16-Civ-A1 Elementary Structural Analysis · Undated paper

Question 7 of 8: Three-hinged frame — reactions and internal-force diagrams

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 7: Three-hinged frame — reactions and internal-force 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. A symmetric trapezoidal frame with pin supports at joint 1 (0, 0) and joint 5 (10, 0). The inward-leaning legs rise 6 m to joint 2 (2.5, 6) and joint 4 (7.5, 6); the horizontal top member 2–4 is 5 m long with an internal hinge at its midpoint, joint 3 (5, 6). Each leg is $\sqrt{2.5^{2} + 6^{2}} = 6.5\text{ m}$ long. Loads: 16 kN/m acting downward over the top member (total 80 kN at $x = 5$), and 10 kN/m acting horizontally leftward on member 4–5 per metre of vertical projection (total 60 kN, resultant at mid-height, $y = 3\text{ m}$).

Find. The four reaction components and the shear and bending-moment diagrams for all three members, with extreme ordinates.

Approach. Four unknown reaction components and four equations (three global plus $M = 0$ at the crown hinge) make the frame determinate. Use the three global equations first, then the hinge condition on the left free body.

1. Determinacy. With $m = 4$, $j = 5$, $r = 4$ and one release, $$\text{DSI} = 3(4) + 4 - 3(5) - 1 = 0$$ so the frame is statically determinate — the classical three-hinged arrangement (two support pins plus a crown hinge).
2. Global vertical and horizontal equilibrium. $$A_y + E_y = 80\ \text{kN}, \qquad A_x + E_x = 60\ \text{kN}$$ where $A$ denotes joint 1 and $E$ joint 5, and both horizontal reactions act to the right to resist the leftward wind.
3. Global moment about joint 1. The 80 kN vertical resultant sits at $x = 5$, and the 60 kN leftward resultant at $y = 3$: $$\sum M_1 = -80(5) + 60(3) + 10E_y = 0$$ $$E_y = \frac{400 - 180}{10} = \boxed{22\ \text{kN} \uparrow}, \qquad A_y = 80 - 22 = \boxed{58\ \text{kN} \uparrow}$$ The wind therefore relieves the leeward support and loads the windward one.
4. Crown-hinge condition. Take the free body to the left of joint 3 (5, 6). It carries the reactions at joint 1 and half the roof load, $16 \times 2.5 = 40\text{ kN}$ acting at $x = 3.75$. Because a hinge carries no moment, $$\sum M_3^{\text{left}} = 6A_x - 5A_y + 40(1.25) = 0$$ $$6A_x = 5(58) - 50 = 240 \;\Rightarrow\; A_x = \boxed{40\ \text{kN} \rightarrow}$$ $$E_x = 60 - 40 = \boxed{20\ \text{kN} \rightarrow}$$ Checking the right free body about the hinge gives $120 + 110 - 50 - 180 = 0$, an independent confirmation.
5. Left leg 1–2. It carries no load along its length, so its axial force, shear and moment gradient are all constant. Resolving the base reaction $(40, 58)$ along and normal to the member direction $(2.5, 6)/6.5$: $$N = \frac{40(2.5) + 58(6)}{6.5} = \boxed{68.92\ \text{kN compression}}, \qquad V = \boxed{14.62\ \text{kN}}$$ The moment grows linearly from zero at the pin to $$M_2 = 40(6) - 58(2.5) = 240 - 145 = \boxed{95\ \text{kN}\cdot\text{m}}$$ which is also $V \times L = 14.62 \times 6.5 = 95$, as it must be.
6. Top member 2–4. Taking moments of the left-hand forces about a section at $x$ (with sagging positive), $$M(x) = 58x - 240 - 8(x - 2.5)^{2}$$ This gives $-95\text{ kN}\cdot\text{m}$ at joint 2, exactly zero at the crown hinge $x = 5$ (the design check), and $-5\text{ kN}\cdot\text{m}$ at joint 4. The shear falls linearly from $+58\text{ kN}$ at joint 2 to $-22\text{ kN}$ at joint 4, vanishing at $x = 6.125\text{ m}$ where the moment peaks: $$M_{\max} = \boxed{+10.13\ \text{kN}\cdot\text{m at } x = 6.125\ \text{m}}$$ The member also carries a constant axial compression of 40 kN — the arch thrust.
7. Right leg 4–5. Measuring $y$ upward from joint 5 and working from below, the section carries the reactions at joint 5 plus the strip of wind load already passed: $$M(y) = 29.17y - 5y^{2}$$ This closes at $-5\text{ kN}\cdot\text{m}$ at joint 4, matching the top member, and peaks where the shear vanishes: $$M_{\max} = \boxed{42.53\ \text{kN}\cdot\text{m at } y = 2.92\ \text{m}}$$ The shear runs from $26.92\text{ kN}$ at joint 5 to $-28.46\text{ kN}$ at joint 4, and the axial compression grows from 12.62 kN at the base to 35.69 kN at the top.
8. Overall check. The largest bending moment anywhere in the frame is the 95 kN·m at the windward knee, joint 2 — not at midspan, because the crown hinge forces the moment there to zero and pushes the peak into the corners.