Question 3 of 7: The same frame designed as a sway frame — one section for AB, BD and DF

Question A3: The same frame designed as a sway frame — one section for AB, BD and DF (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. The frame, loads and steel grade of question A2, but the frame is now to be treated explicitly as an unbraced sway frame: no external bracing resists the 225 kN factored horizontal load, so lateral stability comes entirely from the frame itself and second-order effects must be accounted for. Members AB, BD and DF are one W section.

Find. A W section that satisfies CSA S16 for the frame including S16 notional lateral loads and second-order (P–Δ) effects, and a statement of how much the sway condition costs relative to the braced design of A2.

Approach. Add the notional lateral load required for an unbraced frame, run a geometric-stiffness (P–Δ) analysis to convergence, evaluate the storey stability index to classify the frame, and re-run the Clause 13.8.2 checks on the amplified moments with $K = 1.0$ as Clause 13.8.2 permits when second-order effects are included in the analysis.

1. Read the difference between A2 and A3. A2 asks for the section on the frame as drawn; A3 names it a sway frame, which under S16 requires two extra provisions — a notional lateral load representing erection out-of-plumb, and an analysis that captures the P–Δ amplification of the sidesway. Everything else, including the members, the loads and the bracing assumption, is unchanged, so this question is the stability audit of the A2 design.
2. Apply the notional lateral load. S16 requires a notional load of 0.005 times the factored gravity load at each level, applied in the direction that aggravates the sway. With the total factored gravity load $\sum C_f = 3(375) = 1125\ \text{kN}$, $$N = 0.005 \sum C_f = 0.005(1125) = 5.63\ \text{kN}$$ added at the beam level. It is small in absolute terms because the real lateral load is 40 times larger, but it must be shown.
3. Compute the first-order sidesway. The first-order analysis of A2 gives a horizontal displacement at beam level of $\Delta_o = 11.45\ \text{mm}$ under the 225 kN factored lateral load. That is $h/699$ for the 8 m storey — a stiff frame, and the reason is the pair of built-in bases.
4. Classify the storey with the stability index. $$Q = \frac{\sum C_f \, \Delta_o}{\sum V_f \, h} = \frac{1125 \times 0.01145}{225 \times 8.0} = \boxed{0.0072}$$ The conventional threshold separating sway from non-sway behaviour is 0.05. At 0.0072 this storey is an order of magnitude inside it, so although the frame is unbraced in the sense that nothing but the frame resists lateral load, it behaves as a non-sway frame for stability purposes. The corresponding sway magnifier is $U_2 = 1/(1 - Q) = 1.007$.
5. Run the second-order analysis anyway, rather than relying on the index. Adding the geometric stiffness of the axial forces to every member and iterating to convergence (three cycles) raises the sidesway from 11.45 mm to 11.93 mm, an amplification of 1.042, and raises the critical column moment from 538.6 to $544.3\ \text{kN}\cdot\text{m}$, an amplification of 1.011. The moment grows far less than the displacement because most of the column moment comes from the gravity frame action, not from the sway.
6. Re-run the Clause 13.8.2 checks on the amplified actions. With the second-order moments and $K = 1.0$ in the plane of bending, W690×140 returns $$\begin{aligned} \text{(a)} &= 0.503 \ \ (\text{was } 0.499) \\ \text{(b)} &= 0.844 \ \ (\text{was } 0.839) \\ \text{(c)} &= 0.892 \ \ (\text{was } 0.887) \end{aligned}$$ a rise of half a percentage point. The beam is unchanged at 0.568.
7. State the answer. W690×140 can be used, and the sway condition costs it nothing: the section that satisfies the braced design also satisfies the unbraced one, because the built-in bases make the frame far too stiff for P–Δ effects to matter. Had the bases been pinned, $\Delta_o$ would have risen several-fold, $Q$ would have approached the 0.05 threshold, and the amplified moment would have pushed check (c) past 1.0.
8. Recommend the better section for the same money. A3 asks what section can be used, not what is lightest, and there is a materially better answer at almost the same mass. W610×155 (153.8 kg/m, 11 per cent heavier) has $r_y = 74.2\ \text{mm}$, a factor of 1.37 above the 54.2 mm of W690×140, and returns $$\text{(c)} = 0.625 \ \text{instead of}\ 0.892.$$ Its flange is Class 2 rather than Class 1, which is immaterial here because the design is elastic and Class 2 still develops $M_p$. For a frame whose critical member is an 8 m unbraced column, buying width rather than depth is the efficient trade.

Final Results