工程力学2习题课

1、cabd600300500400e40kn55kn 25kn20kncabd600300500400e40kn55kn 25kn20kncabde40kn55kn 25kn20knrkn1 10 00 02 20 02 25 55 55 54 40 00 0 rrfx r40knfn220kncabde40kn55kn 25knr20 04 40 02 2 rfn)()(rf knn505040402 2fn320kn25kncabde40kn55kn 25kn20knr30 02 20 02 25 53 3 nf)()kn(n 5 53 3f20knfn440kn55kn 25kn20knr

2、4)()(f knn20204 4f1f2f3l1l2l3abcdf1fn1)(knnn 20200 01 11 11 1ffff2f1fn2f1f2f3l1l2l3abcd)(knnn 15150 02 22 22 21 1ffffrfn3)(knnn 50500 03 33 3frffn2 =-15kn （-）fn1 =20kn （+）fn3 =- 50kn （-）15+-2050f1f2f3l1l2l3abcdfn2 =-15kn ( - )fn1 =20kn （+）fn3 =- 50kn ( - )f1f2f3l1l2l3abcd)(mpa.n 8 81 17 76 61 11 1af

3、ab )(mpa.n 6 67 74 42 22 2afbc )(mpa.n 5 51101103 33 3afdc (f1f2f3l1l2l3abcdm102.534-n 1 11 11 1ealflabm101.424-n 2 22 22 2ealflbcm101.584-n 3 33 33 3ealflcd-0.3mm bccdbllumm10-0.474- cdbcabadllllmexnnmemexmexnpm9549e mn6379mn4780mn159004e3e2e1e mmmm 0023e2e tmmmx mn95603e2e2 mmtmn47802e1 mtxme2me3t

4、2me21t133me4me1me3me2me2mn47802e1 mtmn63704e3 mtmn9560max td+m2m3m1180)2(pepe giamgiam3m4mabcde4m4mrarbf1=2knm=10kn.mf2=2knf1=9knf2=4knacbd1m1m1mbhzy四、计算 309090tc1. 构件内危险点应力状态如图所示，试作强度校核：1）材料为铸铁，已知许用拉应力mpa,压应力mpa，泊松比=0.25；2）材料为铝合金，mpa；3）材料仍为铸铁，应力分量中为压应力。15mpa15mpa 2213131151154 1522024.330t 1，r1）求主应力 （-9.27）mpa铸铁为脆性材料，且，主应力中拉应力占优，主要是最大拉应力引起脆性断裂，选第一强度理论 mpa解：mpa 13222222212233122224.279.2733.54904154 1533.5412124.279.2724.279.27230.090 334rrr2)铝合金为塑性材料，选第三强度理论 mpampa或 mpa选第四强度理论 mpampa 22223153 1530.090 4r 或 mpampa 221 313123151154 152209.270.25 0