Question 1 of 7: Plastic design of the steel rigid frame; welded connection at C

Question 1: Plastic design of the steel rigid frame; welded connection at C (12 + 8 = 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. Rectangular portal ABCD, fixed at A and D; column height 8 m, beam span 15 m with the 250 kN load at mid-span (7.5 m each side). Plastic capacities: columns $M_p$, beam $1.5M_p$. Specified loads from Figure 1:

Find. The plastic moment $M_p$ required for collapse under the factored loads, welded I-sections for the beam BC and the column CD that provide it, and a welded moment connection at the knee C.

Approach. Apply the kinematic (upper-bound) method to the three independent mechanisms, take the largest $M_p$ they demand, confirm by a static (lower-bound) check that a bending-moment distribution exists everywhere within capacity, then select Class 1 welded sections and detail the knee so that the hinge forms in the member and not in the weld.

1. Part (a) — count the redundancies and list the mechanisms. A portal fixed at both bases has six reaction components and three equations of statics, so it is indeterminate to the third degree; collapse therefore needs four hinges, and there are $n - r = 5 - 3 = 2$ independent mechanisms plus their combination. The two vertical loads at B and C sit directly over the columns and generate no beam bending, so only the 250 kN mid-span load drives the beam mechanism and only the 80 kN load drives sway.
2. Beam mechanism. Hinges form at B, at mid-span and at C. At B and C the beam offers $1.5M_p$ but the column only $M_p$, so the hinge forms in the column and absorbs $M_p$; the mid-span hinge is in the beam and absorbs $1.5M_p$ through a relative rotation $2\theta$. Equating internal and external work, $$M_p\theta + 1.5M_p(2\theta) + M_p\theta = W\left(\tfrac{L}{2}\right)\theta \;\Rightarrow\; 5M_p\theta = 375 \times 7.5\,\theta$$ $$\boxed{M_p = \frac{2812.5}{5} = 562.5\ \text{kN}\cdot\text{m}}$$
3. Sway mechanism. Hinges form at A, B, C and D, all four in columns of capacity $M_p$, each rotating $\theta$ while the storey drifts $h\theta$: $$4M_p\theta = H h \theta = 120 \times 8\,\theta \;\Rightarrow\; M_p = \frac{960}{4} = 240\ \text{kN}\cdot\text{m}$$ Sway alone is far from critical, as expected for a frame whose horizontal load is only 80 kN against 1250 kN of gravity.
4. Combined mechanism. Superimposing the two cancels the hinge at B, because the rotations there are equal and opposite; the internal work loses $2M_p\theta$: $$(5 + 4 - 2)M_p\theta = 7M_p\theta = 2812.5\theta + 960\theta = 3772.5\theta \;\Rightarrow\; M_p = 538.9\ \text{kN}\cdot\text{m}$$ Combining reduces the demand here, so the beam mechanism governs. Taking the largest of the three, the design value is $M_p = 562.5$ kN·m for the columns and $1.5M_p = 843.75$ kN·m for the beam.
5. Static (lower-bound) check — is that really the collapse load? With hinges at B, mid-span and C the beam is determinate: its end moments are $-562.5$ kN·m and the free moment is $WL/4 = 375 \times 15/4 = 1406.25$ kN·m, so the mid-span value is $1406.25 - 562.5 = 843.75 = 1.5M_p$, exactly at capacity. Sharing the 120 kN storey shear equally between the columns gives base moments $$M_A = M_D = 562.5 - 60 \times 8 = 82.5\ \text{kN}\cdot\text{m} \;<\; M_p$$ A distribution therefore exists that is everywhere within capacity, so the kinematic and static solutions coincide and $M_p = 562.5$ kN·m is the true collapse value.
6. Section requirements. For a Class 1 section $M_r = \phi Z F_y$ with $\phi = 0.90$: $$Z_{\text{col}} \geq \frac{562.5 \times 10^6}{0.9 \times 350} = 1786 \times 10^3\ \text{mm}^3, \qquad Z_{\text{beam}} \geq \frac{843.75 \times 10^6}{0.9 \times 350} = 2679 \times 10^3\ \text{mm}^3$$
7. Select the beam BC. Try a welded I with flanges $240 \times 16$ and web $560 \times 10$, so $d = 592$ mm and $A = 13\,280$ mm². Its plastic modulus is $$Z = b t (d - t) + \frac{w h^2}{4} = 240(16)(576) + \frac{10(560)^2}{4} = 2\,211\,840 + 784\,000 = 2996 \times 10^3\ \text{mm}^3$$ $$\boxed{M_r = 0.9 (2996\times10^3)(350) = 944\ \text{kN}\cdot\text{m} \;\geq\; 843.75\ \text{kN}\cdot\text{m}}$$ Class 1 is satisfied: flange $b/t = 7.2 \le 145/\sqrt{F_y} = 7.75$ and web $h/w = 56.0 \le 1100/\sqrt{F_y} = 58.8$.
8. Select the column CD. Flanges $250 \times 16$, web $370 \times 10$, $d = 402$ mm, $A = 11\,700$ mm²: $$Z = 250(16)(386) + \frac{10(370)^2}{4} = 1\,544\,000 + 342\,250 = 1886 \times 10^3\ \text{mm}^3$$ $$\boxed{M_r = 0.9(1886\times10^3)(350) = 594\ \text{kN}\cdot\text{m} \;\geq\; 562.5\ \text{kN}\cdot\text{m}}$$ with $b/t = 7.5$ and $h/w = 37.0$, again Class 1. Question 2 revisits this member once the 937.5 kN of axial load is admitted.
9. Bracing required by the plastic method. The paper's note about lateral support at joints and load points is not enough. CSA S16 Cl. 13.7 limits the unbraced length beside a hinge to $L_{cr} = (25\,000 + 15\,000\kappa)r_y/F_y$; with $\kappa = 0$, $r_y = 52.7$ mm for the beam and 59.7 mm for the column, $$L_{cr,\text{beam}} = \frac{25\,000 (52.7)}{350} = 3766\ \text{mm}, \qquad L_{cr,\text{col}} = \frac{25\,000(59.7)}{350} = 4264\ \text{mm}$$ so the beam must be braced at its quarter points (3.75 m) and each column at mid-height (4.0 m). State this as part of the design; without it the assumed mechanism cannot form.
10. Part (b) — forces the knee at C must carry. The connection transmits the hinge moment $M_p = 562.5$ kN·m together with the beam end shear $V = W/2 = 187.5$ kN. Taking the moment as a flange couple over the beam depth less one flange, $$T = \frac{M_p}{d - t} = \frac{562.5\times10^6}{592 - 16} = 977\ \text{kN}$$
11. Flange welds. Use complete-joint-penetration groove welds with matching E49XX electrode, so the joint develops the flange itself: $$T_r = \phi A_f F_y = 0.9 (240 \times 16)(350) = 1210\ \text{kN} \;\geq\; 977\ \text{kN}$$ A CJP weld of matching electrode needs no separate weld-metal calculation — it is the base metal that governs, which is exactly what plastic design requires.
12. Web welds. The flange couple already carries the whole moment, so two fillet welds along the 560 mm web need only take the shear. For a 6 mm fillet (the CSA S16 Table 3 minimum against 16 mm plate), with $X_u = 490$ MPa, $\phi_w = 0.67$ and $\theta = 0$, $$V_r = 0.67\phi_w A_w X_u = 0.67(0.67)(0.707 \times 6 \times 2 \times 560)(490) = 1045\ \text{kN} \gg 187.5\ \text{kN}$$ $$\boxed{\text{6 mm fillet welds, both sides of the web, full depth}}$$
13. Panel-zone shear — the check that actually bites. The knee panel must carry $T$ less the column shear, $916.6$ kN. The Question 1 column web offers $$V_r = \phi (0.66F_y) w d_c = 0.9(0.66 \times 350)(10)(402) = 836\ \text{kN} \;<\; 917\ \text{kN}$$ so a diagonal stiffener is needed. Across a panel $402 \times 592$ mm the diagonal is 715.6 mm long and inclined so that $\cos\theta = 402/715.6 = 0.562$; the shortfall of 80.8 kN requires a diagonal force of $80.8/0.562 = 143.8$ kN, i.e. an area of $143.8\times10^3/(0.9 \times 350) = 456$ mm². $$\boxed{\text{diagonal stiffener: pair of } 10 \times 90\ \text{mm plates} = 1800\ \text{mm}^2}$$ With the heavier column adopted in Question 2 ($w = 12$ mm, $d_c = 440$ mm) the panel resistance rises to 1098 kN and the diagonal is no longer needed — a good reason to settle the beam-column check first.
14. Continuity stiffeners. The 977 kN flange force delivered into the column web would yield it locally, since $B_r = \phi_{bi} w (N + 10t)F_y = 0.80(10)(16 + 160)(350) = 493$ kN. A pair of $100 \times 12$ mm stiffeners in line with each beam flange adds $0.9(2400)(350) = 756$ kN, giving $756 + 493 = 1249$ kN ≥ 977 kN.