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The following lecture is adopted from the following books:
1. “Some Basic Problems of the Mathematical Theory of Elasticity” by N.I.
Muskhelishvili
2. “A Treatise on the Mathematical Theory of Elasticity ” by A. E.H. Love
Displacement:
Displacement occurs when particles in a body moves from initial state to a final state. If
the length of a line joining the two particles remains unaltered in the initial and the final
state, then the displacement is called rigid body displacement. If the displacement alters
this length then the final state of the body is said to be in “strained state” and the initial
state is called the “unstrained state”. Let x , y z , be the location of a point occupied by a
particle which in the strained state occupies a location: x + u , y + v z , + w , then
u v w , , are the projections of displacements of the particle. In simple uni-axial extension
along the x axis, displacement of a particle is given by u = ex v , = 0, w = 0 , where e is
the extension. In simple shear along the x axis, the planes parallel to the x axis slide past
each other so that particles in plane parallel to x , y remain in that plane. The
displacement is then given by u = sy v , = 0, w = 0 where s = 2 tan α, v = 0, w = 0.
Homogeneous strain: If the components of displacements are linear function of
coordinates, the strain is called homogeneous strain.
Analysis of Strain:
Deformation in a continuous body is defined as the change in the positions of the points
in the body so that their relative distances are altered. Say a point at ( x , y , z ) in the
undeformed / unstarined state of the body moves to a unique location (
x , y , z ). Then
the co-ordinates
x , y , z be a definite function of the coordinates x , y , z of the same
point before deformation:
x = f 1 (^) x y z , , , y = f 2 (^) x y z , , , z = f 3 x y z , , 1.
where functions f 1 (^) , f 2 (^) , f 3 are assumed to be continuous in the region occupied by the
body. Similarly, the coordinates x , y , z are also continuous functions of
x , y , z.
Affine Transformation :
The transformations of this form are called affine if the coordinates
x , y , z are linear
functions of the coordinates x , y , z
11 12 13
21 22 23
31 32 33
x a x a y a z a
y a x a y a z b
z a x a y a z c
where a 11 Kare constants. Above equations have nontrivial solution for x , y , z , so that
the determinant
11 12 13
21 22 23
31 32 33
a a a
D a a a
a a a
Properties: (a) Inverse affine transformation is also affine, because by solving 1.2 one
obtains x , y , z in terms of
x , y , z :
11 12 13
21 22 23
31 32 33
x b x b y b z
y b x b y b z
z b x b y b z
(b) Points lying on a plane ∏ before transformation lies on a different plane
∏. Say
Ax + By + cz + D = 0 is the equation of plane ∏ , then after the transformation they lie
on a different plane
A x + B y + c z + D = 0 , which is the equation of plane
∏.
(c) Points lying on a straight line, which is the intersection of two different planes 1
and 2 ∏ , will now lie again on a straight line, which is basically the intersection of two
planes
1 ∏ and
2 ∏. Hence, any straight segment is transformed into a straight segment
and any vector to a vector.
Let P =( ξ ,ψ ζ , )
ur
be the vector which transforms to a vector ( )
P = ξ , ψ ,ζ
uur
. If
x 0 , y 0 , z 0 and x , y , z be the end points of vector P =( ξ , ψ ζ, )
ur , then
ξ= x − x 0 , ψ= y = y 0 , ζ = z − z 0
Similarly
ξ = x − x 0 , ψ = y − y 0 , ζ = z − z 0 1.
Where ( )
x = 1 + a 11 x + a 12 y + a 13 z + a and ( )
x 0 (^) = 1 + a 11 (^) x 0 (^) + a 12 (^) y 0 (^) + a 13 0 z + a ,
etc. Subtracting one obtains
11 12 13
21 22 23
31 32 33
a a a
a a a
a a a
a 11 = a 22 = a 33 = 0 and ( a 12 + a 21 ) = ( a 13 + a 31 ) = ( a 23 + a 32 )= 0
which implies that aij = − aji. Then equation 1.8 can be written as,
δξ = q ζ − r ψ , δψ = r ξ − p ζ , δζ = p ψ − q ξ 1.
where p = a (^) 32 = − a (^) 23 , q = a 13 (^) = − a 31 (^) , r = a 21 (^) = − a 12. These quantities are
infinitesimal angles of rotation about the coordinate axes and are called the components
of rotation.
We can introduce following notations:
11 22 33
32 23 13 31 12 21
xx yy zz
yz zy xz zx xy yx
a e a e a e
a a e e a a e e a a e e
e ij are called the components of strain. Furthermore we can introduce the notations:
p = a − a q = a − a r = a − a
Hence we can divide the tensor aij in following symmetric and anti-symmetric parts:
32 13 21
23 31 12
yz zx xy
yz zx xy
a e p a e q a e r
a e p a e q a e r
Using these definitions of a (^) ij in equation 1.8, we have:
xx xy xz
yx yy yz
zx zy zz
e e e q r
e e e r p
e e e p q
δξ ξ ψ ζ ζ ψ
δψ ξ ψ ζ ξ ζ
δζ ξ ψ ζ ψ ξ
So the original affine transformation can be divided into two parts: symmetric and anti-
symmetric.
xx xy xz
yx yy yz
zx zy zz
e e e
e e e
e e e
ξ ψ ζ
ξ ψ ζ
ξ ψ ζ
q r
r p
p q
Geometric meaning of components of strain: Let us go back to equation 1.9, under
new notations it has the form:
2 2 2 2 2 2 xx yy zz xy xz yz P δ P = e ξ + e ψ + e ζ + e ξψ + e ξζ + e ψζ
Consider a vector P ( ξ,0,0)
ur which is parallel to the Ox axis. For this vector:
2 P δ P = e xx ξ which results in (^) xx
P
e P
δ = , i.e. ex (^) x denotes the relative increase in length
of the vectors.
Consider two vectors P 1 ( 0, ψ 1 ,0)
uur
and P 2 ( 0,0, ζ 2 )
uur : which are initially directed
towards y and z , after deformation turn to vectors (^) ( )
P 1 δξ ψ 1 , 1 +δψ 1 ,δζ 1
uur and
2 2 2 2 2
P δξ , δψ ,ζ +δζ
uur
, the angle 2
yz
between which is written as:
1 2 1 1 2 1 2 2
2 2 2 2 2 2 1 1 1 1 2 2 2 2
cos 2
yz
π δξ δξ^ ψ^ δψ^ δψ^ δζ^ ζ^ δζ
⎛ ⎞ +^ +^ +^ +
For a small ε yz ,
⎟=^ = ⎠
⎞ ⎜ ⎝
⎛ − ε yz ε yz
π
2
cos
1 2 1 2 2 1
1 2 2 1
But for P 1 ( 0, ψ 1 ,0)
uur
and P 2 ( 0,0, ζ 2 )
uur we have from 1.15:
2 2 2
1 1 1
yz
zy
e p
e p
Introducing this formula into equation 1.19, one obtains ε yz = e yz + ezy = 2 eyz.
The angle 2 e (^) yz represents the decrease in angle between the two vectors originally in the
directions O y and O z.
O B
B’
C
C
K
K
x
z
y
( ) (^) ( ) ( )
2 1 2 2 2 (^1 2) xx (^1 2) yy (^1 2) zz (^2) yx (^2) zx 2
ds l m n mn nl xylm ds
⎜ ⎟ =^ +^ +^ +^ +^ +^ +^ +^ +
where ε xx Kare the following,
2 2 2 1
xx
u u v w
x x x x
2 2 2
2 2 2
yy
zz
v u v w
y y y y
w u v w
z z z z
∂ ⎪⎛ ∂ ⎞^ ⎛^ ∂ ⎞^ ⎛^ ∂ ⎞ ⎪
yz
zx
xy
w v u u v v w w
y z y z y z y z
w u u u v v w w
x z x z x z x z
v u u u v v w w
x y x y x y x y
We thus obtain the general expressions for the components of strain in terns of the
gradients of displacements.
The left hand side of 1.24 is a constant and the right hand side signifies an ellipsoid. It
has the property that in any direction, the length of its radius is inversely proportional to
ds 1
ds
. Such an ellipsoid is called the reciprocal strain ellipsoid.
Transformation of components of strain:
Consider 1.17, here P δ P is a quadratic with respect to ξ , ψ ζ,. But since P δ P has a
definite meaning, it should be independent of the choice of coordinate axes or of the
transformation of coordinates. In other word, if (^) ' ' , , (^) ' ', x x x y
e K e K are the components of
strain in the new coordinate system and if
' ' '
ξ ,ψ ,ζ are the component of the vector P ,
then we have:
' ' ' ' ' ' ' ' ' ' ' '
2 2 2
'2 '2 '2 ' ' ' ' ' '
xx yy zz xy xz yz
x x y y z z x y x z y z
P P e e e e e e
e e e e e e
Now let’s say 1 1 1 l , m , n are the direction cosines of the x ' axis new coordinate system
with respect to the old one and like wise. We can then develop following table:
' 1 1 1
' 2 2 2
' 3 3 3
x y z
x l m n
y l m n
z l m n
The coordinate of the vector P in the old system can be expressed with respect to the
new one as,
' ' ' 1 1 1
' ' ' 2 2 2
' ' ' 3 3 3
l m n
l m n
l m n
Substituting this into 1.26 and then matching the coefficients for
'2 ' '
ξ , K, ξ ψ,K in the
left and right hand sides,
2 2 2 ' ' 1 1 1 1 1 1 1 1 1
' ' 2 3 2 3 2 3 2 3 3 3
2 3 3 2 2 3 3 2
x x xx yy zz yz zx xy
y z xx yy zz yz
zx xy
e e l e m e n e m n e n l e l m
e e l l e m m e n n e m n m n
e n l n l e l m l m
K
K
Furthermore we can rewrite 1.17 as,
2 2 2 2
P e = exx ξ + e yy ψ + ezz ζ + 2 exy ξψ + 2 exz ξζ + 2 eyz ψζ
where
P
e P
= is the relative increase in length of vector P =( ξ , ψ ζ, )
ur
. Since
2 P e does
not depend on any direction but only its magnitude, hence for every
direction,
2 2 P e = ± c , where c is an arbitrary constant with the dimension of length. If
the starting point of P lies on the origin, then its other end point lies on the surface
2 2 P e = ± c , or
2 2 2 2
e xx ξ + eyy ψ + ezz ζ + 2 e xy ξψ + 2 exz ξζ + 2 e yz ψζ = c
which is called the strain surface.
If the axes are so chosen that they coincide with the principal axes of the surface then,
1.32 takes the form:
2 2 2 2 1 2 3
e ξ + e ψ + e ζ = c , where
1 2 3 e , e , e are the , , xx yy zz e e e
of the new axes. Principal axes are the roots of the equation:
deformations. Now we put them all back to the body. Question is will the adjoining faces
and lines will perfectly match without any gaps, after this exercise. It will almost be
impossible to do that. Hence, it is obvious now that components of strain must satisfy
certain conditions in order to result in deformations without discontinuities.
Mathematically also we see the following: we have,
xx yy zz
xy yz zx
u v w e e e x y z
u v w v w u e e e y x y z x z
⎝ ∂^ ∂^ ⎠ ⎝ ∂^ ∂^ ⎠ ⎝^ ∂^ ∂ ⎠
v u r x y
⎝ ∂^ ∂ ⎠
,
u w q z x
⎛ ∂^ ∂ ⎞
⎝ ∂^ ∂ ⎠
and
w v p y z
⎝ ∂^ ∂ ⎠
where , , , xx xy e K e Kare single valued functions having continuous second order
derivatives. One has six equations and three unknown functions, which implies that the
problem may not have solutions if exx , K, exy ,K are not subjected to additional
conditions.
Let V be simply connected region occupied by a body. In this domain ( x , y z , ) is the
current location of a point at which e (^) xx , K, exy ,K are given and we need to find out the
deformations. Let M 0 ( x 0 , y 0 , z 0 )be any point in V and u 0 , v 0 , w 0 are the components
of displacement at M 0 and p 0 , q 0 , r 0 are the components of rotation. Let M 1 ( x y z , , )be
any other point at V at which we want to find out the components of displacements. Let
M (^) 0 M 1 be a line that joins M (^) 0 and M (^) 1 and lies in V. Then if the partial derivatives
u u u
x y z
are known, then the displacements u 1 (^) , v 1 (^) , w 1 can be found out as,
0 1
1 0
M M
u u u u u dx dy dz x y z
⎝ ∂^ ∂^ ∂ ⎠
∫
where (^) xx , (^) xy , xz
u u u e e r e q x y z
Hence, (^) ( ) ( )
0 1 0 1
1 0 xx xy xz
M M M M
u = u + e dx + e dy + e dz + qdz − rdy ∫ ∫
Let us focus on the second integral:
By partial integration:
( ) (^) ( ( ) ( ))
( ) ( ) (^) (( ) ( ) )
0 1 0 1
1 1
0 0 0 1 0 1
0 1
1 1
1 1 1 1
0 1 0 0 1 0 1 1
M M M M
y z
y z M M M M
M M
qdz rdy rd y y qd z z
r y y y y dr q z z z z dq
q z z r y y y y dr z z dq
∫ ∫
∫ ∫
∫
Now
v u r x y
⎝ ∂^ ∂ ⎠
, so that,
2 2 2 2 2 2
2 2
2 2 2 2 2 2
2 2
2 2
xy (^) xx
yy xy
r v u v u u u e^ e
x (^) x y x (^) x y x y x y x x y
r v u v v u v e^ e
y x y (^) y x y x y (^) y y x x y
r v u
z x z y z
∂ ⎛^ ∂ ∂ ⎞^ ⎛^ ∂ ∂ ∂ ∂ ⎞^ ∂^ ∂
∂ ⎜^ ∂ ∂ ⎟^ ⎜^ ∂ ∂ ∂ ∂ ∂ ∂ ⎟ ∂ ∂
∂ ⎛^ ∂ ∂ ⎞^ ⎛^ ∂ ∂ ∂ ∂ ⎞^ ∂^ ∂
∂ ⎜^ ∂ ∂ ⎟^ ⎜^ ∂ ∂ ∂ ∂ ∂ ∂ ⎟ ∂ ∂
∂ ⎛^ ∂ ∂ ⎞
∂ ⎜^ ∂ ∂ ∂ ∂ ⎟
2 2 2 2
v w u w e yz^ exz
x z y x y z y x x y
⎜ +^ −^ −^ ⎟=^ −
1.39a
And
u w q z x
⎛ ∂^ ∂ ⎞
⎝ ∂^ ∂ ⎠
, so that
2 2 2 2 2 2
2 2
2 2 2 2 2 2
2 2
2
xx xz
xy zy
q u w u u w u e e
x x z (^) x x z z x (^) x z x z x
q u w u v w v e^ e
y z y y x z y z x y x z x z x
q u w
z (^) z x z
∂ ⎛^ ∂ ∂ ⎞^ ⎛^ ∂ ∂ ∂ ∂ ⎞^ ∂ ∂
∂ ⎜^ ∂ ∂ ⎟^ ⎜^ ∂ ∂ ∂ ∂ ∂ ∂ ⎟ ∂ ∂
∂ ⎛^ ∂ ∂ ⎞^ ⎛^ ∂ ∂ ∂ ∂ ⎞^ ∂^ ∂
∂ ⎜^ ∂ ∂ ∂ ∂ ⎟^ ⎜^ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ⎟ ∂ ∂
∂ ⎛^ ∂ ∂ ⎞
∂ ⎜^ ∂ ∂ ⎟
2 2 2 2
2
u w w w e xz ezz
z z x^ x z^ x z^ z^ x
⎜ +^ −^ −^ ⎟=^ −
1.39b
Now substituting these expressions into
2 2 2 2 2 2 2
2 2 2
2 2 2 2 2 2 2
2 2 2
2 2 2 2 2 2 2
2 2 2
yy (^) zz yz (^) xx yz (^) zx xy
zz xx zx yy^ zx xy^ yz
xx yy^ xy^ zz xy^ yz zx
e (^) e e (^) e e (^) e e
z y y z^ y z^ x x y^ x z
e e e e^ e e^ e
x z z x^ z x^ y y z^ y x
e e^ e^ e e^ e e
y x x y^ x y^ z z x^ z
∂ ∂ ∂ ∂^ ∂ ∂^ ∂ ∂ ∂^ ∂ ∂
∂ ∂ ∂ ∂^ ∂ ∂^ ∂
∂ ∂ ∂ ∂^ ∂ ∂^ ∂ ∂ ∂^ ∂ ∂
∂ ∂^ ∂^ ∂ ∂^ ∂ ∂
∂ ∂ ∂ ∂^ ∂ ∂^ ∂ ∂ ∂^ ∂ ∂ y
These equations are called the conditions of compatibility of Barré de Saint-Venant.
