工程力学材料力学练习册答案.ppt

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1、习题解答习题解答轴向拉压轴向拉压P61 34-1(1)F FF F2 2F FF FF FF F11F FF F2 2F FF FN N1 111F F2 2F FF FN N2 22 2F FF FN N3 3F FF FN N1 11122223333F FF FF FN N2 2F FF FF FF FN N3 3 4F3F2FFN图图F FFN图图 Fy=0 FN1=4F Fy=0 FN2=3F Fy=0 FN3=2F Fy=0 FN1=F Fy=0 FN2=0 Fy=0 FN3=FP61 34-1(4)六要素六要素R RF FN N3 3P61 34-2ABCDF F4 4F F5

2、5F F2 2F F1 1F F3 3E 500 N920 N640 N240 NF1=500 NF2=420 NF3=280 NF4=400 NF5=240 NFN 图图s s=AFNs sAB=1.25 MPas sBC=2.3 MPas sCD=1.6 MPas sDE=0.6 MPa应应力力单单位位:MPaP62 34-3F F3 3F F1 1F F2 2ABCD112233F FN N1 1F F3 3 Fx=0 FN1=F3F F3 3F F2 2F FN N2 2F F3 3F F1 1F F2 2F FN N3 3=30 kN Fx=0 FN2=F3+F2=40 kN Fx=

3、0 FN3=F3+F2 F1=30 kN 30kN40kN30kNFN图图s s1=A1FN1A1=410 2 m2=0.75 MPas s2=A2FN2A2=210 2 m2=2 MPas s3=A3FN3A3=410 2 m2=0.75 MPaP62 34-4F FF Fs s s sa a a a =s s s s coscos2 2a a a a2 2s s s st t t ta a a a =sin2sin2a a a aa a=0s s=AFNFN=F=5 kN=50 MPaa a=30 a a=45a a=60a a=90 50 MPa0 37.5 MPa 21.65 MPa

4、25 MPa 25 MPa 12.5 MPa 21.65 MPa00a ans sa at ta as ss ss ss s力的力的单单位位:kNP63 35-1F FA AF FB BF FD DF FC CABCDaaaFNiEAEAF FNi Ni a aD D D Dl li i =D D D Dl li i =S DS DS DS Dl li i 4 kN1 kN3 kN 0.05 mm 0.0125 mm 0.0375 mm0e e 呢呢?e e 与与 D Dl 区区别别?S Se e?P63 35-2l2l2hhABCDEH123F FN N1 1F FN N2 2F F=F F

5、N N3 3FN1=80 kNFN2=40 kNF FN N3 3 =120 kN120 kNE1A1FN1 2hD Dl1=0.4 mmE2A2FN2 2hD Dl2=0.4 mmE3A3FN3 hD Dl3=0.2 mmd dC d dH D Dl1 D Dl2 d dC=D Dl1 d dH=d dC+D Dl3=0.4 mm=0.6 mm(1)(2)d dC d dH D Dl1 D Dl2 D Dl1=0.2 mmD Dl2=0.4 mmD Dl3=0.2 mmd dH=d dC+D Dl3=0.267 mm=0.467 mmd dC=D Dl1+D Dl23132位移分析位移分析!

6、P64 35-3ABCl=4mh=3m5mq=10 kN/mS SMA=0 F FBCBCF FA Ay yF FA Ax xFBC=kN3100FN=(1)斜杆用钢丝索斜杆用钢丝索斜杆用钢丝索斜杆用钢丝索A=n4p pd2s s=AFN s s 取取 n=67或或(2)斜杆用两根等边角钢斜杆用两根等边角钢斜杆用两根等边角钢斜杆用两根等边角钢A=2A1s s=AFN s s 取取 203(A A1 1 =1.13 mm1.13 mm2 2)结论结论!n 66.366A1 1.044 mm2P64 35-4F FACB 45456060F FN N2 2F FN N1 1S SFx=0 S SF

7、y=0 FN1=0.897 FFN2=0.732 Fs s1=A1FN1 s s 1 F 107 kNs s2=A2FN2 s s 2 F 123 kN则则 F =107 kNP65 36-1A AB BF FF FFN=5 kNs s=AFN=25 MPas s=Ee e1E=e e1s s=208 GPag g=e e1e e2=154=0.267P65 36-2aaaA AB BC CD DF FF FN NF FC Cy yF FC Cx xS SMC=0 F FN N =2 2 F Fs s=AFNp pd28F=(1)p pd28F=s sS =37.7 kN若若若若 s s s

8、s =s s s sP Pp p p pd d2 28 8F F=s s s sP P =31.4 kN 31.4 kN(2)p pd28F=s sb =62.8 kN(s s s se )令令 s s=s sS令令 s s=s sbP66 36-3lABCDaaaF FF FN N1 1F FN N2 2 S SMA=0 F FN N1 1 +2+2F FN N2 2 =3 3F FEAFN1 lD Dl1=EAFN2 lD Dl2=代入代入D Dl1 D Dl2 FN2=2FN1联立联立 求解求解FN1=30 kNFN2=60 kN且且s s2=AFN2 s s 取取补补充方程充方程!结论

9、结论结论结论!F FA Ay yF FA Ax xD D D Dl l1 1 =2 2D D D Dl l2 2FN2 FN1A 相同相同s s2 s s1A 6 cm2A =6 cm2P66 36-4F FN N2 2F FN N1 1F FN N3 3123aF F3030D Dl1D Dl2D Dl3S SFx=0 23FN1+FN2=0S SFy=0 21FN2+FN3=FEAFN1aD Dl1=23EAFN2 aD Dl2=13EAFN3 aD Dl3=2321代入代入 D Dl1+D Dl3=D Dl23FN1+FN3=4F2联立联立 求解求解FN1=0.122 kNFN2=0.141 kNFN3=0.930 kN位移分析位移分析!受力图受力图受力图受力图!