Question 1 of 8: Displacement of a Three-Bar Truss

Question 1: Displacement of a Three-Bar Truss (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-ended bars meet at the one free joint B; their far ends A, C and D are held by fixed supports, and the 7000 N load acts along line BC.

Find. Horizontal displacement $u_B$ and vertical displacement $v_B$ of joint B.

Approach. Only joint B can move and three bars run to fixed supports, so the joint is statically indeterminate to the first degree. The cleanest energy route writes the strain energy in the two joint displacements and stationarizes it, giving a $2\times2$ joint-stiffness relation $\mathbf{K}\,\boldsymbol{\delta}=\mathbf{F}$ (Castigliano’s second theorem applied at the joint).

1. Member directions and lengths. With B at the origin, the unit vectors from B toward each support and the lengths are $\hat{n}_{BA}=(-1,0),\ L=0.50\,\text{m}$; $\hat{n}_{BD}=(0,-1),\ L=0.60\,\text{m}$; $\hat{n}_{BC}=(-0.6402,-0.7682),\ L=0.781\,\text{m}$.
2. Axial stiffness of each bar. $k_i=\dfrac{AE}{L_i}$ with $AE=750\times10^{-6}\cdot70\times10^{9}=5.25\times10^{7}\ \text{N}$: $k_{BA}=1.050\times10^{8}$, $k_{BD}=8.750\times10^{7}$, $k_{BC}=6.722\times10^{7}\ \text{N/m}$.
3. Assemble the joint stiffness. Each bar contributes $k_i\,\hat{n}_i\hat{n}_i^{\mathsf T}$, so $$\mathbf{K}=\sum_i k_i\begin{bmatrix}c_i^2&c_is_i\\c_is_i&s_i^2\end{bmatrix}=\begin{bmatrix}13.26&3.306\\3.306&12.72\end{bmatrix}\times10^{7}\ \text{N/m}.$$
4. Load components at B. The unit vector from C to B is $(0.6402,\ 0.7682)$, so $F_x=7000(0.6402)=4481\ \text{N}$ and $F_y=7000(0.7682)=5378\ \text{N}$ (both positive).
5. Solve for the joint displacement. Inverting $\mathbf{K}\{u_B,v_B\}^{\mathsf T}=\{F_x,F_y\}^{\mathsf T}$ gives $$\boxed{u_B = 0.0249\ \text{mm}\ (\rightarrow),\qquad v_B = 0.0358\ \text{mm}\ (\uparrow)}$$
6. Physical check. The load pulls B up and to the right, away from all three supports, so every bar is in tension and B displaces a small amount into the load quadrant — consistent with both components coming out positive.