Displacements of a Fixed–Fixed Stepped Rod: 22-Mec-A7 Advanced Strength of Materials, May 2014, Q1

Question 1: Displacements of a Fixed–Fixed Stepped Rod (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. Three collinear welded rods between two rigid walls (order wall–A–1–B–2–C–3–D–wall), loaded only at the interior joints B and C.

Given data
Rods 1, 3	$L_1=L_3=1.5$ m, $E_1=E_3=50$ GPa, $A_1=A_3=0.01$ m$^2$
Rod 2	$L_2=2.0$ m, $E_2=30$ GPa, $A_2=0.025$ m$^2$
Load at B	$160$ kN to the left ($-x$)
Load at C	$500$ kN to the right ($+x$)

Fig. 1—Three welded rods between rigid walls; interior loads at joints B and C. Positive $x$ points right.

Find. The horizontal displacements $u_B$ and $u_C$ of joints B and C.

Approach. The member is axially statically indeterminate (both ends fixed), so use the displacement (stiffness) method: with $u_A=u_D=0$, write joint equilibrium at B and C in terms of the two unknown displacements through each rod’s axial stiffness $k=EA/L$.

Axial stiffnesses. Each rod behaves as a spring of stiffness $k=EA/L$: $$k_1=k_3=\frac{50\times10^{9}(0.01)}{1.5}=333.3\ \text{MN/m},\qquad k_2=\frac{30\times10^{9}(0.025)}{2.0}=375.0\ \text{MN/m}.$$
Joint equilibrium. Taking tension positive with $N_1=k_1u_B$, $N_2=k_2(u_C-u_B)$, $N_3=-k_3u_C$, equilibrium of the applied loads at B and C gives $$(k_1+k_2)\,u_B-k_2\,u_C=-160\ \text{kN},\qquad -k_2\,u_B+(k_2+k_3)\,u_C=+500\ \text{kN}.$$
Solve the 2×2 system. Substituting the stiffnesses (in MN/m and kN), $$\begin{bmatrix}708.3&-375.0\\-375.0&708.3\end{bmatrix}\!\begin{Bmatrix}u_B\\u_C\end{Bmatrix}=\begin{Bmatrix}-160\\500\end{Bmatrix}\ \text{kN}\;\Rightarrow\;\boxed{u_B=0.205\ \text{mm},\quad u_C=0.815\ \text{mm}}$$ both directed to the right (positive $x$).
Internal-force check. Back-substituting the displacements, $$N_1=68.5\ \text{kN (T)},\quad N_2=228.5\ \text{kN (T)},\quad N_3=271.5\ \text{kN (C)},$$ and the joint balances $N_2-N_1=160$ kN and $N_3-N_2=-500$ kN both close, confirming the solution.

Quantity	Result
Displacement of B, $u_B$	$0.205$ mm (right)
Displacement of C, $u_C$	$0.815$ mm (right)
Rod forces $N_1,N_2,N_3$	$68.5$ (T), $228.5$ (T), $271.5$ (C) kN

Topic: Displacements of an axially loaded statically indeterminate (fixed–fixed) stepped member.

Key relations: Member stiffness $k=EA/L$; bar force $N=k\,\Delta$; joint equilibrium $\sum N+F_{\text{ext}}=0$; wall constraints $u_A=u_D=0$.

Why this works: Because both ends are held, statics alone cannot find the internal forces—there are three unknown bar forces but only two interior joint equations. The displacement method restores solvability by making the joint displacements the primary unknowns: each rod’s force is written as its stiffness times the relative displacement of its ends, so equilibrium becomes two linear equations in $u_B$ and $u_C$. This is the one-dimensional stiffness (finite-element) form, and it automatically satisfies compatibility because the rods share the joint displacements.

Common pitfall: Sign errors on the 160 kN load (it acts to the left, so it enters as $-160$ kN) and forgetting that rod 3 is in compression—its force must come out negative. Also do not assume the load simply “splits” by area; the differing moduli and lengths make the stiffnesses unequal.

Source: Hibbeler, Mechanics of Materials 10th ed., §4.4–4.5 (statically indeterminate axially loaded members); Ugural & Fenster, Advanced Strength and Applied Elasticity Ch. 2.

Question 1 of 8: Displacements of a Fixed–Fixed Stepped Rod

Question 1: Displacements of a Fixed–Fixed Stepped Rod (20 marks)