07-Str-A2 · Undated paper

Question 3 of 7: A3 — HSS member selection for a two-size truss

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

Notes on this paper

Paper format. 07-Str-A2 Elementary Structural Design, May 2019 sitting, 3 hours, closed book with handbooks and textbooks permitted. Seven questions of 20 marks each: Part A (A1–A3, structural steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). The candidate answers two from Part A, two from Part B and one from Part C — all seven are solved here.

Reference texts. CSA S16 Design of Steel Structures and the CISC Handbook of Steel Construction (section properties, Tables 4-4 and 7); CSA A23.3 Design of Concrete Structures with Brzev & Pao, Reinforced Concrete Design: A Practical Approach; CSA O86 Engineering Design in Wood with the Canadian Wood Council Wood Design Manual; NBCC Part 4 for load combinations.

Question 3: A3 — HSS member selection for a two-size truss (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 pin-and-roller planar truss of 9 m span in three 3 m panels, 4 m deep over the two interior top nodes. Three identical loads $P_1$ act: at the two top nodes and at the bottom node under the second top node.

Given data — A3
Quantity	Value
Span / panel length / depth	9 m / 3 m / 4 m
$P_1$ (from the figure)	100 kN live + 180 kN dead
Load points	U1, U2 (top) and L2 (bottom)
Supports	Pin at A, roller at B
Steel	HSS Class C, 350W: $F_y=350$, $F_u=450$ MPa

Find. The single HSS size that best suits every tension member, and the single HSS size that best suits every compression member.

Figure 3 — A3 truss as dimensioned on page 3 of the paper. Nine members, six joints, pin at A and roller at B.

Approach. Confirm determinacy, factor $P_1$, find the reactions, solve every member force by the method of joints, then size one tension member for the largest tie force and one compression member for the largest strut force at its own length.

Confirm the truss is determinate. Counting nine members, six joints and three reaction components, $$m+r=9+3=12=2n=2(6)\quad\checkmark$$ so the method of joints alone will deliver every force.
Factor the panel load. Page 1 states all loads shown are unfactored, so with $1.25D+1.5L$: $$P_{1f}=1.25(180)+1.5(100)=\boxed{375\ \text{kN}}$$
Find the reactions. Taking moments about A, with the loads at $x=3$, 6 and 6 m: $$B_y=\frac{375(3)+375(6)+375(6)}{9}=\frac{5625}{9}=625\ \text{kN},\qquad A_y=3(375)-625=500\ \text{kN}$$ The right-hand reaction is the larger because two of the three loads sit in the right half.
Work the joints from the supports inward. At joint A the diagonal A–U1 rises 4 in 5, so vertical equilibrium gives $F_{AU1}(4/5)+500=0$, hence $F_{AU1}=-625$ kN, and horizontal equilibrium gives $F_{AL1}=+375$ kN. At joint U1, with the 375 kN load applied, $$F_{U1L1}=500-375=+125\ \text{kN},\qquad F_{U1U2}=-375\ \text{kN}$$ Returning to joint L1, $125+F_{L1U2}(4/5)=0$ gives $F_{L1U2}=-156.25$ kN, and then $F_{L1L2}=375+0.6(156.25)=+468.75$ kN. At joint B, $F_{BU2}(4/5)+625=0$ gives $F_{U2B}=-781.25$ kN and $F_{L2B}=+468.75$ kN, which closes against the bottom chord force found from the left — the equilibrium check.
Tabulate the member forces (tension positive).

Factored member forces — A3
Member	Length (m)	Force (kN)	Action
A–L1	3.0	+375.0	tension
L1–L2	3.0	+468.75	tension (max)
L2–B	3.0	+468.75	tension (max)
U1–L1	4.0	+125.0	tension
U2–L2	4.0	+375.0	tension
U1–U2	3.0	−375.0	compression
L1–U2	5.0	−156.25	compression
A–U1	5.0	−625.0	compression
U2–B	5.0	−781.25	compression (max)

Select the tension member. The governing tie force is 468.75 kN in the bottom chord. For a welded HSS with no bolt holes, $A_{ne}=A_g$ and the shear-lag factor is unity, so the two tension limit states are $$T_r=\min\left(\phi A_gF_y,\ 0.85\phi_uA_{ne}F_u\right)=\min\left(0.9A_g(350),\ 0.85(0.75)A_g(450)\right)$$ The fracture branch is the smaller, requiring $A_g\ge468750/286.9=1634$ mm². Adopt HSS 102 × 102 × 4.8 Class C ($A_g=1770$ mm², $r=39.5$ mm): $$T_r=\min(557.6,\ 507.7)=\boxed{508\ \text{kN}\ >\ 468.75\ \text{kN}}$$ Slenderness is trivially satisfied, $L/r=3000/39.5=76 < 300$.
Select the compression member. The governing strut is the 5 m diagonal U2–B at 781.25 kN, with $K=1.0$ for a pin-jointed truss. Trying HSS 152 × 152 × 9.5 Class C ($A=5150$ mm², $r=57.4$ mm): $$\frac{KL}{r}=\frac{5000}{57.4}=87.1,\qquad \lambda=\frac{87.1}{\pi}\sqrt{\frac{350}{200000}}=1.160$$ $$C_r=0.90(5150)(350)\left(1+1.160^{2.68}\right)^{-1/1.34}=\boxed{822\ \text{kN}\ >\ 781.25\ \text{kN}}$$ The wall is not Class 4: $(b-4t)/t=(152-38)/9.53=12.0 < 670/\sqrt{350}=35.8$.
Confirm the selection is the lightest that works. The next lighter square section, HSS 152 × 152 × 8.0 ($A=4400$ mm², $r=57.9$ mm), gives $C_r=709$ kN — 9 % short of the demand. HSS 152 × 152 × 9.5 is therefore the smallest adequate compression member, at a utilisation of 0.95.

Two sections serve the whole truss: HSS 102 × 102 × 4.8 for every tie and HSS 152 × 152 × 9.5 for every strut. The compression member needs almost three times the area of the tension member even though its force is only 67 % larger, which is the whole story of the question — a tie is limited by material strength and a strut by geometry, so the two limit states diverge sharply once $KL/r$ passes about 80.

Final results — A3
Quantity	Value
Factored panel load $P_{1f}$	375 kN
Reactions $A_y$ / $B_y$	500 kN / 625 kN
Maximum tension (L1–L2, L2–B)	468.75 kN
Maximum compression (U2–B, 5 m)	781.25 kN
Tension member	HSS 102 × 102 × 4.8, $T_r=508$ kN
Compression member	HSS 152 × 152 × 9.5, $C_r=822$ kN

Topic. Statically determinate truss analysis followed by tension and compression member selection, CSA S16 Clauses 13.2 and 13.3.

Key relations. Determinacy $m+r=2n$; joint equilibrium $\sum F_x=\sum F_y=0$; tension $T_r=\min(\phi A_gF_y,\ 0.85\phi_uA_{ne}F_u)$; compression $C_r=\phi AF_y(1+\lambda^{2n})^{-1/n}$ with $n=1.34$ for a Class C HSS.

Why this works. Because the truss is determinate, member forces depend only on geometry and load, so the analysis is complete before any section is chosen — sizing cannot feed back into the force distribution. The two-size restriction then reduces the design to two envelope problems: the largest tie force and the largest strut force at its own length. That last qualifier matters. The 3 m top chord carries 375 kN and the 5 m diagonal carries 781 kN, but even if those forces were equal the longer member would need the bigger section, because $C_r$ falls roughly as the inverse square of length in this slenderness range.

Common pitfall. Sizing the compression member on the largest force without checking which member length it occurs in; and, on the tension side, stopping at $\phi A_gF_y$ and never testing net-section fracture, which governs here and is 9 % lower. A third trap is forgetting to factor the loads at all — page 1 says the figure loads are unfactored, and using 280 kN instead of 375 kN understates every member force by 25 %.

Source. CSA S16, Cl. 13.2 and 13.3.1; CISC Handbook of Steel Construction, HSS Class C property and column-load tables.

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