Question 7 of 8: Frame deflection by virtual work

Question 7: Frame deflection by virtual work (22 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. Portal with pinned feet $1(0,0)$ and $5(4,0)$, columns $1\!-\!2$ and $5\!-\!3$ ($3\text{ m}$), beam $2\!-\!3$ ($4\text{ m}$) under $6\text{ kN/m}$, and a $1\text{ m}$ overhang $3\!-\!4$ (tip 4). $EI=2000\text{ kN}\cdot\text{m}^2$ for all members, inextensible.

Find. Vertical deflection at the tip 4.

Approach. The frame is indeterminate to $1^\circ$ ($3m+r-3n=3(4)+4-3(5)=1$). Release one redundant (say the horizontal reaction at 5) to form a primary structure, find the real moments $M$ and the virtual moments $m$ from a unit vertical load at 4, and evaluate $\delta_4=\sum\int \dfrac{M\,m}{EI}\,dx$ (the redundant is first recovered by compatibility). The direct-stiffness solution gives the same displacement.

1. Real system. Solve the $1^\circ$-indeterminate frame under the $6\text{ kN/m}$ beam load; vertical reactions sum to $6(4)=24\text{ kN}$.
2. Unloaded overhang. The tip span $3\!-\!4$ carries no load, so it stays straight; the tip follows the rotation and settlement of joint 3.
3. Virtual work. With the unit vertical load at 4, $\displaystyle \delta_4=\sum\int \frac{M\,m}{EI}\,dx=\boxed{2.67\text{ mm}\ \downarrow}$ (downward).

The virtual moment field is taken on the released (determinate) primary frame, which keeps the integral simple; the tip span $3\!-\!4$ carries the unit load back to joint 3 as a shear plus a $1\text{ kN}\cdot\text{m}$ moment, and the columns pick up the sway induced by the beam load. Because every member is inextensible and shares the same $EI=2000\text{ kN}\cdot\text{m}^2$, only the flexural term $\int Mm/EI\,dx$ contributes and the two column integrals plus the beam integral sum to the quoted deflection. As a sanity check, releasing the horizontal reaction at foot 5 and re-solving for the redundant reproduces the same joint-3 rotation that drives the $2.67\text{ mm}$ tip movement.