Past exam of the solid mechanics course of the University of Oxford / 2011 / 2 / b

Solution

The surface element relation is Nanson's formula. The force on a corresponding pair of current and reference surface elements must agree:

t d a = (S \circ y) n d a = (S \circ y) cof F N d A = TN d A .

T = (S \circ y) cof F = J (S \circ y) F^{- T} .

It measures force per reference area, with a current force direction and a reference normal direction.

In components, differentiate

T_{i A} = S_{ij} (y (x)) (cof F)_{j A}

\partial_{A} (cof Dy)_{j A} = 0

. One proof writes the cofactor matrix as

\frac{1}{2} ε_{jk l} ε_{A BC} y_{k, B} y_{l, C}

; differentiating in

A

produces symmetric second derivatives contracted against antisymmetric epsilon symbols, hence zero. The chain rule consequently gives

\partial_{A} T_{i A} = \frac{\partial S _{ij}}{\partial y _{k}} (y) F_{k A} (cof F)_{j A} = J \frac{\partial S _{ij}}{\partial y _{j}} (y),

since

F_{k A} (cof F)_{j A} = J δ_{kj}

. Thus, with the coordinate composition made explicit,

Div_{x} T = J (div_{y} S) \circ y .

The distinction between reference divergence and current divergence is essential to this identity.

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