Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(Darren Dugan) #1

TENSORS


Further, Poisson’s ratio is defined asσ=−e 22 /e 11 (or−e 33 /e 11 ) and is thus

σ=

(


1


e 11

)


λθ
2 μ

=


(


1


e 11

)(


λ
2 μ

)


Ee 11
3 λ+2μ

=


λ
2(λ+μ)

. (26.49)


Solving (26.48) and (26.49) forλandμgives finally


pij=

σE
(1 +σ)(1− 2 σ)

ekkδij+

E


(1 +σ)

eij.

26.13 Integral theorems for tensors


In chapter 11, we discussed various integral theorems involving vector and scalar


fields. Most notably, we considered the divergence theorem, which states that, for


any vector fielda, ∫


V

∇·adV=


S

a·nˆdS , (26.50)

whereSis the surface enclosing the volumeVandnˆis the outward-pointing unit


normal toSat each point.


Writing (26.50) in subscript notation, we have

V

∂ak
∂xk

dV=


S

aknˆkdS. (26.51)

Although we shall not prove it rigorously, (26.51) can be extended in an obvious


manner to relate integrals oftensor fields, rather than just vector fields, over


volumes and surfaces, with the result


V

∂Tij···k···m
∂xk

dV=


S

Tij···k···mˆnkdS.

This form of the divergence theorem for general tensors can be very useful in


vector calculus manipulations.


A vector fieldasatisfies∇·a=0inside some volumeVanda·ˆn=0on the bound-
ary surface∫ S. By considering the divergence theorem applied toTij =xiaj, show that
VadV=^0.

Applying the divergence theorem toTij=xiajwe find


V

∂Tij
∂xj

dV=


V

∂(xiaj)
∂xj

dV=


S

xiajnˆjdS=0,

sinceajˆnj= 0. By expanding the volume integral we obtain


V

∂(xiaj)
∂xj

dV=


V

∂xi
∂xj

ajdV+


V

xi

∂aj
∂xj

dV

=



V

δijajdV

=



V

aidV=0,

where in going from the first to the second line we used∂xi/∂xj=δijand∂aj/∂xj=0.

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