Question 6 of 7: Check of the steel beam-column AB (14 + 6 marks)

Question 6: Check of the steel beam-column AB (14 + 6 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. Column AB of the Question 3 frame — W840×251, 12 m long, pinned at A — carrying the collapse actions found there: 1230 kN of axial compression with 855 kN·m at the top and zero moment at the base.

Find. Whether W840×251 satisfies all three parts of Clause 13.8.2 — cross-sectional strength, overall in-plane member strength and lateral-torsional buckling strength — and, if not, what the cure is.

Approach. Run the three interaction checks in order. The section was selected in Question 3 purely on plastic modulus, so the informative question is whether its slender weak axis over an unbraced 12 m can carry the axial load at all.

1. Check (a) — cross-sectional strength. Here $C_r$ is taken at zero slenderness and $U_{1x}\ge1.0$: $$C_r=\phi AF_y=0.9(26\,889)(350)=8470\ \text{kN},\qquad M_r=\phi ZF_y=2680\ \text{kN}\cdot\text{m}$$ $$\frac{C_f}{C_r}+\frac{0.85U_{1x}M_{fx}}{M_{rx}} =\frac{1230}{8470}+\frac{0.85(855)}{2680}=0.145+0.271=\boxed{0.42}\ \le\ 1.0$$ The cross-section itself is barely stressed.
2. Check (b) — overall in-plane member strength. The frame sways, and with a pinned base and a joint stiffness ratio $G_B=(I_c/L_c)/(I_b/L_b)=1.59$ the sway alignment chart gives $K\approx2.4$, so $$\frac{KL}{r_x}=\frac{2.4(12\,000)}{343.5}=83.8\ \Rightarrow\ \lambda=1.116,\qquad C_r=8470\left(1+\lambda^{2.68}\right)^{-1/1.34}=4487\ \text{kN}$$ $$\frac{1230}{4487}+\frac{0.85(855)}{2680}=0.274+0.271=\boxed{0.55}\ \le\ 1.0$$ In its own plane the member is fine: the deep section gives it a very large $r_x$.
3. Check (c) — lateral-torsional buckling strength. Out of plane the member is held only at its ends, so $KL=12\,000$ mm against $r_y=61.4$ mm: $$\frac{KL}{r_y}=195.3\ \Rightarrow\ \lambda=2.60,\qquad C_r=\boxed{1185\ \text{kN}}$$ which is less than the applied $C_f=1230$ kN — the member cannot carry the axial load alone, let alone with moment. Completing the check with $\omega_2=1.75$ (moment varying from 855 to zero), $$M_u=\frac{\omega_2\pi}{L}\sqrt{EI_yGJ+\left(\frac{\pi E}{L}\right)^2I_yC_w} =1519\ \text{kN}\cdot\text{m},\qquad M_r=1367\ \text{kN}\cdot\text{m}$$ $$\frac{1230}{1185}+\frac{0.85(855)}{1367}=1.038+0.532=\boxed{1.57}\ >\ 1.0$$
4. Report the verdict. W840×251 as detailed is not satisfactory. It passes the cross-section check at 0.42 and the in-plane check at 0.55, but fails lateral-torsional buckling by 57 per cent. The pattern of the three ratios names the defect precisely: a member that is comfortable on cross-section and in-plane strength yet fails badly out of plane has a slenderness deficiency, not an area deficiency.
5. Give the remedy, and show why upsizing is the wrong one. Adding a single lateral brace at mid-height halves the weak-axis unbraced length: $$\frac{KL}{r_y}=\frac{6000}{61.4}=97.7\ \Rightarrow\ C_r=3711\ \text{kN}$$ and the critical upper segment, with moment falling from 855 to 428 kN·m ($\omega_2=1.30$), gives $M_r=2338$ kN·m, so $$\frac{1230}{3711}+\frac{0.85(855)}{2338}=0.331+0.311=\boxed{0.64}\ \le\ 1.0$$ A brace at mid-height costs almost nothing and takes the member from 1.57 to 0.64. Choosing a heavier deep section instead would not work: the deficiency is in $r_y$ and unbraced length, and any W840 shape has a weak axis of the same order, so the 12 m unbraced length would still govern.
6. Note the consequence for Question 3. Bracing the column changes no strength, so the collapse mechanism, $M_p=1230$ kN·m, and the beam and column selections all stand. What changes is the framing: the design must call for secondary steel at mid-height of both columns, and the same brace should be provided at D, whose column carries 1770 kN with a 1845 kN·m top moment and is therefore the more heavily loaded of the two.