Welded three-rod assembly between rigid walls: 22-Mec-A7 Advanced Strength of Materials, December 2016, Q5

Question 5: Welded three-rod assembly 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.

Q5 — rods 1–2–3 in series between fixed walls at A and D; joints B and C carry 200 kN (left) and 400 kN (right) respectively.

Given. Three axial rods in series, both outer ends built into rigid walls, with axial loads applied at the two interior welds.

Given data
Rods 1, 3	L = 0.5 m, E = 180 GPa, A = 0.008 m² → k = 2.88×10⁹ N/m
Rod 2	L = 1 m, E = 120 GPa, A = 0.015 m² → k = 1.80×10⁹ N/m
Load at B	200 kN leftward (−x)
Load at C	400 kN rightward (+x)

Find. The axial displacements u_B and u_C.

Approach. Walls at A and D are fixed ($u_A=u_D=0$), leaving two free joints. Assemble the axial stiffness matrix for B and C and solve $\mathbf{Kd=F}$.

Bar stiffnesses. $k_1=k_3=EA/L=180\times10^9(0.008)/0.5=2.88\times10^9$ N/m; $k_2=120\times10^9(0.015)/1=1.80\times10^9$ N/m.
Assemble and load. Node B ties rods 1&2, node C ties rods 2&3: $$\begin{bmatrix}k_1+k_2 & -k_2\\ -k_2 & k_2+k_3\end{bmatrix}\begin{Bmatrix}u_B\\ u_C\end{Bmatrix}=\begin{Bmatrix}-200\\ 400\end{Bmatrix}\text{kN}.$$
Solve. With $k_1+k_2=4.68\times10^9$ and $k_2+k_3=4.68\times10^9$, $$\boxed{u_B=-0.0116\ \text{mm},\qquad u_C=+0.0810\ \text{mm}.}$$ B moves slightly left (toward its own load), while C is pulled well to the right.

Final results — Q5
Joint	Displacement
B	−0.0116 mm (left)
C	+0.0810 mm (right)

Topic: Statically indeterminate axial assembly by the stiffness method.

Key relations: Axial bar stiffness $k=EA/L$; global $\mathbf{Kd=F}$ with fixed-wall constraints $u_A=u_D=0$.

Why this works: Two fixed walls make the series of rods one degree statically indeterminate by equilibrium alone, but the stiffness method handles it directly: each rod is a spring $EA/L$, the interior joints are the only unknowns, and compatibility (shared weld displacements) is built into the assembled matrix. Solving the 2×2 system yields both displacements, from which each rod’s force follows as $k\,\Delta u$.

Common pitfall: Sign errors on the applied loads — the 200 kN acts to the left, so it enters the force vector as negative; flipping it reverses $u_B$.

Source: Hibbeler, Mechanics of Materials §4.4–4.6 (indeterminate axial members); Boresi & Schmidt Ch. 5.

Question 5 of 8: Welded three-rod assembly between rigid walls

Question 5: Welded three-rod assembly between rigid walls (20 marks)