Joint displacement of a three-element truss (Castigliano / virtual work): 22-Mec-A7 Advanced Strength of Materials, December 2018, Q7

Question 7: Joint displacement of a three-element truss (Castigliano / virtual work) (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 pin-jointed members meet at the loaded joint B; their far ends A, C, D are pinned to supports.

Given data (from the figure)
Geometry	$a=0.75$ m (horizontal A–B), $b=0.90$ m (vertical B–D)
Member AB	horizontal, $L=0.75$ m
Member BD	vertical, $L=0.90$ m
Member CB	diagonal, $L=\sqrt{0.75^2+0.90^2}=1.172$ m
Section area	$A=5\text{ cm}^2=5\times10^{-4}$ m$^2$ (all)
Modulus $E$	200 GPa
Load at B	16 000 N, 30° above horizontal, up-and-right

Three-element truss: joint B connects horizontal AB, vertical BD and diagonal CB to fixed supports A, C, D. Load P = 16 kN acts 30° above horizontal.

Find. Horizontal ($u$) and vertical ($v$) displacement of joint B.

Approach. Joint B is the only free node; each member ties it to a fixed support, so B behaves like a mass on three axial springs. Assemble the 2×2 joint stiffness $\mathbf K=\sum(EA/L)\,\hat{\mathbf n}\hat{\mathbf n}^{\mathsf T}$ and solve $\mathbf K\{u,v\}=\mathbf P$. This is equivalent to a unit-load (Castigliano) evaluation but collapses the bookkeeping.

Unit vectors from B to each support. Taking B at the origin, right/up positive: to A $(-1,0)$, to D $(0,-1)$, to C $(-0.640,-0.768)$.
Member axial stiffnesses. $EA=200\times10^9\times5\times10^{-4}=1.0\times10^{8}$ N; then $EA/L$ = $1.333\times10^{8}$ (AB), $1.111\times10^{8}$ (BD), $8.536\times10^{7}$ N/m (CB).
Assemble K. Summing $(EA/L)\,\hat{\mathbf n}\hat{\mathbf n}^{\mathsf T}$, $$\mathbf K=\begin{bmatrix}1.683\times10^{8} & 4.198\times10^{7}\\[2pt] 4.198\times10^{7} & 1.615\times10^{8}\end{bmatrix}\ \tfrac{\text{N}}{\text{m}}.$$
Load vector and solve. $\mathbf P=(16\,000\cos30^\circ,\,16\,000\sin30^\circ)=(13\,856,\,8\,000)$ N. Then $$\{u,v\}=\mathbf K^{-1}\mathbf P\;\Rightarrow\;\boxed{u_B=0.075\text{ mm (right)},\quad v_B=0.030\text{ mm (up)}}.$$

Final results — Question 7
Component	Value
Horizontal $u_B$	0.075 mm (rightward)
Vertical $v_B$	0.030 mm (upward)

Topic. Displacement of a loaded joint in a determinate pin-jointed truss.

Key relations. Member force–displacement $\delta=NL/EA$; joint stiffness $\mathbf K=\sum(EA/L)\,\hat{\mathbf n}\hat{\mathbf n}^{\mathsf T}$; Castigliano $\Delta=\sum \dfrac{N\,n\,L}{EA}$.

Why this works. Because every member runs from the single free joint to a fixed support, the truss is a parallel spring set: each member resists motion of B along its own axis with stiffness $EA/L$, and the outer-product $\hat{\mathbf n}\hat{\mathbf n}^{\mathsf T}$ projects that stiffness into $x$–$y$. Summing gives the joint stiffness matrix, whose inverse maps the applied load to the joint displacement — identical to a Castigliano/unit-load calculation but without solving member forces first.

Common pitfall. Mis-reading the load angle or its sense (here up-and-right, both components positive), or using member lengths in cm — keep SI. The diagonal length must come from $\sqrt{a^2+b^2}$, not $a$ or $b$ alone.

Source. Hibbeler, Mechanics of Materials, §14.7–14.8 (Castigliano's theorem for trusses); Boresi & Schmidt, Ch. 5.

Question 7 of 8: Joint displacement of a three-element truss (Castigliano / virtual work)

Question 7: Joint displacement of a three-element truss (Castigliano / virtual work) (20 marks)