第四章弯曲应力 4-1试求图示各梁中指定截面上的剪力和弯矩。 解:(a)Fs-1=0 2kN.m 5kN M,,=-2kN.m M2-2=-2-5×2=-12kNm (b) FR=2kN RB=3kN 5kN F SI-I =F,=+2kN 2×3=6kNm M2-2=FRB×2=6kN·m (c) FR=4kN Fn=4kN 10 FSL=+4kN M1-1=+4kN F=+4kN 2.5m M2-2=-4×1.5=-6KN.m 20×2 (d) FRA 二=13.33kN FRB=6.67kN FH=153、(20+10)×1 =-1.67kN HIm H=13.3x110×12 10×1 23 13.33-3.33-5.0=5.0kN.m M (e)FR/≈FgB4a FsI 回5- 4 M 0 -M
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(f)F=24.31kN 16.5kN/m Fs1_1=3×2+65×1=125kN ttt进 B M1=-3×2×2-65×1×=-152kNm Fs2=Fs-1-FRB=12.5-2431=-118kN 15.2kN.m (g)FR= FRR=40kN 80kN Fsu1=40-10×1=30kN Y M1-1=40×1-80-10×1×=-45kNm F 2 M,,=40×2-20×1-10×2×=-80=-40kNm (h)f=f=g0x 20 q04 40a 0,M qo 4-2试写出下列各梁的剪力方程和弯矩方程,并作剪力图和弯矩图。 解:(a)0≤x q(x) 9()=9 F g(x)x qo q(r) qo 40
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b)0≤x≤1m时 Fs(x)=+30+15x=30+15 M(x)=-30x-7.5y 1≤x≤3m时 Fs(x)=30+15=45 FSmax = 45kN M=-127.5kNm 7.5kN-m (c) FRA=FRB=49.5kN 0≤x≤4m时 中H 4m F5(x)=49.5 ≤x≤8m时 F5(x)=49.5-3x-7 49.5kN 174kN·m A=174kN-m d 0.6kN RB=1.4KN 0≤x≤8m Fs(x)=06-0.2x M(x)=0.6x-0.1x 8≤x<10 F(x)=0.6-0.2x 0.2 M(x)=0.6 4 2 4+0.6x-0.1x 1.4kN 2. 4kN. m
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20kN (e)0≤x≤2m时, 6kN-m 20KN Fs(x)=-2kN M(x)=6 2≤x≤3m时, FS(x)=-22kN M(x)=6-2x-20(x-2) M=20kN.m 6kN.m 4 kN-m (f)AB段:F5(x) 8 段:F5(x)=q M(x) q (g)AB段内:F5(x)=-F M(x=-Fx F BC段内:F5(x)=+ FSmax F/2 FSmax = F F/2 L
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40kN (h)AB段内:F5(x)=0 M()=-30kNm BC段内:F5(x)=30kN M(x)=-30+30x CD段内:F5(x)=-10kN M(x)=+10x +f=30kN 15KN.m F.=10kN 30kN.m 4-3试利用荷载集度、剪力和弯矩间的微分关系作下列各梁的剪力图和弯矩图。 15k NH-1OKN - m 5kN 4kN M/3a
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250kN 10kN/m 40kN 545kN-n 4-4试作下列具有中间铰的梁的剪力图和弯矩图。 A + B D /2 B D
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FM=912 q12/2 ↑大 a 过B 4-5试根据弯矩、剪力与荷载集度之间的微分关系指出图示剪力图和弯矩图的错误。 解: (a)正确: (b)正确: (c)正确: B
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4-6已知简支梁的剪力图如图所示。试作梁的弯矩图和荷载图。已知梁上没有集中力偶作 用。 2kN.m 05 48kN ut 0.25KNm 4kN.m =2kN/m 十十口 18KN 14KN 4-7试根据图示简支梁的弯矩图作出梁的剪力图与荷载图。 10kN OkN OkN.m 10kN. m 40kN. m lOkN 20kN
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4-8试用叠加法作图示各梁的弯矩图。 F7/2 30kN. m 20kN 松 2 m 20kN 20kN-m 15kN t太 iiiiti3B 10kN O N 15kN-m OkN-m 0.045 10kN-m
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4-9选择适当的方法,试作图示各梁的剪力图和弯矩图。 40kN m AATTIIIITITE
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