第十一章弹性力学的变分原理

§11§111数学上的变分法：求解泛函的极值方弹性力学中的变分法y=f(x)I＝I[f(yyAy(x)BaxbL=

1+(bb

)2dxbab

1+(y')2bI[y(x)]=b

f(x,

dxy(x)2、函数的微分

=y'(x)dx记dyyxyx若dyx)很小，但dyy'y(*)并不小，称为零阶接近度；若dyx)很小，dy称为一阶接近度；

y'也很小，接近度阶数越高，则两条曲线越接近(dy)=y(x)-y(xd(y)=(y)' d（=（y） bI=af(x,y,y)dxDf=f(x,y+dy,y+dy)-f(x,y,=

dy

y¢f的一阶变分，表示为df=¶fdy+¶fdy bbb=bb

f(x,y+dy,y(x)+dyx-

f(x,y,y)dx=a[f(x,y+b

y,y+dy)-f(x,y,y)]dx

f(dy及dy的高阶项b定义泛函IbdI

a(df又dI（a

f(x,y,ybbabab

f(x,y,y'

df(x,y,y')dxaa则称函数y(x)在x＝x０

dy=0当

x=

0时，称yyx0)取极值的必要条件为dy0 y'(x)= y''(x)<

yx)y''(x)

0y''(x)>y''(x)=

yx)为极小泛函IIyx)]在yyox)处有极值则在yyox)处必有dI0o当dIyx)]yyxo

0称Iy

为驻值泛函IIyx)]取得极值的必要条件充分条件

I d2I0d2I I d2

¶ ¢¶aI=[(dya

+d

f讨论

I=

f(x,y,y')dxbb

y(b)=dI

b(

dy

y)dx

b b=ab

dy)dx d

-

)d y¢

y¢=b[

-

(¶f)]dydx= dy是任意的，¶f-

(¶f

y¢ bb==

f(x,y,y'边界条件任

-d(¶f)= y'

x=a=

x=b=

()§112应变能与余应变

dW:VedVe=fiduidv+ fiδuidv+σijnjδuids =fiδuidv+(σijδui),jdv VV=fiδuidv+(σij,jδui+sijδui,j)dvVV=(σij,j

+fi)δuidv+sijδui,jdvVVe=σijδεijdvV密度)为ve Ve=vedv 得

sij vesijijsij

=ijve的表达式设初始状态sijij受载后sij，ij在状态1的应变能密度ve

t1σ*δ s、 s、

e*:0fite(0£t£ s*:0fi

(0£t£\

δε*

tσεt t0

0=1σ

ve(e)

s0σvc(s)

ε(s)d0

Vc=vcdvVvε+vc=sijijvε=

=1sjj2§§113e=1 + x˛

i, ji=ui x˛Su与ui对应的sij

+fi= x˛s =t x˛

与真实位移ui对应的ij－－真实应变与真实应变ij对应的sij－－真实应力uikui

=1(uk

+ x˛

i,

jii

=ui

x˛Suk比 uk与u的区 uk只对应于续的位移场，但不一定对应于一个平衡的应力状态，即与uk iiiiss

+f= x˛nsns

=t

x˛Ss比ss与s的区 微小位移，记为dui有 =u+u =e+ ui

=1 i,

+duj

x˛=与ui对应的应变称为虚应

x˛有：ssss dsij,j=ds =

x˛x˛ sijnj= fukdv V

tukds

esesV证明：ss是静力容许\fukdv=-s uk ij, s s = S

snukds

i,ju u

=-tuk

+V

ee fukdv V

tukds

esesV说明由（b）、 fukdv=-tukds+s'ek =-tukds+s'=-tukds+s' =-tukds+s'nukds-=-tukds+s'nukds-uk ij, +f)ukdv

-t)ukds= +f= x˛\ ij, \ nj =ti

x˛由（a）、 设：ui--则：uku+du

=e+

+du

i, j=

(su fukdv V

tukds=

sesess= fi(ui+dui)dv+ti(ui+dui =sijV

+dijfiduidv+tiduids=sijdij fukdv V

tukds=

sesei设：uk--iss=s+ 则sij, =0(V sijn=0(sssijnjfiuidv+tiuids+(ti+dti )ijV\dtiuids=dsijeij

(2) t(1)(s t(2)(s u(1)(s u(2)(s

,e,,e,

(2),e,s,e,sf(1)u(2)dv+t(1)u(2)ds=f(1)u(2)dv+t(1)u(2)ds=(1)e(2) f(2)u(1)dv+t(2)u(1)dsf(2)u(1)dv+t(2)u(1)ds=(2)e(1)

(2) s ss sfi dv+ ds=fiiiSVdv+S §§11－4fiduidv+tiduids=sijdijdv sdve=sijijsv\v

=

fidv

tiids

=

fidv

ssdui微\外力大小和方向在i

则：dV=d fudv+tudss s

=-

fiuidv

s则：d(Ve+Vp)s定义：J=Ve+Vp 总势J(u)=1sedv

fudv-tiu d2J>

dJ=sijdijdv-fiduidv-ti =sijdui,jdv-fiduidv-ti =-(si,j+fi)duidv+sijnjduids-ti =-(si,j+fi)duidv+(sijnj-ti J＝0

+fi= x˛

n

=ti

x˛所以变分问题J＝0 sij,

fi0J(ui)=ve(ij)dv-fiuidv-ti uuifiui+ui=uk +ij =iJ(ui+ui)=ve(ij +ijV -fi(ui+ui)dv-ti(ui+ui DJdJ1d2J1d2 DJ=J(ui+ui)-J(ui)[ve(ij +ij)-ve(ij)]dv-fiduidv-ti =dvdv+ d2vdv-fiduidv-tidu 2 e e2=dJ+12V

d2v

dede

ijdij sij 得：s ed2v =[（）d+2Gdee =2+2Gdede>d2d2J>§§11－5

=s

=ve(eij)+vc(sij)=sijij

+¶vc

=sde

+eds

s

(

-e

-s)de=s

dsij

\e=eijdsijdv=

u u

dvcdv=

VVc=suitidsu3ui是边界Ｓu上给定的已知函数，uJc=Vc-su

JcdJc(sij)=dsV=[eijdsij

+(uidsij),j-ui,jdsij]dv-=(ij-ui,j)dsijdv+uidsijnjds- =V

-1 i,

-ui)dsijn\

=1( i,

+uj

i) x˛=

x˛ =0的方程为几何方程Ritzu=u0v=

+Amum+Bmvmw=

+Cmwmm

fiuidv+

ssdu=umdAmmdv=vmdBmmdw=wmdCmme e

(

d

dC

mm mm

sfuidv+s

=(

dA+¶VedB

dC)

C (fxumdAm+fyvmdBm+fzwmdCm)dv (txumdAm+tyvmdBm+tzm)ds=

=fxumdxdydzV=fyvmdxdydzV

txumdstyvmdsCm

=fV

wmdxdydz

tzwmdsdVe=sijdijdv=sijdui,jdv =sijnjduids-sij,jduidv 由:dJdV

fdudv-tduds=ss

(sij,j

+fi)duidv+(ti-sijnj)duids=(sij,j

+fi)duidv=(¶s

+

+ +

)ud

( yx+

+

+fy)vmdBmdvm (

+

+

)w

dv=

+

+ +

)udxdydz=V V ( yx+

+

+fy)vmdxdydz= (

+

+

)wdxdydz=V Vw=asinplw(０)＝０，J

d2w =Ve-P(2)=20

dx2

)dx- 2

J

p4EIa=4l3

-3由最小势能原理δＪ＝０，

a=p4

w2Plw=

sinpp4 = = =u=x(A1+A2x+A3yv=y(B1+B2x+B3y

u=1 v=计算veVe

(

+

2(1-u2)0 =2(1-

(A2+B2+2uAB J=

(A2+B2+2uAB2(