Question 8 of 8: Three welded rods between rigid walls (indeterminate axial)

Question 8: Three welded rods between rigid walls (indeterminate axial)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Chain A–B–C–D fixed at walls A and D; a 300 kN force at B (leftward) and a 500 kN force at C (rightward) — directions read from the figure.

Find. the horizontal displacements of B and C.

[Figure not reproduced. See the official exam paper.]

Approach. The structure is singly indeterminate; assemble the two nodal stiffness equations for the free nodes B and C (walls give $u_A=u_D=0$) and solve.

1. Rod stiffnesses. $k=EA/L$ gives $$k_1=k_3=\frac{(140\,000)(50\,000)}{1500}=4.667\times10^{6}\ \tfrac{\text{N}}{\text{mm}},\qquad k_2=\frac{(80\,000)(90\,000)}{1500}=4.800\times10^{6}\ \tfrac{\text{N}}{\text{mm}}.$$
2. Nodal equilibrium. Taking rightward positive, with $F_B=-300$ kN and $F_C=+500$ kN, $$\begin{aligned}(k_1+k_2)u_B-k_2u_C&=-300\,000\\-k_2u_B+(k_2+k_3)u_C&=+500\,000\end{aligned}$$
3. Solve the 2×2 system. $$u_B=\boxed{-0.0066\ \text{mm}}\ (\text{i.e. }6.6\ \mu\text{m left}),\qquad u_C=\boxed{+0.0495\ \text{mm}}\ (49.5\ \mu\text{m right}).$$
4. Equilibrium check. The recovered rod forces satisfy the node balances $N_2-N_1+F_B=0$ at B and $N_3-N_2+F_C=0$ at C, confirming the solution. The displacements are small because the given cross-sections are large ($\ge500\ \text{cm}^2$).