Displacements of a Pin-Jointed Truss Joint: 22-Mec-A7 Advanced Strength of Materials, May 2015, Q1

Question 1: Displacements of a Pin-Jointed Truss Joint (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-bar truss with the free (loaded) joint B restrained by three pinned members meeting at B.

Given data
Horizontal load at B, P	38 000 N (→)
Member area, A	9 cm² = 900 mm²
Panel width, a	80 cm = 0.80 m
Panel height, b	100 cm = 1.00 m
Young’s modulus, E	180 GPa

Find. The horizontal displacement u and vertical displacement v of joint B.

Truss: member AB (horizontal), CB (diagonal) and DB (vertical) all pinned to fixed supports; only joint B is free.

Approach. Joint B carries three members but offers only two equilibrium equations, so the truss is statically indeterminate to the first degree; solve by the direct-stiffness (displacement) method using the two unknown displacements (u, v) at B.

Set geometry and direction cosines. With A(0, 0), B(0.8, 0), C(0, −1.0), D(0.8, −1.0) m, the unit vectors from each support toward B are $$\text{AB}:(c,s)=(1,0),\ L=0.80;\quad \text{CB}:(0.625,0.781),\ L=1.281;\quad \text{DB}:(0,1),\ L=1.00.$$ The axial rigidity is $EA=180\times10^{9}\times900\times10^{-6}=1.62\times10^{8}\ \text{N}$.
Assemble the joint stiffness. Each bar contributes $k_i=\dfrac{EA}{L_i}\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$, so $$K=\sum_i k_i=\begin{bmatrix}2.519\times10^{8}&6.171\times10^{7}\\[2pt]6.171\times10^{7}&2.391\times10^{8}\end{bmatrix}\ \text{N/m}.$$
Solve the joint equilibrium. With the applied load $\{F\}=\{38\,000,\,0\}^{\mathsf T}$ N, solving $K\{u,v\}^{\mathsf T}=\{F\}$ gives $$\boxed{u=+0.161\ \text{mm (right)},\qquad v=-0.0416\ \text{mm (down)}.}$$ The horizontal displacement dominates because member AB is aligned with the load; the small downward v arises from the flexibility of the inclined member CB.

Final results — Question 1
Quantity	Value
Horizontal displacement u of B	+0.161 mm (in the load direction)
Vertical displacement v of B	−0.0416 mm (downward)

Topic: Statically indeterminate pin-jointed trusses by the stiffness method.

Key relations: element stiffness $k=\tfrac{EA}{L}[c^2,cs;cs,s^2]$; global $K\{d\}=\{F\}$; bar force $N=\tfrac{EA}{L}(c\,u+s\,v)$.

Why this works: With one free joint the two nodal displacements are the only unknowns. Every bar behaves as a linear spring whose stiffness projects onto the global axes through its direction cosines. Summing these projected stiffnesses enforces compatibility (all bars share the joint displacement) automatically, while the matrix equation enforces equilibrium—so a single 2×2 solve handles the redundancy that a force method would need a compatibility equation for.

Common pitfall: forgetting that stiffness depends on $c^2$ and $s^2$, so the sign of the direction cosine is irrelevant—what matters is measuring the angle consistently from the global x-axis. Mixing degrees/radians or dropping the $10^{-6}$ in the area is the usual numerical slip.

Source: Hibbeler, Mechanics of Materials, Ch. 9 (statically indeterminate axial members); Ugural & Fenster §7 (matrix methods).

Question 1 of 8: Displacements of a Pin-Jointed Truss Joint

Question 1: Displacements of a Pin-Jointed Truss Joint (20 marks)