Three-hinged gabled frame: 07-Str-A1, May 2014, Q7

Question 7: Three-hinged gabled frame (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 gabled frame with unequal legs and unequal rafter slopes, pinned at both feet and hinged at the crown. Taking the level of the right-hand support as the datum and the line of the left column as $x = 0$:

Question 7 — node coordinates and loading
Node	Coordinates (m)	Comment
①	$(0,\ 3)$	pin support, 3 m above the datum
②	$(0,\ 6)$	head of the 3 m left column
③	$(4,\ 9)$	crown, internal hinge
④	$(10,\ 4.5)$	head of the 4.5 m right column
⑤	$(10,\ 0)$	pin support

The rafters carry 10 kN/m over their horizontal projection, that is over the full 10 m span, giving a total of 100 kN.

Find. All four reaction components, and shear, axial force and bending moment diagrams for each of the four members with the maximum and minimum ordinates labelled.

Question 7 — the three-hinged gabled frame. Pins at 1 and 5 and the crown hinge at 3 give four reaction components against three equilibrium equations plus one condition equation, so the frame is determinate. Reactions shown in blue.

Approach. Four unknown reactions are resolved by the three global equilibrium equations plus the condition that the bending moment vanishes at the crown hinge, then the internal forces follow by cutting each member and resolving the running resultant along and normal to it.

Global moment equation. Taking moments about the left pin ①, the 100 kN resultant acts at $x = 5$ m and the right reaction acts 10 m to the right and 3 m below: $$-500 + 10\,V_{B} + 3\,H_{B} = 0$$ where $V_{B}$ is positive upward and $H_{B}$ positive to the right.
Condition at the crown. Moments about the hinge ③ for the right-hand portion only, which carries 60 kN of the load at $x = 7$ m and is 6 m to the right of and 9 m below the crown: $$-180 + 6\,V_{B} + 9\,H_{B} = 0$$
Solve the pair. Eliminating $V_{B}$, $$7.2\,H_{B} = -120 \;\Rightarrow\; H_{B} = \boxed{-16.67\ \text{kN, i.e. } 16.67\ \text{kN} \leftarrow}$$ $$V_{B} = 50 - 0.3(-16.67) = \boxed{55.0\ \text{kN} \uparrow}$$ and global equilibrium then gives $V_{A} = 45.0$ kN upward with $H_{A} = +16.67$ kN, that is 16.67 kN to the right. Both horizontal reactions point inward, which is the arch thrust one expects.
Check the left half. Moments about the crown for the left portion, which carries 40 kN at $x = 2$ m: $$80 - 4(45.0) + 6(16.67) = 80 - 180 + 100 = 0 \ \checkmark$$
Left column ①–②. It carries no load along its length, so the axial force is 45.0 kN compression, the shear is a constant 16.67 kN, and the moment grows linearly to $$M_{2} = 16.67(3) = \boxed{50.0\ \text{kN}\cdot\text{m}}$$ at the knee ②, putting the outside face in tension.
Left rafter ②–③. Cutting at horizontal station $x$ and taking moments of the left-hand forces, $$M(x) = 5x^{2} - 32.5x + 50$$ which is $+50.0$ at ②, zero at $x = 2.5$ m and again at the crown, and reaches its extreme where $dM/dx = 0$, at $x = 3.25$ m: $$M_{\min} = 5(3.25)^{2} - 32.5(3.25) + 50 = -2.81\ \text{kN}\cdot\text{m}$$ Resolving the running resultant $(16.67,\ 45.0 - 10x)$ along the rafter unit vector $(0.8,\,0.6)$ and its normal, the axial force falls from 40.3 kN to 16.3 kN compression and the shear from $+26.0$ kN to $-6.0$ kN, crossing zero at the same $x = 3.25$ m.
Right rafter ③–④. The same construction from the right-hand pin gives $$M(x) = 5x^{2} - 57.5x + 150$$ zero at the crown and at $x = 7.5$ m, with $$M_{\min} = -15.31\ \text{kN}\cdot\text{m} \text{ at } x = 5.75\ \text{m}, \qquad M_{4} = \boxed{+75.0\ \text{kN}\cdot\text{m}}$$ Its unit vector is $(0.8,\,-0.6)$, so the axial force runs from 10.3 kN to 46.3 kN compression and the shear from $+14.0$ kN to $-34.0$ kN.
Right column ④–⑤. Again unloaded along its length: 55.0 kN compression, a constant 16.67 kN shear, and a moment that grows from zero at the pin to $16.67(4.5) = 75.0$ kN·m at the knee ④, matching the rafter value and closing the solution.

Question 7 — bending moment along the frame, plotted against the horizontal projection. Positive ordinates put the outside face of the frame in tension. The two knee moments dominate; the rafters carry very little moment because the arch thrust cancels most of the gravity moment.

The result worth carrying away is how small the rafter moments are. A simply supported 10 m beam under the same 10 kN/m would carry 125 kN·m at mid-span; here the largest rafter moment is 15.3 kN·m, because the 16.67 kN horizontal thrust generated by the three-hinged geometry cancels most of the gravity moment. What the frame pays for that is the pair of knee moments and the horizontal reactions its foundations must resist.

Question 7 — reactions and governing ordinates
Member or point	Axial	Shear	Moment
Reaction at ①	45.0 kN up, 16.67 kN to the right
Reaction at ⑤	55.0 kN up, 16.67 kN to the left
Column ①–②	45.0 kN C	16.67 kN constant	0 to $+50.0$ kN·m
Rafter ②–③	40.3 → 16.3 kN C	$+26.0$ → $-6.0$ kN	$+50.0$; min $-2.81$ at $x = 3.25$ m; 0 at the crown
Rafter ③–④	10.3 → 46.3 kN C	$+14.0$ → $-34.0$ kN	0 at the crown; min $-15.31$ at $x = 5.75$ m; $+75.0$ at ④
Column ④–⑤	55.0 kN C	16.67 kN constant	$+75.0$ kN·m to 0

Topic. The three-hinged gabled frame: reactions, arch thrust and internal force diagrams.

Key relations. Four reaction components, three equilibrium equations and one condition equation $\sum M_{\text{hinge}} = 0$ taken over one side only. For an inclined member, resolve the running resultant $\mathbf{F} = (H,\ V - wx)$ along the member unit vector for axial force and along its normal for shear; the moment extreme sits where that normal component vanishes.

Why this works. Adding a hinge to a two-pinned frame supplies exactly the fourth equation needed, so the structure is determinate however unsymmetric it is — and this frame is deliberately unsymmetric, with legs of 3 m and 4.5 m and rafter runs of 4 m and 6 m, so no shortcut based on symmetry is available. The horizontal reactions are equal and opposite because no horizontal load acts, and their common value is set entirely by the requirement that the moment vanish at the crown; that thrust is what converts a beam-like bending problem into an arch-like one.

Common pitfall. Taking moments about the crown for the whole structure rather than for one side. The condition equation only says that the internal moment at the hinge is zero, which is a statement about the forces on one side of it; applied to the whole frame it degenerates into the global moment equation and yields nothing. A second frequent error is treating the 10 kN/m as acting along the sloping rafter length; the drawing shows the load block over the horizontal projection, so the total is $10 \times 10 = 100$ kN and not $10 \times 12.5$ kN.

Source. Hibbeler, Structural Analysis 10th ed., §5.3 and §5.5; Kassimali, Structural Analysis 6th ed., §5.6 and §6.4.

Question 7 of 8: Three-hinged gabled frame

Question 7: Three-hinged gabled frame (22 marks)