Question 8 of 8: Three-Hinged Frame — Reactions and Diagrams

Question 8: Three-Hinged Frame — Reactions and Diagrams (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.

Find. All four reaction components, and the shear force and bending moment diagrams for every member with the extreme ordinates labelled.

Approach. Two pins and one internal hinge make this a classic three-hinged frame: $3(4) + 4 - 3(5) - 1 = 0$, determinate. Use the three global equilibrium equations plus the condition that the bending moment vanishes at the hinge, taken on the shorter (right-hand) free body.

1. Reduce the distributed load and set out the equations. The uniform load resultant is $W = 2(11) = 22.0\ \text{kN}$ acting at $x = 6.5\ \text{m}$. Vertical and horizontal equilibrium give $$V_1 + V_5 = 32.5 + 22.0 = 54.5\ \text{kN},\qquad H_1 + H_5 = 0$$
2. Take moments about node 1 for the whole frame. With node 1 at $(0,1)$ and node 5 at $(14.5,0)$, $$14.5\,V_5 + H_5 = 32.5(6) + 22.0(6.5) = 338$$ the $H_5$ term appearing because node 5 sits 1 m below node 1.
3. Apply the hinge condition to the right-hand free body. Everything to the right of node 3 is unloaded, so the only force on that free body besides the hinge reaction is the support reaction at node 5. Taking moments about node 3 at $(12,6)$, $$2.5\,V_5 + 6\,H_5 = 0 \;\Longrightarrow\; H_5 = -\tfrac{5}{12}V_5$$
4. Solve for the reactions. Substituting, $$\left(14.5 - \tfrac{5}{12}\right)V_5 = 338 \;\Longrightarrow\; \tfrac{169}{12}V_5 = 338$$ $$\boxed{V_5 = 24.0\ \text{kN}\ \uparrow,\quad H_5 = 10.0\ \text{kN}\leftarrow, \quad V_1 = 30.5\ \text{kN}\ \uparrow,\quad H_1 = 10.0\ \text{kN}\rightarrow}$$
5. Work out the beam moments. Along the beam at height $y = 6$, taking the left-hand free body, $$M(x) = 30.5\,x - 50.0 - (x-1)^2 - 32.5\,\langle x-6 \rangle$$ where the last term applies only beyond the point load. This gives $M = -19.5\ \text{kN}\cdot\text{m}$ at node 2 (hogging, produced by the inclined leg), a zero at $x = 1.653\ \text{m}$, the sagging peak $$M(6) = 183 - 50 - 25 = +108.0\ \text{kN}\cdot\text{m}$$ directly under the point load, and zero again at the hinge, as required.
6. Trace the beam shear. $V = 30.5\ \text{kN}$ at node 2, falling under the uniform load to $V(6^-) = 30.5 - 10 = 20.5\ \text{kN}$; the point load drops it to $V(6^+) = -12.0\ \text{kN}$ and the uniform load carries it on to $V = -24.0\ \text{kN}$ at the hinge. Since the shear changes sign exactly at the point load, that is where the moment peaks — there is no interior stationary point.
7. Finish the short beam 3–4 and the two legs. The hinge passes a force of $(10.0,\ -24.0)\ \text{kN}$ to the right-hand portion, so member 3–4 carries a constant shear of $24.0\ \text{kN}$ and its moment grows linearly from zero at the hinge to $$M_4 = -24.0(1.5) = -36.0\ \text{kN}\cdot\text{m}$$ Leg 4–5 carries only the reaction at node 5. Resolving that reaction normal to the leg gives a constant transverse shear of $5.918\ \text{kN}$ and an axial compression of $25.317\ \text{kN}$, and the moment runs linearly from $36.0\ \text{kN}\cdot\text{m}$ at node 4 to zero at the pin; the check is $5.918 \times 6.083 = 36.0$. Leg 1–2 likewise carries a transverse shear of $3.824\ \text{kN}$, an axial compression of $31.869\ \text{kN}$ and a moment from zero at the pin to $19.5\ \text{kN}\cdot\text{m}$ at node 2, with $3.824 \times 5.099 = 19.5$.
8. Collect the extreme ordinates. $$\boxed{M_{\max}^{+} = +108.0\ \text{kN}\cdot\text{m}\ \text{at } x = 6\ \text{m};\quad M_{\max}^{-} = -36.0\ \text{kN}\cdot\text{m}\ \text{at node 4}}$$ $$\boxed{V_{\max}^{+} = +30.5\ \text{kN}\ \text{at node 2};\quad V_{\max}^{-} = -24.0\ \text{kN}\ \text{from the hinge to node 4}}$$