Question 7 of 8: Trapezoidal frame with a crown hinge

Question 7: Trapezoidal frame with a crown hinge (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 trapezoidal frame with pinned bases 10 m apart at nodes 1 (0, 0) and 5 (10, 0), and a horizontal top reach 5 m long between nodes 2 (2.5, 6) and 4 (7.5, 6), so each inclined leg spans 2.5 m horizontally over a 6 m rise and is 6.5 m long. An internal hinge is drawn at node 3, the mid-point of the top reach. A downward uniform load of 16 kN/m runs over the whole top reach 2–4, and a horizontal uniform load of 10 kN/m acts leftwards on member 4–5 over its 6 m vertical projection.

Find. The four reaction components, and the shear and bending-moment diagrams for each member with the extreme ordinates labelled.

Approach. Four unknown reaction components against three equations plus the hinge condition, so the frame is determinate: take global moments for the vertical reactions, then moments about the hinge for one half to split the horizontal thrust, and finally walk the member chain accumulating shear and moment.

1. Resolve the applied loading. The vertical load totals $16(5) = 80$ kN acting at $x = 5$ m, the centre of the top reach. The horizontal load totals $10(6) = 60$ kN acting leftwards, and because it is distributed over the vertical projection of the leg its resultant sits at mid-height, $y = 3$ m. Determinacy is $r = 4$ against $3 + 1 = 4$ equations.
2. Global moments about node 1 give the vertical reactions. A horizontal force contributes a moment equal to force times height, so $$\sum M_1 = 0:\qquad 10\,E_y - 80(5) + 60(3) = 0$$ $$E_y = \frac{400 - 180}{10} = \boxed{22.0\ \text{kN up}},\qquad A_y = 80 - 22 = \boxed{58.0\ \text{kN up}}$$ Horizontal equilibrium fixes only the sum, $A_x + E_x = 60$ kN.
3. The hinge condition splits the thrust. Take the free body to the right of node 3 — the half reach 3–4, the leg 4–5 and the pin at node 5 — and set the moment about the hinge to zero. That free body carries $16(2.5) = 40$ kN of vertical load at $x = 6.25$ m and the whole 60 kN horizontal load at $y = 3$ m: $$\sum M_3 = 0:\qquad 22(5) + 6E_x - 40(1.25) - 60(3) = 0$$ $$E_x = \frac{50 + 180 - 110}{6} = \boxed{20.0\ \text{kN}},\qquad A_x = 60 - 20 = \boxed{40.0\ \text{kN}}$$ both acting to the right. Repeating the calculation on the left-hand free body returns zero identically, which is the check that the vertical reactions were right.
4. Leg 1–2 carries no transverse load, so its diagrams are straight. Resolving the reaction $(40, 58)$ along and normal to the leg's unit vector $(2.5, 6)/6.5$, $$N = -(40)(0.3846) - (58)(0.9231) = -68.9\ \text{kN} \ \ (\text{compression}),\qquad V = -14.6\ \text{kN}.$$ The shear is constant, so the moment grows linearly from zero at the pin to $14.6(6.5) = 95.0$ kN·m at the knee. Confirming it directly, the moment of the reaction about node 2 is $58(2.5) - 40(6) = -95$ kN·m, hogging: $$\boxed{M_2 = -95.0\ \text{kN}\cdot\text{m}}$$
5. Top reach 2–4. Measuring $x$ from node 1's vertical, $$M(x) = 58x - 40(6) - 8(x - 2.5)^{2},\qquad V(x) = 58 - 16(x - 2.5).$$ The shear runs from $+58$ kN at node 2 through $+18$ kN at the hinge to $-22$ kN at node 4. The moment starts at $-95$ kN·m, passes through exactly zero at the hinge — the arithmetic check on the whole solution — peaks at the point of zero shear, $$x = 2.5 + \frac{58}{16} = 6.125\ \text{m},\qquad M_{\max} = \boxed{+10.1\ \text{kN}\cdot\text{m}}$$ and closes at $-5.0$ kN·m at node 4.
6. Leg 4–5 carries the horizontal load. Working up from the pin, at height $y$ the moment is $$M(y) = \left(\frac{22(2.5)}{6} + 20\right)y - 5y^{2} = 29.17y - 5y^{2},$$ which is zero at the pin, reaches its peak where $\mathrm{d}M/\mathrm{d}y = 0$, i.e. at $y = 2.92$ m, $$M_{\max} = \boxed{+42.5\ \text{kN}\cdot\text{m}}$$ and closes on $-5.0$ kN·m at node 4, matching the top reach across the rigid knee. The corresponding developed shear runs from $+28.5$ kN at node 4 to $-26.9$ kN at the pin, passing through zero at the same height.
7. Name the extremes member by member. The governing hogging moment is the 95.0 kN·m at knee 2, and the governing sagging moment is the 42.5 kN·m in the windward leg — four times the 10.1 kN·m peak in the beam. That inversion is the point of the question: the horizontal load, not the gravity load, controls this frame, and a candidate who sketches the diagrams by analogy with a simple portal will put the peak in the wrong member.