Part (a): $ABCD$ is a rigid L-shaped member, 20 mm in thickness, supported by a pinned connection at $C$ and two axial force members $BF$ and $DE$.

Member $BF$ has length 260 mm and diameter 32 mm.

Member $DE$ has length 175 mm and diameter 16 mm.

Both steel members have modulus of elasticity $E=207\,\text{GPa}$ and coefficient of thermal expansion $11.7\times 10^{-6}/^\circ{}C$.

At $5^\circ{}C,\; ABC$ is horizontal and there is no stress in either of the steel axial members.

Determine the vertical deflection at $A$ when the temperature of the bars $BF$ and $DE$ is increased to $25^\circ{}C$.

Try each of the following steps before viewing the solution ;)
1. Take moments about $C$ to find a relationship between the forces in $BF$ and $DE$.
Assuming both members are in tension ... \begin{align} \Sigma M_C &= F_{DE}(175\,\text{mm})-F_{BF}(300\,\text{mm}) = 0\\\\ F_{DE} &= 1.7143 F_{BF} \end{align}
1. From the geometry of the deformation, find a relationship between $\delta_{BF}$ and $\delta_{DE}$
Let the size of the displacement at $B$ be $BB'$ and the size of the displacement at $D$ be $DD'$. Then, by similar triangles, \begin{align*} \frac{BB'}{300\,\text{mm}} &= \frac{DD'}{175\,\text{mm}} \\\\ BB' &= 1.7143 DD' \end{align*} Suppose that member $BF$ gets longer (which means that $DE$ must get shorter). Then, $\delta_{BF} = BB' \text{ and } \delta_{DE} = -DD'$ The negative sign is important! It follows from the two results above that: $\delta_{BF} = - 1.7413\delta_{DE}$
1. Use the two results $F_{DE} = 1.7143 F_{BF}$ and $\delta_{BF} = - 1.7143\delta_{DE}$ to find the forces in the two axial members.
\begin{align} \delta_{BF} &= \frac{F_{BF}(260\,\text{mm})}{\frac{\pi}{4}(32\,\text{mm})^2\times 207\thinspace 000\,\mathsf{N/mm^2}} + 11.7\times 10^{-6}/^\circ{}C\times (260\,\text{mm})\cdot 20^\circ{}C\\\\ &= 1.5618\times 10^{-6}\,F_{BF}\,\text{mm} + 0.06084\,\text{mm}\\\\ \delta_{DE} &= \frac{F_{DE}(175\,\text{mm})}{\frac{\pi}{4}(16\,\text{mm})^2\times 207\thinspace 000\,\mathsf{N/mm^2}} + 11.7\times 10^{-6}/^\circ{}C\times (175\,\text{mm})\cdot 20^\circ{}C\\\\ &= 4.2047\times 10^{-6}\,F_{DE}\,\text{mm} + 0.04095\,\text{mm}\\\\ &= 4.2047\times 10^{-6}\,(1.7143F_{BF})\,\text{mm} + 0.04095\,\text{mm}\\\\ &= 7.2081\times 10^{-6}\,F_{BF}\,\text{mm} + 0.04095\,\text{mm}\\\\ \end{align} From $\delta_{BF} = - 1.7143\delta_{DE}:\\$ \begin{align*} 1.5618\times 10^{-6}\,F_{BF} + 0.06084 &= -1.7143\left[ 7.2081\times 10^{-6}\,F_{BF} + 0.04095 \right]\\\\ &= -12.357\times 10^{-6}\,F_{BF} - 0.070201\\\\ 13.919\times 10^{-6}\,F_{BF} &= -0.13104\\\\ F_{BF} & = -9414.7\,\text{N}\\\\ F_{DE} & = -16140\,\text{N}\\\\ \end{align*} Note that the forces are in newtons since the modulus of elasticity was expressed in N/mm$^2$ above.
1. Now that we have $F_{BF}$, find the deformation in $BF$:
\begin{align*} \delta_{BF} &= \frac{-9414.7\,\text{N}\times 260\,\text{mm}}{\frac{\pi}{4}(32\,\text{mm})\cdot 207\thinspace 000\,\mathsf{N/mm^2}}+\left(11.7\times 10^{-6}/^\circ{}C\right)\left(260\,\text{mm}\right)\left( 20^\circ{}C\right)\\\\ &= 0.046137\,\text{mm} \end{align*}
1. Find the deflection at $A$.
Using similar triangles, \begin{align*} d_A &= \frac{300\,\text{mm}+110\,\text{mm}}{300\,\text{mm}}\delta_{F}\\\\ &= 0.063053\,\text{mm} \end{align*} The deflection is downwards since $\delta_{BF}$ is positive, i.e. $BF$ gets longer.

Part (b): The shear strength of the bolt at $C$ is 260 MPa and a factor of safety of 2.5 is required with respect to shear. Determine the minimum diameter for the bolt at $C$ and the bearing stress (using the diameter just found) between the bolt at $C$ and the member $ABCD$.
1. Find the force at $C$
For member $ABCD$, \begin{align*} \Sigma F_x &= C_x+F_{DE} = 0\\\\ \Rightarrow C_x &= 16.140\,\text{kN}\\\\ \Sigma F_y &= C_y+F_{BF} = 0\\\\ \Rightarrow C_x &= 9.4147\,\text{kN}\\\\ \end{align*} Then \begin{align*} C &= \sqrt{\left(9.4147\,\text{kN}\right)^2 + \left(16.140\,\text{kN}\right)^2}\\\\ &= 18.685\,\text{kN} \end{align*}
1. Find the required diameter of the bolt at $C$
\begin{align*} \tau_{allow} &= \frac{\tau_f}{\text{FS}}\\\\ &= 104\,\text{MPa}\\\\ \text{and } \tau &= \frac{V}{\frac{\pi}{4}d^2}\\\\ \Rightarrow d &= \sqrt{\frac{4V}{\pi\tau}}\\\\ &= \sqrt{\frac{4\times 18685\,\text{N}}{\pi(104\,\mathsf{N/mm^2})}}\\\\ &= 15.125\,\text{mm}\\\\ \end{align*}
1. Finally, the bearing stress
\begin{align*} \sigma_b &= \frac{F}{td}\\\\ &= \frac{18625\,\text{N}}{20\,\text{mm}\times15.125\text{ mm}}\\\\ &= 61.769\,\mathsf{N/mm^2}\\\\ &\approx 61.8\text{ MPa} \end{align*}