Question 4 of 7: Plastic design of the steel rigid frame

Question 4: Plastic design of the steel rigid frame

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 of Figure 2: columns AB and ED, each 8 m and built in at the base; beam BD spanning 16 m with a real hinge at its mid-point C. Unfactored loads are 500 kN at each joint (B and D) and 300 kN at 4 m from each column. All members share one plastic moment $M_p$; $F_y = 350$ MPa. Lateral support exists at all joints and load points, i.e. at 4 m centres along the beam and at the ends only of each column.

Find. (a) an adequate rolled section; (b) a welded knee at D.

[Figure not reproduced: Figure 2 as printed. C is a genuine hinge in the beam, not a plastic hinge, so the frame is twice redundant. See the official exam paper.]

Approach. Exploit the symmetry to make the beam moments determinate, confirm the collapse load by virtual work, select the lightest Class 1 section, then check the columns as beam-columns and design the knee.

1. Part (a) — count the redundants. Two built-in bases give six reactions and three equations of global equilibrium, so a portal with rigid joints is three times redundant. The hinge at C supplies the condition $M_C = 0$ and reduces this to two. Collapse therefore needs three hinges, one of which the frame already has.
2. Use symmetry to make the beam determinate. Structure and loading are both symmetric about C, so the shear there vanishes: $V_C = 0$, and the hinge gives $M_C = 0$. Cutting at C and taking the left half as a free body, there is no load between $x = 4$ m and $x = 8$ m, so the moment is zero over that stretch; only the 300 kN load acts to the left of it. Taking moments about B, $$M_B = P_i \, a = 300 \times 4 = 1200 \text{ kN}\cdot\text{m} \text{ (unfactored)}$$ The two 500 kN loads sit directly over the columns and contribute nothing. This moment is fixed by statics alone — it does not depend on the section chosen.
3. Confirm the mechanism by virtual work. Impose a rotation $\theta$ at each of B and D. The beam halves rotate about those points, the real hinge at C opens, and the point at $x = 4$ m drops $\delta = a\theta = 4\theta$. With the factored loads $$W_{ext} = 2 \times 450 \times 4\theta = 3600\theta, \qquad W_{int} = M_p (\theta_B + \theta_D) = 2 M_p \theta$$ so $M_p = 1800$ kN·m — the same answer, which confirms that upper and lower bounds coincide and the solution is exact. Restricting the mechanism to one half of the beam (hinges at B, at 4 m and at C) gives only 600 kN·m and is therefore not critical.
4. Add the frame self weight. Taking the section that emerges below (W840x193, 193 kg/m, $= 1.89$ kN/m factored at 1.25) and repeating the free-body calculation adds $w L_{half}^2/2 = 75.7$ kN·m, so $$\boxed{M_p = 1800 + 75.7 = 1875.7 \text{ kN}\cdot\text{m}}$$ An independent stiffness analysis of the whole frame with the hinge released returns exactly 1875.7 kN·m at B and D, together with a column axial load of 1218.9 kN and a base moment of 777.7 kN·m.
The collapse mechanism and the factored moment field. The beam moment is zero between the two interior loads.
5. Select the section on plastic modulus and class. Plastic design under Cl 13.5 requires a Class 1 section throughout, so the flange must satisfy $b/2t \le 145/\sqrt{F_y} = 7.75$ and the web $h/w \le 1100/\sqrt{F_y} = 58.8$. The required modulus is $$Z_x \ge \frac{M_p}{\phi F_y} = \frac{1875.7 \times 10^{6}}{0.9 \times 350} = 5955 \times 10^{3} \text{ mm}^{3}$$ W840×176 has ample modulus at $6706 \times 10^3$ mm$^3$ and is the lightest shape that clears it — but its $b/2t = 7.77$ makes it Class 2 by two hundredths, and a Class 2 section is not admissible in plastic design. That disqualification is a clause check, not an arithmetic one, and is easy to miss.
6. Check the columns, which govern the choice. The joint delivers the full $M_p$ into the column top, so each column carries $C_f = 1218.9$ kN together with $M_f = 1875.7$ kN·m at D and 777.7 kN·m at the base — single curvature, $\kappa = -0.415$, hence $\omega_1 = 0.766$ and $\omega_2 = 1.358$. With lateral support only at the ends, the unbraced length out of plane is the full 8 m and Cl 13.8.2(c) returns 1.530 for W840x193 — a clear failure, and every candidate section in the W690 to W840 range fails the same check unbraced. The deficiency is slenderness, not area: $r_y = 60.8$ mm gives $KL/r_y = 131$. One lateral brace at mid-height halves it, and then $$\begin{gathered}\text{(a) } 0.832, \\ \text{(b) } 0.845, \\ \text{(c) } 0.927\end{gathered}$$ all satisfy Cl 13.8.2. Adopt W840x193 throughout, with one lateral brace at mid-height of each column. It is the lightest Class 1 shape that survives; the lighter W760×173 reaches 1.02 on the cross-sectional check alone.
7. Check the beam. Braced at 4 m centres and with $\omega_2 = 1.746$ over the critical B-to-load segment, Cl 13.6 returns $M_r = 2368$ kN·m — the full $\phi M_p$, because lateral torsional buckling is not critical at that spacing. Including the 137.2 kN of axial thrust the beam-column ratios are 0.691 and 0.700. The end shear is 468.9 kN against $V_r = 0.66 \phi d w F_y = 2567$ kN, a utilisation of 0.183.
8. Part (b) — size the flange forces at the knee. The moment arriving at D is the full $M_p$. Resolving it into a flange couple, $$F_f = \frac{M_f}{d - t} = \frac{1875.7 \times 10^{6}}{840 - 21.7} = 2292 \text{ kN}$$ A complete-joint-penetration groove weld across each beam flange, plus a CJP weld or full-strength fillets at the web, develops the whole section and needs no calculation of weld size beyond specifying matching electrodes.
9. Check the panel zone. The web of the corner must carry that flange force less the column shear: $$V_{f,panel} = F_f - V_{col} = 2292 - 137.2 = 2155 \text{ kN}$$ against $V_r = 0.55 \phi w d F_y = 2139$ kN, a ratio of 1.007. The classical plastic-design expression gives the same verdict: the required panel thickness is $$w_{req} = \frac{\sqrt{3}\, M_p}{\phi\, d_b d_c F_y} = 14.62 \text{ mm}$$ against the 14.7 mm supplied. The panel is at its limit with no reserve, so provide a pair of diagonal stiffeners 135 × 20 mm across the knee; they raise the panel resistance to about 3300 kN and take the ratio to 0.65.
10. Add the continuity stiffeners. The column web alone resists only $B_r = \phi w (t + 10w) F_y = 781 \text{ kN}$ against a flange force of 2292 kN, so transverse stiffeners in line with each beam flange are mandatory. The shortfall needs $A_{st} = (F_f - B_r)/(\phi F_y) = 4797$ mm$^2$; two plates 135 × 20 mm supply 5400 mm$^2$ and fit within the $(292 - 14.7)/2 = 139$ mm available each side of the web.