Past exam of the elasticity and plasticity course of the University of Oxford / 2010 / 3

Table of contents
- a 3
  - Solution a
- b 3
  - Solution b
- c 3
  - Solution c

a

Solution

Differentiate the displacement components. The shear strains involving

z

vanish because

u_{z} = - w_{x}

and

v_{z} = - w_{y}

. The full infinitesimal strain tensor is

E = z - a - b 0 - b - c 0 00 \frac{ν ( a + c )}{1 - ν} .

In particular

tr E = (2 ν - 1) (a + c) z / (1 - ν)

. Substituting the Lamé coefficients into the constitutive law gives

τ = - \frac{E z}{1 - ν ^{2}} a + ν c (1 - ν) b 0 (1 - ν) b c + ν a 0 000 .

For example

τ_{zz} = λ tr E + 2 μ e_{zz} = 0

: the thickness-direction strain supplies exactly the plane-stress Poisson correction. Also

τ_{x y} = - E b z / (1 + ν)

, confirming the off-diagonal factor.

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 .

c

Solution

Use the same first variation on the half-plane, with no restrictions on the edge traces of

η

. Write

M = w_{xx} + ν w_{yy}

. Two

x

integrations of the

M η_{xx}

term produce the boundary term

- M η_{x} + M_{x} η

at

x = 0

. The mixed term, after one integration in

x

and one along the edge in

y

, contributes

2 (1 - ν) w_{x yy} η

. All other boundary terms vanish at infinity. Hence

δ U = D \iint_{x > 0} η \nabla^{4} w d x d y - D \int_{- \infty}^{\infty} {η_{x} (w_{xx} + ν w_{yy}) - η [w_{xxx} + (2 - ν) w_{x yy}]}_{x = 0} d y .

The bulk equation makes its area term zero. Stationarity therefore gives exactly

\int_{- \infty}^{\infty} {η_{x} (w_{xx} + ν w_{yy}) - η [w_{xxx} + (2 - ν) w_{x yy}]}_{x = 0} d y = 0.

The two edge traces can be prescribed independently: use

η (x, y) = χ (x) [f (y) + xg (y)]

with

χ = 1

near zero and compactly supported smooth

f, g

. The fundamental lemma on the edge then gives

w_{xx} + ν w_{yy} = 0, w_{xxx} + (2 - ν) w_{x yy} = 0 (x = 0) .

They are zero bending moment and zero effective transverse shear at a free straight edge. The second includes the edge derivative of the twisting moment; omitting the mixed contribution would give the wrong coefficient.

 Ancestors (10)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version