Problem 5.1. Uniform temperature change

Normal stress in the bar, σ [N/mm²]=

Follow the steps of Problem 3.1.

Step 1. Calculate the elongation of the bar from the temperature change removing the left support.
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Eq.(5-1)

$\epsilon =\alpha ∆T=1.2\times {10}^{–5}\times 50=6\times {10}^{–4} (\sigma =0)$

Eq. (5-2)

$∆L=L\epsilon =600\times 6\times {10}^{–4}=0.360 \mathrm{mm}$

Step 2. Displacement is restricted by the left support. Determine the force which would cause the restricted displacement at the end of the cantilever (the difference between the displacement of the free end and the distance to the support).

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Step 3. Give the stress in the bar.

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Follow the steps of Problem 3.1.

First the we remove the left support. The elongation of the bar from the temperature is the following.

Eq.(5-1)

$\epsilon =\alpha ∆T=1.2\times {10}^{–5}\times 50=6\times {10}^{–4} (\sigma =0)$

Eq.(5-2)

$∆L=L\epsilon =600\times 6\times {10}^{–4}=0.360 \mathrm{mm}$

This displacement is restricted by the left support. Now the force is determined which would cause the restricted displacement at the end of the cantilever (the difference between the displacement of the free end and the distance to the support is ΔL – 0.2 mm).  The normal force applied at the end is calculated from$u=∆L–0.2=\frac{N}{EA}L \to N=\left(∆L–0.2\right)\frac{EA}{L}=\left(0.36–0.2\right)\frac{210\times 1000\times {20}^2\pi /4}{600}=17.59 \mathrm{kN}$

Superposition of the above two cases results in 0.2 mm elongation at the end of the rod, stress arises from the compressive force only:

$\mathit{\sigma }=\frac{N}{A}=\frac{17.6\times {10}^3}{{20}^2\pi /4}\mathbf{=}\mathbf{56}\mathbf{.}\mathbf{0}\mathbf{ }\frac{\mathbf{N}}{{\mathbf{mm}}^{\mathbf{2}}}\mathbf{ }\mathbf{\left(}\mathbf{compression}\mathbf{\right)}$