Question 1 of 8: Displacement of a Three-Bar Truss by an Energy Method

Question 1: Displacement of a Three-Bar Truss by an Energy Method (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 bars (BA, BD, BC) meet at the single free joint B; their far ends A, C and D are pinned to fixed supports.

Find. Horizontal displacement $u$ and vertical displacement $v$ 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 is to write the strain energy in terms of the two joint displacements and stationarize it, which yields 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. Taking C as origin, the unit vectors from B toward each support and the lengths are $\hat{n}_{BA}=(-1,0),\,L=1.0\,\text{m}$; $\hat{n}_{BD}=(0,-1),\,L=0.75\,\text{m}$; $\hat{n}_{BC}=(-0.8,-0.6),\,L=1.25\,\text{m}$.
2. Axial stiffness of each bar. $k_i=\dfrac{AE}{L_i}$: $k_{BA}=3.50\times10^{7}$, $k_{BD}=4.667\times10^{7}$, $k_{BC}=2.80\times10^{7}\ \text{N/m}$.
3. Assemble the joint stiffness. Each bar contributes $k_i\,\hat{n}_i\hat{n}_i^{\mathsf T}$, so $$\mathbf{K}=\sum_i k_i\begin{bmatrix}c_i^2&c_is_i\\c_is_i&s_i^2\end{bmatrix}=\begin{bmatrix}5.292&1.344\\1.344&5.675\end{bmatrix}\times10^{7}\ \text{N/m}.$$
4. Load components at B. $F_x=10\,000\cos30^\circ=8660\ \text{N}$, $F_y=+10\,000\sin30^\circ=5000\ \text{N}$ (upward).
5. Solve for the joint displacement. Inverting the $2\times2$ system $\mathbf{K}\{u,v\}^{\mathsf T}=\{F_x,F_y\}^{\mathsf T}$ gives $$\boxed{u = 0.150\ \text{mm}\ (\rightarrow),\qquad v = 0.0525\ \text{mm}\ (\uparrow)}$$
6. Physical check. The load pulls B up and to the right — away from all three supports — so every bar is in tension ($N_{BA}=5.26$, $N_{BC}=4.25$, $N_{BD}=2.45\ \text{kN}$), consistent with a small positive displacement in the load direction.