Question 3.16: A rectangular beam, width b and depth d has a temperature va...
A rectangular beam, width b and depth d has a temperature variation given by:
T = T_{o} \left\{1 – \frac{4\gamma ^{2}}{d^{2}} \right\}
There is no restraint or applied loading (i.e. P = M = 0). Obtain the stress distribution.
Axial force equilibrium (equation (3.94))
0 = E \bar{\varepsilon }\cdot bd – E\alpha \int_{\frac{-d}{2} }^{d/2}{T_{o} \left\{1 – \frac{4\gamma ^{2}}{d^{2}} \right\} (bd\gamma )}
∴ \bar{\varepsilon } = \frac{\alpha }{d} T_{o} \int_{\frac{-d}{2} }^{d/2}{T_{o} \left\{1 – \frac{4\gamma ^{2}}{d^{2}} \right\} d\gamma }
i.e. \bar{\varepsilon } = \frac{\alpha }{d} T_{o} \left[\gamma – \frac{4\gamma ^{3}}{3d^{2}}\right]_{-d/2}^{d/2}
∴ \bar{\varepsilon } = \frac{2}{3} \alpha \cdot T_{o}
With M = O we can obtain 1/R from the moment equilibrium (equation (3.95)) but from symmetry we can see that (1/R) = O.
Using equation (3.93)
= E \left(\frac{2}{3}\alpha T_{o} + O – \alpha T_{o} \left(1 – \frac{4\gamma ^{2}}{d^{2}} \right) \right)
= E\alpha T_{o} \left(\frac{4\gamma ^{2}}{d^{2}} – \frac{1}{3} \right)
At y = O, \sigma _{x} = \frac{- E\alpha T_{o}}{3}
At = \frac{\pm d}{2}, \sigma _{x} = E\alpha T_{o} \left(1 – \frac{1}{3} \right) = \frac{2E\alpha T_{o}}{3}
\sigma _{x} = 0 When \frac{4\gamma ^{2}}{d^{2}} = \frac{1}{3} i.e. at γ = ±0.287 d
This is the stress distribution away from the ends.
At the ends, \sigma _{x} = 0 and there is a gradual transition between the two.
