Question 5 of 7: Plastic design of the steel frame and the welded corner at B

Question 5: Plastic design of the steel frame and the welded corner at B (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.

Find. The required plastic moment, a section that supplies it, and the details of the welded knee at B.

Approach. Enumerate the independent mechanisms, combine them to find the least upper bound, confirm the answer by a lower-bound equilibrium check, then size a Class 1 welded I-section and design the corner to transfer the member plastic moment.

1. Part (a) — independent mechanisms. A pin-based portal has four reaction components and three equations of equilibrium, so it is indeterminate to the first degree and collapses when two hinges form. The two independent mechanisms are$$\text{beam: } P_f\frac{L}{2}\theta = 4M_p\theta \Rightarrow M_p = \frac{750(7.5)}{4} = 1406.25\ \text{kN}\cdot\text{m}$$$$\text{sway: } Hh\theta = 2M_p\theta \Rightarrow M_p = \frac{120(10)}{2} = 600\ \text{kN}\cdot\text{m}$$The 900 kN loads at B and C do no work in either mechanism — they sit directly over the columns and translate horizontally in the sway mechanism, vertically not at all in the beam mechanism.
2. Combine and cancel the hinge at B. Adding the two mechanisms makes the hinge rotations at B equal and opposite, so that hinge disappears and $2M_p\theta$ of internal work is recovered. The remaining hinges are at mid-span and at C, each rotating $2\theta$:$$P_f\frac{L}{2}\theta + Hh\theta = 4M_p\theta \Rightarrow \boxed{M_p = \frac{750(7.5) + 120(10)}{4} = 1706.25\ \text{kN}\cdot\text{m}}$$This is 21 % above the beam mechanism alone and nearly three times the sway mechanism; stopping at the independent mechanisms would under-design the frame badly.
The governing combined mechanism. The hinge at B cancels, leaving hinges at mid-span and at the top of the leeward column.
3. Confirm by the lower-bound theorem. An upper-bound mechanism is only a bound until a statically admissible moment field is exhibited. Vertical equilibrium and moments about A give $V_D = 1355$ and $V_A = 1195\ \text{kN}$ (these are determinate). Setting the moment at C equal to $M_p$ fixes $H_D = -M_p/h = -170.6\ \text{kN}$, hence $H_A = -120 - H_D = 50.6\ \text{kN}$, and$$M_{mid} = 7.5V_A - hH_A - 900(7.5) = 1706.25\ \text{kN}\cdot\text{m} = M_p\ \checkmark$$$$M_B = |hH_A| = 506.25\ \text{kN}\cdot\text{m} < M_p\ \checkmark$$Both hinge sections reach $M_p$, nowhere is $M_p$ exceeded, and equilibrium holds, so $1706.25\ \text{kN}\cdot\text{m}$ is the exact collapse value, not merely an upper bound.
4. Select the section. The requirement is $Z \ge M_p/(\phi F_y) = 1706.25\times10^6/(0.9 \times 350) = 5.42\times10^6\ \text{mm}^3$. A welded I with a $710 \times 14$ web and $280 \times 20$ flanges ($d = 750\ \text{mm}$) gives$$Z = b t_f(d - t_f) + \frac{t_wh_w^2}{4} = 5.852\times10^6\ \text{mm}^3, \qquad \boxed{M_r = \phi ZF_y = 1843.5 > 1706.25\ \text{kN}\cdot\text{m}}$$Both elements are Class 1, as plastic design requires: the flange ratio is $(280-14)/2/20 = 6.65 < 145/\sqrt{350} = 7.75$ and the web ratio is $710/14 = 50.7 < 1100/\sqrt{350} = 58.8$. The member weighs $166\ \text{kg/m}$.
Welded section supplying the required plastic moment.
5. Bracing at the plastic hinges (S16 Cl 13.7). A hinge must be able to rotate without buckling laterally, so the code limits the laterally unsupported length adjacent to it to $L_{cr} = (25\,000 + 15\,000\kappa)r_y/F_y$. With $r_y = \sqrt{I_y/A} = 58.9\ \text{mm}$ and $\kappa = 0$ at the mid-span hinge,$$L_{cr} = \frac{25\,000(58.9)}{350} = 4207\ \text{mm}$$The stated bracing at load points and joints leaves $7500\ \text{mm}$ unbraced, which is not enough. Add lateral bracing at the quarter points of BC, reducing the unbraced length to $3750\ \text{mm} < 4207\ \text{mm}$. This is a real design finding, not a formality: without it the frame would fail by lateral torsional buckling before the assumed mechanism could develop, and the plastic analysis above would be invalid.
6. Part (b) — forces at the welded corner. A plastic-design connection must develop the member rather than merely the applied moment, because the hinge is expected to rotate. The flange force at full yield is$$T_f = b t_f F_y = 280(20)(350) = 1960\ \text{kN}$$Take the flanges continuous through the joint with complete-joint-penetration groove welds: a CJP weld with matching electrode develops the full plate, so no weld sizing calculation is needed, and it is the only detail that reliably survives hinge rotation. The web is joined by fillet welds sized for the web shear.
7. Web panel shear and the diagonal stiffener. The knee panel must carry the whole member moment as a shear couple across a panel of depth $d - t_f = 730\ \text{mm}$:$$V_{panel} = \frac{M_p}{d - t_f} = \frac{2048.3\times10^3}{730} = 2805.9\ \text{kN}$$while the unstiffened panel resists only$$V_r = 0.55\phi d_c w F_y = 0.55(0.9)(730)(14)(350) = 1770.6\ \text{kN}$$The panel is 58 % short, so a diagonal stiffener is required. Across a square panel the diagonal runs at $45^{\circ}$, so it carries$$F_{diag} = \frac{2805.9 - 1770.6}{\cos 45^{\circ}} = 1464.1\ \text{kN}, \qquad A_{req} = \frac{F_{diag}}{\phi F_y} = 4648\ \text{mm}^2$$$$\boxed{\text{provide two }200 \times 12\ \text{plates } (4800\ \text{mm}^2), \text{ one each side of the web}}$$Add horizontal continuity plates opposite each beam flange to carry the 1960 kN flange force into the column web, groove welded to the column flanges.