## Related questions with answers

In the steel bars $BE$ and $AD$ each have a $6 \times 18-\mathrm{mm}$ cross section. Knowing that $E=200 \mathrm{~GPa}$, determine the deflections of points $A,B$, and $C$ of the rigid bar $ABC$., the $3.2-\mathrm{kN}$ force caused point $C$ to deflect to the right. Using $\alpha=11.7 \times 10^{-6} /{ }^{\circ} \mathrm{C}$, determine (a) the overall change in temperature that causes point $C$ to return to its original position, (b) the corresponding total deflection of points $A$ and $B$.

Solution

VerifiedLet us list down the given data from Problem 9-77.

Given:

- Cross-sectional area: $A_{\text{BE}} = A_{\text{AD}} = 6\times18\ \text{mm}$
- Modulus of elasticity: $E = 200\ \text{GPa} = 200\times10^{3}\ \text{MPa}$
- Length of AB: $L_{\text{AB}} = 75\ \text{mm}$
- Length of BC: $L_{\text{BC}} = 300\ \text{mm}$
- Length of EB: $L_{\text{EB}} = 400\ \text{mm}$
- Length of AD: $L_{\text{AD}} = 400\ \text{mm}$
- Load: $P = 3.2\ \text{kN} = 3200\ \text{N}$
- Thermal coefficient: $\alpha = 11.7 \times 10^{-6}\ ^{\circ}\text{C}^{-1}$

Required:

a.) Change in temperature needed to revert C to its initial point, $\Delta T$

b.) Deflection at A and B due to this change in temperature, $\delta_{\text{A}}$ and $\delta_{\text{B}}$

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