Question 7 of 8: Displacement of a Three-Member Truss Joint

Question 7: Displacement of a Three-Member 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. Members BA (horizontal, length $a=1.0$ m), BD (vertical, length $b=1.2$ m) and BC (diagonal) meet at the loaded free joint B; supports at A, C, D. $A=12$ cm$^2$, $E=180$ GPa, $P=58000$ N horizontal.

Find. The horizontal ($u$) and vertical ($v$) displacements of joint B.

Approach. Only joint B is free (2 DOF); with three members it is one degree statically indeterminate, so assemble the 2×2 joint stiffness $\mathbf K=\sum k_i(\hat n_i\hat n_i^{\mathsf T})$ and solve $\mathbf K\{u,v\}^{\mathsf T}=\{P,0\}$.

1. Member data. Unit vectors from B: BA $(-1,0)$, $L=1000$ mm; BD $(0,-1)$, $L=1200$ mm; BC $(-0.640,-0.768)$, $L=1562$ mm. Stiffness $k=EA/L$ with $EA=180000\cdot1200=2.16\times10^8$ N gives $k_{BA}=216$, $k_{BD}=180$, $k_{BC}=138.3$ kN/mm.
2. Assemble. Each member adds $k\begin{bmatrix}c^2&cs\\cs&s^2\end{bmatrix}$: $$\mathbf K=\begin{bmatrix}272.7&68.0\\68.0&261.6\end{bmatrix}\ \text{kN/mm}.$$
3. Solve. $\mathbf K\{u,v\}^{\mathsf T}=\{58,0\}^{\mathsf T}$ kN gives $\boxed{u=0.227\ \text{mm (right)},\quad v=-0.059\ \text{mm (down)}}$.
4. Check. The horizontal load produces mostly horizontal movement; the small downward $v$ comes from the diagonal BC coupling the two directions through its off-diagonal stiffness.