《理论力学》课程教学资源（英文复习）Review Problems For Theoretical Mechanics

Review ProblemsFor Theoretical MechanicsEditedbyTheDepartmentofMechanics2019.11

Review Problems For Theoretical Mechanics Edited by The Department of Mechanics 2019.11

StaticsEp1.Known:P.a,theweightandfrictionof eachmember arenot counted:Findtheconstraint reactionatAandBSOLUTION:Firstly, take the whole as the research objectanditsforcediagramisshowninthefigure(a)P(b)(c)(a)Ep1[Analysis]: There are altogether four unknown forces in this force diagram, all ofwhich are required constraint forces. There are only three independent equilibriumequations, so it is impossible to solve all of them. But by taking the moment of A or BWwe can solve for Fay or Fgy first.1Zms(F)=0From-2axFAy-Pxa=0weget:Fay23ZY=0FAy +FBy -P=0we get:FByTaketherodEC,and itsforce isshown inFigure3.1 (c),fromZmc(F)=0FResin45°xa-Pxa=0FRE=/2PWe obtainTake bar AED, and its force diagram is shown in Fig.3.1 (b), from2a×FAx-2a×F4y-2a×FRE=0Zm,(F)=0PWe getFAx：2For the overall force figure 3.1 (a), from

Statics Ep 1. Known: P, a, the weight and friction of each member are not counted; Find the constraint reaction at A and B. SOLUTION: Firstly, take the whole as the research object, and its force diagram is shown in the figure (a). [Analysis]: There are altogether four unknown forces in this force diagram, all of which are required constraint forces. There are only three independent equilibrium equations, so it is impossible to solve all of them. But by taking the moment of A or B, we can solve for FAY or FBY first. From mB (F) = 0  − 2a  FAY − P  a = 0 we get： FAY P 2 1 = − Y = 0 FAY + FBY − P = 0 we get： FBY P 2 3 = Take the rod EC, and its force is shown in Figure 3.1 (c), from mC (F) = 0  FRE  sin 45  a − P a = 0  We obtain FRE  = 2P Take bar AED, and its force diagram is shown in Fig. 3.1 (b), from mD (F) = 0  2a FAx − 2a FAy − 2a FRE = 0 We get 2 P FAx = For the overall force figure 3.1 (a), from

ZX=0FAx + Fx = 0PFBx = -we get2Exercise: Known: P, a, the weight and frictionof each member are not counted;Calculatethe constraint reaction atA,BandE.EAB工R力a-Qa--a→人→EP补DUMD:1534777772:

 X = 0 FAx + FBx = 0 we get 2 P FBx = − Exercise: Known: P, a, the weight and friction of each member are not counted; Calculate the constraint reaction at A, B and E

Ep2.Known:P=10kN, q=2kN/m, M=2kN·m, a=1m, the weight andfriction of eachmemberarenotcounted;CalculatetheconstraintreactionatASOLUTION:Firstly,takethewholeastheresearchobject,and its stress is shown inFigure2 (a)fLFoyFosForCPP1DDcEM-BFFMFFaB9F(a)(b)(c)Ep 21ZX=0FAx +=x×3q×3a-qa=02We getFx =-7kNTake thebar CE,and itsforcediagram is shown inFIG.3.2 (c),fromZm.(F)=0F,cos45°xa+Pxa-M=0We getF,cos45°=-8kNZY=0FromFesin45°-F,=0Fg, =-8kNgetTake the component BCD.and its force diagram is shown in Figure 3.2 (b),froma=0Zm,(F)=0Fxa+Fc,xa-qax2GetFB=9kNForthewhole,fromZY=0FA, +Fs =0GetFAy=-9kN

Ep 2. Known：P=10kN，q=2kN/m, M=2kN·m, ɑ=1m, the weight and friction of each member are not counted; Calculate the constraint reaction at A. SOLUTION: Firstly, take the whole as the research object, and its stress is shown in Figure 2 (a).  X = 0 3 3 0 2 1 FAx +  q  a − qa = We get FAx = −7kN Take the bar CE, and its force diagram is shown in FIG. 3.2 (c), from m (F) = 0 c  FE cos45  a + P a − M = 0  We get FE cos45 = −8kN  From Y = 0 FE sin 45 − FCy  = 0  get FCy  = −8kN Take the component BCD, and its force diagram is shown in Figure 3.2 (b), from mD (F) = 0  0 2  +  −  = a FB a FCy a qa Get FB = 9kN For the whole, from Y = 0 FAy + FB = 0 Get FAy = −9kN Ep 2

1aFromZm(F)=0MA+aFB-M+qax=×3q×3axa=022M,=1kN·mgetBA?Exercise: Known: AB=BC=CD=a. The material weight is P,and the weight of pulley and each bar is not counted. Find theAconstraintreactionatES45

From mA (F) = 0  3 3 0 2 1 2 + − +  −  q  a  a = a M A aFB M qa get M A = 1kN  m Exercise：Known：AB=BC=CD=ɑ. The material weight is P, and the weight of pulley and each bar is not counted. Find the constraint reaction at E

Ep 3. Known: P, 1, R, Each bar and pulley shall be excluded; Find the constraintreactionatfixedendAFaFM0(b)(c)Ep 3SOLUTION:TaketheCDrod(includingthepulley)astheresearchobject,anditsforceisshowninFigure(b),fromZm,(F)=0FcB×21+F(1+R)-PR= 0PWe getFcB =2Take rod AB and its force diagram is shown in Figure (c), fromZX=0FA+FBC+F=0PWe getFx =2FromZY=0FA,=0Zm,(F)=0MA-F(1+R)-FBc×21=0We getM =PRExercise:Known:g=500N/m,P=2000N,M=500Nm,a=lm,Not countingtheweight of eachpoleCalculate:theforceofABbaronCDbaratBE

Ep 3. Known: P, l, R, Each bar and pulley shall be excluded; Find the constraint reaction at fixed end A. SOLUTION: Take the CD rod (including the pulley) as the research object, and its force is shown in Figure (b), from mD (F) = 0  FCB  2l + FT (l + R) − PR = 0 We get 2 P FCB = − Take rod AB and its force diagram is shown in Figure (c), from  X = 0 FAx + FBC + FT = 0 We get 2 P FAx = − From Y = 0 FAy = 0 mA (F) = 0  MA − FT (l + R) − FBC 2l = 0 We get M A = PR Exercise: Known: q = 500N / m,P = 2000N,M = 500Nm,a =1m, Not counting the weight of each pole. Calculate: the force of AB bar on CD bar at B. Ep 3

Ep4.Known:P,a, M=Pa,Notcounting the weight of eachpoleFind the constraint reaction at support ACand D.BMDSOLUTION:Firstly, BC bar is taken as the object ofstudy, and its force diagram is shown inFigure(c),fromZmc(F)=0-F×2a+Pa-M=0We getFby = 0CB7M公FFAFAxa20T(b)(a)FrFeFcFeNFcFtyMpCBMDFix(d)(c)Ep 4Finally, take the CD rod as the research object, and the force is shown in Figure (d),PZX=0-Fc+FDx=0getFpx=2ZY=0Fd, = PFdy-Fo, =0get

Ep 4. Known: P ,a,M = Pa ，Not counting the weight of each pole. Find the constraint reaction at support A and D. SOLUTION: Firstly, BC bar is taken as the object of study, and its force diagram is shown in Figure (c), from mC (F) = 0  − FBy  2a + Pa − M = 0 We get FBy  = 0 取折杆 AB 为研究对象，受力如图（b）所示，由 Y = 0 FAy − P − FBy = 0 得 FAy = P mB (F) = 0  − FAy 2a + Pa + FAx 2a = 0 得 2 P FAx =  X = 0 FAx + FBx = 0 得 2 P FBx = − 取 BC 为研究对象，受力图（c），由  X = 0 − FBx  + FCx  = 0 得 2 P FCx  = − Y = 0 FBy  + FCy  − P = 0 得 FCy  = P Finally, take the CD rod as the research object, and the force is shown in Figure (d),  X = 0 − FCx + FDx = 0 get 2 P FDx = − Y = 0 FDy − FCy = 0 get FDy = P Ep 4

Zmp(F)=0M,+2aFα+2aF,=0get Mp =-PaExercise:Known:AB=2BC=2CD=2m,q=2000N/m,XDM=500Nm, Each rod is homogeneous and itsweight per unit length is 500N/m.Calculate: Constraint reaction at A.WM29

mD (F) = 0  MD + 2aFCx + 2aFcy = 0 get M Pa D = − Exercise: Known: AB=2BC=2CD=2m，q=2000N/m， M=500Nm, Each rod is homogeneous and its weight per unit length is 500N/m. Calculate: Constraint reaction at A

Ep 5.Determine thex andycomponents ofeach of theforces shown.U28mm84mm96 mm50N80mm29NO51N90 mm-48mm-SOLUTIONo.Compute the following distances:OA=(84)2+(80)2SON=116 mmOB= (28)2 +(96)2.51N=100mmOC=/(48)2 +(90)2=102mm84F,=+(29 N)F,=+21.0N29-NForce:11680F, = +(29 N)-F, =+20.0 N11628F=-(50 N)F,=-14.00N50-NForce:10096F, =+(50 N),F, =+48.0 N10048F=+(51N),F,=+24.0N51-NForce:10290F, = (51 N)F, =-45.0 N102

Ep 5

Ep 6.Knowing that α =35°, determine the resultant of the threeforces shown.aa30%100N200N150NSOLUTIONF,=+(100N)c0s35°=+81.915N100-N Force:F, = -(100 N)sin35°= -57.358 N150-NForce:F,=+(150N)cos65°=+63.393NF, = (150 N)sin 65°= 135.946 NF,=(200N)cos35°=163.830N200-NForce:F, = (200 N)sin35°= 114.715 NForcex Comp. (N)yComp. (N)100N+81.91557.358150N+63.393135.946200N163.830114.715R, =18.522R, =308.02R=Ri+R,jR,=-18.522j=(18.522 N)i + (308.02 N)jRtanα=R.308.0218.522Ry= -308.02jα=86.5590K308.02 NR=R=309N786.6°sin 86.559

Ep 6