Question 7 of 8: Horizontal deflection by virtual work

Question 7 — Horizontal deflection by virtual work [22]

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. A determinate portal (pin at 1, roller at 4). $20$ kN↓ at the beam midspan. $EI=2.0\times10^{5}$ kN·m² throughout.

Find. Horizontal deflection $\delta_{3h}$ at joint 3.

Approach. $\delta_{3h}=\displaystyle\int\frac{M\,m}{EI}\,ds$ with $M$ from the real load and $m$ from a unit horizontal load at 3. Both columns turn out to carry no product, so only the beam integral remains.

1. Real reactions. No horizontal load ⇒ $H_1=0$; $V_1=V_4=10$ kN. With $H_1=0$ both columns have zero real moment; the beam is a simple $20$ m span, $M_{\text{real}}$ a triangle peaking $-100$ kN·m at midspan (our sign: tension top).
2. Virtual system (unit → at 3). $H_1=-1$, $V_1=-0.5$, $V_4=+0.5$. Column 1–2 has $m=-y$ (0 to $-10$); the beam has $m=0.5x-10$ (from $-10$ at 2 to $0$ at 3); column 3–4 has $m=0$.
3. Only the beam contributes ($M_{\text{real}}=0$ in column 1–2; $m=0$ in column 3–4):$$\delta_{3h}=\frac{1}{EI}\int_0^{20} M_{\text{real}}(x)\,m(x)\,dx,\quad M_{\text{real}}=\begin{cases}-10x,&0\le x\le10\\10x-200,&10\le x\le20\end{cases},\ m=0.5x-10.$$
4. Evaluate. $\displaystyle\int_0^{20}M_{\text{real}}\,m\,dx=5000$ kN²·m³, so $$\delta_{3h}=\frac{5000}{2.0\times10^{5}}=0.025\text{ m}=\boxed{25\text{ mm}}.$$