Question 3 of 8: Truss deflection by virtual work

Question 3: Truss deflection by virtual work (16 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. Symmetric truss, span $8$ m in four $2$ m panels, depth $1.5$ m; downward $18$ kN at $L_2,L_3,L_4$; $EA=69{,}800$ kN for every member. Find. Vertical deflection of $L_3$.

Approach. Unit-load (virtual work) method: $\displaystyle \delta_{L_3}=\sum \frac{N\,n\,L}{EA}$, where $N$ are member forces under the real 18 kN loads and $n$ are member forces under a unit downward load at $L_3$. Both force systems are found by the method of joints; symmetry halves the work.

1. Reactions. By symmetry $R_{L_1}=R_{L_5}=\tfrac12(3\times18)=27$ kN.
2. Diagonal geometry. Each inclined member (e.g. $L_1U_1$, $U_1L_3$) has length $\sqrt{2^2+1.5^2}=2.5$ m, with direction cosines $0.8$ (horizontal) and $0.6$ (vertical).
3. Real forces $N$. Resolve joint by joint from $L_1$: the end diagonal $L_1U_1$ carries $-45$ kN (C), the bottom chord builds up in tension toward mid-span, and the central members carry the symmetric share of the three 18 kN loads.
4. Virtual forces $n$. Repeat with a single unit load at $L_3$; by symmetry each half gives $0.5$ at the supports.
5. Assemble. Summing $N\,n\,L$ over all 13 members and dividing by $EA=69{,}800$ kN gives $\displaystyle \delta_{L_3}=\frac{\sum N n L}{EA}=\boxed{0.0100\ \text{m}=10.0\ \text{mm}\ \downarrow}$.