07-Str-A1 · December 2018

Question 8 of 8: Three-Hinged Gabled Frame — Reactions and Diagrams

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2018 — 07-Str-A1 Elementary Structural Analysis; 3 hours, CLOSED BOOK (an approved Casio or Sharp calculator is permitted). Six questions constitute a complete paper: answer all of Questions 1–5, then one only of Questions 6, 7 or 8. Marks are shown in the left margin (6, 18, 18, 18, 18, then 22 for the optional question). Every question is worked below, including all three alternatives, because the full set is the more useful study resource.

Reference texts. R. C. Hibbeler, Structural Analysis (Ch. 2 determinacy and stability; Ch. 4 influence lines; Ch. 6–9 deflections and virtual work; Ch. 12 moment distribution); A. Kassimali, Structural Analysis (Ch. 3, 6, 8, 13, 16); K. Leet, C.-M. Uang and A. Gilbert, Fundamentals of Structural Analysis; J. C. McCormac, Structural Analysis: Using Classical and Matrix Methods. For Canadian practice the companion documents are the National Building Code of Canada (Part 4 load combinations) and CSA S16 Design of Steel Structures; this paper is pure analysis, so no design code is invoked in the answers.

Check — two figure readings that the printed figures leave open, both stated where they are used. (i) In Question 2(c) the hatched plane under the upper roller is drawn parallel to the inclined member, so the reaction is taken normal to the member. That reading makes every result an integer (R = 28 kN, shears 32 / −4 / −28 kN, axial forces 25 / 10 / 0 kN), whereas a vertical-plane reading gives nothing round; and because the two readings differ only by a force acting along the member axis, the shear and moment diagrams are identical either way. (ii) In Question 7 both wall supports are drawn as triangles on hatching. The truss is analysable only if one of them releases a component; taking the pin at U₂ and the horizontal-only roller at L₃ gives δ = 29.0 mm, and the same value follows if the thin vertical line between the two wall joints is a dimension witness line rather than a member. The alternative (pin at L₃) would give 31.8 mm.

Question 8: Three-Hinged Gabled Frame — Reactions and Diagrams (22 marks, optional)

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.

Item	Value
Node coordinates	1(0, 0), 2(0, 7), 3(12, 12), 4(24, 7), 5(24, 0) m
Supports	pins at nodes 1 and 5; rigid knees at 2 and 4; internal hinge at the peak, node 3
Rafter geometry	12 m run over 5 m rise — each rafter is exactly 13 m long
Wind load on column 1-2	4.26 kN/m horizontal, left to right, over the full 7 m ⇒ 29.82 kN at $y = 3.5$ m
Gravity load on rafter 2-3	8.45 kN/m on the horizontal projection over 12 m ⇒ 101.4 kN at $x = 6$ m

Find. The four reaction components and the shear and bending moment diagrams for all four members, with extreme ordinates.

Question 8: a three-hinged gabled frame — pins at nodes 1 and 5, an internal hinge at the peak, and load on the left half only.

Approach. Four reaction components against three global equations plus one hinge condition: determinate. Because the right half of the frame carries no load at all, the resultant reaction at node 5 must act along the line joining node 5 to the hinge, which supplies the fourth equation in one line and immediately fixes the right-hand half of both diagrams.

Confirm determinacy and identify the shortcut. With $m = 4$, $j = 5$, $r = 2 + 2 = 4$ and $c = 1$, $$i = 3(4) + 4 - 3(5) - 1 = 0$$ Members 3-4 and 4-5 carry no applied load, so the free body 3-4-5 is acted on by exactly two forces: the reaction at node 5 and whatever passes through the hinge. A two-force body requires those forces to be collinear, so the node-5 reaction acts along the line from $(24,\ 0)$ to $(12,\ 12)$ — direction $(-1,\ 1)$, i.e. at 45 degrees.
Global moments about node 1 give the vertical reaction at node 5. The wind resultant is 29.82 kN acting rightwards at $y = 3.5$ m and the gravity resultant is 101.4 kN acting down at $x = 6$ m, so $$24E_y = (29.82)(3.5) + (101.4)(6) = 104.37 + 608.40 = 712.77$$ $$E_y = \boxed{29.70\ \text{kN}\ \uparrow}$$
Use the 45-degree line to get the horizontal component, then close the equilibrium. Since the reaction is along $(-1,\ 1)$, $$E_x = -E_y = 29.70\ \text{kN acting leftwards} \qquad |E| = 29.70\sqrt{2} = 42.00\ \text{kN}$$ Vertical and horizontal equilibrium of the whole frame then give $$A_y = 101.4 - 29.70 = \boxed{71.70\ \text{kN}\ \uparrow} \qquad A_x = -29.82 + 29.70 = \boxed{0.12\ \text{kN leftwards}}$$ The horizontal reaction at node 1 is essentially zero — the two load intensities in this paper are very nearly the pair that would make it vanish exactly, so the wind thrust is carried almost entirely by the far pin. Report the 0.12 kN rather than rounding it away, because it shows the equilibrium closes.
The unloaded right half, done in two lines. The 42.00 kN reaction resolved along and normal to the right rafter (unit vectors $(0.9231,\ -0.3846)$ and $(0.3846,\ 0.9231)$) gives a constant 38.84 kN of axial compression and a constant 15.99 kN of shear in member 3-4, and the moment grows linearly from zero at the hinge to $$M_4 = (15.99)(13) = (29.70)(7) = \boxed{207.9\ \text{kN}\cdot\text{m}}$$ The right column carries that same 207.9 kN·m at the knee, falling linearly to zero at the pin, with a constant 29.70 kN shear and 29.70 kN of axial compression. Two independent routes to $M_4$ — via the rafter and via the column — agree, which is the check.
The left column, where the wind acts. Measuring $y$ upwards from the pin, the moment of everything below the section is $$M(y) = A_x y + \tfrac{1}{2}(4.26)y^2 = -0.121y + 2.13y^2$$ so $M$ is zero at the pin, dips to a negligible $-0.002$ kN·m at $y = 0.028$ m, and reaches $$M_2 = -0.85 + 104.37 = \boxed{103.5\ \text{kN}\cdot\text{m}}$$ at the knee. The shear runs linearly from 0.12 kN at the base to 29.70 kN at the knee, and the axial compression is a constant 71.70 kN.
The loaded left rafter. Taking moments of the pin reaction, the wind resultant and the part of the gravity load to the left of the section, the sagging moment as a function of horizontal distance $x$ from node 2 is $$M(x) = -4.225x^2 + 59.327x - 103.521 \ \ \text{kN}\cdot\text{m}$$ which correctly returns $-103.5$ kN·m at the knee (matching the column, as a rigid joint demands) and exactly zero at the hinge. It changes sign at $x = 2.04$ m and peaks where the shear vanishes: $$x = \frac{59.327}{2(4.225)} = 7.021\ \text{m}, \qquad M_{\max} = \boxed{+104.7\ \text{kN}\cdot\text{m}}$$ which is 7.61 m along the rafter from node 2. The transverse shear falls linearly from $+54.76$ kN at the knee to $-38.84$ kN at the hinge, and the axial compression from 54.99 kN to 15.99 kN. The two rafters exchange, through the hinge, a force of $(29.70,\ -29.70)$ kN — a purely 45-degree thrust, which is why the left rafter's shear at the hinge equals the right rafter's axial force and vice versa.

Question 8: developed shear and bending moment diagrams, abscissa measured along the member chain 1-2-3-4-5 (total 40 m). The moment is exactly zero at the crown hinge and the two panels satisfy V = dM/ds throughout.

Quantity	Value
Reaction at node 1 (pin)	$A_x = 0.12$ kN leftwards, $A_y = 71.70$ kN ↑
Reaction at node 5 (pin)	$E_x = 29.70$ kN leftwards, $E_y = 29.70$ kN ↑ (resultant 42.00 kN at 45°, through the hinge)
Column 1-2	V = 0.12 to 29.70 kN; M = 0 at the pin to 103.5 kN·m at the knee; N = 71.70 kN compression
Rafter 2-3	V = +54.76 to −38.84 kN; M = −103.5 kN·m at node 2, zero at $x = 2.04$ m, +104.7 kN·m at $x = 7.02$ m, zero at the hinge; N = 54.99 to 15.99 kN compression
Rafter 3-4	V = 15.99 kN constant; M = 0 at the hinge to 207.9 kN·m at node 4; N = 38.84 kN compression
Column 4-5	V = 29.70 kN constant; M = 207.9 kN·m at node 4 to 0 at the pin; N = 29.70 kN compression
Largest moment anywhere	207.9 kN·m at knee 4 (the unloaded half)

Topic. Three-hinged frames and the two-force-member shortcut

Key relations. Determinacy $i = 3m + r - 3j - c$ with $c = 1$ for the crown hinge. A free body acted on by only two forces has those forces equal, opposite and collinear. Distributed load on the horizontal projection of a rafter: resultant $wL_h$ at mid-projection. Peak moment where the transverse shear vanishes; on a rafter loaded over its projection this is still at the horizontal centroid position.

Why this works. The crown hinge does two things at once: it supplies the fourth equation that makes a two-pin gabled frame determinate, and it isolates the unloaded half as a two-force body. That second consequence is the real economy here — instead of writing a moment equation about the hinge, you can read the direction of the far reaction straight off the geometry (45 degrees, because node 5 to node 3 is 12 across and 12 up), and one global moment equation then finishes the reactions. The unloaded half also produces the largest bending moment in the frame, 207.9 kN·m at knee 4 against 104.7 kN·m under the load, because that reaction has a long lever arm about the knee.

Common pitfall. Reading the distributed rafter load as acting per metre of rafter (13 m) instead of per metre of projection (12 m); here that would inflate the resultant from 101.4 kN to 109.85 kN and change every reaction. The second trap is discarding the very small horizontal reaction at node 1 as a rounding artefact — it is a genuine 0.12 kN, and reporting it is what demonstrates that the horizontal equilibrium has actually been checked.

Source. Hibbeler, Structural Analysis, §5.3 (three-hinged frames and arches) and §5.1–5.2; Kassimali, §6.4; McCormac, Ch. 4.

Closing note on paper strategy (not examinable)

Questions 6, 7 and 8 are alternatives worth 22 marks each, and they are deliberately not equivalent in effort. Question 8 is the shortest of the three once the two-force-member observation is made — the whole right-hand half of both diagrams follows from one 45-degree line — and it needs no stiffness data at all. Question 7 is next: two joint walks and a seven-row table, with three of the seven members contributing nothing. Question 6 carries the most bookkeeping, though the modified-stiffness treatment shown above reduces it to a single balancing operation and is far quicker than the classical iterative table. A candidate who recognises the releases early can finish any of the three inside the 30 minutes the mark allocation implies.

← Question 7 Back to the paper →