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CONTINUUM MECHANICS for ENGINEERS

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In this equation, the first term on the right-hand side gives the change<br />

occurring in the property at position x, known as the local rate of change; the<br />

second term results from the particles changing position in space and is<br />

referred to as the convective rate of change. Since by Eq 4.2-12 the velocity is<br />

defined as v = dx/dt (or v k = dx k/dt), Eq 4.5-3 may be written as<br />

d<br />

∂<br />

∂<br />

[ Pij... ( x, t)] = [ Pij... ( x, t)]<br />

+ [ Pij... ( x,<br />

t)] v<br />

dt ∂ t<br />

∂ x<br />

(4.5-4)<br />

from which we deduce the material derivative operator <strong>for</strong> properties expressed<br />

in the spatial description<br />

d<br />

= vk<br />

dt t x<br />

∂<br />

∂ +<br />

∂<br />

∂<br />

d<br />

or = (4.5-5)<br />

dt t<br />

∂<br />

+ v⋅ ∂<br />

The first <strong>for</strong>m of Eq 4.5-5 is <strong>for</strong> rectangular Cartesian coordinates, while the<br />

second <strong>for</strong>m is coordinate-free. The del operator ( )<br />

will always indicate<br />

partial derivatives with respect to the spatial variables unless specifically<br />

stated.<br />

Example 4.5-1<br />

Let a certain motion of a continuum be given by the component equations,<br />

x 1 = X 1e –t , x 2 = X 2e t , x 3 = X 3 + X 2(e –t – 1)<br />

and let the temperature field of the body be given by the spatial description,<br />

θ = e –t (x 1 – 2x 2 + 3x 3)<br />

Determine the velocity field in spatial <strong>for</strong>m, and using that, compute the<br />

material derivative dθ/dt of the temperature field.<br />

Solution<br />

Note again here that the initial configuration serves as the reference configuration<br />

so that Eq 4.2-9 is satisfied. When Eq 4.5-2 is used, the velocity<br />

components in material <strong>for</strong>m are readily determined to be<br />

v 1 = –X 1e –t , v 2 = X 2e t, v 3 = –X 2e –t<br />

Also, the motion equations may be inverted directly to give<br />

k<br />

X 1 = x 1e t , X = x 2e –t , X 3 = x 3 – x 2(e –2t – e –t )<br />

k<br />

k

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