Question 4 of 8: Three Welded Rods Between Rigid Walls

Question 4: Three Welded Rods Between Rigid Walls (20 marks)

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. A three-rod series bar fixed at both walls (A and D), with a 200 kN leftward force at B and a 350 kN rightward force at C.

Find. The axial displacements of the interior joints B and C.

Approach. Treat each rod as an axial spring $k=EA/L$ and assemble the two equilibrium equations for the free joints B and C (walls impose $u_A=u_D=0$), then solve the 2×2 system.

1. Spring stiffnesses. $k_1=k_3=\dfrac{70000\cdot11000}{1000}=770$ kN/mm; $k_2=\dfrac{100000\cdot17000}{1800}=944.4$ kN/mm.
2. Joint equilibrium (rightward positive). At B: $(k_1+k_2)u_B-k_2u_C=-200$; at C: $-k_2u_B+(k_2+k_3)u_C=+350$ (kN, mm).
3. System. $$\begin{bmatrix}1714.4&-944.4\\-944.4&1714.4\end{bmatrix}\begin{Bmatrix}u_B\\u_C\end{Bmatrix}=\begin{Bmatrix}-200\\+350\end{Bmatrix}.$$
4. Solve. $\boxed{u_B=-0.0060\ \text{mm},\quad u_C=+0.201\ \text{mm}}$ — B barely moves (a hair to the left) while C moves right.
5. Check with rod forces. $N_1=k_1u_B=-4.6$ kN (compression), $N_2=k_2(u_C-u_B)=+195.4$ kN (tension), $N_3=k_3(0-u_C)=-154.6$ kN (compression). Joint B: $N_2-N_1=195.4+4.6=200$ kN balances the leftward 200 kN; joint C: $N_2-N_3=195.4+154.6=350$ kN balances the rightward 350 kN. The wall reactions are 4.6 kN at A and 154.6 kN at D.