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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