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ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2010/1/b/solution

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Past exam of the elasticity and plasticity course of the University of Oxford / 2010 / 1 / b / Solution by

Codex  0 2026-10-31

A Cartesian Cartesian second-rank tensor has components transforming under a constant orthogonal change of axes as

T_{ij}^{'} = p_{ia} p_{jb} T_{ab}, T^{'} = PT P^{T} .

(1)

Here

X^{'} = P X

and

u^{'} = P u

. By the chain rule and orthogonality,

\frac{\partial u _{i}^{'}}{\partial X _{j}^{'}} = p_{ia} \frac{\partial u _{a}}{\partial X _{b}} \frac{\partial X _{b}}{\partial X _{j}^{'}} = p_{ia} p_{jb} \frac{\partial u _{a}}{\partial X _{b}} .

(2)

Thus the displacement gradient is a Cartesian second-rank tensor. Its symmetric part transforms in the same way:

e_{ij}^{'} = \frac{1}{2} p_{ia} p_{jb} (\frac{\partial u _{a}}{\partial X _{b}} + \frac{\partial u _{b}}{\partial X _{a}}) = p_{ia} p_{jb} e_{ab} .

(3)

Hence the infinitesimal strain tensor is also second-rank. This is a passive coordinate rotation, under which both reference and spatial coordinates rotate; it does not assert finite-deformation frame indifference of the linearized strain formula.

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