同濟(jì)大學(xué) 材料力學(xué) 習(xí)題解答6(練習(xí)冊P86-P94)

材料力學(xué)習(xí)題解答(六)P86

46-1FCFBlCA0.5lBMx=FCFB

=

m

=0

l

MM(x)

=(l

-

x)

l

M彎矩方程撓曲線近似微分方程EI

=(

x

-

l)dx2

d2y

l

M積分EIq

=(

x

-

l

)2

+

C2l

MEIy

=(

x

-

l

)3

+

Cx

+

D6l

M邊界條件x

=0,y=0：x

=l,y=0：D

=Ml26

1撓曲線方程y(x)=[(

x

-

l

)3-

l

2x

+

l

3]6EIl

M轉(zhuǎn)角方程q(x)=[3(

x

-

l

)2

-

l

2]6EIl

MA處撓度yA

=16EIMl

2x

=：2

lB處轉(zhuǎn)角x

=l

：qB

=-6EIMl

ymax31x

=(1-

)l

討論：q

=0：令ymax

=9EIMl

230.06250.06415誤差2.6％

C

=-

Ml6

13l1-積分法坐標(biāo)P86

46-2(b)ACaMBlx彎矩方程ACBCM1(x)=-

MM2(x)=0撓曲線近似微分方程EI

=

Mdx2

d2y1EI

=0dx2

d2y2EI=

Mx+

C1dx

dy1EI=

C2dx

dy2ACBCEIy1=

Mx2

+

C1x

+D121EIy2=

C2x

+

D2邊界條件x

=0,q1=0：C1=0x

=0,y1=0：D1=0q2(x)=EIMa2EIMay2(x)=

(2x

-

a)連續(xù)條件=

qC=

yCx

=

a,q2=

q1x

=

a,

y2=

y1=EIMa2EIMa2=C2=

MaD2

=

-

Ma221qB=EIMa2EIMayB=(2l

-

a)積分q1(x)=EIMx2EIMx2y1(x)=P87

46-4FAaACaFBDBaqxFA

=

qaFB

=

qa

mA

=0

Fy

=0

4

3

4

9彎矩方程ABBCM1(x)=

qax

-

qx22

143M2(x)=-(3a

-

x)22

q撓曲線近似微分方程EI

=

qx2

-qaxdx2

d2y12

143EI

=

(3a

-

x)2dx2

d2y22

qEI=

qx3

-qax

2

+

C1dx

dy16

183EI=

-

(3a

-

x)3+

C2dx

dy26

qEIy1=

qx4

-qax3

+

C1x

+D124

1243EIy2=

(3a

-

x)4+

C2x

+

D224

q邊界條件x

=0,

y1=0：D1=0x

=2a,

y1=0：q1(x)=

(4x3

-9ax2+

4a3)24EIq24EIqy1(x)=

(x4-3ax3+4a

3x)連續(xù)條件=

qBx

=2a,q2=

q1x

=2a,

y2=

yB=

0：=

0：C2=qa361D2

=

-

qa483q2(x)=

[a3

-

(3a

-

x)3]6EIq24EIqy2(x)=

[(3a-x)4+4a3x-9a4]C1=6qa

3qA=6EIqa312EIqa4yD=qB=

08EIqa4yC=P88

47-1(b)BM(qa2)CF(qa)qAaaP224表19-1序號1

2EIM(2a)2yCM=

-qCM=

-EIM(2a)EI2qa4=

-=

-EI2qa3P224表19-1序號3

6EI5qa4==2EIqa3yCF=yBF+

qBF·aqCF=

qBF=2EIFa23EIFa3=

+

qBF·aP224表19-1序號5

24EI7qa4=yCq=yBq+

qBq·aqCq=

qBq=6EIqa38EIqa4=

+

qBq·ayC=yCM+yCF+yCqqC=qCM+qCF+qCq8EI7qa4=

-3EI4qa3=

-P88

47-2(a)AC0.5lB0.5lM(Fl)FP225表19-1序號8

yCM=

0qAM=

-24EIMlP225表19-1序號9

48EIFl3yCF=qAF=16EIFl2yC=yCM+yCFqA=qAM+qAF=

qBM=

-

qBF48EI5Fl2=

-qB=qBM+qBF48EIFl2=48EIFl3==

-24EIFl2FP89

47-3(c)ABCqF(qa)aaP225表19-1序號11

yCq=qBq·aqCq