西安建筑科技大学：《材料力学》课程习题与答案_第2章 轴向拉伸和压缩

第1次作业: (+) F F (c)3F F 第2次作业: 2-3 F=-20kN. F=-10kN. F=10kN 10KN F 20×103N A =-100MPa 200mm 10kN F 10×103N 20KN =-33.3MPa A 300 10×10N A. 400mm225MPa 2-19 FN=2×(F+P)=60kN A=2×405.8=811.6mm a=x=60×10=739MPa 8116 AD杆 300×2 300×10 FNE =300N24=x ]170 =1764.7mm,A1=8.82cm 选择80×80×6(9397cm2) AB杆

第 1 次作业： 第 2 次作业： 2-3 FN1 = −20kN, FN2 = −10kN, FN3 =10kN MPa mm N A FN 100 200 20 10 2 3 1 1 1 = − −   = = MPa mm N A FN 33.3 300 10 10 2 3 2 2 2 = − −   = = MPa mm N A FN 25 400 10 10 2 3 3 3 3 =   = = 2-19 73.9 [ ] 811.6 60 10 2 405.8 811.6 2 ( ) 60 3 2  =    = = =  = =  + = MPa A F A mm F F P kN N N 2-21 AD 杆： 80 80 6(9.397 ) 1764.7 , 8.82 170 300 10 [ ] 300 ,2 2 300 2 2 2 1 2 3 1 1 1 cm mm A cm F F k N A N N   = =  = = =  = 选择  AB 杆： (a) F (b) 2F (+) F (+) (-) (+) F (c) 3F (d) 2F F (+) F (+) (-) 2F 10kN (+) (-) 10kN 20kN

FN=2×300=600kN,2A=FN2_600×103 15294mm,A1=1765cm 选择100×100×101926lcm2) 第3次作业 FN 3.1M23,N33 3EA 3EA3EA 3EA F=F==10kN.F,=0 M1=△1=FM1=10×103×100 EA=210×X10°×100=0467m a=0467m(→),δ,=0467m(↓) d(x)=d+42-d (x)=2(d1+22,x) 4 Fdx 4Fl d(△/)= M=[d(△) (d rEd, d2

100 100 10(19.261 ) 1529.4 , 17.65 170 600 10 [ ] 2 300 600 ,2 2 2 1 2 3 2 1 2 cm mm A cm F F k N A N N   = =  =  = = = 选择  第 3 次作业： 2-7 EA Fl EA Fl EA Fl EA Fl EA l F EA l F EA l F l N N N 3 3 3 3 3 3 3 1 2 3 + = − = +  = + + ( ) 3 =  = → EA Fl u l D 2-12 mm EA F l l l k N F F F F N N N N 0.467 210 10 100 10 10 1000 10 , 0 2 3 3 1 1 2 1 2 3 =      =  = = = = = = = 0.467mm(→), = 0.467mm()  Cx  Cy 2-15 1 2 0 2 1 2 1 2 1 2 1 2 1 1 4 , ( ) ( ) 4 ( ) ( ) 4 ( ) , ( ) Ed d Fl l d l x l d d E d Fdx d l x l d d x A x d l d d d x d l     =  = − +  = − = + − = +  F F F