Question 2 of 7: Rigid-jointed frame — one wide-flange section for beam and columns

Question A2: Rigid-jointed frame — one wide-flange section for beam and 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, used for all five members, satisfies S16 Clause 13.8.2 as a beam-column under the second-order forces of this unbraced frame.

Approach. Factor the loads, solve the once-redundant frame by the stiffness method, repeat the solution with the geometric stiffness of the columns included so that P-delta effects are captured, and then check the beam and both columns as beam-columns with the unbraced lengths the frame actually provides.

1. Factor the loads. $$P_{1f} = 1.25(180) + 1.5(100) = 375\ \text{kN}, \quad P_{2f} = 1.25(360) + 1.5(200) = 750\ \text{kN}$$ $$H_f = 1.5(50) = 75\ \text{kN}, \quad \sum C_f = 2(375) + 750 = 1500\ \text{kN}$$ The frame has no bracing, so S16 Clause 8.4.2 adds a notional lateral load of $0.005(1500) = 7.5\ \text{kN}$ at deck level, in the same direction as the real horizontal load.
2. Count the redundancy and set up the analysis. Two pinned bases give four reaction components against three equations of equilibrium, so the frame is once redundant. It was solved with a plane-frame stiffness model of five prismatic members (A–B, B–C, C–D, D–E, E–F) carrying axial, shear and flexural stiffness, with the member self weight applied as a distributed load through consistent fixed-end forces.
3. First-order results. For the trial section W840x176 ($I_x = 2420\times10^{6}\ \text{mm}^4$) the first-order solution gives $$M_B = 1126\ \text{kN}\!\cdot\!\text{m}, \quad M_C = 1650\ \text{kN}\!\cdot\!\text{m}, \quad \Delta = 61.0\ \text{mm}$$ The beam moment is close to the simply supported value $P_{2f}L/4 = 2250\ \text{kN}\!\cdot\!\text{m}$ reduced by the average end restraint, which is the check that the analysis is behaving.
4. Add second-order effects. A 12 m pin-based column is very flexible, so the P-delta contribution is not negligible: with $\sum C_f = 1500\ \text{kN}$ acting through a 61 mm sway the secondary moment is of the order of 90 kN·m. The model was therefore re-solved with the consistent geometric stiffness matrix of each member formed from its own axial force and iterated to convergence, with the notional load included: $$\Delta_{2nd} = 70.7\ \text{mm}, \quad \text{amplification} = 70.7/61.0 = 1.16$$ $$\boxed{M_B = 1191\ \text{kN}\!\cdot\!\text{m},\ \ M_C = 1673\ \text{kN}\!\cdot\!\text{m},\ \ M_D = 45\ \text{kN}\!\cdot\!\text{m}}$$ The corresponding reactions are $A_y = 883.9$, $F_y = 684.3$, $A_x = 94.1$ and $F_x = 11.6\ \text{kN}$; the sum of the vertical reactions reproduces the applied load plus the member self weight, and the base shears sum to the applied 82.5 kN.
5. Note where the horizontal load actually goes. The 12 m column carries 94.1 kN of base shear and the 8 m column only 11.6 kN, even though the shorter column is the stiffer of the two. The reason is that the frame is asymmetric: the gravity load alone sways it, and that gravity sway happens to oppose the lateral load in the short column. Designing the short column for its share of a symmetric-frame shear would be wrong by a factor of five.
6. Establish the unbraced lengths. The paper states no bracing, so the working assumption is lateral support at the joints and at the load points only — B, C and D on the beam and the ends of each column. That gives $L_b = 6\ \text{m}$ for each beam segment, 12 m for column A–B and 8 m for column D–F. Because the second-order analysis has already been performed, $K = 1.0$ is used for in-plane strength and $U_{1x} = 1.0$ in checks (b) and (c) of Clause 13.8.2.
7. Check the beam segment B–C. The moment runs from $-1191$ at B to $+1673$ at C, so $\kappa = 0.71$ in double curvature and $\omega_2 = 1.75 + 1.05(0.71) + 0.3(0.71)^{2} = 2.65$, capped at 2.5. With $Z_x = 6.71\times10^{6}\ \text{mm}^3$ the section is Class 1 ($b_f/2t_f = 7.77 \le 8.37$; $h/w = 57.0 \le 60.0$ allowing for the small axial force): $$\phi M_p = 0.9(6.71\times10^{6})(300) = 1810.6\ \text{kN}\!\cdot\!\text{m}$$ $$\frac{C_f}{C_r} + 0.85\frac{M_f}{M_r} = \frac{94.1}{2774} + 0.85\frac{1673}{1810.6} = 0.819$$
8. Check column A–B, which is where the design is decided. Carrying $C_f = 883.9\ \text{kN}$ and $M_f = 1191\ \text{kN}\!\cdot\!\text{m}$ over an unbraced 12 m, the weak-axis slenderness is $$\frac{KL}{r_y} = \frac{12\,000}{59.4} = 202 \ \Rightarrow\ C_r = 907\ \text{kN}, \quad M_{r,LTB} = 942.8\ \text{kN}\!\cdot\!\text{m}$$ $$\frac{883.9}{907} + 0.85\frac{1191}{942.8} = 2.05 \ \gg 1.0 \quad \text{(fails)}$$ This is a slenderness deficiency, not an area deficiency: checks (a) and (b), which use the strong axis, return only 0.71 and 0.72. No commercially available W section fixes it, because every shape deep enough to carry 1673 kN·m in the beam has $r_y$ of the same order.
9. Supply the remedy and re-check. A single lateral brace at mid-height of column A–B halves both the unbraced length and the effective length about the weak axis: $$\frac{KL}{r_y} = \frac{6000}{59.4} = 101 \ \Rightarrow\ C_r = 2774\ \text{kN}, \quad M_{r,LTB} = 1735.2\ \text{kN}\!\cdot\!\text{m}$$ $$\boxed{\frac{883.9}{2774} + 0.85\frac{1191}{1735.2} = 0.902 \le 1.0}$$ Column D–F, carrying 667 kN with only 45 kN·m over 8 m, returns 0.390 and needs no brace. Bracing changes no strength, so the frame analysis and the moments above all stand.
10. Check shear and serviceability drift. The largest beam shear is 470.5 kN against $V_r = \phi A_w F_s = 2083\ \text{kN}$, so shear is nowhere near critical. Under the 50 kN lateral service load alone the frame drifts 24.1 mm, that is $h/498$ for the 12 m column, inside the customary $h/400$ limit. Note that most of the 70.7 mm factored drift comes from gravity-induced sway of the asymmetric frame rather than from the lateral load.