A rectangular beam (again b × d ), but with: [latex]T = T_{o} × \frac {2γ}{d}[/latex] It is constrained so that [latex]\bar{\epsilon } = 0[/latex] and 1/R = 0. Determine the stresses and restraints.

Axial force equilibrium (equation (3.94)) [latex]P = E \bar{\varepsilon } A - E\alpha \int_{A}^{}{TdA}[/latex] [latex]\int_{A}^{}{TdA} = \frac{2T_{o}b}{d} \int_{-d/2}^{d/2}{\gamma d\gamma } = \frac{2T_{o}b}{d} \left[\frac{\gamma ^{2}}{2} \right] _{-d/2}^{d/2} = 0[/latex] Also, [latex]\bar{\varepsilon } = 0[/latex] ∴ P = 0 Moment equilibrium (equation (3.95)) [latex] M = \frac{EI}{R} - E\alpha \int_{A}^{}{T\gamma dA}[/latex] [latex]\int_{A}^{}{T\gamma dA} = \frac{2T_{o}b}{d} \int_{-d/2}^{d/2}{\gamma^{2} d A } = \frac{2T_{o}b}{d} \left[\frac{\gamma ^{3}}{3} \right] _{-d/2}^{d/2}[/latex] [latex]= \frac{2T_{o}}{d} \left[\left(\frac{d^{3}}{24} \right) - \left(\frac{- d^{3}}{24} \right) \right] = \frac{T_{o}bd^{2}}{6}[/latex] Also, 1/R = 0, ∴ [latex]M = \frac{- E\alpha bd^{2}T_{o}}{6} [/latex] Using equation (3.93) (with [latex] \bar{\varepsilon } = (1/R = 0)[/latex]) [latex] \sigma_{x} = \left(\bar{\varepsilon } + \frac{\gamma }{R} - \alpha T \right) = - E\alpha \cdot T[/latex] ∴ [latex] \sigma_{x} = - E\alpha \cdot T_{o} \frac{2\gamma }{d} [/latex] i.e. [latex] \sigma_{x} = \frac{- 2E\alpha T_{o}}{d}\gamma [/latex]

Question 3.17: A rectangular beam (again b × d ), but with: T = To × 2γ/d I...

An Introduction to Mechanical Engineering

A rectangular beam (again b × d ), but with: $T = T_{o} × \frac {2γ}{d}$ It is constrained so that $\bar{\epsilon } = 0$ and 1/R = 0. Determine the stresses and restraints.

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Axial force equilibrium (equation (3.94))

P = E \bar{\varepsilon } A  –  E\alpha \int_{A}^{}{TdA}

\int_{A}^{}{TdA} = \frac{2T_{o}b}{d} \int_{-d/2}^{d/2}{\gamma d\gamma } = \frac{2T_{o}b}{d} \left[\frac{\gamma ^{2}}{2} \right] _{-d/2}^{d/2} = 0

Also,

$\bar{\varepsilon } = 0$ ∴ P = 0

Moment equilibrium (equation (3.95))

M = \frac{EI}{R}  –  E\alpha \int_{A}^{}{T\gamma dA}

\int_{A}^{}{T\gamma dA} = \frac{2T_{o}b}{d} \int_{-d/2}^{d/2}{\gamma^{2} d A } = \frac{2T_{o}b}{d} \left[\frac{\gamma ^{3}}{3} \right] _{-d/2}^{d/2}

= \frac{2T_{o}}{d} \left[\left(\frac{d^{3}}{24} \right)  –  \left(\frac{-   d^{3}}{24} \right) \right] = \frac{T_{o}bd^{2}}{6}

Also,

1/R = 0, ∴ $M = \frac{- E\alpha bd^{2}T_{o}}{6}$

Using equation (3.93) (with $\bar{\varepsilon } = (1/R = 0)$ )

$\sigma_{x} = \left(\bar{\varepsilon } + \frac{\gamma }{R} – \alpha T \right) = – E\alpha \cdot T$
∴ $\sigma_{x} = – E\alpha \cdot T_{o} \frac{2\gamma }{d}$

i.e.