Three welded rods between rigid walls: 22-Mec-A7 Advanced Strength of Materials, December 2018, Q5

Question 5: 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-segment axial member A–B–C–D built in at both walls (A and D fixed); loads act at the interior joints B and C.

Given data (from the figure)
Rod 1 (A–B)	$L_1=1.5$ m, $E_1=50$ GPa, $A_1=0.01$ m$^2$
Rod 2 (B–C)	$L_2=2$ m, $E_2=30$ GPa, $A_2=0.025$ m$^2$
Rod 3 (C–D)	$L_3=1.5$ m, $E_3=50$ GPa, $A_3=0.01$ m$^2$
Force at B	150 kN to the left
Force at C	400 kN to the right

Three rods welded in series between rigid walls A and D; 150 kN acts leftward at B, 400 kN rightward at C.

Find. The axial displacements $u_B$ and $u_C$ of the interior joints.

Approach. Model each rod as an axial spring of stiffness $k=EA/L$; assemble the 2×2 stiffness system for the two free joints (walls fixed) and solve for the joint displacements. Take rightward as positive.

Segment stiffnesses. $$k_1=\frac{E_1A_1}{L_1}=\frac{(50\times10^9)(0.01)}{1.5}=3.333\times10^{8}\ \tfrac{\text{N}}{\text{m}},$$ $$k_2=\frac{E_2A_2}{L_2}=\frac{(30\times10^9)(0.025)}{2}=3.750\times10^{8}\ \tfrac{\text{N}}{\text{m}},\qquad k_3=k_1=3.333\times10^{8}\ \tfrac{\text{N}}{\text{m}}.$$
Assemble the free-joint system. With $u_A=u_D=0$, equilibrium of joints B and C gives $$\begin{aligned}(k_1+k_2)\,u_B-k_2\,u_C&=F_B=-150\text{ kN},\\ -k_2\,u_B+(k_2+k_3)\,u_C&=F_C=+400\text{ kN}.\end{aligned}$$
Solve the 2×2 system. Substituting the stiffnesses (in $10^{8}$ N/m: $7.083,\,-3.750,\,7.083$), $$u_B=\boxed{+0.121\text{ mm}},\qquad u_C=\boxed{+0.629\text{ mm}}\quad(\text{both to the right}).$$ Even though the 150 kN load at B points left, the much larger 400 kN pull at C drags both interior joints rightward.

Final results — Question 5
Joint	Displacement
$u_B$	+0.121 mm (right)
$u_C$	+0.629 mm (right)

Topic. Statically indeterminate axial bar system (fixed–fixed) solved by the stiffness (displacement) method.

Key relations. Spring stiffness $k=EA/L$; joint equilibrium $\sum k(\Delta u)=F$; nodal system $\mathbf{K}\mathbf{u}=\mathbf{F}$.

Why this works. Each uniform rod behaves as a linear axial spring, and welding them in series while fixing both ends makes the assembly one degree indeterminate by force methods — but the displacement method handles it directly: the two interior joint displacements are the unknowns, and equilibrium at each joint gives exactly two equations. The fixed walls simply drop the end nodes from the system.

Common pitfall. Getting a load sign wrong — the figure shows 150 kN left and 400 kN right (the text description is unreliable). Also mixing the two different moduli: rod 2 uses $E_2=30$ GPa, not 50 GPa.

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

Question 5 of 8: Three welded rods between rigid walls

Question 5: Three welded rods between rigid walls (20 marks)