Question 6 of 8: Horizontal deflection of a joint on a determinate frame (alternative 1 of 3)

Question 6: Horizontal deflection of a joint on a determinate frame (alternative 1 of 3) (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-shaped frame: a 6 m column from the pin at joint 1 up to joint 2, and a 12 m beam from joint 2 to the roller at joint 3. A horizontal load of 36 kN is applied at joint 2, and EI = 16.0 × 10⁵ kN.m² for both members, which are inextensible, so only bending contributes to the displacement.

Find. The horizontal deflection of joint 2, in millimetres, with its direction.

Approach. Find the real bending moment diagram, apply a horizontal unit load at joint 2 to obtain the virtual diagram, and integrate $\int M m \, ds / EI$ over both members; because the unit load acts at the same point and in the same direction as the real load, $m = M/36$ and the integral collapses to a single evaluation.

1. Reactions under the real load. The roller at joint 3 takes vertical force only, so horizontal equilibrium sends the whole 36 kN back through the pin. Taking moments about joint 1, $$36(6) = V_3(12) \;\Rightarrow\; V_3 = 18.0\ \text{kN}\ (\uparrow)$$ $$\begin{aligned} H_1 &= 36.0\ \text{kN}\ (\leftarrow), \\ V_1 &= 18.0\ \text{kN}\ (\downarrow) \end{aligned}$$
2. Real bending moment diagram. On the column, measuring $y$ upward from the pin, the only force below the section is the pair at joint 1, of which only the horizontal component has a lever arm, so $M = 36y$, rising from zero at the pin to $$M_2 = 36(6) = \boxed{216\ \text{kN}\cdot\text{m at the knee}}$$ On the beam, measuring $s$ back from the roller, $M = 18s$, which reaches the same 216 kN.m at joint 2 — the check that the knee is in equilibrium. Both diagrams are straight lines, which makes the integration elementary.
3. Virtual system. Remove the 36 kN and apply 1 kN horizontally at joint 2 in the direction whose displacement is wanted. The structure and the load position are unchanged, so every internal action simply scales: $$\begin{aligned} m &= \frac{M}{36} \;\Rightarrow\; m = y \ \text{on the column}, \\ m &= 0.5 s \ \text{on the beam} \end{aligned}$$
4. Apply the unit-load equation. For a frame in which axial and shear deformations are neglected, $$1 \cdot \Delta = \int \frac{M m}{EI}\,ds$$ and with $EI$ constant it factors out. Integrating each straight-line product separately, $$\int_0^{6} (36y)(y)\,dy = 36\left[\frac{y^3}{3}\right]_0^{6} = 36(72) = 2592\ \text{kN}^2\text{m}^3$$ $$\int_0^{12} (18s)(0.5s)\,ds = 9\left[\frac{s^3}{3}\right]_0^{12} = 9(576) = 5184\ \text{kN}^2\text{m}^3$$
5. Combine and convert. The beam contributes twice as much as the column, because although its moment is smaller its length is doubled and the integral grows with the cube: $$\Delta_2 = \frac{2592 + 5184}{16.0 \times 10^{5}} = \frac{7776}{16.0 \times 10^{5}} = 4.86 \times 10^{-3}\ \text{m} = \boxed{4.86\ \text{mm}}$$ The sign is positive, so joint 2 moves in the direction of the unit load, that is horizontally in the same direction as the applied 36 kN. Both members are straight and the diagrams triangular, so the same answer follows from the standard product integral $\int M m \, ds = \tfrac{1}{3} M_{max} m_{max} L$ applied to each member, $\tfrac{1}{3}(216)(6)(6) + \tfrac{1}{3}(216)(6)(12) = 7776$, divided by EI.