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ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2009/2/b/iii/solution

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Past exam of the elasticity and plasticity course of the University of Oxford / 2009 / 2 / b / iii / Solution by

Codex  0 2026-10-31

Multiplication of the preceding displacement formula by

- ik

gives

2 μ v_{x} = P [- ik y e^{- q y} - 2 i (1 - ν) \frac{k}{q} e^{- q y}] .

For the supplied kernels,

f_{1, x} = - ik e^{- q y}

and

f_{2} = i (k / q) e^{- q y}

. The convolution theorem therefore gives

2 μ v_{x} = P * [y f_{1, x} - 2 (1 - ν) f_{2}] .

Since

f_{1, x} (s, y) = - 2 ys / [π (s^{2} + y^{2})^{2}]

and

f_{2} (s, y) = s / [π (s^{2} + y^{2})]

, combine the two fractions and divide by two:

μ v_{x} (x, y) = - \frac{1}{π} \int_{- \infty}^{\infty} \frac{( 1 - ν ) s ^{2} + ( 2 - ν ) y ^{2}}{( s ^{2} + y ^{2} ) ^{2}} P (x - s) s d s .

The sign is tied to the positive forward Fourier phase used here. This is a displacement derivative, so an additive displacement reference drops out. At

y > 0

the kernel is regular near

s = 0

; the usual decay/integrability of the applied load controls the tails. A boundary-limit singular integral, if wanted at

y = 0

, would require its own principal-value interpretation.

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