Question 7 of 8: Three-hinged gabled frame

Question 7: Three-hinged gabled 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 gabled frame ABCDE with a hinge at the apex C and pinned supports at A and E, whose bases are at different levels.

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

Approach. Four unknown reaction components against three equations of global equilibrium; the apex hinge supplies the fourth condition, $\sum M_{C} = 0$ taken for the limb on one side of it. Solve the pair simultaneously, then cut each member to obtain the internal forces.

1. Global equilibrium. Vertically, $A_{y} + E_{y} = 108$ kN. Horizontally, $A_{x} + E_{x} = 213$ kN. Taking moments about A, and remembering that the 213 kN load acts 11 m above A and that E is 7 m above A, $$24E_{y} - 7E_{x} = 108(18) - 213(11) = 1944 - 2343 = -399$$
2. Condition at the apex hinge. Taking moments about C for the right-hand limb CDE, which carries the roof load, the 213 kN load and the reaction at E, $$12E_{y} + 9E_{x} = 108(6) + 213(5) = 648 + 1065 = 1713$$ The lever arms are the horizontal and vertical offsets of E from C, namely 12 m and 9 m.
3. Solve the pair. Eliminating $E_{y}$, $$E_{x} = \boxed{153\ \text{kN}}, \qquad E_{y} = \boxed{28\ \text{kN}}$$ and back-substitution into the global equations gives $$A_{x} = 213 - 153 = \boxed{60\ \text{kN}}, \qquad A_{y} = 108 - 28 = \boxed{80\ \text{kN}}$$ all four acting in the positive sense, so both horizontal reactions push to the right and both vertical reactions push up. The check is $\sum M_{C}$ for the left limb, which carries no load at all: $16(60) - 12(80) = 960 - 960 = 0$, exactly as the hinge requires.
4. Column AB. The reaction at A is the only force on this member, so the axial thrust is a constant 80 kN compression, the shear a constant 60 kN, and the moment grows linearly: $$M_{B} = 60(11) = \boxed{660\ \text{kN}\cdot\text{m}}$$ putting the inside (right-hand) face in tension.
5. Rafter BC. No load acts between B and C, so the moment falls linearly from $660\ \text{kN}\cdot\text{m}$ at B to zero at the apex hinge. With a member length of $\sqrt{12^{2}+5^{2}} = 13$ m, the shear is a constant $660/13 = 50.8$ kN and the thrust is $$N = \tfrac{12}{13}(60) + \tfrac{5}{13}(80) = 86.2\ \text{kN compression}$$
6. Rafter CD. Measuring $u$ horizontally from the apex, the moment is $$M(u) = -105u + 4.5u^{2}\ \text{kN}\cdot\text{m}$$ which is zero at the hinge and $-612\ \text{kN}\cdot\text{m}$ at D. Its extreme is at $u = 105/9 = 11.67$ m, where $$|M|_{\max} = \boxed{612.5\ \text{kN}\cdot\text{m}}$$ so the peak is only marginally larger than the value at the knee. The shear runs from $-96.9$ kN at the apex to $+2.8$ kN at D.
7. Column DE. Working up from the pin at E, the moment grows linearly to $$M_{D} = 153(4) = \boxed{612\ \text{kN}\cdot\text{m}}$$ which matches the value obtained from the rafter side — the 213 kN load at D is a force and not a couple, so the moment must be continuous there. The shear is a constant 153 kN and the axial force 28 kN compression.