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时间:2019-09-21
《工程力学多媒体课室习题参考答案》由会员上传分享,免费在线阅读,更多相关内容在行业资料-天天文库。
1、工程力学多媒体课室习题参考答案第一章1-1FR=669.5N,Ð(FR,i)=34.901-2略1-3SMA=-180kN.m,不会倾倒。1-4略1-5略1-6略第二章2-1F¢R=466.6N,Mo=21.44N.m,FR=466.6N,d=45.96mm2-2F¢R=150N,Mo=900N.m,FR=150N,d=-6mm2-3F¢R=8270N,Mo=6121N.m,FR=8270N,Ð(FR,i)=267.60,x=-0.763m(在O点左侧)2-4FR=608N,Ð(FR,i)=96.30,x=0.488m(在O点左侧)2-5(a)yc=24cm;(b)xc=11cm第三章3
2、-1(a)FAx=3Ö2F/4(®),FAy=Ö2F/4(),FNB=F/2(b)FAx=0,FAy=5qa/4-F/2-m/2a,FNB=-qa/4+3F/2+m/2a3-2(a)FAx=qatana/8(®),FAy=7qa/8(),MA=3qa2/4FNC=qa/8cosa,FBx=qatana/8,FBy=3qa/8(b)FAx=0,FAy=3m/4-9q/2,FNB=4q-m/2,FND=m/4+q/2,FCx=0,FCy=-m/4+3q/2(c)FAx=38.6kN(®),FAy=90kN(),FBx=38.6kN(¬),FBy=90kN(),FCx=38.6kN,FCy=0
3、3-3F=15kN,Fmin=12kN,方向于OB垂直3-4FCE=F2/2cosa-F1/2sina3-5(a)FNA=FNB=m/l,(b)FNA=FNB=m/l,(c)FNA=FNB=m/lcosa,(d)FNA=FNB=Ö2m/l3-6FNA=FNC=m/2Ö2a3-7FNE=Ö2F,FAx=F-6qa,FAy=2F(),MA=5Fa+18qa23-8FCx=33.8kN,FCy=0,FAB=33.75kN3-9FT=200N,FBz=FBx=0,FAx=86.6kN,FAy=150kN,FAz=100kN3-10FT2=4kN,Ft2=2kN,FAx=-6.375kN,FAz=
4、1.3kN,FBx=-4.125kN,FBz=3.9kN第四章4-1F=sin(a+jm)G/cos(a-jm),当q=jm时,Fmin=Gsin(a+jm)4-2(1)F1=1.492kN,F2=1.508kN(2)FT1=26.06kN,FT2=20.93kN4-3s=0.456l4-4b《110mm4-5F=Gr(a/fS-b)/Rl第五章5-1(a)FN1=-30kN,FN2=0kN,FN3=60kN;(b)FN1=-20kN,FN2=0kN,FN3=20kN;(c)FN1=20kN,FN2=-20kN,FN3=40kN;(d)FN1=-25kN,FN2=0kN,FN3=10kN
5、。5-2(a)êFNïmax=3kN;(b)êFNïmax=10kN;(c)êFNïmax=FP;(d)êFNïmax=5kN;(e)êFNïmax=FP;(a)êFNïmax=2FP;(a)êFNïmax=FP+Alrg。5-3(a)êTïmax=2Me;(b)êTïmax=2Me;(c)êTïmax=30kN.m。5-4T1=954.9N.m5-5(a)FS1=0,FS2=-FP,FS3=0,M1=M2=FPa,M3=0(b)FS1=FS2=-qa,FS3=0,M1=M2=-qa2/2,M3=0(c)FS1=-q0a/2,FS2=q0a/12,M1=M2=-qa2/6(d)FS1=-
6、qa,FS2=-3qa/2,M1=-qa2/2,M2=-2qa2(e)FS1=1333N,FS2=-667N,M1=267N.m,M2=333N.m(f)FS1=FS2=-100N,FS3=-200N,M1=-20N.mM2=M3=-40N.m5-6(a)êFSïmax=2ql,êMïmax=3ql2/2;(b)êFSïmax=ql,êMïmax=3ql2/2;(c)êFSïmax=5FP/3,êMïmax=5FPa/3;(d)êFSïmax=2qa,êMïmax=5ql2/2;(e)êFSïmax=2FP,êMïmax=FPa;(f)êFSïmax=FP,êMïmax=FPa;(g)ê
7、FSïmax=2qa,êMïmax=qa2;(h)êFSïmax=3qa/8,êMïmax=9qa2/128;(i)êFSïmax=3Me/2a,êMïmax=3Me/2;(j)êFSïmax=qa,êMïmax=qa2/2;(k)êFSïmax=5qa/8,êMïmax=qa2/8;(l)êFSïmax=3.5FP,êMïmax=2.5FPa;(m)êFSïmax=qa,êMïmax=qa2;(n)êFSïmax=qa/2,êMï
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