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