741" Advanced Mechanics of Solids

[, ' 1r]1

A,x,r= R,

fly=

, A,x*Lu- , LY+A,u, Lz+L,u-,rx = A;7-, tt y = -- AJ-, ,r, =

Ls,

Substituting for As' from Eq. (2.28) and for Lu,, Lur, Lu"from Eq. (2.7a)-(2.7c)

, I l(. 0u,\ du, dr* In'' = 1 a s*L[.t.7fJ n- * 7; ", * ai "' )

, t ldu,nt = yr, gpol ^

,,

y axis: Ly + Lu,= %- * *( ,*41^ 'l

n, * 0!^, p (2.30)' dx \ ay) dz

z axis: Az t Lu,= 4. * + 4, nu *( t* 4-) *' dx*'0Y -/'\'' dz)*

Av .--Lz,a_ - -A.s ' ' A^s

*(r*u2\n *u!, ,,f\ dv ) ' dz ''-] e'2et

, I lau- du- .(,.dr.\ lil'= 1,,spala|'-*dr "*l'* u;)")The value of e"n is obtained using Eq. (2.20).

ig]il0; CUBICAL DILATATIONConsider a point A with coordinates (x, y, z) and a neighbouring point B withcoordinates (x + Lt,y + Ly, z+ Lz). After deformat!.on, the points A and B move to

A' andB' with coordinates

A':(x+ux,y+ur,z*ur)B' ; (x+ Ax + u,* Lu,,y + Ay + ur+ Lur, z+ Lz+ ur+ Lur)

where a,, u, and u, are the components of diplacement of point A, and from Eqs(2.7a)-(2.7c)

Lu,= d!, n*+d!, ny+(*dx oy oz

. 0u- du" 0u,A,u,= -] Ax + -;-r Ly + --z- Lz

'dxdy'dz. du, du- du-LU_ =--=-! A-r + -=l Ar' + -=! A:"dxdy'dz

The displaced segement A'B' wlll have the.following components along the x, yand z axes:

t/

/'zFig.2.S

Consider now an infinitesimr(Fig. 2.5). When the body ur

becomes an oblique paralleleIdentifying PQ of Fig.2.5 w.Eqs (2.30) the projections of

alongx axis: I

alongy u*ir, !:

along z axis: !.a

Hence, one can successivel.Az = }l,P.S ( Ax = Ay = 0) an<

x, y and z axes as

P(/.(duxaxts: ll+_[/,

y axis: dY"0x

z axis: du, ,

0x

The volume of the right paralAz. The volume of the defonrformula from analytic geometl

V'=V+A,V=

Az + Arr-

r- 1a.., Au. from Eq. (2.1a)-(2.1c)

Analysis of Strain 75

^€-.:71.." V

zFig,2,5 Deformation of right parallelepiped

Consider now an infinitesimal rectangular parallelepiped with sides M, Ly and Az(Fig. 2.5). When the body undergoes deformation, the right paralleiepiped PQRSbecomes an oblique parallelepiped P'Q'R'8.Identifying P Q of F ig. 2.5 urilh AB of Eqs (2.30), one has Ay = At = 0. Then, fromEqs (2"30) the projections of P'Q'wil1 be

/:\along x a*;s:

It . #)*

alongy u*ir, d!.'

6*ox

along z axis: 4i2 Lr6x

Hence, one can successively identify AB with PQ (Ly - Az = 0), PR (A{ =Lz = O),P,S ( Ar = Al = 0) and get the components of P'Q', P'R' andP'9 along the

x, y and z axes as

(2.29)

; r and a neighbouring Point B with

brmation, the Points A and B move to

;r-.i* L,z*ur,+Lur)iplacement of Point A, and from Eqs

:

af'

following comPonents along the x, Y

du.,: lr' * -=-l- Azr dz.. du-

z axls: ------L lltCr

6u, ^,

(t*4)*dv \ cr)

0u.. I-1..+-=:n. Idz 'l

ou, 1lt I

-fl-

|dz "ll

du- \ |1+---ln. Idz ) " )

3u- At,+ .l nz-dz

. du\- 1+-=-i-l&',. dz )

Pg

xaxis: ?:.*)*

P,S

dtt, p

6u$' - ^-MCZ

y axis: lt * t

PR'

Cu" L,0y .'

cut \rtI

a1 l

(2.30) The volume of the right parallelepiped before deformation is equal to V = Ax LyAz. The volume of the deformed parallelepiped is obtained from the well-knownformula from analytic geometry as

V'=V+LV:D.A.xLyLz

76 Advanced

where D is the

Mechanics of Solids

following determinant:

D:

}ttr.

0x

['. %) olu a!,I d*) dy 0:

% [,.L) Y'dx i ay) D:

Y' [' ',-4-)dy ( d:l

(2.3r)

lgiifr;tii CIIANGETWO Ltr

Let PQ be a line elerelement with directi,

VL

RIIl/iA,1 l)-olP

zy'Fig.2.6 Change it

two line

g69, B'=

-(l+r

+(+l

In particular, if the tbefore strain, then a

cos 0'= (I+ ton

Now (90' - 0') re

by a, then

0'=or cos 9'=since rr is small. ThPR. If we represent

T*0, at P: cos 0' =

Example2.T The

u = k(x2

At a point P(2, 2, 3ing direction cosine

PQ'. n*r= n,.

Determine the angl<

If we assume that the strains are small quantities such that their squares andproducts can be negelected, the above determinant becomes

- 0u - 0u, Ou-D=I+ ='+ = +dr 0y D.

-lrc ac rc-r'cx\rLt,vtc-___ (2.32)

Hence, the new volume according to the linear strain theory will be

V'=V+ LV= (1+ e,,+€.rr+ err) LxLy Lz (2.33)

The volumetric strain is defineci as

A= + =t*rt"*t. (2.34)I'Therefore, according to the iinear theory, the volumetric strain, also known ascubical dilatation, is equal to the sum of three linear strains.

Example 2.5 The folloving state of strain exists at a point P

I o.o: -0.04 s I

r,,|=|-;;; ;;; -;"Il o -oo2 o l

In the direction PQ having direction cosines n,-Q.9, n,=0 and n. : 0.8,determine €oo.

Solution From Eq. (2.20)

epe=0.02 (0.36) + 0.06 (0) + 0 (0.64) - 0.0,1(0) - 0.02 (0) + 0 (0.48)= 0.007

Example 2.6 [n Exarmple 2.5. whar i,s the t-ttbical tlilatation at point p?

Solution From Eq. (2.34)

A=t"r*tr*t.=0.02+0.06+0=0.08