Three-hinged frame: reactions and diagrams (22 marks): 16-Civ-A1 Elementary Structural Analysis, December 2019, Q7

Question 7 — Three-hinged frame: reactions and 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 frame with joints at 1 $(0,0)$, 2 $(0,4)$, 3 $(12,9)$ and 4 $(15.75,0)$, all in metres. Joints 1 and 4 are pins and joint 3 is an internal hinge. Member 1–2 is a 4 m column, member 2–3 rises 5 m over a 12 m horizontal run (length 13 m) and member 3–4 falls 9 m over a 3.75 m run (length 9.75 m). A load of 3.9 kN/m acts downward on member 2–3 per metre of horizontal projection, and 3.9 kN/m acts horizontally on member 3–4 per metre of vertical projection.

Find. The four reaction components and the shear and moment diagrams for all three members.

Q7 — three-hinged frame: pins at 1 and 4 and a crown hinge at 3. The 3.9 kN/m loads act per metre of horizontal projection (member 2–3) and per metre of vertical projection (member 3–4).

Approach. Two pins and one hinge give four unknowns against three overall equations plus zero moment at the hinge, so the frame is determinate. Resolve the projected loads into resultants, use overall equilibrium, then close with the hinge condition applied to the right-hand free body.

Load resultants. Because the loads are specified per metre of projection, $$W_v = 3.9(12) = 46.8\ \text{kN downward at } x = 6\ \text{m}$$ $$W_h = 3.9(9) = 35.1\ \text{kN horizontal at } y = 4.5\ \text{m}$$ Note the levers: the vertical resultant acts at mid-span, not at mid-length of the sloping member, and likewise for the horizontal one.
Overall moments about joint 1. $$\sum M_1 = -46.8(6) + 35.1(4.5) + 15.75B_y = 0$$ $$15.75B_y = 280.8 - 157.95 = 122.85 \;\Rightarrow\; B_y = \boxed{7.8\ \text{kN}\ \uparrow}$$ and vertical equilibrium gives $A_y = 46.8 - 7.8 = \boxed{39.0\ \text{kN}\ \uparrow}$.
Hinge condition on the right free body. Taking moments about joint 3 for member 3–4 plus its share of the horizontal load, $$\sum M_3 = B_y(15.75-12) - B_x(0-9) - 35.1(4.5) = 0$$ $$9B_x = 157.95 - 29.25 = 128.7 \;\Rightarrow\; B_x = \boxed{14.3\ \text{kN}}$$ and horizontal equilibrium gives $A_x = 35.1 - 14.3 = \boxed{20.8\ \text{kN}}$. As a check, the moment of the left free body about the hinge is $39.0(12) - 20.8(9) - 46.8(6) = 468 - 187.2 - 280.8 = 0$.
Column 1–2. It carries no transverse load, so its shear is the constant 20.8 kN and its moment grows linearly from zero at the pin to $$M_2 = -20.8(4) = \boxed{-83.2\ \text{kN}\cdot\text{m}}$$ at the knee, with a constant axial compression of 39.0 kN.
Member 2–3. At a horizontal distance $x$ from the column line the section sits at height $4 + 5x/12$, so $$M(x) = 39.0x - 20.8\left(4 + \frac{5x}{12}\right) - 1.95x^{2} = 30.333x - 83.2 - 1.95x^{2}$$ This starts at $-83.2$ kN·m (matching the column, as a rigid knee requires), crosses zero, and peaks where $30.333 - 3.9x = 0$, that is at $x = 7.78\ \text{m}$: $$M_{\max} = \boxed{+34.8\ \text{kN}\cdot\text{m}}$$ falling back to exactly zero at the crown hinge.
Member 3–4. Measuring $y$ up from the pin at joint 4, the free body below the section gives $$M(y) = -17.55y + 1.95y^{2}$$ which vanishes at both ends — zero at the pin and zero at the hinge — and reaches its extreme at mid-height: $$M(4.5) = \boxed{-39.5\ \text{kN}\cdot\text{m}}$$ The whole of this member therefore hogs, with the largest value halfway up.

Question 7 — results
Reaction at joint 1	$A_x = 20.8$ kN, $A_y = 39.0$ kN
Reaction at joint 4	$B_x = 14.3$ kN, $B_y = 7.8$ kN
Column 1–2	$M$: 0 to −83.2 kN·m; $V = 20.8$ kN constant
Member 2–3	$M$: −83.2 to +34.8 kN·m (at $x = 7.78$ m) to 0
Member 3–4	$M$: 0 to −39.5 kN·m (mid-height) to 0
Moment at the crown hinge	0 (the defining check)

Topic. Determinate three-hinged frames with loads specified on projected areas.

Key relations. Four unknowns are closed by $\sum F_x = \sum F_y = \sum M = 0$ plus $\sum M_{\text{hinge}} = 0$ taken for one side only; a load given per metre of horizontal projection has resultant $w \times (\text{span})$ acting at mid-span, and one given per metre of vertical projection has resultant $w \times (\text{height})$ at mid-height.

Why this works. The crown hinge cannot transmit moment, so writing moment equilibrium for everything on one side of it introduces no new unknowns beyond the two reaction components at that side’s support — which is precisely the fourth equation needed. Taking the unloaded or less loaded side keeps the algebra short. Projected loading is the normal idealisation for snow and wind: snow accumulates on plan area, so it is quoted per horizontal metre, while wind pressure acts on the elevation, so it is quoted per vertical metre.

Common pitfall. Multiplying a projected load by the sloping member length instead of its projection, which inflates the resultant by the ratio 13/12 here and corrupts every subsequent number. The second trap is taking the hinge moment for the whole structure rather than one free body — that just reproduces overall equilibrium and yields no new information. Finally, check that the moment closes to exactly zero at the crown from both sides; it is the cheapest confirmation available.

Source. Hibbeler, Structural Analysis 10th ed., §5.2 and Ch. 4 for the diagrams; Kassimali 6th ed., §5.6–5.7.

Question 7 of 8: Three-hinged frame: reactions and diagrams (22 marks)

Question 7 — Three-hinged frame: reactions and diagrams (22 marks)