工程力学材料力学练习册答案.ppt

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,= 30 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 mm,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 = 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