Question 1 of 7: Plastic design of the rigid steel frame and its welded corner

Question 1: Plastic design of the rigid steel frame and its welded corner (12 + 8 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 of Figure 1: a single-bay portal on pinned bases at A and F, both columns 8 m high, beam C–D–E spanning $7 + 7 = 14$ m. Point B lies on the windward column 5 m above A.

Find. (a) the plastic moment the frame must develop and the rolled sections that supply it; (b) the plate work, stiffening and welds that make joint E capable of transmitting that plastic moment.

[Figure not reproduced: Figure 1 as printed: pinned portal, equal 8 m columns, 14 m beam. The 80 kN lateral load acts at B, 5 m above the windward base. See the official exam paper.]

Approach. Factor the loads, count the redundancy to fix how many hinges a mechanism needs, enumerate the independent mechanisms and their combination, take the largest $M_p$, confirm it with the static (lower-bound) theorem, then select Class 1 sections and check the column that hosts the hinge as a beam-column before designing the knee for the beam flange force and the panel shear.

1. Factor the loads. With the single load factor set out above, $$\begin{gathered}P_C = P_E = 1.5(400) = 600\ \text{kN}, \\ P_D = 1.5(500) = 750\ \text{kN}, \\ H = 1.5(80) = 120\ \text{kN}.\end{gathered}$$
2. Vertical reactions follow from statics alone. The frame is determinate vertically, so taking moments about A, $$V_F = \frac{P_D L_1 + P_E L + H h_B}{L} = \frac{750(7) + 600(14) + 120(5)}{14} = 1017.86\ \text{kN},$$ and $V_A = 600 + 750 + 600 - 1017.86 = 932.14$ kN. These do not depend on the redundant, so they are fixed for every mechanism considered below.
3. Establish how many hinges a mechanism needs. With three members, four joints and two pinned bases the degree of static indeterminacy is $3(3) + 4 - 3(4) = 1$, so collapse requires $1 + 1 = 2$ hinges. The critical sections are C, D and E (the bases are pinned and cannot host a hinge), giving $3 - 1 = 2$ independent mechanisms.
4. Beam mechanism. Hinges at C, D and E with a rotation $\theta$ at each end and $2\theta$ under the load give $$4M_p\theta = P_D\!\left(\tfrac{L}{2}\right)\theta \;\Rightarrow\; M_p = \frac{750(7)}{4} = 1312.5\ \text{kN}\cdot\text{m}.$$
5. Sway mechanism. Hinges at C and E, both columns rotating about their pins through $\theta$; the lateral load rides at 5 m, not at the eaves, so it travels only $h_B\theta$: $$2M_p\theta = H h_B \theta \;\Rightarrow\; M_p = \frac{120(5)}{2} = 300\ \text{kN}\cdot\text{m}.$$ This is small and cannot govern on its own, but it is not negligible in combination.
6. Combined mechanism — the hinge at C cancels. Adding the two and removing the hinge at C leaves hinges at D and E only, which is exactly the two-hinge count the frame requires. The kink at D is $2\theta$ and the kink at E is also $2\theta$, so $$4M_p\theta = P_D\!\left(\tfrac{L}{2}\right)\theta + H h_B\theta \;\Rightarrow\; M_p = \frac{5250 + 600}{4} = \boxed{1462.5\ \text{kN}\cdot\text{m}}$$ — 11.4 per cent above the beam mechanism. Stopping at the independent mechanisms would under-design the frame.
7. Confirm with the lower-bound theorem. Taking the horizontal reaction $H_A$ as the redundant, the three critical moments are $M_C = -8H_A - 360$, $M_D = 1965 - 8H_A$ and $M_E = M_D - 2925$ (kN·m). Minimising $\max|M|$ gives $8H_A = 502.5$, i.e. $H_A = 62.81$ kN and $H_F = -182.81$ kN, whence $$\begin{gathered}M_C = -862.5, \\ M_D = +1462.5, \\ M_E = -1462.5\ \text{kN}\cdot \text{m}.\end{gathered}$$ Every section satisfies $|M| \le M_p$ and two reach it, so the static and kinematic bounds coincide and $M_p = 1462.5$ kN·m is exact, not an upper bound.
8. Select the beam. Plastic design (Cl 8.5) requires Class 1 sections throughout, so $$Z_{req} = \frac{M_p}{\phi F_y} = \frac{1462.5\times10^6}{0.90(350)} = 4.64\times10^{6}\ \text{mm}^3 .$$ W690×152 supplies $Z_x = 4.94\times10^{6}$ mm$^3$, i.e. $\phi M_p = 1556$ kN·m, and its flange ratio $b/2t = 254/(2\times21.1) = 6.02$ clears the Class 1 limit $145/\sqrt{350} = 7.75$. Adopt W690×152 for the beam C–D–E.
9. The columns are not a repeat of the beam. The leeward column F–E hosts the plastic hinge, carries the larger axial force $C_f = 1017.9$ kN, and — because "joints and load points" places lateral support only at F and E — is unbraced out of plane over the full 8 m. Checking W690×152 on Cl 13.8.2(c) with $\omega_2 = 1.75$ gives $C_r = 1427$ kN and $M_r = 1169$ kN·m, so $$\frac{1017.9}{1427} + \frac{0.85(1462.5)}{1169} = 1.78 \;\gt\; 1.0 .$$ The beam section fails badly as a column; its weak-axis radius of gyration ($r_y = 54.9$ mm) is simply too small for an 8 m storey.
10. Adopt a wider-flange column. W610×217 has almost the same mass class but $r_y = 76.9$ mm and $Z_x = 6.81\times10^{6}$ mm$^3$ ($\phi M_p = 2144$ kN·m). Over 8 m it gives $C_r = 3491$ kN and $M_u = 3676$ kN·m, hence $M_r = 2018$ kN·m, so $$\frac{1017.9}{3491} + \frac{0.85(1462.5)}{2018} = 0.907 \;\lt\; 1.0 .$$ Its flange ratio is $b/2t = 5.92 \lt 7.75$ and its web $h/w = 34.7$ clears the Class 1 limit $\frac{1100}{\sqrt{350}}\!\left(1 - 0.39\frac{C_f}{\phi C_y}\right) = 56.1$. Adopt W610×217 for both columns — the same section each side, which is what would be detailed in practice.

Part (b) now has a definite demand: joint E must carry the full plastic moment of the beam into the column without the panel yielding in shear or the column flanges dishing.

1. Part (b) — the beam flanges deliver a couple. At the hinge the beam moment is resolved into equal and opposite flange forces separated by $d - t$: $$F_f = \frac{M_p}{d_b - t} = \frac{1462.5\times10^6}{688 - 21.1} = \boxed{2193\ \text{kN}}.$$
2. Net shear on the panel zone. The column shear at E relieves part of that force; from the collapse state $V_{col} = |H_F| = 182.8$ kN, so $$V_p = F_f - V_{col} = 2193 - 182.8 = 2010\ \text{kN}.$$ The column web resists it over its full depth (Cl 13.4.1.1, a stocky web): $$V_r = \phi(0.66F_y)d_c w = 0.90(231)(628)(16.5)/10^3 = 2154\ \text{kN},$$ a utilisation of $0.93$, so the web alone is adequate in shear.
3. The plastic-design knee rule is more demanding. For a corner required to develop $M_p$ the classical requirement on the panel thickness is $$w_{req} = \frac{\sqrt{3}\,M_p}{F_y d_b d_c} = \frac{1.732(1462.5\times10^6)}{350(666.9)(600.3)} = 18.1\ \text{mm} \;\gt\; 16.5\ \text{mm}.$$ The shortfall is carried by a diagonal stiffener running corner to corner at $\theta = \arctan(666.9/600.3) = 48.0^\circ$: $$A_{st} = \frac{(w_{req} - w)\,d_c}{\cos\theta} = \frac{(18.08 - 16.5)(600.3)}{0.669} = 1416\ \text{mm}^2 .$$ Provide 2 – 100 × 12 plates ($2400$ mm$^2$), one each side of the web.
4. Continuity plates opposite the beam flanges. Without them the column web must take $F_f$ on a bearing length of $10k$, giving $$B_r = 0.80\,w(10k)F_y = 0.80(16.5)(10)(39.7)(350)/10^3 = 1834\ \text{kN} \;\lt\; 2193\ \text{kN}.$$ The balance needs $A_{cp} = (2193 - 1834)\times10^3/(0.90\times350) = 1139$ mm$^2$; provide 2 – 140 × 16 plates ($4480$ mm$^2$) in line with each beam flange, which also anchors the diagonal.
5. Welds. The flange couple is developed by complete-joint- penetration groove welds at both beam flanges — matching electrode (E49xx), so no calculation of size is required. The beam web carries the end shear $V = V_F - P_E = 417.9$ kN; with $X_u = 490$ MPa and a fillet on each side over the 645.8 mm web depth the required leg is only $$D = \frac{417.9\times10^3}{2(0.67)(0.90)(490)(0.707)(645.8)} = 1.5\ \text{mm},$$ so the 8 mm minimum fillet for a 27.7 mm thick part (S16 Table 4) governs.