16-Civ-A1 Elementary Structural Analysis · Undated paper

Question 3 of 8: Member forces in a Pratt truss

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

Notes on this paper

Paper format. Engineers Canada / EGBC national examination 16-Civ-A1 Elementary Structural Analysis, 3 hours, CLOSED BOOK. Six questions constitute a complete paper: answer Q1–Q5 and one of Q6/Q7/Q8. For completeness all eight questions are worked here.

Reference texts: R. C. Hibbeler, Structural Analysis (10th ed., Pearson) — determinacy and stability (Ch. 2), method of joints and sections (Ch. 3), shear and bending-moment diagrams (Ch. 4), influence lines and moving loads (Ch. 6), virtual-work deflections (Ch. 8–9), slope-deflection and moment distribution (Ch. 10–11); A. Kassimali, Structural Analysis (6th ed., Cengage) — internal hinges, compound (Gerber) beams, three-hinged frames, Maxwell’s law of reciprocal deflections.

Sign convention. Upward reactions positive; sagging bending moment positive (tension on the underside of a beam), hogging negative; member axial force tension positive (T), compression negative (C). Shear is positive when the resultant of the forces on the left of a section acts upward.

Check: figure reading. All geometry, loads and support symbols used below were read directly from the printed figures. Where a support symbol governs the answer it is called out explicitly in the Given.

Question 3: Member forces in a Pratt truss (16 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. Bottom-chord joints L₁ to L₆ at 4 m centres (span 20 m); top-chord joints U₁ to U₄ directly above L₂ to L₅ at a height of 3 m. Verticals L₂U₁, L₃U₂, L₄U₃, L₅U₄; diagonals L₁U₁, L₂U₂, L₃U₃, U₃L₅ and U₄L₆. Downward 30 kN loads at L₂, L₃ and L₄. Pin at L₁, roller at L₆. Each sloping web member has length $\sqrt{4^{2} + 3^{2}} = 5\text{ m}$, so its direction cosines are $4/5$ horizontally and $3/5$ vertically.

Find. The axial force in the four listed members, with tension or compression stated.

Q3 — the truss. Panel length 4 m, height 3 m; the sloping web members are 3–4–5 triangles.

Approach. Confirm determinacy, find the reactions from global equilibrium, then take a vertical section through the panel L₂–L₃ for the chord and diagonal, and isolate joints L₁ and L₂ for the end diagonal and the vertical.

Determinacy. With $m = 17$ bars, $j = 10$ pins and $r = 3$, $$m + r - 2j = 17 + 3 - 20 = 0$$ so the truss is statically determinate and the method of joints and sections suffices.
Reactions. Taking moments about L₁, $$R_{L6}(20) = 30(4) + 30(8) + 30(12) = 720$$ $$R_{L6} = \boxed{36\ \text{kN} \uparrow}, \qquad R_{L1} = 90 - 36 = \boxed{54\ \text{kN} \uparrow}$$ There is no horizontal load, so the pin carries no thrust.
End diagonal L₁–U₁, by joint L₁. Only two bars meet the end pin, the bottom chord (horizontal) and the diagonal rising at 3 in 5, so vertical equilibrium alone gives the diagonal: $$\sum F_y = 54 + F_{L1U1}\left(\tfrac{3}{5}\right) = 0$$ $$F_{L1U1} = -54 \times \tfrac{5}{3} = \boxed{90\ \text{kN compression}}$$ A negative result means the bar pushes back on the joint, which is compression — as expected for the top-going end diagonal of a simply supported truss.
Bottom chord L₂–L₃, by sections. Cut vertically between L₂ and L₃: the cut crosses the top chord U₁U₂, the diagonal L₂U₂ and the bottom chord. Taking moments of the left free body about U₂ (at $x = 8\text{ m}$, $y = 3\text{ m}$) eliminates the other two: $$\sum M_{U2} = 54(8) - 30(4) - F_{L2L3}(3) = 0$$ $$F_{L2L3} = \frac{432 - 120}{3} = \boxed{104\ \text{kN tension}}$$
Diagonal L₂–U₂, from the same section. Vertical equilibrium of that left free body involves only the diagonal, because both chords are horizontal: $$\sum F_y = 54 - 30 + F_{L2U2}\left(\tfrac{3}{5}\right) = 0$$ $$F_{L2U2} = -24 \times \tfrac{5}{3} = \boxed{40\ \text{kN compression}}$$ The net panel shear of 24 kN is carried by the diagonal alone.
Vertical L₂–U₁, by joint L₂. Four bars and the 30 kN load meet here; of the four only the vertical and the diagonal have vertical components: $$\sum F_y = F_{L2U1} + F_{L2U2}\left(\tfrac{3}{5}\right) - 30 = 0$$ $$F_{L2U1} = 30 + 40\left(\tfrac{3}{5}\right) = \boxed{54\ \text{kN tension}}$$ This hanger carries the panel load plus the vertical component thrown at it by the compression diagonal.
Independent check at joint L₂. Horizontal equilibrium was not used above, so it is a free check. With $F_{L1L2} = 72\text{ kN}$ tension (from horizontal equilibrium at L₁): $$-72 + 104 + (-40)\left(\tfrac{4}{5}\right) = -72 + 104 - 32 = 0 \quad\checkmark$$

Member	Force	Sense
L₁–U₁	90 kN	Compression
L₂–L₃	104 kN	Tension
L₂–U₂	40 kN	Compression
L₂–U₁	54 kN	Tension
Reaction at L₁	54 kN	upward
Reaction at L₆	36 kN	upward

Topic. Axial forces in a statically determinate planar truss by the method of joints and the method of sections.

Key relations. Determinacy $m + r - 2j = 0$; joint equilibrium $\sum F_x = \sum F_y = 0$ at every pin; for a section cutting three non-concurrent bars, moments about the point where two of them meet isolate the third; for parallel chords the panel shear is carried entirely by the diagonal, $F_{\text{diag}} = -V_{\text{panel}}/\sin\theta$.

Why this works. Pin-jointed bars carry axial force only, so a free body cut through the truss is a rigid body loaded by three unknown bar forces and the known external loads — three equilibrium equations, three unknowns. Choosing the moment centre at the intersection of two unknowns turns a 3×3 solve into a single line of arithmetic; choosing $\sum F_y$ when both chords are horizontal does the same for the diagonal. The method of joints is the degenerate case where the “section” encircles a single pin, which is why an end pin with only two bars can be solved immediately.

Common pitfall. Mis-reading the web layout. In this truss the diagonals run L₂U₂ and L₃U₃ (rising to the right) but U₃L₅ and U₄L₆ (falling to the right) — a Pratt arrangement that reverses at the centre. If you assume a uniform diagonal direction the panel-shear sign flips and L₂U₂ comes out in tension. Second pitfall: forgetting the diagonal’s vertical component at L₂, which gives 30 kN instead of 54 kN for the hanger.

Source. Hibbeler, Structural Analysis, 10th ed., §3.4 (method of joints) and §3.6 (method of sections); Kassimali, Structural Analysis, 6th ed., Ch. 4.

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