFEA 10(3.3.11)Equations 3.3.10 and 3.3.11 again lead to Equation 3.3.9.Wecandiscussthesimilaritiesbetweenthetrussandthebeamfurther.We regard the truss as a beam having the section shown in Figure 3.3.8. As will be dis-cussed in Chapter 4, the moment of inertia of the section is I Ay = 22(3.3.12)Substituting y = 5 into Equation 3.3.11, we getI A = 50 . Equations 3.3.9 and 3.3.12 lead to =MEI (2.4.6)which we obtained in Section 2.6 Chapter 2 (Equation 3.4.6).In addition to the fexural deformation shown in Figure 3.3.7b, panel ABDE is distorted (Figure 3.3.7c) because the diagonal member AE is subjected to a tensile force 2F. You can simulate the distortion if you provide GOYA-T with the boundary conditionshowninFigure 3.3.7d.Asimilardistortionalsooccursinbeams.Itis yyAAfigure3.3.8 Section.6861X_C003.indd 189 2/7/08 4:23:36 PM190Understanding Structures: An Introduction to Structural Analysiscalled shear deformation or shear distortion (note that the axial force in the diago-nal member is related to the shear force in the corresponding beam) (Figure 3.3.9). However, the effect of the shear deformation (or distortion) on the total defection is small in shallow trusses or beams such as those shown in Figures 3.3.1 or 3.3.4.ExerciseChoose any numbers i and j to set the loads shown in Figure 3.3.8. Draw fgures simi-lar to Figure 3.3.1 for the truss shown in Figure 3.3.8. Also, draw the deformed shape. Compare your result with those obtained by a friend for the same problem. 1010 + i (N) 1010 + j (N).You have made many virtual experiments with beams using GOYA. In this section, you will test an actual beam to review Chapters 2 and 3.You need the following:1.Two pieces of wood 1/8 in. thick and 1/4 in. wide, with length of 2 ft. In this section, we call them beams. You can buy them at a hardware shop. Do not buy balsa.2.A ruler longer than 8 in.3.A small plastic bag.4.Two binder clips.5.Two desks.6.A kitchen scale.7.A friend to help you.AU: Figure 3.3.9 not cited in text. Ten-tatively cited here. OK?AU: Figure 3.3.9 not cited in text. Ten-tatively cited here. OK?(a) Deformation of ABDE (b) Simulation using GOYA-TBDAE2 F2 F2 Ffigure3.3.9 Shear deformation.6861X_C003.indd 190 2/7/08 4:23:38 PMMoment and Defections in Simply Supported Beams191test 1: MeasUring YOUngs MODUlUs fOr wOODWeshallmeasureYoungsmodulusforwoodusingEquation3.1.7developedin Section 3.1.Step 1: Mark one of the beams as shown in Figure 3.4.1a and fx the plastic bag at point C using a binder clip.Step 2: Fix the ruler to the other beam using another binder clip as shown in Figure 3.4.1b.Step3:Layapenciloneachdeskatadistanceof20in.asshownin Figure 3.4.2. Place the marked beam on the pencils.Step 4: Take the other beam and place the end without the ruler on the foor. Make sure it is vertical. Measure the initial location of point C.Step 5: Hang the plastic bag at midspan of the beam. Place objects (mar-bles,sand,etc.)inthebagsothatthedefectionreachesapproximately2 in. Measure the defection at point C (vCC in Figure 3.4.2) as accurately as you can.Step 6: Weigh the plastic bag with its contents using the kitchen scale.Step 7: Recall the equation for the defection of a simply supported beamvFLEI= 348, (3.1.7)DRulerBinder clip(a) Beam for test (b) Beam for measurementABC10 in5 in5 in1/8 in1/4 infigure3.4.1 Mark one of the beams and attach a ruler to the other beam.DeskF 10 in 10 inPencilPencilDeskABCDvCCfigure3.4.2 Measure defection.6861X_C003.indd 191 2/7/08 4:23:41 PM192Understanding Structures: An Introduction to Structural Analysisand calculate Youngs modulus E by substituting the measured defection v, the load F, the beam length L, and the moment of inertiaI bh =312 / . The unit of the load should be stated in lbf.Step8:CheckyourresultnotingthatYoungsmodulusofwood(except balsa)isusuallybetween1000and1800ksi.Ifyourresultisoutofthis range, reexamine your calculation and/or your measurements.test 2: testing the eqUatiOn fOr the DeflecteD shaPe Of a siMPlY sUPPOrteD BeaMIn this test, we shall check Equation 3.1.5, which determines the defected shape of the beam, using the same equipment as in Test 1.Step1:CalculatethedefectionvBCshowninFigure 3.4.3.Thenotation vBCindicatesthatthedefectionismeasuredatpointBandtheforceis appliedatpointC.UseEquation3.1.5andthevalueforEdeterminedin Test 1.Step2:MeasurethedefectionvBCandcompareitwiththecalculation result.test 3: testing the reciPrOcitY theOreM Using a siMPlY sUPPOrteD BeaMInthistest,weshallinvestigatethereciprocitytheoremdiscoveredbyScottish physicistJamesClerkMaxwell(18311879),whoiswellknownforhiscontribu-tions to electromagnetism.Step1:AssumethattheweightisatthelocationshowninFigure 3.4.4. Calculate the defection at point A vCB using Equation 3.1.17. This should be equal to the calculated defection vBC in Figure 3.4.3. This equality is a demonstration of Maxwells reciprocity theorem.Step2:DothetestshowninFigure 3.4.4tomeasurethedefectionvCB. Compare it with the calculated value.Prediction is very important in civil engineering. Car designers can test the safety of their car by test-ing many prototypes before actually selling the car. On the other hand, civil engineers can rarely test their design using a full-scale model; they need to predict structural behavior through calculation.10 inDesk DeskBvBCF CD5 in 5 inAfigure3.4.3 Measure defection between support and load.6861X_C003.indd 192 2/7/08 4:23:43 PMMoment and Defections in Simply Supported Beams193The reciprocity theorem is applicable to any structure. Figure 3.4.5 shows two other examples in which vBA (the displacement of point B on a structure caused by a load F acting at point A) is always equal to vAB (the displacement of point A caused by the same amount of load F acting at point B).test 4: testing the eqUatiOn fOr the DeflecteD shaPe Of a cantilever BeaMUse the same beam and the same weight you used for Test 3.Step 1: Place the beam at the edge of a desk and press it down as shown in Figure 3.4.6 with a stiff book or pencil case. Press it frmly. Otherwise, the beamwoulddeform,asshowninFigure 3.4.7,andyouwouldnotobtain the correct boundary condition. (The correct boundary condition is that the beam has zero slope and zero defection at the edge of the table or that we have a fxed end at C). Measure the defections vAA and vBA at locations A and B, respectively.Step 2: Calculate the defections vAA and vBA for this test using the equa-tions in Chapter 2. Do they agree with the measurements?AU: Insertion OK?AU: Insertion OK?DeskDeskF BvCBADC10 in 5 in 5 infigure3.4.4 Move the load to the left string.AAABA(a) Simple beam (b) Simple beam with projectionvABFvBAFFFvABvBABBBfigure3.4.5 The reciprocity theorem.6861X_C003.indd 193 2/7/08 4:23:46 PM194Understanding Structures: An Introduction to Structural Analysistest 5: testing the reciPrOcitY theOreM Using a cantilever BeaMStep 1: Hang the load at location B as shown in Figure 3.4.8 and measure the defections vAB and vBB at locations A and B, respectively.Step 2: Calculate the defections vAB and vBB for this test using the equa-tions in Chapter 2. Do they agree with your measurements?Step 3: Study your results. Do they conform to the reciprocity theorem?PencaseDeskBAF vBAPressrmlyC5 in 5 invAAfigure3.4.6 Defections at loaded point and between support and load.PencaseDeskBAC5 in 5 infigure3.4.7 If you press the pen case insuffciently .PencaseDeskBAvBBF vAB5 in 5 infigure3.4.8 Move the load.6861X_C003.indd 194 2/7/08 4:23:49 PMMoment and Defections in Simply Supported Beams195test 6: investigating the effects Of MOMent Of inertiaStep 1: Rotate the beam around its longitudinal axis by 90 so that the beam hastheshorterdimensionofitscrosssectionparalleltothedesksurface (Figure 3.4.9).Step 2: Measure the defection at point A.Step3:Calculatethedefectionandcompareitwiththemeasurement. Compare it also with the result obtained in Test 4.test 7: sensing a cOUPleStep 1: Hold the wood beam with your fngers at midspan.Step 2: Ask a friend to push one end up and the other end down with the same force (Figure 3.4.10). You will sense the twist or couple that is required to resist the moment generated by equal and opposite forces applied at the ends of the beam (Figure 3.4.11a). You will also see that the defection of the beam is antisymmetrical about midspan.Step 3: Try holding the beam at different points away from its middle as shown in Figure 3.4.11b. Observe the defected shape.DeskABF 10 inCfigure3.4.9 Rotate the beam around its longitudinal axis by 90.figure3.4.10 Sense a couple.6861X_C003.indd 195 2/7/08 4:23:51 PM196Understanding Structures: An Introduction to Structural Analysis3.5 probLeMS(Neglect self-weight in all the problems. Assume that all the beams are prismatic.)3.1Find an incorrect statement among the following fve statements concern-ing the simply supported beam in Figure 3.5.1.1.The shear force at point B is larger than that at point D.2.The bending moment at point B is the same as that at point D.3.The bending moment reaches maximum at point C.4.The defection reaches maximum at point C.5.The slope reaches maximum at point A.3.2Whichsetsofloadsyieldthebending-momentdiagrams(1)(5)inFig-ure 3.5.2? Select the correct answer from among (a)(e) in Figure 3.5.2.L/2 L/2M/2M/22L/3 L/3M/3MM/L M/LMM/L M/L2M/3figure3.4.11 Reactions.L L 2L 2L6LFACB DEfigure3.5.1(4)(2) (3)(1)(5)(a) (b) (c) (d) (e)figure3.5.2AU: Caption? AU: Caption?AU: Caption? AU: Caption?6861X_C003.indd 196 2/7/08 4:23:54 PMMoment and Defections in Simply Supported Beams1973.3ThewidthandtheheightofthebeaminFigure 3.5.3are100mmand 180 mm, respectively. What is the maximum bending stress in the beam? Select the correct answer from among (1)(5).(1) 50 N/mm2(2) 100 N/mm2(3) 200 N/mm2(4) 300 N/mm2(5) 400 N/mm23.4Consider a continuous beam subjected to a uniformly distributed load and supported as shown in Figure 3.5.4a. What is the ratio of the reaction at A (RA) to that at B (RB)? Select the correct answer from among (a)(e). (Hint: The defections due to a uniformly distributed load and a concentric load are shown in Figure 3.5.4b.): (a) 1/2(b) 1/3(c) 2/5(d) 3/5 (e) 3/103.5ThebeamshowninFigure 3.5.5haszerobendingmomentatpointA. Find the correct ratio of P to wL from among (1)(5).AU: Provide answers.AU: Provide answers.w = 12 kN/m6.0 mfigure3.5.3wL/2 L/25wL4384EIFL/2 L/2FL348EIRBL/2 L/2wRARC = RA(a) Continuous beam (b) Deectionsfigure3.5.4AU: Caption? AU: Caption?AU: Caption? AU: Caption?6861X_C003.indd 197 2/7/08 4:23:56 PM198Understanding Structures: An Introduction to Structural Analysisthe correct answer from among (1)(5), where I is the moment of inertia of the beam section.(1) FLEI38(2) FLEI33(3) FLEI32(4) 233FLEI(5) 563FLEI3.7What is the defection at point A of the beam shown in Figure 3.5.7? Select the correct answer from among (1)(5), where I is the moment of inertia of the beam section. (Hint: use the result in Problem 3.6)(1) 233FLEI (2) 563FLEI (3) FLEI3 (4) 433FLEI (5) 533FLEI3.8Asimplysupportedbeamissubjectedtouniformlydistributedcou-plesasshowninFigure 3.5.8.Thebeamwidthanddepthare80mm each, and Youngs modulus is 1000 N/mm2. Determine the moment dia-gram, the defected shape, and the maximum defection. (Hint: Determine themomentdiagramforthecasesshowninFigure 3.5.9a,b,wherethe AU: Corrected as 3.8 OK?AU: Corrected as 3.8 OK?LFLAfigure3.5.6AU: Caption? AU: Caption?LFLALfigure3.5.7AU: Caption? AU: Caption?AU: Caption? AU: Caption?LFL L LwF : wL1 : 31 : 21 : 12 : 13 : 1A(1)(2)(3)(4)(5)figure3.5.56861X_C003.indd 198 2/7/08 4:24:05 PMMoment and Defections in Simply Supported Beams199uniformlydistributedcouplesarerepresentedbyaconcentratedcouple 1010=100kN-mandtwoconcentratedcouples510=50kN-m, respectively. Recall that these couples are represented by pairs of horizon-tal forces, as shown in Figure 3.5.9 c, d. You can simulate each case using GOYA-S, as shown in Figure 3.5.9 e, f.)10 m10 kN-m/mfigure3.5.8 Uniformly distributed couples.1 m100 kN-m(a) Concentrated couple50 kN-m(b) Two couples1 m100 kN(e) A pair of vertical forces50 kN(f) Two pairs of vertical forces100 kN1 m50 kN1 m100 kN 50 kN(c) A pair of horizontal forces (d) Two pairs of horizontal forces1 m50 kN50 kN 50 kN 100 kNfigure3.5.9 Hint.6861X_C003.indd 199 2/7/08 4:24:07 PM6861X_C003.indd 200 2/7/08 4:24:07 PM2014Bending and Shear Stresses4.1 FirstMoMentIn Chapters 2 and 3 we considered beams with rectangular sections and learned that1.The stress, s, in the extreme fber of the section, is determined using the section modulus MZ. (2.5.7)2.The curvature of the beam can be determined using the moment of inertia (second moment) MEI. (2.6.4)3.Integrating the curvature obtained by Step 2, we can determine the defec-tion of the beam.These procedures apply not only for rectangular sections but also for others such as I sections and tubes. In this chapter, we shall consider sections that are not rectangu-lar. First, we need to defne the frst moment of a section.Start GOYA-I to reach the window in Figure 4.1.1a. Each square, surrounded by the grid lines, measures 10 10 mm. The fgure shows a rectangular section with b30 mm and h40 mm. Click the Bending stress button to get the applet shown in Figure 4.1.1b. This applet shows the distribution of stress for a bending moment of 10 103 N-mm. You can change the magnitude of the moment using the sliding bar. The red and blue colors indicate compressive and tensile stresses, respectively.Click the six squares indicated by the arrows in Figure 4.1.1a. You will get an inverted T-section, as shown in Figure 4.1.2a,b. As you click, you will notice that the red line changes its position. The red line indicates the position where the bending stress is zero, as shown in Figure 4.1.2c. It is called the neutral axis. To calculate the position of the axis, we shall show that the frst moment defned by the following We call this the frst moment because Equation 4.1.1 includes the frst order of the coordinate y. If it is replaced with the zero order of the coordinate y, i.e., y01, we obtain the area of the section A y dA dA 0 (4.1.2) The second moment will be defned similarly in the following section.Au: Please check the numbring of fgures cited or Display maths also.Au: Please check the numbring of fgures cited or Display maths also.6861X_C004.indd 201 2/7/08 2:25:41 PM202Understanding Structures: An Introduction to Structural Analysis(a) Section (b) Bending stressy0ydAhy0ydAh(a) Beam (b) Section (c) Stress distribution(d) Cut the beam at the middle (e) Dene the coordinate y from the bottomFigure4.1.2 Beam of inverse-T section.6861X_C004.indd 202 2/7/08 2:25:44 PMBending and Shear Stresses203equation should be zero about the neutral axis. S y dAyh y 000 (4.1.1)wherey0isthedistancefromthebottomofthesectiontotheneutralaxis,andy isthedistancefromtheneutralaxistoaninfnitesimalsectiondA,asshownin Figure 4.1.2b.To derive the previous equation, let us again assume a linear strain distribution over the depth of the section: y (2.5.2)wherefisthecurvatureatthesection.Stressatanylevelinthesectionmaybe expressed using Youngs modulus, E, as E (1.2.3)or we may relate it to curvature using Equation 2.5.2. E y (2.5.3)Thisequationindicatesthatthestressisdistributedlinearlyoverthedepthofthe section, as shown in Figure 4.1.2c. The axial force on the section is the product of the area dA and the stress s: P dA E y dA ES (4.1.3)Theaxialforceinabeamsubjectedtotransverseloadiszero(P ES 0) because of the equilibrium of forces along the beam axis (Figure 4.1.2d). We there-fore get S0 (Equation 4.1.1). If we redefne the y-coordinate as distance from the bottom of the section (Figure 4.1.2e), Equation 4.1.1 can be rewritten as S y y dAh ( )000(4.1.4)We can expand this equation as ( ) y y dA ydA y dA ydA y Ah h h h 00 000 000andobtainthefollowingequationthatismoreusefulforcalculatingy0than Equation 4.1.1. yy dAAh00

(4.1.5)6861X_C004.indd 203 2/7/08 2:25:51 PM204Understanding Structures: An Introduction to Structural AnalysisExample 4.1.1Determinethelocationoftheneutralaxisy0fortheinverse-Tsectionshownin Figure 4.1.3a.SolutionWe shall use Equation 4.1.5 for defning the coordinate y as distance from the bottom (Figure 4.1.3). Noting that the infnitesimal area isdA ady 3in the wide part of the section (0 y a) and dAa dy in the narrow part (a y 3 a), we have ydA y ady yady a ah aaa + + 0 043 33 1 5 7 5 9 . . aa3The area of the total section is A a a a a a + 3 3 62Substituting this in Equation 4.1.5, we have yydAAaaah0032961 5 .Example 4.1.2Calculatethelocationoftheneutralaxisy0forthetriangularsectionshownin Figure 4.1.4.dydy(a) Section (b) Area of innitesimal sections3aay0a a aya3adA = 3a dydA = a dyFigure4.1.3 Inverted-T section.bdy dAyhf(y)Figure4.1.4 Triangular section.6861X_C004.indd 204 2/7/08 2:25:56 PMBending and Shear Stresses205SolutionWeuseEquation4.1.5again.Thewidthoftheinfnitesimalareaf(y)shownin Figure 4.1.4 can be defned as f ybh yh( )

Because the infnitesimal area is dAf(y) dy, ydA y f y dybhy h y dybhh h h 0 0 026( ) ( )We get yydAAbhbhhh00262 3

//indicating that the neutral axis of a triangle crosses its centroid (i.e., center of gravity).Determining the CentroiD of a SeCtion by experimentWe need the following:1.A sheet of cardboard (preferably with a grid printed on it)2.A thread, a needle, a pair of scissors, a ruler, and a calculatorIn this test, we shall investigate if the neutral axis of any section, as in the case of a triangle, crosses its centroid.1.Draw any section using GOYA-I.2.Cut the cardboard to have the same shape as the section.3.Calculate the location of the neutral axis y0 and check to see if the result agreeswiththeredlineinthescreen.Drawthelineonthecardboard model (line AB in Figure 4.1.5).TreadABy0x0Figure4.1.5 Hang by thread.6861X_C004.indd 205 2/7/08 2:25:59 PM206Understanding Structures: An Introduction to Structural Analysis4.Punch a hole on line AB near the edge of the cardboard model. Hang it using the thread as shown in Figure 4.1.5. If your calculation of y0 is cor-rect, the line AB will be vertical. This is the proof that the neutral axis AB crosses the centroid of the board.5.Calculate the location of the neutral axis x0 assuming that the section in Figure 4.1.5 is bent around the horizontal axis. Hang the board as shown in Figure 4.1.6. Make certain again that the line CD is vertical. This is the proof that the neutral axis CD crosses the centroid of the board.6.Punch another hole in the cardboard model. The hole must not be on lines AB or CD (Figure 4.1.7). Check to see if the extension of the thread (the brokenlineinFigure 4.1.7)crossestheintersectionpointoftheneutral axes. The broken line represents the neutral axis when the section is bent around the broken line.A BCDy0x0Figure4.1.6Au: provide CaptionAu: provide CaptionABDCFigure4.1.7Au: provide CaptionAu: provide Caption6861X_C004.indd 206 2/7/08 2:26:01 PMBending and Shear Stresses2077.Sticktheneedleandthreadintothecardboardmodelasshownin Figure 4.1.8, and make certain that the cardboard is horizontal. This is the proof that the stuck point is the centroid of the board.Let us examine the reason why the line AB in Figure 4.1.5 is vertical. We shall use Figure 4.1.9,wherethey-coordinateismeasuredfromtheneutralaxis.Iftisthe thickness of cardboard and r its density, the moment around the hole caused by the slice dA is dM t dA y ( ) Integrating the moment over the whole body gives M dM t y dA t S where S is the frst moment defned by Equation 4.1.1. Because S0, we get M0, which means that the body is in equilibrium in terms of moment and does not rotate.NeedleTreadFigure4.1.8 Hang by needle and thread. AU: Caption? AU: Caption?dA . t . dAy0ytFigure4.1.9 Moment around the hole.6861X_C004.indd 207 2/7/08 2:26:08 PM208Understanding Structures: An Introduction to Structural AnalysisIf you conduct an integration over both the x and y directions, you can demon-strate the reason why the board is horizontal in Figure 4.1.8.Coffee breakThefrstpersonwhodiscoveredthecomputationmethodforthecenterofgravity wasaGreekphysicist,Archimedes(BC287212).Itisanimportantextensionof his famous principle of the lever. Note that we also used this principle in reference to Figure 4.1.9. Archimedes also invented integral calculus, which is indispensable to the computation of the center of gravity. E. T. Bell describes him as the greatest scientist in antiquity in the book Men in Mathematics. The inventors of differen-tial calculus, the counterpart of integral calculus, were Isaac Newton and Gottfried W. von Leibnitz, who lived in the 17th century.4.2 secondMoMentandsectionModulusIn the preceding section, we defned the frst moment as S y dA (4.1.1)whereyisthedistancebetweentheinfnitesimalsectiondAandtheneutralaxis (Figure 4.2.1). In this section, we shall defne the second moment (or the moment of inertia) by replacing y with y2 in Equation 4.1.1: I y dA2 (4.2.1)InSection2.5,Chapter2,welearnedthatthebendingmomentistheintegralof the axial force of the infnitesimal section s.dA multiplied by the distance from the neutral axis y: M y dA (2.5.3) Bell, Eric Temple, 1937. Men in Mathematics, Touchstone (Simon & Schuster), New York.ydyy yhbyymax(a) Rectangle (b) H (vertical) (c) H (horizontal) (d) General shapedA dydAdy dy dA dAFigure4.2.1 Various sections and infnitesimal segments.6861X_C004.indd 208 2/7/08 2:26:11 PMBending and Shear Stresses209Thisequationappliestoallkindsofsections.InSection2.6welearnedthatthe stress is proportional to the curvature f and the distance from the neutral axis y, as expressed in Equation 2.6.3. Ey (2.6.3)Substituting this into Equation 2.5.3, M E y dA EI 2 (4.2.2)The moment of inertia plays an important role in relating the bending moment to the curvature (curvature is a measure of how or at what rate the beam bends). For the rectangular section in Figure 4.2.1a, dAb.dy and I y dA y b dybhhh 2 232212 // (4.2.3)as we learned in Section 2.6.Figure 4.2.2showstheinitialwindowofGOYA-I.Aswasstatedearlier,each squaremeasures1010mm.Thedigitsintheright-handcolumnshowthecon-tribution of each row to the moment of inertia. For example, the contribution of the uppermost row is I y dy 210203 430 70 10 mmas listed in the column of numbers that appear in Figure 4.2.2. If we are interested in obtaining an approximate value, we can state the contribution of these three squares Figure4.2.2 Window of GOYA-I.6861X_C004.indd 209 2/7/08 2:26:15 PM210Understanding Structures: An Introduction to Structural Analysisto the moment of inertia as I y dA ( )2 2 3 415 30 10 67 5 10 . mmwhereyis the distance from the neutral axis to the centroid of these squares. The total moment of inertia is shown at the bottom of the column (I70 + 10 + 10 + 70160 103 mm4).Press Ctrl + N three times to create four windows. In these windows, draw the four sections shown in Figure 4.2.3. All the sections have the same area, A2000 mm2, but very different moments of inertia, I, ranging from 90 to 1907 103 mm4. The large differences are caused primarily by the different contributions of the extreme rows (I y dA 2 in Figure 4.2.3). For the section in Figure 4.2.3a, the average distance totheextremerowsisassmallasy15 mm,butforthesectioninFigure 4.2.3d the average distance is as large asy35 mm. For the section in Figure 4.2.3c, the average distance to the extreme rows (or squares) is large but the area dA is small. The expression M EI /indicates that the beam with the section in Figure 4.2.3d will have a smaller curvature and, therefore, smaller defection for a given load over a given span than that of the other sections.Substituting M EI /into Ey, yI M (4.2.4)Press the Bending stress button in the windows showing the sections in Figure 4.2.3 and obtain the stress distributions shown in Figure 4.2.4. Note that the stresses vary linearly with the distance from the neutral axis y. If we defne the distance between the edge of the section and the neutral axis ymax, as shown in Figure 4.2.4, the maxi-mum stress in the section smax is maxmax

yIM (4.2.5)Ifthebeamismadeofbrittlematerialwithstrengthsf,itwillfailatthebending moment MIyf f

max (4.2.6)(a) I = 267 (b) I = 417(c) I = 577(d) I = 1907I = 117y = 15 mmI = 182I = 163I = 863y = 35 mmFigure4.2.3 Moment of inertia (unit: 103 mm4).6861X_C004.indd 210 2/7/08 2:26:24 PMBending and Shear Stresses211We call the following coeffcient as the section modulus. ZIy

max (4.2.7)In GOYA-I, Z is indicated at the bottom of the window. The section modulus of a rectangle is ZIybh h bh

max3 212 2 6 (4.2.8)as we learned in Section 2.5, Chapter 2. Now, we can rewrite Equation 4.2.6 as M Zf f (4.2.9)Inotherwords,thestrengthofabeamisproportionaltoitssectionmodulus. Because the section modulus of the section in Figure 4.2.4d is much larger than that of the section in Figure 4.2.4c, an I-shaped section should be positioned as shown in Figure 4.2.4d to effciently resist bending.Technical termsfanges and web:Figure 4.2.5showsthetypicalsectionofasteelI-beam.Structuralengineers call the strips in the top and bottom as fanges, and the vertical plate as a web, which y(d) Vertical H (c) Horizontal HmaxmaxymaxymaxFigure4.2.4 Stress distribution.FlangeWebFlangeFigure4.2.5 Steel I-beam.6861X_C004.indd 211 2/7/08 2:26:28 PM212Understanding Structures: An Introduction to Structural Analysismay look similar to the skin (web) that joins the toes of swans. Flanges are typically thicker than webs, as shown in the fgure to resist bending moment effectively.Example 4.2.1Calculate the section modulus of the section shown in Figure 4.2.6.SolutionFirst,weevaluatethemomentofinertiaasthesumofthreeparts(top,middle,and bottom): I y dAy ady y ady y aaaaa + + 223423325 5 ddya a a aaa + + 434 4 4 4185354318534243Noting ymax4a, we have ZIyaaa

max42434 341063Because the section considered is symmetrical about its neutral axis, we can shorten thecalculationprocessbypartitioningthesectionasshowninFigure 4.2.7arect-angular section of 8a 5a minus two sections of 6a 2a. RecallingI bh 312 /for a rectangular section, we obtain the same result. Ia a a aa a ( ) ( ) ( ) ( ) 5 81222 612640321633 34 4

42434a : OKa3a3aaa 2a 2aX XFigure4.2.6AU: Caption? AU: Caption?6861X_C004.indd 212 2/7/08 2:26:32 PMBending and Shear Stresses213However, we should not use this technique for calculating the section modulus because ymax of the outer rectangle (4a) is different from that of the inner ones (3a). Za a a aa a ( ) ( ) ( ) ( ) 5 8622 661603723882 23 33310633 3a a < : NG!Also, we cannot use this shortcut for calculating the moment of inertia of a section that is not symmetrical about the horizontal axis (Figure 4.2.8) because the neutral axes of the partitioned sections are different from each other. This technique is valid only for themomentofinertiaofasectionsymmetricalaboutthebendingaxis.Thecorrect moment of inertia of the section in Figure 4.2.8 is I y ady y adyaaaa + 21 50 520 52 53 3 2..... 55 5 25 8 54 4 4a a a +. .If you use the shortcut, you will get an incorrect answer: Ia a a aa a ( ) ( ) ( ) ( ). .3 412231216 4 5 113 34 455 8 54 4a a > . : NG!=8a5a8a 2x5a6a2a6aymax = 4aymax = 3a2x3a3aa3a4a3aa1.5aFigure4.2.8 Never do this because neutral axes are different.6861X_C004.indd 213 2/7/08 2:26:37 PM214Understanding Structures: An Introduction to Structural AnalysisExample 4.2.2Calculatethemomentofinertiaandthesectionmodulusofatriangularsection (Figure 4.2.9).SolutionConsider an infnitesimal slice of thickness dy and width f(y), as shown in Figure 4.2.9. The dimension f(y) can be expressed as 2323h y f y h b f yyhb

_,

_,

: ( ) : ( ) or Because the area of the slice is f(y) dy, we have I y dA yyhbdybhhh

_,

2 232 3 323 36 // (4.2.10)Noting that ymax2h/3, we obtain ZIy hbhbh

max/33622 3 24 (4.2.11)Boththemomentofinertiaandthesectionmodulusforthetriangularsectionare smaller than those of the rectangular section with width b and depth h. That does not surprise us.Example 4.2.3Calculate the moment of inertia of a circular section with a radius of R.SolutionWe defne the angle between the neutral axis and the edge of the slice, q, as shown in Figure 4.2.10a. The width of the slice dA varies, as f y R ( ) cos2 Figure 4.2.10b shows the segment defned by dq. Noting that dqis so small that the arc length, R dq, approximates the chord length, we obtain Figure 4.2.10c, which shows in detail how we express dy in terms of R dq and cosq: dy Rd cosbdy dAhh/32h/3 y2h/3yf(y)Figure4.2.9 Triangular section.6861X_C004.indd 214 2/7/08 2:26:42 PMBending and Shear Stresses215The area of the slice in Figure 4.2.10a is dA f y dy R d ( ) cos 22 2 Noting that y R sin as shown in Figure 4.2.10a, the moment of inertia, I, is I ydA R dR 2 4 2 22242222sin cossin// ddRdR ( ) ////cos22 422 441 44 (4.2.12)Letuscomparetheprecedingresultwiththemomentofinertiaofasquaresection having the same area, i.e., h R2 2 where h denotes the side dimension of the square. Substituting the previous equation into Equation 4.2.12, IR h h4 4 44 4 12 56 .showing that the moment of inertia of a circular section is similar to that of the square section (I h 412 / ) having the same area.Example 4.2.4Buildingcolumnsorbridgepiersmaybesubjectedtobendingmomentbothinthe x-andy-directionbyearthquakeorstormeffects.Assumethatthetubesectionof Figure 4.2.11aissubjectedtobendingmomentsofM Mx y 50 106N-mm and computethemaximumstressinthesection.Thistypeofcolumnisoftenusedin bridges. This is equivalent to a bending moment ofabout the inclined axis in Figure 4.2.11b. See Figure 4.2.11c showing the vector summation of Mx and My. As we learned in Chapter 2, the bending moment itself is not a vector. However, if we cut the member and consider the forces at the cut, we can treat the moment acting on the cut as a vector. You will remember that we did that for axial forces.dyR . cos R . cosRdRdy = R . sindARd(a) Whole section (b) Segment dened by dRddy = Rd . cos(c) Detail to express dyf(y)Figure4.2.10 Circular section.6861X_C004.indd 215 2/7/08 2:26:49 PM216Understanding Structures: An Introduction to Structural AnalysisSolutionThemomentofinertiaaroundthex-axisisobtainedbysubtractingthemomentof inertia of the inner rectangle (400 500) from that of the outer one (500 600): Ix 500 60012400 5001248 3 103 38 4. mmThe corresponding section modulus is ZIyxx max..48 3 1030016 1 1086 3mmThe maximum stress caused by the bending moment ofMx 50 106N-mm is xxxMZ

50 1016 1 103 11662.. N/mmThe stress distribution is shown in Figure 4.2.12a.600505050050 50 400500MxMxMy(b)(a)CompressionMyM(c)TensionM = 502 106 N-mmFigure4.2.11 Tube section. = 3.50 N/mm2Tension = 3.11 N/mm2Compression = 6.61 N/mm2 = 6.61 N/mm2MxMyTension = 3.11 N/mm2Compression = -3.50 N/mm2TensionCompressionM(b) (a) (c)Figure4.2.12 Stress distribution.6861X_C004.indd 216 2/7/08 2:26:54 PMBending and Shear Stresses217The moment of inertia around the y-axis is Iy 600 50012500 4001235 8 103 38 4. mmIy is smaller than Ix because of the smaller height (h600 mm). The corresponding section modulus is ZIyyy max..35 8 1025014 3 1086 3mmThe maximum stress caused by the bending moment ofMy 50 106N-mm is yyyMZ

50 1014 3 103 50662.. N/mmThe stress distribution is shown in Figure 4.2.12b.The stress caused by the simultaneous bending moments ofM Mx y 50 106N-mmis shown in Figure 4.2.12c. The maximum stress is ++ x y3 11 3 50 6 61 . . . N/mm2andoccursatthecornerswherethemomentMxandMycausestressesofthesame sense (tension or compression).Determine the optimum proportionS for a SeCtion reSiSting momentWe wish to design a beam section that can resist a bending moment of +50 103 N-mm(usingGOYA-I)withthelimitationthatneitheritsheightnoritswidth should exceed 80 mm. Note that the positive sign of the bending moment indicates that the bottom fber will be in tension. Assume that we can use a material with a compressivestrengthof2N/mm2andatensilestrengthof1N/mm2.Designthe section so that the area (the number of squares) is minimized. The section should becontinuous,asshowninFigure 4.2.13a,andshouldnotbediscontinuous,as showninFigure 4.2.13b.Itisdiffculttofndthebestsolutiondirectly.Doitby trial and error.(a) (b)Figure4.2.13 Examples.6861X_C004.indd 217 2/7/08 2:27:00 PM218Understanding Structures: An Introduction to Structural Analysis4.3 constructionandtestoFastyroFoaMBeaMWe need the following:1.A Styrofoam sheet, 2 2 ft, with a thickness of 1/2 in.2.A knife3.Adhesive for the Styrofoam4.A weight of approximately 5 lb, such as marble or sand5.A small plastic bag6.A piece of wood 2 ft long7.A ruler longer than 8 in.8.A binder clip9.A kitchen scaleInthissection,wewillmakearectangulartubebeam(Figure 4.3.1)usingaSty-rofoamsheet. Topreventfailureatthemiddleofthetubebeam,thejointshallbe strengthened,asshowninFigure 4.3.2,usingtheremainingmaterial.Thebeam shallbelocatedbetweendesksplaced40in.apart,asshowninFigure 4.3.3,and designed to fail if the load applied at midspan reaches a value of F4 lbf. The beam height (h) shall be 2 in. The beam width shall be determined based on the following assumptions.Assumption 1: The tensile strength of Styrofoam is ;sf 30 psi Its compressive strength is higher.Assumption 2: The beam will break at either end of the strengthened part (Figure 4.3.3), where the bending moment is MFL 22 18 36 lbf-in.Therefore, the width of the beam shall be designed to satisfyMfZsf36 1bf-in 2 inb24 in24 inGlueFigure4.3.1 Rectangular tube beam.6861X_C004.indd 218 2/7/08 2:27:02 PMBending and Shear Stresses219To make optimum use of the beam, we shall also evaluate the Youngs modulus of the Styrofoam as follows:1.Fix the ruler to the wood using a binder clip as we did in Section 3.4.2.Hang the plastic bag on the beam.3.Place objects (marbles, sand, etc.) into the plastic bag so that the defection reaches approximately 2 in.4.Measure the defection at midspan.5.Determine the weight.6.Assumethatthestrengthenedpartisrigid.Thedefectionofthebeam shall be similar to that of the cantilever beam shown in Figure 4.3.4, from which you can calculate Youngs modulus.2 inb4 inFigure4.3.2 Strengthen the joint.F = 4 lbfF/24 in L = 18 in4 in 4 inBreaks at either endBar with ruler F/2L = 18 in40 inFigure4.3.3 Beam under load.6861X_C004.indd 219 2/7/08 2:27:04 PM220Understanding Structures: An Introduction to Structural AnalysisAfter the calculation, increase the weight gradually until the beam fails. Then com-pute the strength of your Styrofoam from the test data using f fM Z/ (i.e., assum-ing that Styrofoam is a brittle material).Wehaveconsideredtensileandcompressivestresses(normalstresses)causedby bending moment. In this section, we investigate stresses caused by shear.Figure 4.4.1 shows a body on a desk pushed by a force V. If the force is smaller than a certain threshold value, the body will not move because of the friction between VL = 18 inDeectionFigure4.3.4 Equivalent cantilever beam.VV(a) Solid on desk (b) FrictionA = V/AFigure4.4.1 Shear stress on desk.6861X_C004.indd 220 2/7/08 2:27:07 PMBending and Shear Stresses221thebodyandthedesksurface.Wedefnetheaverageshearstressonthecontact surface between the base and the desk as the ratio of the force, V, to the area of the contact surface, A: VA (4.4.1)We can defne the unit shear stress on a beam section similarly by dividing the shear force on a section by the area of the section. However, the actual distribution of shear stress is not necessarily uniform in either case considered. In this section we exam-ine shear-stress distribution on a beam section.Before tackling the problem of shear-stress distribution over a beam section, we need to develop a set of defnitions for stresses. Figure 4.4.2a shows an infnitesimal cube that is subjected to four stresses acting on the top face ADEH and side face ABEF. Axis x is perpendicular to face ABEF, and axis y is perpendicular to face ADEH.We focus on four stresses (also see Table 4.4.1):s: the normal stress acting in the direction of the x-axis on face ABEFsy: the normal stress acting in the direction of the y-axis on face ADEHtxy: the shear stress acting in the direction of the y-axis on face ABEFtyx: the shear stress acting in the direction of the x-axis on face ADEHThestressessxandsyarecallednormalstressesbecausetheyactinadirection perpendicular to the faces of the cube (Table 4.4.1). The stresses txy and tyx, acting in xyxyDCBFEHxyzxxyyABCD ABCDyx(a) Innitesimal cubexy(b) Normal stress (c) Shear stressxyvxxyyxxyAFigure4.4.2 Stresses on an infnitesimal cube.Direction of ForceDirectionof facexxyy Shear stressNormal stressxyyx (= xy)xy1AU: Table title not provided.AU: Table title not provided.6861X_C004.indd 221 2/7/08 2:27:10 PM222Understanding Structures: An Introduction to Structural Analysisdirections parallel to the faces of the cube, are called shear stresses. Generally, there are normal as well as shear stresses on the face ABCD; but for simplicity we shall ignore them in order to be able to show them in two dimensions (Figure 4.4.2b,c). In Figure 4.4.2b, note that the equilibrium of forces requires a stress of sx on face DC and a stress of sy on face BC in the opposite direction to the normal stresses on faces ABandAD.InFigure 4.4.2c,similarly,txyandtyxactonfacesDCandBC.The equilibrium condition for moment leads to the following important equation. xy yx

(4.4.2)Let us move to the problem of the beam shown in Figure 4.4.3. We assume that the beam has a rectangular section. The left and the right faces of the slice dx are sub-jected to bending moments M and M + dM, respectively. Figure 4.4.4a shows the dis-tribution of normal stresses caused by the bending moment. Normal stresses st and st + dst denote the tensile stresses at the bottom edge of the slice. They are related to bending moments M and M + dM. tMZMbh 26 / (4.4.3) t tdM dMZM dMbh+ +

+26 / (4.4.4)dxM + dM MVVhFigure4.4.3 Simply supported beam.(a) Stress on slice dx (b) Cut the slice at the middle (c) Shear stressctt + dtc + dc =dx dxbhdxhh/2tt + dt32VbhFigure4.4.4 Equilibrium of slice dx.6861X_C004.indd 222 2/7/08 2:27:15 PMBending and Shear Stresses223CuttingthebeamsliceinFigure 4.4.4aatmid-heightresultsinthefree-bodydia-gramshowninFigure 4.4.4b.Thevolumesofthetriangularprismsrepresentthe tensile forces acting on the two sides of the beam slice. They are Tbh Mht432 (4.4.5) T dTbhdM dMht t+++432( ) (4.4.6)The difference between the tensile forces, dT, is balanced by the shear force on the upper face of the free body in Figure 4.4.4b. The required force is the shear stress, t, multiplied by the area of the face, b.dx. b dx dTdMh32 (4.4.7)Rearranging, 32dM dxbh/ (4.4.8)Recalling that VdM/dx, we have the following important equation for shear stress at the neutral axis of a beam with a rectangular section: 32Vbh (4.4.9)Equation 4.4.9 indicates that the shear stress at the middle height of the section is 1.5 times the average shear stress, V/bh. Though the shear stress in Figure 4.4.4b is in the horizontal direction, the equilibrium of moments in the infnitesimal square, shown inFigure 4.4.4c,requiresthesameshearstressintheverticaldirection(Equation 4.4.2). You might be surprised to fnd shear stresses in the horizontal direction; but if you cut the beam horizontally, as shown in Figure 4.4.5, you will see the need for glue (or shear strength) to prevent slip.Cutting the slice dx at a distance y from mid-height, we obtain Figure 4.4.6. The volume of the solid body on the left of the slice represents the tensile force on the left face. Tb yhhyMhht t +

_,

_,

_,

222623 222

1]1yFigure4.4.5 Beam cut horizontally.6861X_C004.indd 223 2/7/08 2:27:21 PM224Understanding Structures: An Introduction to Structural AnalysisThe tensile force on the right side, T + dT, can be obtained similarly. These results and dT b dx /( )lead to

_,

1]162322Vbhhy (4.4.10)This equation indicates that the shear stress distribution is parabolic (Figure 4.4.7). Shear stress reaches its maximum at mid-height (y0) and is zero at the top and bot-tom (yh/2) of the section. The latter statement is intuitively obvious because the top and bottom faces of the beam are unrestrained. We cannot justify stresses acting on those surfaces. To satisfy equilibrium, the force represented by the volume of the curved solid representing the distribution of shear stress on the section (Figure 4.4.7) must be equal to the shear force in the beam.Wecansimilarlycalculatetheshearstressdistributionforsectionswithnon-rectangular shapes. Consider the section in Figure 4.4.8a. Let us compute the shear stress at distance y1 below the neutral axis. Recall that the normal stress caused by the bending moment at position y from the neutral axis is yI M (4.2.4)dxybt t + dtNeutral axisFigure4.4.6 Cut the slice at y.hb = 32VbhFigure4.4.7 Shear stress distribution.6861X_C004.indd 224 2/7/08 2:27:25 PMBending and Shear Stresses225whereIisthemomentofinertia.Integratingthisfromthebottom(y0)tothetar-get (y1), we obtain the tensile force acting on the left face of the body, as shown in Figure 4.4.8b. T dAyIMdAyyyy 0101 (4.4.11)The tensile force on the right face is obtained similarly. T dT d dAyIM dM dAyyyy++ + ( ) ( ) 0101 (4.4.12)From equilibrium in the horizontal direction, b dx dTyIdMdAyy101 (4.4.13)where b1 is the width of the beam at y1. The preceding equation results in Vb Iy dAyy1 01 (4.4.14)Note that the integral in this equation is the frst moment of the section from the bot-tom to y1. Thus, we may rewrite it as follows: Sb I V11 (4.4.15)where S1 is the frst moment.Example 4.4.1Compute the shear stress at the neutral axis of the I-section in Figure 4.4.9a subjected to a shear force V.Neutral axis(a) Section (b) Stressdxb1y1y1ydAy0dAb1T + dTTFigure4.4.8 Shear stress in general section.6861X_C004.indd 225 2/7/08 2:27:30 PM226Understanding Structures: An Introduction to Structural AnalysisSolutionThe moment of inertia of the section is obtained in accordance with Figure 4.4.9b: I 112300 700 300 13 700 2 24 195 13 3[ ( ) ( ) ] 007 4mmThe frst moment from the bottom to the neutral axis is S ydy ydy + 300 13 3 12 1035032632606 3. mmThus, the shear stress at the neutral axis is Sb I V V116753 12 1013 195 1012 3 10..Wecancomputetheshearstressesattheotherlocationssimilarlyasshownin Figure 4.4.9c.Notethattheshearstresschangesabruptlyattheborderbetweenthe fangeandtheweb(about23times,inthiscase)becausethewidth(b1inEquation 4.4.15) changes abruptly from 300 mm to 13 mm (about 1/23).If we simply divide the shear force V by the area of the web, we get VAVVweb13 700 2 2411 8 105( ).Notethatthisapproximationisveryclosetothemaximumshearstressshownin Figure 4.4.9c,indicatingthatthewebresistsalmostalltheshearforceandthatthe contribution of the fanges to shear resistance is negligible. You may be surprised to see such an abrupt change in stress. In fact, Equation 4.4.15 is an approximate determination of the shear-stress distribution. An improved formation leads to continuous (but rapid) change at the border, though Equation 4.4.15 is suffciently accurate for structural design.(a) Section (c) Shear stress 105V(b) Moment of inertia7001324243000.40.49.69.612.3Figure4.4.9 H-shaped section.6861X_C004.indd 226 2/7/08 2:27:34 PMBending and Shear Stresses2274.5 ProBleMs4.1A beam with a square section made of a brittle material is to be subjected toamomentactingaboutanaxisparalleltoanedge(Figure 4.5.1a)or about a diagonal axis (Figure 4.5.1b). About which axis is it (1) stronger and (2) stiffer?4.2A steel pipe has an outer radius of 500 mm, an inner radius of 480 mm, and a thickness of 20 mm (Figure 4.5.2a). Calculate the ratio of the moment ofinertiaofthepipetothatofthesolidcircularsectionwiththesame cross-sectional area. (Hint: the moment of inertia of a circular section with radius r is I = pr4/4.) Which of the fve options listed here is the correct one?1.Approximately 52.Approximately 153.Approximately 254.Approximately 355.Approximately 454.3CalculatetheratioofthemomentofinertiaoftheI-sectioninExample 4.4.1 (or Figure 4.5.3a) to that of the solid square section with the same cross-sectionalarea.Whichofthefveoptionslistedhereisthecorrect one?aaaa(a) (b)Figure4.5.1 Square section.?(a) Pipe (b) Equivalentcircular section20500480Figure4.5.2AU: Caption? AU: Caption?6861X_C004.indd 227 2/7/08 2:27:36 PM228Understanding Structures: An Introduction to Structural Analysis1.Approximately 52.Approximately 153.Approximately 254.Approximately 355.Approximately 454.4Aforceof1500Nisappliedtoacantilevercolumnwitharectangular section as shown in Figure 4.5.24. Select the correct set of shear stresses at the points A, B, and C from among the options listed in Table 4.5.1. The stress unit is N/mm2.AU: Should this be Figure 4.5.3?AU: Should this be Figure 4.5.3?A00551012345B1015101510C005510a a120 kNSection a-aA B C150150200 200Table 4-5-1OptionsFigure4.5.3 Cantilever column.6861X_C004.indd 228 2/7/08 2:27:37 PM2936Buckling6.1 SimplemodelSInSection1.7Chapter1,wedevelopedanexpressiontodeterminethebuckling load for a wooden column (Equation 1.7.1). In Chapter 6, we shall derive a similar but more general expression based on what we learned in Chapter 2. In this section, weshallapproachthegeneralexpressionbyusingsimplemodelsofthebuckling phenomenon.Before we do that, we shall conduct a simple test to obtain a feel for the physical phenomenon of buckling.1.Take a wooden stick approximately 2-ft long with a 1/8-in. (~3.2 mm) square section.2.Insert the stick into a small block of Styrofoam as shown in Figure 6.1.1a.3.Attach a binder clip to the top of the stick (Figure 6.1.1b).4.Place several small magnets onto the binder clip carefully until the stick startstobendasshowninFigure 6.1.1c.Bendingmayoccursuddenly without warning, or slowly, depending on the straightness of the stick and the arrangement of the weights.What you have observed is another example of a buckling column albeit on a small scale.LeonhardEuler(17071783),aSwissmathematicianandphysicist,derived thefollowingrelationshipbetweenthebucklingloadforaconcentricallyloaded (axisofloadalignedwiththeaxisofthecolumn)columnanditsgeometricaland materialproperties. AnexampleoftheEulerequationapplicabletoacantilever column subjected to an axial load is reproduced in Equation 6.1.1. It should apply to our experiment. PEILEILcr 22 242 47 . (6.1.1)where E is Youngs modulus, I is the moment of inertia, and L is the free length of the stick. Buckling can be very dangerous in a structure. Structural designers need to understand well the buckling mechanism so they can prevent it in structures they design.Figure 6.1.2a shows a cantilever column subjected to a horizontal load F at its freeend.Weignoretheself-weightofthecolumn.Anexpressionforthelateral defection at the free end of such a column was developed in Section 2.8, Chapter 2. vFLEI

33 (2.8.14)Au: Please check the numbring of fgures cited or Display maths also.Au: Please check the numbring of fgures cited or Display maths also.6861X_C006.indd 293 2/7/08 4:31:36 PM294 UnderstandingStructures:AnIntroductiontoStructuralAnalysisIn order to help us understand the buckling phenomenon, we set up an analog column (Figure 6.1.2a). The analog column is rigid throughout its height. It is supported on a pin at the base and its free end is maintained in position by a horizontal, linearly elastic spring attached to the end of the column and a fxed point. In effect, the fex-ibility of the entire column is concentrated in the spring.Inspection of Figure 6.1.2 will reveal that the spring is analogous to the fexural stiffness of the cantilever column. The column resists the load F because of its fex-ural stiffness. The analogous column resists the load F with the help of the spring. L(b) (c)MagnetsBinderclipStyrofoam(a)Push verticallyWoodFigure6.1.1 Experiment.v vFLFF = k . vk(b) Rigid columnwith a spring(a) Cantilever columnFigure6.1.2 System with horizontal spring.6861X_C006.indd 294 2/7/08 4:31:38 PMBuckling 295Without the spring, it would topple over. To improve the analogy, we make the stiff-ness of the spring, k: kEIL

33 (6.1.2)Note that Fkv is the same as Equation 2.8.14. Even though the lateral stiffness of theanalogcolumnisprovidedbyadifferentmechanism,itmimicsthecantilever column successfully.What would happen if you push the column in Figure 6.1.2b at the top? If you could push the column at its exact cross-sectional center with a load that is exactly vertical, and if the column is perfectly straight and isotropic throughout its length, the column would remain as it is except for a small amount of shortening. However, such a setup is virtually impossible under practical conditions. The vertical is almost always eccentric with respect to the resistance axis of the column.Atthistime,westopandintroducethedefnitionofeccentricity.Itistheper-pendicular distance between the axes of column resistance and applied load. We can illustrate it in two dimensions as depicted in Figure 6.1.3a. The axis of column resis-tance may be represented by the centerline of the column in Figure 6.1.3a (assuming that the column is straight and the column section and material are uniform through-out). YouwillnotethattheaxialverticalloadPactsatasmalldistance,e,tothe rightofthecenterlineofthecolumnatthepointofapplication.Thedistance,e, is the eccentricity of the applied vertical load with respect to the center of column resistance.We return to the analogous column (Figure 6.1.3a), which is loaded with an axial load P at eccentricity e. Because of the eccentricity, the axial load generates a clock-wise moment at the top equal to Pe. In response to the applied moment, the analogous 0.1L 0.2Lve/L = 0.001e/L = 0.0120kPPk . vkk . vveek . Lk . LPPLP(a) Light load (b) Heavy load (c) Load-deection relationshipBAFigure6.1.3 System with horizontal spring.6861X_C006.indd 295 2/7/08 4:31:40 PM296 UnderstandingStructures:AnIntroductiontoStructuralAnalysiscolumntendstorotateclockwise.Thistendencyisresistedbythespring.Taking moments about the free end of the analogous column, P v e kv L + ( ) (6.1.3)RearrangingEquation6.1.3todefnethedefection,v,atthetopoftheanalogous column, vPekL P

(6.1.4)ThesolidlineandthebrokenlineinFigure 6.1.3cshowtherelationshipbetween Pandvforcasesoflargeeccentricity(e/L0.01)andsmalleccentricity (e/L0.001),respectively.Forthecaseoflargeeccentricity,thedisplacementv increases gradually as the load P increases. For the case of small eccentricity, the dis-placement v increases dramatically when the load P approaches the value of kL because thedenominatorinEquation6.1.4approacheszero.Thisphenomenonissimilarto what we observed when we loaded the wooden stick in the experiment (Figure 6.1.1). We call PcrkL the buckling load. If we substitute Equation 6.1.2, we obtain P kLEILcr

32 (6.1.5)This equation is similar to Equation 6.1.1 (the exact solution) except that the coeff-cient 3 is 20% larger than 2.47. Note that this equation is independent of the strength of the spring. The way we have set it up with our assumptions, the spring does not fail. The buckling strength of the analogous column is determined by the stiffness of the spring. If we remove the load, the moment P (v + e) will disappear and the spring will push back. The defection at the top of the analogous column will return to zero. Therefore,wecallthisfailuremechanismelasticbucklingasitisreferredtoin engineering jargon even though it should be called buckling in the range of linearly elastic response.Exercise 6.1.1InGOYA-U1,youcanfndasystemwithk0.5N/mmandL200mm.Fillin Table 6.1.1forthethreecases:e0.1mm,e1mm,ande1mm.Checkyour results using GOYA-U1.Figure 6.1.4a shows another simple model with a rotational spring at the bottom of a rigid column. The stiffness of the spring is assumed as KEIL

32 (6.1.6)If q is small enough so thatv L L sin , FKq/L is equivalent to Equation 2.8.14. Figure 6.1.4b defnes the eccentricity. Figure 6.1.4c shows the deformation caused by the axial force. Moment equilibrium around the spring leads to P L e K +( ) (6.1.7)6861X_C006.indd 296 2/7/08 4:31:44 PMBuckling 297or PPeLKL (6.1.8)from which PKLEILcr

32 (6.1.9)Exercise 6.1.2In GOYA-U2, you can fnd a system with K20 103 N-mm and L200 mm. Fill in Table 6.1.2 for the case of e1 mm in terms of radians and degrees. Check your results using GOYA-U2.Table6.1.1loadversusdefectionloadP(N)defectionv(mm)e0.1mm e0.1mm e0.1mm02040608095KM = K . v = L . LeFPPM = K . L . q eK KLPP(b) Light load (c) Heavy load (a) Horizontal forceFigure6.1.4 System with rotational spring.6861X_C006.indd 297 2/7/08 4:31:47 PM298 UnderstandingStructures:AnIntroductiontoStructuralAnalysisNow we use a two-spring model that we had considered in Section 2.7, Chapter 2, to evaluatethedefectionofacantileverbeam.Figure 6.1.5ashowsthemodel.Recall that each spring represents the fexural deformation of a length of beam equal to L/2. The relationship between the bending moments (MA and MB) and the rotations of the springs (aA and aB) are M K M K KEILA A B B and where2 (6.1.10)The free-body diagrams shown in Figure 6.1.5b,c lead to P v e M KA A A + ( ) (6.1.11) P v v e M KA B B B + + ( ) (6.1.12)Table6.1.2loadversusrotationloadP(N)rotationq(e0.1mm)unit:rad. unit:deg.02040608095PPP . eMAMBMA = KaAeaBPePaBeP PL/2eL/4L/4AaAvAvB(c) Free-bodyABvAvAvB(d) Free-body (e) Bending moment (a) Model (b) Deformed shapeABABMB = KaBFigure6.1.5 Two-spring model.6861X_C006.indd 298 2/7/08 4:31:50 PMBuckling 299Figure 6.1.5d shows the moment distribution along the height. We assume that aA and aB are small enough (sin A A andsin B B ). From Figure 6.1.5b, we obtain vLA A B + ( ) 4 (6.1.13) vLB B 2 (6.1.14)Substituting Equations 6.1.13 and 6.1.14 into Equations 6.1.11 ad 6.1.12, PLe KA B A + +

1]1( ) 4 (6.1.15) PL Le KA B B B + + +

1]1( ) 4 2 (6.1.16)If we solve these equations in terms of aA and aB, we obtain AK PLPePL KPL K

+( )( )8 48 82 2 (6.1.17) BKPePL KPL K

+88 82 2( ) (6.1.18)Equations 6.1.17 and 6.1.18 have the same denominators. They will be zero if PKLEIL

+ 2 22 12 342. (6.1.19)or PKLEIL

2 22 113 72. (6.1.20)Figure 6.1.6 shows the relationship between the load P and the rotations of the springs (aA and aB) assuming EI4 106 N/mm2, L200 mm (or K40,000 N-mm), and 0501001502002500 1 2 3 4 5 6 7 8Rotation (deg)BALoad (N)Figure6.1.6 Load-rotational relationship.6861X_C006.indd 299 2/7/08 4:31:58 PM300 UnderstandingStructures:AnIntroductiontoStructuralAnalysise1 mm. We can see that the rotations increase dramatically as the load approaches K234 N (the value given by Equation 6.1.19). Note that Equation 6.1.19 is similar to Equation 6.1.1. The error for the approximate solution is only 5%.Exercise 6.1.3In GOYA-U3, you can fnd a two-spring model with L200 mm. Fill in Table 6.1.3 for the case of EI4 106 N/mm2 and e1 mm. Check your results using GOYA-U3.Example 6.1.1Figure 6.1.7ashowstheplanofa110-storyskyscraper.Eachfoorissupportedby 76columnswithcrosssectionsshowninFigure 6.1.7b.Thecolumnsaresteelwith a Youngs modulus of 30,000 ksi. The weight of foor per unit area is 200 lbf/ft2. The story height is 15 ft (Figure 6.1.7c). Assume that the beams are much stiffer than the columns. Estimate the safety factor of the structure against buckling for the following twocases:(Inthisapplication,thesafetyfactorisdefnedasthebucklingstrength divided by the axial load.)Case 1: Structure as it is (the solid line in Figure 6.1.7c).Case 2: All the beams supporting the second and third foors are destroyed because of fre (the solid line in Figure 6.1.7d). Furthermore, Youngs modulus of steel is reduced to 7,500 ksi because of the high temperature.Hint: Calculate the horizontal stiffness against the force F shown in Figure 6.1.7c,d. Use the method shown in Figure 6.1.3.SolutionThe total weight of the building is W 110 60 1000 8 80 102 8. lbfWe ignore the possibility of rotation of the building about its vertical axis.Table6.1.3loadmomentdefectionloadP(N)bendingmomentdefectionv(mm) Ma (N-m) Mb (N-m)0501001502002206861X_C006.indd 300 2/7/08 4:32:00 PMBuckling 301The moment of inertia of the column section is I 2412161222 2004 44, inRecalling Equation 5.5.4, the horizontal stiffness for Case 1 is kEIH

12 76 12 30 000 10 22 20013 12333( , ) ,( ) 1 60 108. lbf/inThe critical load for Case 1 is P kHcr 1 60 10 13 12 2 50 108 10. ( ) . lbfThe safety factor is PWcr

2 50 108 80 1028 4108...200 ft200 ft13 ft13 ft13 ft13 ft13 ft13 ft13 ft13 ft13 ft13 ft13 ft13 ft24 in24 in16 inFF(a) Floor plan (b) Section(c) Before re (d) After reFigure6.1.7 A skyscraper.6861X_C006.indd 301 2/7/08 4:32:04 PM302 UnderstandingStructures:AnIntroductiontoStructuralAnalysisThe structure is quite safe against buckling. On the other hand, the horizontal stiffness for Case 2 is kEIH

12 76 12 7 500 10 22 2003 13 1233( , ) ,( )3361 48 10 . lbf/inThe critical load for Case 1 is P kHcr 1 48 10 3 13 12 6 93 106 8. ( ) . lbfThe safety factor is PWcr

6 93 108 80 100 7988...The structure should collapse as shown by the broken line in Figure 6.1.7d. Collapse is caused by the gravity force, not by a horizontal force. Note that Pcr is proportional to EI/H2. Now that E is 1/4 and H is three times their values before the fre, the safety factor is (1/4) (1/3)21/36 of that in Case 1.Example 6.1.2AcolumnisloadedincompressionasshowninFigure 6.1.8a.Estimatethebuck-ling load assuming that the eccentricity e is small enough. (Hint: the column buckles asshownbythebrokenlinesinFigure 6.1.8a.Usethetwo-springmodelshownin Figure 6.1.8.)PM = KaePeeP . eMMPePPLPeP PL/2eL/4L/4(b) Model (c) Deformed shapeABABAav(d) Free-bodyv(e) Bending momenta(a) ColumnFigure6.1.8 Simply compressed column.6861X_C006.indd 302 2/7/08 4:32:07 PMBuckling 303SolutionFigure 6.1.8c s