 Solution

ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2010/3/b/solution

Contract the tensors, counting both off-diagonal shear terms:

W = \frac{1}{2} τ_{ij} e_{ij} = \frac{E z ^{2}}{2 ( 1 - ν ^{2} )} {a^{2} + 2 ν a c + c^{2} + 2 (1 - ν) b^{2}} .

Since

\int_{- h /2}^{h /2} z^{2} d z = h^{3} /12

, integration gives

\int_{- h /2}^{h /2} W d z = \frac{D}{2} {a^{2} + 2 ν a c + c^{2} + 2 (1 - ν) b^{2}}, D = \frac{E h ^{3}}{12 ( 1 - ν ^{2} )} .

For the general midplane deflection, expand the energy density as

w_{xx}^{2} + 2 ν w_{xx} w_{yy} + w_{yy}^{2} + 2 (1 - ν) w_{x y}^{2}

w + ε η

gives

δ U = D \iint_{Ω} {(w_{xx} + ν w_{yy}) η_{xx} + (w_{yy} + ν w_{xx}) η_{yy} + 2 (1 - ν) w_{x y} η_{x y}} d x d y .

For clamped variations,

η = \partial_{n} η = 0

on the boundary, so integrating twice transfers all derivatives to

w

without a boundary contribution. The bulk coefficient is

w_{xxxx} + 2 ν w_{xx yy} + w_{yyyy} + 2 (1 - ν) w_{xx yy} = (\partial_{x}^{2} + \partial_{y}^{2})^{2} w .

Stationarity for all compactly supported variations and the fundamental lemma therefore give

\nabla^{4} w = 0 in the plate interior .

 New to topics? Read the docs here!