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1、材料力学,习题解答,轴向拉压,P61 34-1(1),4F,3F,2F,FN图,Fy=0,FN1=4F,Fy=0,FN2=3F,Fy=0,FN3=2F,Fy=0,FN1=F,Fy=0,FN2=0,Fy=0,FN3=F,P61 34-1(4),六要素,P61 34-2,A,B,C,D,E,500 N,920 N,640 N,240 N,F1=500 N,F2=420 N,F3=280 N,F4=400 N,F5=240 N,FN 图,sAB=1.25 MPa,sBC=2.3 MPa,sCD=1.6 MPa,sDE=0.6 MPa,应力单位,:MPa,P62 34-3,Fx=0,FN1=F3,=3
2、0 kN,Fx=0,FN2=F3+F2,=40 kN,Fx=0,FN3=F3+F2 F1,=30 kN,A1=410 2 m2,=0.75 MPa,A2=210 2 m2,=2 MPa,A3=410 2 m2,=0.75 MPa,P62 34-4,sa=s cos2a,a=0,FN=F,=5 kN,=50 MPa,a=30,a=45,a=60,a=90,50 MPa,0,37.5 MPa,21.65 MPa,25 MPa,25 MPa,12.5 MPa,21.65 MPa,0,0,a,n,力的单位,:kN,P63 35-1,FNi,Dli=S Dli,4 kN,1 kN,3 kN,0.05 m
3、m,0.0125 mm,0.0375 mm,0,e 呢?,e 与 Dl 区别?,Se?,P63 35-2,=FN3,FN1=80 kN,FN2=40 kN,FN3=120 kN,=0.4 mm,=0.4 mm,=0.2 mm,dC=Dl1,dH=dC+Dl3,=0.4 mm,=0.6 mm,(1),(2),Dl1=0.2 mm,Dl2=0.4 mm,Dl3=0.2 mm,dH=dC+Dl3,=0.267 mm,=0.467 mm,位移分析!,P64 35-3,A,B,C,SMA=0,FN=,(1)斜杆用钢丝索,s,取 n=67,或,(2)斜杆用两根等边角钢,A=2A1,s,取 203,(A1=
4、1.13 mm2),结论!,n 66.3,66,A1 1.044 mm2,P64 35-4,A,C,B,45,60,SFx=0,SFy=0,FN1=0.897 F,FN2=0.732 F,s 1,F 107 kN,s 2,F 123 kN,则 F=107 kN,P65 36-1,A,B,FN=5 kN,=25 MPa,s=Ee1,=208 GPa,=0.267,P65 36-2,A,B,C,D,SMC=0,FN=2 F,(1),=37.7 kN,若 s=sP,=31.4 kN,(2),=62.8 kN,(se),令 s=sS,令 s=sb,P66 36-3,A,B,C,D,SMA=0,FN1+2FN2=3F,代入,FN2=2FN1,联立 求解,FN1=30 kN,FN2=60 kN,且,s,取,补充方程!,结论!,Dl1=2Dl2,FN2 FN1,A 相同,s2 s1,A 6 cm2,A=6 cm2,P66 36-4,1,2,3,30,SFx=0,SFy=0,3FN1+FN3=4F2,联立 求解,FN1=0.122 kN,FN2=0.141 kN,FN3=0.930 kN,位移分析!,受力图!,
