工程力学理论虚位移原理

δw

n δx+ δy+Fδz)=S S

例4图⽰桁架，各杆⻓度例4图⽰桁架，各杆⻓度F2=15kN。试求:杆力FDE,E DEABC lF1) +

lll )δf

F)

3+ l3

)δf=

δrB=lδfA=l δfA= +15+

+

)lδf= FDE=-

A

3 3(2lF1) +(2lF2) -(ICFCB) CIcos =EC= AE=2lcos30=CI= EI=IC=2E =3lδfA=E

3δfA= FCB=例5：例5：拱架结构,F1=2kN，F2=1kN，试求:拱架D、CAEGAEGHOBCD l

DD解：1.求(lF1) +(lF2)δfGHD-(2lFDy)δfGHD= =δfO= FDyAEAEGHOBC Dll求(lF1) +(lF2) +(2lFDx) A

FDxE

G

ll

求(lF1) +(lF2) +FCδyC=2lδfO=2lδfABE=δrE=δrG=δrC=2l δyC=lF1lδfO+F2lδfO+FClδfO= FC=-lCl例 在图所⽰桁架中,已知AB=BC=CA=l,AD=DC= FiδFiδ=解：1. - +PδrDcos45=δrBcos30=δrCcos δrD=δrCcos

=

P

l例6A已知 yBFDyBFDPACxδw=1(Fxiδxi+Fyiδyi)=FDy=F- FBy=-yDl

yB=lsinq22δyD δ 2(-P+F lxC=2lcosq2=1

δq=

sinq12sinq2δq2 sinq1δq1 2 q1=45,q2=60 F=2.37P212 = q

δ

= =

δ

+ +

δ

+…+

δ

j=1

δq¶ ¶ S

riδw=

δri=S

¶qδq

=S(S

¶q Q=

ri

j k

j (i=1,…,n； (SFi δw=SQjδqji=1 j=1

Qj =

+

+

¶zi

Qj

S

[Fix

+Fiy

+Fiz¶zi δw=Q1δq1+Q2δq2++Qkδ取⼀组除δqj= =Qjδq =

δδjkSδw=S

δqj= ,试求维持平衡所需力F的大小 ,1 y=1

=l

=

=lcosq δy1=-2sinqq q

=4lsinqδ

(Fixδxi+(Fixδxi+Fiyδyi)=x δq„ F=Pq ¶q-F

-Pq平衡：Qq= F=P解：自由度：2，取广义坐标 2 2δ

=-M

-(

δ-(lsinf1P)δf1+(l

δQ=δ

=-M-(2

F F

2

PQ2

2

F)

f2f

δF=

M=

Ｂ

Qj

S

[Fix

+Fiy

+Fiz¶ziQ=P

+F¶xB-

1 1Q2= +

+F -M

f2flyC1=

yC

=l

+l

xB=lsinf1+lsin\Q=-P3lsinf+Flcosf-M Q=-Plsinf+Flcosf 1

=12

2

2-PδyC1+PδyC2+FδxB-Mδf1= yC1=

l

δ

2

=l

+ 2C 2

δyC2=-lsinf1δf1-

2 xB=lsinf1+lsin

3

δf1+l lP2lsinf1+Flcosf1-M)δf1+(-P2sinf2+Flcosf2)δf2=因:δf10δf2F=1Ptan M=

§16-4 =-¶V =-¶V n n

+

+

¶zi=-(

+

+

¶zi)=-¶V

jQ=-¶V=j

¶V= yPClllkOlVP(2lcosqh)yPClllkOl2¶V=

即

B cosq=P

q=arccos

,

当θ=0 当θarccosP

PPPPP ¶V=

2 2

=P(2lcosq+h)+1k(2lsin -2Pl+4kl2cosqsinq=sinq=0,q1=cosq=P

=arccosP 22

l当q0,-2Pl4kl20.则2klP2当q

P

-k

4kl2 解：解：FR

Fiy

FMO=Mxi+Myj+M

δr+M0 =Fixδx+Fiyδy+Fizδz+Mxδfx+Myδfy+Mj δQ1=δ

Q2=δ

=Fiy= Q3=δ

=Fiz=Q4

Q=δ δf

Q=δ

=Mzδ⻓=4。梯=0.，∠=6°体重=80A C P1 C

2

-P1

-

l

δw= 2

+

lcosfδq-

2lsinfδq=

1 s

δf= δw=(Pl+

+Pl-

2ltanf)cosfδf=1 2sinf1 h (1

) =P1l(2fstanf-1)sin