Question 2 of 7: A2: One wide-flange section for the whole frame of Figure 2
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 07-Str-A5 Advanced Structural Design, printed footer 07-Str-A5/May 2019. Three hours, "closed book" with handbooks and textbooks permitted. Seven questions of equal value (20 marks each) in three parts: A1–A3 (do two), B1–B3 (do two), C1 (do one) — five solutions make a complete paper. All seven are solved here, because the set is a study resource rather than an exam attempt.
Design data given on page 1. Solutions to CAN/CSA S16 (steel), CAN/CSA A23.3 (concrete) and CAN/CSA O86 (timber). All loads shown on the figures are unfactored. All structural steel is G40.21 300W — $F_y = 300$ MPa, $F_u = 450$ MPa, $E = 200\,000$ MPa, $G = 77\,000$ MPa. All reinforcement is 400W — $f_y = 400$ MPa. Concrete strength is stated per question. Steel sections are W-shapes unless noted otherwise.
Check — load combinations. The figures split the loads into DEAD, LIVE, SNOW and WIND, so NBCC 2020 Table 4.1.3.2 is applied literally. Throughout Part A the governing case is $1.25D + 1.5L$ (case 2 with no companion snow or wind on the figure). For Figures 3 the governing case is $1.25D + 1.5S + 0.4W$ (case 3), with $1.25D + 1.4W + 0.5S$ and $1.4D$ checked and shown not to govern. The "LIVE/SNOW" label on W1 of Figure 3 is read as a single 3 kN/m specified roof load that may be either occupancy live or balanced snow, and the triangular W2 is read as the additional snow-drift surcharge that its shape implies.
Question 2 — A2: One wide-flange section for the whole frame of Figure 2 (20 marks)
Given. An unbraced single-bay frame with pinned bases at A and F, rigid beam-to-column connections at B and D, a 12 m left column, a 12 m beam and an 8 m right column (so that base F sits 4 m above base A), carrying three vertical point loads and one horizontal load at the deck.
Given data — Figure 2
Quantity
Symbol
Value
Column A–B / beam B–C–D / column D–F
—
12 m / 6 m + 6 m / 8 m
Supports and joints
—
A and F pinned; B and D rigid connections
Vertical load at B and at D (dead / live)
$P_1$
180 / 100 kN
Vertical load at C (dead / live)
$P_2$
360 / 200 kN
Horizontal load at D, acting to the left (live)
$H$
50 kN
Steel
$F_y$
300 MPa
Find. The lightest single W-shape that serves simultaneously as the beam and as both columns of this sway frame.
[Figure not reproduced: Figure 2 as printed: base F sits 4 m above base A, so the two columns differ in both height and stiffness. See the official exam paper.]
Approach. Run a first-order elastic frame analysis, amplify it for sway with a P-Δ iteration (CSA S16 Cl. 8.4.3), then check the more heavily loaded column against all three beam-column equations of Cl. 13.8.2 and the beam against Cl. 13.6.
Factor the loads. Case 2 of NBCC Table 4.1.3.2 again governs, with the horizontal load part of the live-load group:
$$P_{1f} = 375\ \text{kN},\qquad P_{2f} = 750\ \text{kN},\qquad H_f = 1.5(50) = 75\ \text{kN}$$
Analyse the frame. The frame has four reaction components against three equations, so it is once statically indeterminate; a direct-stiffness solution with nodes at A(0, 0), B(0, 12), C(6, 12), D(12, 12) and F(12, 4) is used. The first-order sway of the deck is 67.4 mm to the left.
Amplify for stability effects. With 1 500 kN of gravity riding on a 67 mm sway, second-order effects are not negligible. Iterating the equivalent lateral force $\sum C_{fi}\Delta/h_i$ until it converges gives an effective storey shear of 86.2 kN and
$$\Delta = 73.5\ \text{mm},\qquad U_2 = \frac{86.2}{75.0} = 1.149$$
Extract the design actions. The converged second-order analysis gives, for the tall left column and the beam,
$$\boxed{C_f = 839.4\ \text{kN},\quad M_{f,\text{col}} = 1\,150.4\ \text{kN}\cdot\text{m}\ \text{at B},\quad M_{f,\text{beam}} = 1\,636.2\ \text{kN}\cdot\text{m}\ \text{at C}}$$
with a beam shear of 464.4 kN at B and 660.6 kN of axial load in the short right column. The two column axial loads sum to 1 500 kN, as they must.
Try W610×241. $A = 30\,800$ mm$^2$, $Z_x = 7\,650\times10^3$ mm$^3$, $I_y = 184\times10^6$ mm$^4$, $r_x = 264$ mm, $r_y = 77.3$ mm. Flange $b/2t = 5.31$ and web $h/w = 32.0$ put it in Class 1 even with the axial load present, so
$$\phi C_y = \phi AF_y = 8\,316\ \text{kN},\qquad \phi M_p = \phi Z_xF_y = 2\,065.5\ \text{kN}\cdot\text{m}$$
Compression resistances for the column. Taking $K = 1.0$ in plane (legitimate once a second-order analysis has been done) and $K = 1.0$ out of plane over the full 12 m, Cl. 13.3.1 with $n = 1.34$ gives
$$C_r = \phi AF_y\left(1 + \lambda^{2n}\right)^{-1/n}: \qquad C_{rx} = 7\,205\ \text{kN},\qquad C_{ry} = 2\,012\ \text{kN}$$
the weak-axis value reflecting $\lambda = 1.914$ at $KL/r_y = 155$.
Lateral–torsional resistance of the column. The column moment runs linearly from zero at the pinned base to 1 150.4 kN$\cdot$m at B, so $\omega_2 = 1.746$ and, over $L = 12\,000$ mm, $M_u = 2\,507$ kN$\cdot$m, hence $M_r = 1\,766.4$ kN$\cdot$m.
Beam-column checks, Cl. 13.8.2. All three inequalities are evaluated with $U_{1x} = 1.0$, as Cl. 13.8.4 permits for a sway frame analysed to second order:
$$\frac{C_f}{C_r} + \frac{0.85\,U_{1x}M_{fx}}{M_{rx}} \le 1.0$$
giving 0.574 for cross-sectional strength, 0.590 for overall member strength and
$$\boxed{\frac{839.4}{2\,012} + \frac{0.85(1\,150.4)}{1\,766.4} = 0.417 + 0.554 = 0.971}$$
for the lateral–torsional case, which governs.
Check the beam. Taking lateral restraint at the three load points, the critical 6 m segment is C–D, whose moment falls from 1 636.2 kN$\cdot$m to 77.2 kN$\cdot$m; $\omega_2 = 1.797$ raises $M_u$ far above $0.67M_p$, so $M_r$ reaches the full $\phi M_p = 2\,065.5$ kN$\cdot$m. With 95.9 kN of axial load in the beam the interaction ratio is 0.692, and shear is trivial: $V_r = 2\,025$ kN against 464 kN.
Confirm nothing lighter works. W610×195, W610×217 and W690×217 return 1.283, 1.107 and 1.004 respectively on the Cl. 13.8.2(c) check — all fail, the last only marginally. W610×241 is therefore the lightest adequate shape.
Check — two assumptions worth writing on the answer paper. First, the columns are assumed unbraced out of plane over their full height and the beam braced only at B, C and D; a real building would brace the columns at mid-height by wall framing, which would immediately relieve the governing 0.971 check. Second, the frame's serviceability sway under unfactored dead plus live load is 47.1 mm, or h/254, against the h/500 that NBCC commentary suggests for lateral drift. The frame satisfies strength, but a single-bay pin-based portal of this proportion is too flexible; the practical recommendation is to fix the bases or add a bracing bay rather than to keep increasing the section.
