辽宁工业大学：《材料力学》课程教学资源（作业试题）第23次作业

第23次作业: 3-1(b) F2× F 7FI 2E2(2)2× d- hErd 2E 3-1(d) F F+x dx 14F2l E 0 2EA 3兀Ed 3-3(b) x Bale 17q2l dx+ x 2EI0(8 8 15360EI 第24次作业: 3-6(a) Fx am x M(x) 2 aF 2 ov 2c.Fxx F1S OFEl2248EⅠ

第 23 次作业： 3-1(b) 2 2 2 2 2 2 8 7 4 2 4 4 (2 ) 2 4 3 E d F l d E l F d E l F V     =  +  = 3-1(d) 2 2 0 2 3 14 2 Ed F l EA x dx l F F V l   =       + =  3-3(b) EI q l dx qlx qx dx qlx EI V l l 15360 17 8 2 3 2 8 1 2 5 2 2 0 2 2 2 0 =                  + −      =    第 24 次作业： 3-6(a) EI Fl dx Fx x F EI V x F Fx M M x l y 2 2 48 2 2 , 2 ( ) 3 2 0 = =    = =   =  

3-7(c 水平位移 AB: M(x)=Fr aM aM BC: M(x)=Fa+ Fx+Fx OF F (a+x)xc ra= 5Fa3 2EⅠ I2EI 垂直位移 AB: M(x)=Fr o aF aM BC: M(x)=Fa+Fx aF F Fx.xdx+ El 26/(a+xa13F3 LeL A截面转角 AB: M(x)=Fx+M aM aM aM BC: M(x=Fa+ Fx+M iFa Fx×lax+ (a+x)×lax El EI 4EⅠ

3-7(c) 水平位移 ( ) 12 5 ( ) 2 : ( ) , : ( ) , 0 3 0  = + = → =   = + + =   =  EI Fa a x xdx EI F x F M BC M x Fa Fx F x F M AB M x Fx l x f f f 垂直位移 ( ) 12 13 ( ) 2 . : ( ) ' , : ( ) , 3 0 0  = + + =  =   = + =   =   EI Fa a x adx EI F Fx xdx EI F a F M BC M x Fa F x x F M AB M x Fx l l x A 截面转角 ( ) 4 5 ( ) 1 2 1 : ( ) , 1 : ( ) , 1 2 0 0  =  + +  =  =   = + + =   = +   EI Fa a x dx EI F Fx dx EI F M M BC M x Fa Fx M M M AB M x Fx M l l x f f f f

B截面转角 AB: M(r)=Ar aM 0 aM aM BC: M(x)=Fa+ Fx+MraMS F BFa (a+x)×lax 2El 4EⅠ 3-8(b) AB相对线位移 M(r)=Fr aM OF Fa am a M(x)= 2F2 F F 7Fa 2 x×xax+ El I2EI C截面相对转角 aM M(x)=Fx+m om fa O M(x) m oFa 6=2 x×1dx+ E Fa xld 4EⅠ

B 截面转角 ( ) 4 3 ( ) 1 2 : ( ) , 1 : ( ) , 0 2 0  = +  =  =   = + + =   =  EI Fa a x dx EI F M M BC M x Fa Fx M M M AB M x Fx l x f f f 3-8(b) AB 相对线位移： EI Fa dx Fa a x xdx EI F a F Fa M M x x F M M x Fx a a A B 12 7 2 2 2 2 , 2 ( ) ( ) , 3 0 2 0 =       =   +  =   = =   =   C 截面相对转角： EI Fa dx Fa x dx EI F m M m Fa M x m M M x Fx m a a 4 5 1 2 2 1 , 1 2 ( ) ( ) , 1 2 0 2 0 =      =   +  =   = + =   = +   

第25次作业: 3-12(c M BC: M()=F,aE=X aM CD: M(x)=Fbl, aM DA: M(x)=FB/-Fx NpB(Fx×x+FBlx+(FB1-Fx)×l=0 4F/ FI 0 3eI BEI 3 FR 32 3-12(e) aM 1: M(x)=Fcx aM M(x)=Fca, OF 3: M(x)=Fca-Fr OM =a OF EILJCFCXXxdx+ Fax adx+l(Fca- Fx)xadx=0 7F 0 BEI 2ET 3F (个),F BF (),FAx=F(-)

第 25 次作业： 3-12(c) ( ) 32 3 0 3 8 4 ( ( ) 0 1 : ( ) , : ( ) , : ( ) , 3 3 2 0 2 0 0 =  − = =        =  +  + −  =   = − =   = =   =    F F EI Fl EI F l F x xdx F l ldx F l Fx ldx EI l F M DA M x F l Fx l F M CD M x F l x F M BC M x F x B B l B l B l B B B B B B B B 3-12(e) ( ), ( ) 14 3 ( ), 14 3 0 3 2 7 ( ( ) 0 1 3: ( ) , 2 : ( ) , 1: ( ) , 3 3 0 0 0 =  =  =  − = =      =  +  + −  =   = − =   = =   =    F F F F F F EI Fa EI F a F x xdx F a adx F a Fx adx EI a F M M x F a Fx a F M M x F a x F M M x F x B A y A x C a C a C a C C C C C C C C

3-13(a) 1: M(x)=2-g OM OM 2 aX1 aX 2: M(x)=-Xx+x.gaaM_OM 8 OX1 (-x)dx|=0 El )×1dx+(-x1x+X2 )×1dx|=0 E X 3EI 2EI 16El 3X,a go 2EⅠ2EI48EI 12

3-13(a) , , 1 8 2 : ( ) , 0, 1 2 1: ( ) 1 2 2 1 2 1 2 2 2 =   = −   = − + − =   =   = − X M x X qa M M x X x X X M X qx M M x X ) 1 0 8 ) 1 ( 2 ( 1 )( ) 0 8 ( 1 2 0 0 2 1 2 2 2 2 0 2 1 1 2 =       = −  + − + −  =       = − + − −    a a a dx qa dx X x X qx X EI x dx qa X x X EI 0 48 7 2 3 2 0 3 2 16 3 2 2 1 2 4 2 3 1 − + − = − + = EI qa EI X a EI X a EI qa EI X a EI X a 72 5 12 2 2 1 ql X ql X = = −