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=
d det(S + U )
=0

d
’1
= det(S + U )
(det S)2 d =0

= ’ det(S ’1 )(S ’T : U )
An alternative route can be followed to reach the same result by us-
ing the chain rule Equation (2.105c) and both Equations (2.119) and
(2.124) to give,
D det(S ’1 )[U ] = det(S ’1 )(S T : DS ’1 [U ])
= ’ det(S ’1 )(S T : (S ’1 U S ’1 ))
= ’ det(S ’1 )(S ’T : U )




2.4 TENSOR ANALYSIS

Section 2.2 dealt with constant vectors and tensors. In contrast, such items
in continuum mechanics invariably change from point to point throughout
a problem domain. The resulting magnitudes are known as ¬elds in a three-
dimensional Cartesian space and can be of a scalar, vector, or tensor nature.
Examples of scalar ¬elds are the temperature or density of a body. Alter-
35
2.4 TENSOR ANALYSIS



natively, the velocity of the body particles would constitute a vector ¬eld
and the stresses a tensor ¬eld. The study of these quantities requires opera-
tions such as di¬erentiation and integration, which are the subject of tensor
analysis.


2.4.1 THE GRADIENT AND DIVERGENCE OPERATORS
Consider ¬rst a scalar ¬eld, that is, a function f (x) that varies throughout
a three-dimensional space. At a given point x0 , the change in f in the
direction of an arbitrary incremental vector u is given by a vector f (x0 )
known as the gradient of f at x0 , which is de¬ned in terms of the directional
derivative as,
f (x0 ) · u = Df (x0 )[u] (2.125)
The components of the gradient can be obtained by using the de¬nition of
the directional derivative (2.101) in the above equation to give,
d
f (x0 ) · u = f (x0 + u)
d =0
3
d(x0,i + ui )
‚f
=
‚xi d
=0 =0
±=1
3
‚f
= ui (2.126)
‚xi xi =x0,i
±=1

Hence the components of the gradient are the partial derivatives of the
function f in each of the three spatial direction as,
3
‚f
f= (2.127)
ei
‚xi
±=1

For obvious reasons, the following alternative notation is frequently used,
‚f
f= (2.128)
‚x
The gradient of a vector ¬eld v at a point x0 is a second-order tensor
v(x0 ) that maps an arbitrary vector u into the directional derivative of v
at x0 in the direction of u as,
v(x0 ) u = Dv(x0 )[u] (2.129)
A procedure identical to that employed in Equation (2.126) shows that the
components of this gradient tensor are simply the partial derivatives of the
36 MATHEMATICAL PRELIMINARIES



vector components, thereby leading to the following expression and useful
alternative notation,
3
‚vi ‚v
v= ei — ej ; v= (2.130)
‚xj ‚x
±,β=1

The trace of the gradient of a vector ¬eld de¬nes the divergence of such
a ¬eld as a scalar, ·v, which can be variously written as,
3
‚vi
·v = tr v = v:I= (2.131)
‚xi
±=1

Similarly to Equation (2.129), the gradient of a second-order tensor S
at x0 is a third-order tensor S(x0 ), which maps an arbitrary vector u to
the directional derivative of S at x0 as,

S(x0 ) u = DS(x0 )[u] (2.132)

Moreover, the components of S are again the partial derivatives of the
components of S and consequently,
3
‚Sij ‚S
S= ei — ej — ek ; S= (2.133)
‚xk ‚x
i,j,k=1

Additionally, the divergence of a second-order tensor S is the vector ·S,
which results from the double contraction of the gradient S with the
identity tensor as,
3
‚Sij
·S = S:I= (2.134)
ei
‚xj
±,β=1

Finally, the following useful properties of the gradient and divergence
are a result of the product rule,

(f v) = f v+v— f (2.135a)
·(f v) = f · v + v · f (2.135b)
(v · w) = ( v)T w + ( w)T v (2.135c)
·(v — w) = v · w + ( v)w (2.135d)
·(S T v) = S : v+v· ·S (2.135e)
·(f S) = f · S + S f (2.135f)
(f S) = f S+S— f (2.135g)
37
2.4 TENSOR ANALYSIS




EXAMPLE 2.10: Proof of Equation (2.135e)
Any one of Equations (2.135a“g) can be easily proved in component
form with the help of the product rule. For example using Equations
(2.131) and (2.134) gives (2.135e) as,
3

T
·(S v) = (Sij vi )
‚xj
i,j=1
3
‚Sij
‚vi
= Sij + vi
‚xj ‚xj
i,j=1
=S : v+v· ·S



2.4.2 INTEGRATION THEOREMS
Many derivations in continuum mechanics are dependant upon the ability to
relate the integration of ¬elds over general volumes to the integration over
the boundary of such volumes. For this purpose, consider a volume V with
boundary surface ‚V and let n be the unit normal to this surface as shown
in Figure 2.7. All integration theorems can be derived from a basic equation
giving the integral of the gradient of a scalar ¬eld f as,

f dV = f n dA (2.136)
V ‚V
Proof of this equation can be found in any standard text on calculus.
Expressions similar to Equation (2.136) can be obtained for any given
vector or tensor ¬eld v by simply using Equation (2.136) on each of the
components of v to give,

v dV = v — n dA (2.137)
V ‚V
A more familiar expression is obtained by taking the trace of the above
equation to give the Gauss or divergence theorem for a vector ¬eld v as,

·v dV = v · n dA (2.138)
V ‚V

Similarly, taking the trace of Equation (2.137) when v is replaced by a
second-order tensor S and noting that, as a result of Equation (2.72c),
(S — n) : I = Sn, gives,

·S dV = Sn dA (2.139)
V ‚V
38 MATHEMATICAL PRELIMINARIES



x3

n




dA ‚V




x1 x2
V



FIGURE 2.7 General volume and element of area.



EXAMPLE 2.11: Volume of a Three-Dimensional body
The volume of a three-dimensional body is evaluated by the integral,

V= dV
V
Using Equation (2.138) it is possible and often useful to rewrite this
volume in terms of an area integral. For this purpose note ¬rst that
the divergence of the function v(x) = x/3 is 1 and therefore Equa-
tion (2.138) gives,
1
V= x · n dA
3 ‚V



Exercises
1. The second-order tensor P maps any vector u to its projection on a plane
passing through the origin and with unit normal a. Show that:
Pij = δij ’ ai aj ; P = I ’a—a
Show that the invariants of P are IP = IIP = 2, IIIP = 0, and ¬nd the
eigenvalues and eigenvectors of P .
2. Using a procedure similar to that employed in Equations (2.41“42), ob-
tain transformation equations for the components of third- and fourth-
order tensors in two sets of bases ei and ei that are related by the 3-D
transformation tensor Q with components Qij = ei · ej .
39
2.4 TENSOR ANALYSIS



3. Given any second-order tensor S linearize the expression S 2 = SS in the
direction of an increment U .
4. Consider a functional I that when applied to the function y(x) gives the
integral:
b
I(y(x)) = f (x, y, y ) dx
a
where f is a general expression involving x, y(x) and the derivative
y (x) = dy/dx. Show that the function y(x) that renders the above
functional stationary and satis¬es the boundary conditions y(a) = ya
and y(b) = yb is the solution of the following Euler“Lagrange di¬erential
equation:
d ‚f ‚f
=0

dx ‚y ‚y
5. Prove Equations (2.135a“g) following the procedure shown in Exam-
ple 2.10.
6. Show that the volume of a closed 3-D body V is variously given as,

V= nx dA = ny dA = nz dA
‚V ‚V ‚V
where nx , ny and nz are the x, y and z components of the unit normal n.
CHAPTER THREE

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