19991207

Applied Mathematics and Mechanics (English Edition, Vol. 20, No. 12, Dee 1999)

Published by SU, Shanghai, China

Article ID: 0253-4827(1999) 12-1350-08

DOUBLE-MOMENT OF SPACIAL C U R V E D B A R S

WITH CLOSED THIN-WALL CROSS-SECTION *

Zhu Yuchun ( ~ I ~ ) , Zhang Peiyuan ( ~ : L ~ ) , Yan Bo (j~z ~ )

(Department of Engineering Mechanics, Chongqing University,

Chongqing 400044, P R China)

(Communicated by Zhang Ruqing)

Abstract: In this paper, the double-moment of thin-wall, cross-section spacial

curved bars of anisotropic materials is discussed, and a general solving method for

this type of problems as "well as the concrete double-moment formula of planary

curved bars subjected to action of vertical loads are given out.

Key words: closed thin-wall cross-section; spatial curved bar; double-moment

CLC number: TB125 Document code: A

Introduct ion

Although the constrained twist problems of thin-wall straight bars of isotropic and/or aniso-

tropic materials, with closed or open cross-sections, have been well solved [z-3] , for spatial

curved bars of this type of problems there has not been any satisfactory solution up to now. This

problem is urgently needed to be thoroughly studied in engineering structures, especially in bridge

structures with box cross-section curved beams. In this paper a double-moment solution method

for spacial curved bars of anisotropic materials with closed thin-wall cross-sections is given out.

Suppose the tangential , normal , and subnormal unit vectors of a spatial curve l are respec -

o. tively e , , e~, and e 6 . The Frenet formula, for

ra

Fig. a, and r,,

enough smooth curve, is

es = k0en, en = - koe~ + r 0 e b , 1 ) J eb = -- r0 en ,

here ( ) = d( ) / d s , s , ko and r0 are respecuve-

ly arc coordinate, curvature and torsion ~4] .

Choose two orthogonal fixed directions 01

and 01 71 through point 01 on planary surface F .

Let point 01 move along curve l , with surface F

always perpendicular to l , the orbit of surface F

then forms a spacial t a r . The cross angle between direction 01 ~ and normal e n is represented as

0, which is generally a function of s . If the unit vectors of 01 ~ and 01 q are represented by ee

* Received date: 1998-01-14; Revised date: 1999-04-19

1350

Abstract

The one-dimensional problem of the motion of a rigid flying plate under explosive attack has an analytic solution only when the polytropic index of detonation products equals to three. In general, a numerical analysis is required. In this paper, however, by utilizing the "weak" shock behavior of the reflection shock in the explosive products, and applying the small parameter pur- terbation method, an analytic, first-order approximate solution is obtained for the problem of flying plate driven by various high explosives with polytropic indices other than but nearly equal to three. Final velocities of flying plate obtained agree very well with numerical results by computers. Thus an analytic formula with two parameters of high explosive (i.e. detonation velocity and polytropic index) for estimation of the velocity of flying plate is established.

1. Introduct ion

Explosive driven flying-plate technique ffmds its important use in the study of behavior of materials under intense impulsive loading, shock synthesis of diamonds, and explosive welding and cladding of metals. The method of estimation of flyor velocity and the way of raising it are questions of common interest.

Under the assumptions of one-dimensional plane detonation and rigid flying plate, the normal approach of solving the problem of motion of flyor is to solve the following system of equations governing the flow field of detonation products behind the flyor (Fig. I):

ap +u_~_xp + au --ff =o,

au au 1 y =0,

aS as a--T =o,

p =p(p, s),

(i.0

293

where p, p, S, u are pressure, density, specific entropy and particle velocity of detonation products respectively, with the trajectory R of reflected shock of detonation wave D as a boundary and the trajectory F of flyor as another boundary. Both are unknown; the position of R and the state parameters on it are governed by the flow field I of central rarefaction wave behind the detonation wave D and by initial stage of motion of flyor also; the position of F and the state parameters of products

Zhu Yuchun, Zhang Peiyuan and Yan Bo 1351

and e n , then

e e = encosO + ebsinO, (2) J ev = - e. sinO + e 6cosO,

from equation (1) the following expressions are obtained

d., = kee e - k v % , ]

ee = - kee, + r e ~ , l (3)

~ = k~e s - t e e ,

in whichk e = k0cos0, k~ = k0sin0, r = r0 + 0.

Assume curve l is the axis of a bar, and surface F is cross-section of the bar. Denote the

wall thickness as h , the arc crordinate of the wall central line as t . If h is far smaller than the di-

mension of cross-section, the constitutive equations can be approximately written as(see Fig. 1 )

a, = Ee, , r,, = G)',t, (4)

here E , G are elastic modulas E52 .

1 I n t e r n a l F o r c e s , E q u i l i b r i u m E q u a t i o n s a n d G e o m e t r y E q u a t i o n s

Simplifing stress vectors to point 01 on cross-section F , the principal force vector N and

moment M can be obtanied, of which components are respectively denoted by N, , N: , N~ and

M~ , M e , M v , so

N : N~e~ + N~e e + N~e~ , M = M~e~ + Mee e + M~e~ ,

where N, is axial force, N e and N~ are shear forces, M, is torgue, M e and Mr are bending mo-

ments. The external forces and moments per unit length on axis are indicated by p and m as

p = pses + pee~ + p v e v , m = m,e , + mee t + rove v .

The equilibrium equations are

d { N } - [ K ] ' { N } + { p } = {0}, = } , (5)

d { M } - [ K ] - { M } - [ H I - { N } + { m } {0},

here

�9 { N } = [ N , Ne

{p} = [p~ pc

0 k e

[ K ] = - k e 0

- k ~ - r

The general solution is (let { m } = {0}) [rJ

NT/] T , { M } = [M~

p,1 IT, {m} = [m,

- k v

~" , [ H ] =

0

M~ M,1 ]T ,

/75~ /n~ ] T,

0 .

- 1

Abstract


1. Introduct ion




au au 1 y =0,

aS as a--T =o,

p =p(p, s),

(i.0

293


1352 Double-Moment of Spacial Curved Bars

{N}= [A] " ({Q~ " {p}ds)' t

{ M } = [ A ] . { { R o } + f:[A] r . [HI . [ A ] . ( { Q o } + { Q " } ) a s } ,

(6)

here { Qo } and { Ro } are integration constants, { Q" } = - fo [ A iT {p }ds . If the base vectors

of spacial fixed right-handed rectangular coordinate system are e , , er and e~, then

I e,.e, e,.ey e,.e,] [ A ] = e e ' e . ee 'ey e e ' e , . ] (7)

e 0"ex er �9 ey e v. e z

In order to derive geometry "equation, we introduce multiplier 3 u , , du e , 3u~, ~ 9 , , 89e ,

c?~v. Using [ 3us 3ue 3% ] and [ ~79, ~oe ~gv ] to multiply both ends of equations (5 ) ,

and integrating the equation obtained after summation of the two equations in ( 5 ) , in the range

(0 , l ) , we have

f o [ ( N , - keN e + kvN ~ + + + rN~ + p , ) ~ u , (ae pe )$ue +

(IVr - k~N, + rN e + p~)~u~ + (M, - keM e + k~M v + m , ) ~ 9 , +

( M e . + keM s - rM~ - N~ + me)~ge + (M~ - kvM , + rM e + N e + m,~)~9,t]ds =

[Ns~us]g + [Ne~ue]to + [ N ~ u ~ ] ~ + [M,~gs]o t + [Me~ge]g + [ M ~ 9 ~ ] 0 t, (8)

h e r e [ J ] ~ = ( J ) , = ~ - ( J ) , = o a n d

e = a, - keu e + kvu v, Ye = ae + keu, - r u r - 9~, ]

Yv = av - kvu, + rue + 9e , x, = 9s - kec, oe + k v g v ' l (9)

xe = 9e + keg, - rg~, x~ = 9~ - kvg, + rge .

In equation ( 8 ) , e , Ye, Y~, x , , xe, xv are respectively general strains corresponding to general

stresses N, , N e , Nr M,, M e , M, 1, and u, , u e, ur 9s, 9e , 9v are general displacements

corresponding to loads p , , Pe, P~, ms, me, m,1 �9 The boundary conditions are

Ns or u s , N e or u e ,N~ or % ,M, or 9, ,Me or 9e ,M~ or 9~. (10)

Equation (9) is usually referred to as geometry equation, and can be rewritten as

�9 ~ r [K].(r (,~t = ~0t, ~0t,} (11) d { u } - [ K ] ' { u } - [ H ] ' { r { e } =

where {r [*, 9, 9~] T, {u}= [u, ue uv] T,

Abstract


1. Introduct ion




au au 1 y =0,

aS as a--T =o,

p =p(p, s),

(i.0

293


Zhu Yuchun, Zhang Peiyuan and Yah Bo 1353

So the general solution of the geometry equation is

{~a} = [A] . ( { Igo}+ {u/* }) ,

{u}= [a].{{V0}+ flEA]'.<{=}+ [ H ] ' [ A ] ( { I g o } + { I t * }))'ds}, in which { I/to } and { U0 } are integration constants, { I/t* } = F [ A IT. { lr } d s .

30

(12)

2 Double-Moment and Equivalent Constitutive Equations

Double-moment is not considered in the discussion above. The distributions of strain e, and

y , , , similar to [ 1,5 ] , on cross-section of non-free twist spacial thin-wall bar including double-

moment are

~, = ~ + ~ - &~ - x ~ , 1

?',t x, ro - ? 'esina + 7, /cosa - x(r o - e ) , l (13)

here e , ~'e, ~'~, ~:,, xe , xv are the same as ( 8 ) . For closed thin-wall cross-section bar, general

strain x is introduced for considering constrained twist, w.~ , r 0 , r A , e are defined as Is] ( see

Fig. 2)

a'A = ~ f :edt ' e = g'2/( Gh~ d~hh) ~i * - - , W A = r a d t , I] = Wa I , = t . (14)

The determination of coordinate system O l ~r/, point A and the start point P0 for the calculation of

arc length must satisfy following conditions

~E~hdt = ~Er2hdt = O, fE~Ghdt = 0

and E~w~hdt = Eriwahdt = O, Ew~hdt = O.

This coordinate system is referred to as physical centroid principal axises sys tem, WA is physical

principal sector coordinate.

Stresses after introducing internal forces

represented as

N~ = ~a=hdt,N, = - fr~tsinahdt,

N,~ = ~r,tcosahdt,

Ms = ~ r J o h d t ,

M e = ~a~r]hdt,

M~ = - ~a,~hdt,

M B = ~a~w~ hdt,

are

(15) eo(= =O)

Fig. 2 Cross-section properties

Abstract


1. Introduct ion




au au 1 y =0,

aS as a--T =o,

p =p(p, s),

(i.0

293


1354 Double-Moment of Spaeial Curved Bars

where Ms is double-moment. We denote

M,B = ~ G , ( r o - e ) h d t . (16)

It can be easily proved that

o (G3G + G,3X, , )hd t .ds = [N,~e + Ne~7 $ + Nv~7,1 + M,~x, +

Mz3xe + M ~ x v + (MB - M , B ) ~ x ] d s - [ M B ~ x ] o t . ( 1 7 )

On the other hand, the variation of strain energy is

~U = 3 f : ~ 1.~_( Er + Gx2 , )hd t .ds =

f / { D l l e ~ e + D22)'$~)'~ + D33~'r/~:Xv + [D44x, - (D44 - Do)x]3xs + o

Dssxe~xe + D66xv3x ~ + [(D44 - Do)(X - x,) - D w x ] ~ x } d s +

[D77x~x]~ + f i { - ~ C r a s i n a h d t ( G ~ ) ' e + Xe~x,) +

~Gracosahdt (x , 37,~ + 7 , ~ 3 . , ) - ~ G s i n a c o s a h d t ( x , ~ X , 1 + ? ' ~ ? ' , ) +

G(r a - e ) s i n a h d t ( 7 e 3 x + x~)'e ) - f G( r A - e ) c o s a h d t ( ) ' ~ 3 x - x~)'r ) }ds .

(18)

If the cross-section shape of the bar and the distribution of G are symetrical about axises O l ~ and

Ol 7], the last integration part is identically equal to zero. Let the right terms of (17) and (18) be

equal to each other, the equivalent constitutive equations of the bar are obtained as follows

N, = D I I e , N ~ = Dz27e, N,j = D337 ~ , M, = D 4 4 x , - ( D 4 4 - Do)Z , }

Me Dssx~, M, 1 = D66x, l , Ms = - D w g , M,B = (D44 - D o ) ( G - x ) , (19)

here

o,I ~ : Csin2ohd, ~ ={Cco:ohdt

D77 = Ema hd t , D O = 1"22 dt

where D O is anti-twist stiffness of cross-section for free twist bar, having relation

fGroehd t = ~GeZhdt = Do .

The minimum potential energy principle of the entile bar can be written as

8 U - 8 { u } + { m } T ' { ~ } ) d s = O. (21)

Equilibrium equation (5) described with general displacements can be derived out and the varia-

Abstract


1. Introduct ion




au au 1 y =0,

aS as a--T =o,

p =p(p, s),

(i.0

293



tional equation with general displacements is

MB - M,B = 0. (22)

Besides, boundary condition (10) and the following boundary condition can also be obtained

[Ms~x]o t = 0. (23)

3 General Method Determining Double -Moment

Equations for static problems of spatial curved bar with closed thin-wall cross-sections in-

clude: variational equation (21) , geometry equation ( 11 ) , and equivalent constitutive equation

(19) . The general solving steps are as follows:

First, from the 4th, 7th, 8th expressions of (19) and formula ( 2 2 ) , we establish the

equation of x and x, :

D44x, - (D44 - Do)tO - M, = O, ( - D77x) - (D4,, - D o ) ( X , - x ) = O. (24)

If Dkk ( k = 1 , 2 , - - ' , 7) is constant, the solutions of x and x , , from above equations are

D44 - Do x = c l c h a s + c2shas + x * , tc~ - D44 x + - , , , , M ' ' (25)

here x * is a specific solution of equation (24) ,

a = a / ( D 4 4 - Do) " Do~D** O77.

Thus, { r } and { t } can be expressed with { N } , { M} and integration constants cl and c2. Now

the unsolved problem is to determine integration constants { Qo }, { Ro }, { ~0 }, { Uo } as well

as c l and c2.

4 D o u b l e - M o m e n t o f P l a n e C u r v e d B a r U n d e r V e r t i c a l D i s t r i b u t e d L o a d s

r = 0 and 0 = 0 in formula (3) is just the situation of plane curved bar (see Fig. 3) . Fix

the origin of the rectangular coordinate system at the end of the bar ( s = 0 ) , the axis of the bar

being on the plane O x y . The acting loads are

{ m } : { 0 } , { p } = [0 0 q]T

If the axis of the bar is a circle with radium a , we have

f l = s 1 - - , k ~ a a

x : asinfl , y = a ( 1 - cosfl) .

Denote D = 077D44/(D44 - D o ) , solution ( 2 5 ) of

equation (24) becomes

1 [ Mo, cosfl + = c l c h a s + c2shas + D ( k 2 + a2 )

Moesinfl - No,~a[cosf l D ( k 2 + a2) ] Do +

q a Z [ s i n f l s D ( k 2 + a 2 ) ] } a O o '

axis of the bar

Fig. 3 Axis of plane curve

(26)

Abstract


1. Introduct ion




au au 1 y =0,

aS as a--T =o,

p =p(p, s),

(i.0

293


1356 Double-Moment of Spacial Curved Bars

here

/C s - -

D44 - D o 1 O~..4 (c1chot$ + c 2 s h a s ) + ~ - ~ { ( 1 + f ) Mo, c o s f + (1 + f ) M o e s i n fl -

[ ( l + f ) c o s f l + (1 + f D ( k z + a 2 )

]}, f = (D44 - D o ) 2 / [ D 4 4 D 7 7 k 2 + a 2 ) ] .

Using equations (6) and ( 1 2 ) , we obtain

Ms = Moseosf + Moesin/3 + No,lY - q a ( s - a s in f l ) ,

M e = - Mo, sin fl + Moeeosfl + NoTx - q a y ,

N~ = N o T - q s ,

(x , eosfl - x e s i n f ) d s + q~ = r176 + r + cosfl o

(x , sin~? + xecos f l )d s , sinfl o

(27)

l c, = q a [ D ( - ~ +-a2 ) - ,

qa 2 . k 3 1 c2 = - c l t h a l + D ( k 2 + aZ ) a --$ " cha-----~"

The double-moment given by the 7th expression of equation (19) is

(28)

(29)

-- qo2[(l__ .in /sho -s aa k 2 + a 2 j ~ 1 ~ +

~ s

( x s c o s p - xes inf l )ds + q~e = - r + r - sinfl o

( x s s i n f + ~ e c o s f ) d s , C O S ~ 0

(x~cosfl + xes in f l )ds + uT = UoT + r - r x + y T d s + s i n f o

cost? o ( x , sinfl + xecosf l )ds d s ,

w h e r e M 0 , , Moe, No T are the values of M, , M e , N T a t t h e e n d s = 0, and r r UoTare

the values of f , , q~e, uT at s = 0. x e and 7 T are described with M e and N T by ( 1 9 ) . N , , N e ,

M T , u , , u e , ~o~ are all identically equal to zero.

If the bar l is fixed at s = 0, and free at s = l , the boundary conditions are

s = 0 ( p = 0 ) , Uo~ = 0, r = r = 0, x = 0 (with double-moment) ,

s = l ( p = f i t ) , N~ = 0, M, = M e = 0, x = 0 (without double-moment) .

The integration constants determined by aforementioned conditions are

( q " ) l

No~ = q l , Mo~ = qal 1 - - - T s l n f l / ,

Moe = - qa : (1 - cosf l t ) ,

Abstract


1. Introduct ion




au au 1 y =0,

aS as a--T =o,

p =p(p, s),

(i.0

293



(k 2 + a2)a2(chal - 1 + k2 + a2Me , (30)

and the double-moment at fixed end of the bar is

D~-_ D O qa 2 asinflt / M,o = D44 [ ( / - k2-V-~--~a2 ] �9 thal +

+ ~ (1 - ~o~51 (31)

References:

[ 1 ] Gianelige G U, Banuofuco R G. Statics of Elastical Thin Walled Bars [ M] . Hu Hai- chang, Xie Bomin Trasl. Beijing: Science Press, 1955. (Chinese version)

[ 2 ] Chen Tieyun, Chen Bozhen. Band, Twist and Stability of Open Thin Walled Bars [ M ].

National Defence Press, 1965. (in Chinese)

[3] Vlasov V Z. Elastical Thin Walled Bars [ M ]. Moscow: National Science Academic Press, 1963. (in Russian)

[4] Wu Daren. Differential Geometry [ M ] . Beijing: People 's Education Press, 1959. (in Chinese)

[ 5 ] Luo Zudao, Li Huijian. Anisotropic Material Mechanics [ J ] . Shanghai : Shanghai Jiaotong University Press, 1994. (in Chinese)

[ 6 ] Xiong Hanwei, Zhang Peiyuan. Finite element analysis of spatial curved bods[ J ] . Journal

of Chongqing University, 1997,20(4) :31 - 36. (in Chinese)

Abstract


1. Introduct ion




au au 1 y =0,

aS as a--T =o,

p =p(p, s),

(i.0

293


Documents

19991207