Question 6 of 8: Three-Hinged Gable Frame

Question 6: Three-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. A frame pinned to the ground at node 1 (0, 0) and at node 4 (13 m, 0), with node 2 at (3 m, 4 m) and node 3 at (10 m, 4 m); the horizontal dimension string reads 3 m + 7 m + 3 m and the height is 4 m. Node 2 carries a small open circle and is an internal hinge; node 3 has no circle and is a rigid corner. A uniformly distributed load of 13 kN/m acts on the horizontal projection, its block running from directly above node 2 to directly above node 4 — that is, over members 2-3 and 3-4 and not over member 1-2.

Find. The four reaction components, and the shear and bending moment diagrams for members 2-3 and 3-4 with maximum and minimum ordinates labelled.

Approach. Recognise member 1-2 as a two-force member, because it is pinned at both ends and unloaded between them; the reaction at node 1 must then act along the line 1-2, which supplies the fourth equation the two pinned bases need.

1. Count the unknowns and find the fourth equation. Two pins give four reaction components against three equations of overall equilibrium, so one condition equation is needed. The internal hinge at node 2 supplies it, and in this frame it does so in the most convenient possible form: member 1-2 carries no load between the pin at node 1 and the hinge at node 2, so it is a two-force member and the reaction at node 1 is directed along 1-2, in the ratio $3 : 4$.
2. Resolve the applied load. The ruling covers the horizontal projection from $x = 3$ m to $x = 13$ m, so $$W = (13\ \text{kN/m})(10\ \text{m}) = 130\ \text{kN}$$ acting through $\bar{x} = (3 + 13)/2 = 8$ m.
3. Take moments about node 4 for the whole frame. Writing the force in member 1-2 as $S$, its components at node 1 are $\left( 0.6S,\ 0.8S \right)$. The vertical component has a lever arm of 13 m about node 4 and the horizontal component has none, node 1 and node 4 being at the same level, so $$13\left( 0.8S \right) = (130)(13 - 8) = 650$$ $$S = \boxed{62.5\ \text{kN}}$$
4. Write down the reactions. $$H_1 = 0.6S = \boxed{37.5\ \text{kN} \rightarrow}, \qquad V_1 = 0.8S = \boxed{50\ \text{kN} \uparrow}$$ $$V_4 = 130 - 50 = \boxed{80\ \text{kN} \uparrow}, \qquad H_4 = \boxed{37.5\ \text{kN} \leftarrow}$$ Every figure is round, and the hinge condition is satisfied identically because $S$ acts along 1-2 and therefore has no moment about node 2.
5. Shear and moment on member 2-3. This member is horizontal, so its shear is the net vertical force to the left: $$V(x) = 50 - 13(x - 3)\ \text{kN}, \qquad 3 \le x \le 10\ \text{m}$$ running from $+50$ kN at the hinge to $50 - 91 = -41$ kN at the corner. The bending moment starts at zero, as a hinge requires, and is $M(x) = 50(x - 3) - 6.5(x - 3)^{2}$.
6. Peak sagging moment on 2-3. Zero shear occurs $50/13 = 3.846$ m past the hinge, at $x = 6.85$ m, where $$M_{\max} = \frac{V_2^{2}}{2w} = \frac{(50)^{2}}{2(13)} = \boxed{+96.15\ \text{kN}\cdot\text{m}}$$ and at the corner $$M_3 = (50)(7) - 6.5(7)^{2} = 350 - 318.5 = \boxed{+31.5\ \text{kN}\cdot\text{m}}$$
7. Shear on member 3-4. This leg falls 4 m over 3 m, so its axis is $(0.6,\ -0.8)$ and the normal to it is $(0.8,\ 0.6)$. The resultant to the left of a section at horizontal position $x$ is $\left( 37.5,\ 50 - 13(x - 3) \right)$, so $$V = (0.8)(37.5) + (0.6)\left[ 50 - 13(x - 3) \right]$$ which is $+5.40$ kN just below the corner and $-18.0$ kN just above the pin at node 4, crossing zero at $x = 10.69$ m.
8. Bending moment on member 3-4. Taking the free body to the right of the section and including the moment of the horizontal reaction about the sloping axis gives $M_3 = 31.5\ \text{kN}\cdot\text{m}$ at the corner — continuous with member 2-3, as a rigid corner requires — rising to $$M_{\max} = \boxed{+34.62\ \text{kN}\cdot\text{m}} \quad \text{at } x = 10.69\ \text{m}$$ and falling to zero at the pin. Both members are therefore in sagging throughout; there are no negative moment segments, and the minimum ordinates are the zeros at the hinge and at support 4.