Question 6 of 8: Horizontal deflection of joint 2 of an L-frame

Question 6: Horizontal deflection of joint 2 of an L-frame (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. An L-frame: pinned at joint 1, a 6 m column 1–2, a 12 m beam 2–3, roller at joint 3. A 36 kN horizontal load acts at joint 2. Flexural rigidity $EI=16.0\times10^{5}$ kN·m²; axial deformation neglected.

Find. The horizontal deflection of joint 2 (in the direction of the 36 kN load).

Approach. This determinate frame deflects in flexure only, so use virtual work: $\delta=\displaystyle\int \frac{M\,m}{EI}\,ds$, with $M$ the real bending moment and $m$ the moment from a unit horizontal load at joint 2.

1. Reactions (real). $\sum F_x$: $H_1=36$ kN (←). $\sum M_1$: $12V_3=36(6)\Rightarrow V_3=18$ kN ↑, $V_1=18$ kN ↓.
2. Real moments. Column: $M(y)=36y$ (0 at the pin, $216$ at joint 2). Beam: $M(x)=18(12-x)$ (216 at joint 2, 0 at the roller).
3. Unit (virtual) system. A 1 kN horizontal load at joint 2 gives $m_{col}(y)=y$ (0→6) and $m_{beam}(x)=0.5(12-x)$ (6→0).
4. Integrate. Column: $\displaystyle\int_0^6 (36y)(y)\,dy=36\cdot\frac{6^3}{3}=2592$. Beam: $\displaystyle\int_0^{12} 18(12-x)\cdot0.5(12-x)\,dx=9\cdot\frac{12^3}{3}=5184$. Sum $=7776$.
5. Deflection. $\delta_{2H}=\dfrac{7776}{EI}=\dfrac{7776}{16.0\times10^{5}}$ m $\Rightarrow$ ==**$\delta_{2H}=4.86$ mm**== (to the right, i.e. in the load direction).

Both member integrals are of the form $\int(\text{linear})^2\,ds$ because the bending moment varies linearly along each member; the column contributes 2592 and the beam 5184, so the beam — twice as long — dominates the response even though its peak moment (216 kN·m at joint 2) equals the column's. The positive result confirms joint 2 moves in the same sense as the applied 36 kN, as physical intuition demands: the column bends as a vertical cantilever off the pin while the beam rotates about the roller, and the two rotations add at the knee. Axial shortening is neglected because both members are slender and the frame carries the horizontal thrust chiefly in flexure, so a flexure-only virtual-work integral is accurate to well within engineering tolerance here.