Question 8 of 8: Frame deflection by virtual work

Question 8: Frame deflection by virtual work (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. Trapezoidal frame: pin (1)$=(0,0)$, roller (4)$=(7.2,0)$; corners (2)$=(1.6,1.2)$, (3)$=(5.6,1.2)$; top span (2)–(3) is $4\text{ m}$; $12\text{ kN}$ point loads down at (2) and (3); $EI=4000\text{ kN}\cdot\text{m}^2$, flexural strain only.

Find. Vertical deflection at the mid-point of (2)–(3).

Approach. The frame is determinate (pin + roller). Apply a unit vertical load at the mid-point of (2)–(3) and evaluate $\delta=\dfrac{1}{EI}\int M\,m\,ds$ over all members.

1. Real moments. By symmetry $R_{(1)}=R_{(4)}=12\text{ kN}$, $H=0$. Each leg carries a linear moment $0\to12(1.6)=19.2\text{ kN}\cdot\text{m}$; the shear in the top span is $12-12=0$, so span (2)–(3) is in uniform moment $M=19.2\text{ kN}\cdot\text{m}$.
2. Virtual moments. The unit load at mid-span gives legs $0\to0.8$ and a triangular $m$ on (2)–(3) peaking at $1.8$ at the centre ($0.8$ at each end).
3. Integrate. Top span: $\int M\,m\,ds=19.2\times5.2=99.84$; two legs: $2\times\tfrac{19.2(0.8)(2)}{3}=20.48$. Hence $$\delta=\frac{99.84+20.48}{4000}=0.03008\text{ m}.$$

It is worth pausing on why the top span carries pure moment. A vertical cut anywhere in (2)–(3) has, to its left, the $12\text{ kN}$ reaction at (1) and the $12\text{ kN}$ load at (2); these are equal and opposite, so the transverse shear is exactly zero and the bending moment holds constant at $12(1.6)=19.2\text{ kN}\cdot\text{m}$ across the whole span. That single observation collapses the span integral $\int M\,m\,ds$ into $19.2$ times the area under the triangular virtual diagram, and the only remaining work is the two short legs. Because the loads are symmetric about the mid-span, the virtual diagram is symmetric too, which is what makes its area a clean $5.2$.