Question 6 of 8: Reactions and diagrams for members 2–3 and 3–4 of a hinged gable frame (20 marks)

Question 6: Reactions and diagrams for members 2–3 and 3–4 of a hinged gable frame (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. Pin at node 1 (0,0); rafter up to node 2 (3,4), which is an internal hinge (the open circle in the source figure); horizontal chord to node 3 (10,4), a rigid corner; rafter down to a pin at node 4 (13,0). UDL 13 kN/m on the horizontal chord 2–3.

Find. Reactions, and shear/moment diagrams for members 2–3 and 3–4.

Approach. Member 1–2 has a pin at each end and no load along its length, so it is a two-force (purely axial) member: its reaction at node 1 must lie along the line 1–2. That one geometric fact, combined with global equilibrium of the whole frame, is enough to solve all four reaction components without ever needing an indeterminate method.

1. Direction of the axial force in 1–2. $\tan\theta = 4/3$, so the reaction at node 1 satisfies $R_{1x}/R_{1y}=3/4$; write $R_1=P(0.6,\,0.8)$ for unknown magnitude $P$.
2. Moment equilibrium about node 4 for the whole frame (UDL resultant $=13(7)=91\text{ kN}$ at $x=6.5$): solving $\sum M_4=0$ for $P$ gives $P=\boxed{56.875\text{ kN}}$, hence $R_1=(34.125,\ 45.5)\text{ kN}$.
3. Remaining reactions from global equilibrium. $R_4=(-34.125,\ 45.5)\text{ kN}$ — checked: $\sum F_x=0$, $\sum F_y=91.0-91.0=0$, and an independent moment check about node 1 also closes to zero.
4. Member 1–2 is confirmed two-force: axial force $=P=\boxed{56.875\text{ kN, compression}}$, $V=0$, $M=0$ along its whole length (no transverse load and pinned/hinged at both ends).
5. Member 2–3 (horizontal, UDL 13 kN/m, $M=0$ at the hinge end 2). This behaves exactly like a simply-supported beam of span 7 m carrying the same UDL: $V$ runs $+45.5\to-45.5\text{ kN}$ (linear, crossing zero at midspan $x=3.5\text{ m}$); $M_{\max}=\dfrac{wL^2}{8}=\dfrac{13(7)^2}{8}=\boxed{79.625\text{ kN}\cdot\text{m}}$ at midspan, falling back to $M=0$ at node 3 — the rigid corner carries zero moment here because the far leg (3–4) also turns out to be two-force.
6. Member 3–4. Resolving $R_4$ along the rafter's own axis shows the transverse component is zero: $V=0$, $M=0$ throughout — member 3–4 is itself two-force, carrying $56.875\text{ kN}$ compression (equal to member 1–2 by the frame's left–right symmetry), which is exactly the axial thrust needed to close the horizontal-equilibrium of joint 3.