Question 2 of 7: Figure 2 frame — one wide-flange section for the beam and both columns

Question A2: Figure 2 frame — one wide-flange section for the beam and both columns (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.

Find. The lightest W section that satisfies CSA S16 for every member of the frame when one section is used throughout, treating the frame as restrained against sidesway.

[Figure not reproduced: Figure A2.1 — Figure 2 of the paper redrawn. Built-in bases at A and F (the paper draws plain hatch there, not the triangle it uses for a pin in Figure 1). P1 = 180 kN dead + 100 kN live at B, C and D; P2 = 150 kN live at E. See the official exam paper.]

Approach. Because every member is the same section the elastic analysis is independent of which section that is, so the frame is solved once by the direct-stiffness method; the beam is then checked in flexure under Clause 13.6 and each column under the three Clause 13.8.2 beam-column interaction equations, and the lightest shape that clears all of them is selected.

1. Factor the loads. $$P_{1f} = 1.25(180) + 1.5(100) = 225 + 150 = 375\ \text{kN}\ \text{(down, at B, C and D)}$$ $$P_{2f} = 1.5(150) = 225\ \text{kN}\ \text{(horizontal, at E)}$$ P2 carries no dead component, so it attracts the full live-load factor.
2. Note that the analysis does not depend on the section. A frame in which every member has the same $EI$ and $EA$ distributes moment purely by geometry: the stiffness terms cancel out of the compatibility equations. Solving the frame with W610×113, W690×140 and W610×155 in turn changes the computed end moments by less than 0.2 per cent, so one analysis serves for all candidates. This is what makes the question tractable by hand in 20 marks.
3. Solve the frame. The frame has six reaction components and three equations of equilibrium, so it is three times redundant; a direct-stiffness solution with the four nodes B, C, D and E free gives the factored member end actions tabulated below. Equilibrium confirms the result: the base shears sum to $119.0 + 106.0 = 225\ \text{kN} = P_{2f}$ and the base reactions to $577.5 + 547.5 = 1125\ \text{kN} = 3P_{1f}$.
4. Identify the design actions. The largest moment in the frame is the sagging moment at the beam mid-span node C, $$\boxed{M_f = 676.3\ \text{kN}\cdot\text{m} \ \text{at C}}$$ and the most heavily worked member is column A–B, which carries $C_f = 577.5\ \text{kN}$ together with $M_f = 538.6\ \text{kN}\cdot\text{m}$ at its top and $414.8\ \text{kN}\cdot\text{m}$ at its base. Note that the loads at B and at D sit directly over the columns, so they enter the columns as axial force and bend the beam only through the frame action.
5. Set the required plastic modulus from the beam. $$Z_x \ge \frac{M_f}{\phi F_y} = \frac{676.3\times10^6}{0.9 \times 300} = 2.505\times10^6\ \text{mm}^3$$ That eliminates everything below about W530×123 — but it is not the governing requirement, as the next step shows.
6. Fix the bracing assumption before checking the columns. Lateral support is taken at the joints and at the points of load application. On the beam that means braces at B, C and D, giving 6 m unbraced segments. On column D–E–F it means a brace at E where P2 is applied, giving two 4 m segments. Column A–B has no load between its ends, so it is unbraced over its full 8 m — and that single fact controls the whole selection.
7. Check the beam. For W690×140 the segment C–D runs from 676.3 sagging at C to 358.9 hogging at D, so the moment diagram crosses zero and the gradient factor is large: $\omega_2 = 2.310$. Clause 13.6(a) over the 6 m segment then gives $M_r = 1188.3\ \text{kN}\cdot\text{m}$, so $$\frac{M_f}{M_r} = \frac{676.3}{1188.3} = 0.569 \quad \checkmark$$ and the beam shear ratio is only 0.114. The beam is not the problem.
8. Check column A–B under Clause 13.8.2. The two end moments have opposite sign, so the member is in double curvature: $\kappa = +414.8/538.6 = +0.770$, hence $\omega_1 = 0.6 - 0.4\kappa = 0.29$, taken at its floor of 0.40, and $U_{1x} = \omega_1/(1 - C_f/C_{ex})$ is governed by its own lower limit of 1.0. The same reversing diagram gives $\omega_2 = 2.406$ for the buckling check. With $r_y = 54.2\ \text{mm}$ over 8 m, $KL/r_y = 148$, which is slender: $C_r$ falls from 4756 kN at zero slenderness to 1252 kN.
9. Evaluate the three interaction equations. Using $C_f/C_r + 0.85\,U_1 M_f/M_r \le 1.0$ with the appropriate $C_r$ and $M_r$ in each case: $$\begin{aligned} \text{(a) cross-section} &: \frac{577.5}{4756} + \frac{0.85(538.6)}{1211.6} = 0.499 \\ \text{(b) member} &: \frac{577.5}{1252} + \frac{0.85(538.6)}{1211.6} = 0.839 \\ \text{(c) lateral-torsional} &: \frac{577.5}{1252} + \frac{0.85(538.6)}{1075.7} = \boxed{0.887} \end{aligned}$$ Check (c) governs, at 0.887. Column D–E–F, braced at E, returns only 0.429 and 0.392 for its two 4 m segments — less than half the demand on A–B despite carrying almost the same axial load.
10. Confirm that the next lighter shape fails. W690×125 satisfies the beam requirement ($\phi M_p = 1066\ \text{kN}\cdot\text{m}$) and even scrapes through check (b) at 0.966, but its $r_y = 52.8\ \text{mm}$ leaves too little buckling resistance over the 8 m storey and check (c) returns 1.032 — a genuine failure. W610×140, at 139.0 kg/m, passes at 0.982 but is both heavier and more highly worked than W690×140 at 138.3 kg/m. The answer is therefore W690×140.
11. Sanity-check where the material is going. The beam is at 0.57, the lower column at 0.43 and the critical column at 0.89. In a one-section frame that spread is unavoidable and is the price of the fabrication constraint; the section is chosen by the buckling resistance of the tallest unbraced column, not by the largest moment in the frame.

The complete set of factored member end actions from the analysis is:

Final Results