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 = 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. The design starts from the same pinned-base portal of Figure 2, now in steel of $F_y = 350$ MPa, with a single plastic moment capacity $M_p$ common to beam and columns.

Find. The plastic moment required by the governing collapse mechanism, a member section that provides it while remaining Class 1, and the details of the welded knee at B.

Approach. Enumerate the independent mechanisms and their combination by the work equation, take the largest $M_p$, select plates that give the required $Z$ and remain Class 1 with the column axial load present, reduce $M_p$ for that axial load, then size the panel-zone stiffening at the knee.

1. Note which loads do work. The 900 kN loads act at B and C, directly over the columns. In a beam mechanism those joints do not move vertically, and in a sway mechanism they move only horizontally, so neither load does external work in any mechanism. They matter only as column axial force. Only the 750 kN mid-span load, the 120 kN horizontal load and the beam self-weight enter the work equations.
2. Beam mechanism. Hinges at B, E and C with beam rotation $\theta$ give an internal work of $4M_p\theta$ and an external work of $P_f(L/2)\theta + w_fL^2\theta/4$. With a trial factored self-weight $w_f = 2.383$ kN/m, $$M_p = \frac{750(7.5) + 2.383(15)^2/4}{4} = 1439.8\ \text{kN}\cdot\text{m}.$$
3. Sway mechanism. With pinned bases the columns rotate about A and D, so hinges form only at B and C: $$M_p = \frac{H_f h}{2} = \frac{120(10)}{2} = 600.0\ \text{kN}\cdot\text{m}.$$
4. Combined mechanism. Adding the two and cancelling the hinge at B, where the rotations oppose, removes $2M_p\theta$ of internal work while both external works are retained: $$M_p = \frac{750(7.5) + 2.383(15)^2/4 + 120(10)}{6 - 2} = \boxed{1739.8\ \text{kN}\cdot\text{m}}$$ This is the largest of the three and therefore governs. A statical check confirms it is admissible: $H_D = M_p/h = 174.0$ kN, so $H_A = 120 - 174.0 = -54.0$ kN and the moment at B is $|M_B| = 539.8$ kN·m, safely below $M_p$.
5. Select the section. The required plastic modulus is $Z_{req} = M_p/(\phi F_y) = 5.52 \times 10^6$ mm³. Take a welded I-section with flanges $340 \times 22$ mm and a web $700 \times 14$ mm, $d = 744$ mm, giving $$A = 24\,760\ \text{mm}^2, \quad Z = 7.116 \times 10^6\ \text{mm}^3, \quad M_r = \phi Z F_y = \boxed{2241\ \text{kN}\cdot\text{m}} \; > \; 1739.8.$$ Flange projection $b_{el}/t = 163/22 = 7.41 \le 145/\sqrt{F_y} = 7.75$, and web $h/w = 50.0$ against the Class 1 limit for combined flexure and compression, $\left(1100/\sqrt{F_y}\right)\!\left(1 - 0.65C_f/\phi C_y\right) = 51.9$. The section is Class 1 with the column axial load present, as plastic design requires.
6. Reduce $M_p$ for column axial load. The collapse statics give beam shears of 312.9 kN at B and 472.9 kN at C, so the columns carry $C_f = 1236.7$ kN (AB) and $1396.7$ kN (CD). With $C_y = AF_y = 8666$ kN, $$M_{pc} = 1.18M_p\left(1 - \frac{C_f}{C_y}\right) = 1.18(2241)\left(1 - 0.161\right) = 2219\ \text{kN}\cdot\text{m} \; > \; 1739.8\ \text{kN}\cdot\text{m},$$ so the reduced capacity is still adequate.
7. Fix the bracing required by the hinges. CSA S16 Cl. 13.7 limits the unbraced length adjacent to a plastic hinge. For the beam segment next to the mid-span hinge, $\kappa = -539.8/1739.8 = -0.31$ and $$L_{cr} = \frac{25\,000 + 15\,000\kappa}{F_y}\,r_y = \frac{20\,350}{350}(76.3) = 4437\ \text{mm};$$ for the column hinge, $\kappa = 0$ gives $L_{cr} = 5452$ mm. Bracing "at load locations" alone would leave 7.5 m in the beam and 10 m in the columns, which is not enough. Provide additional lateral bracing at the quarter points of BC (3.75 m) and at the third points of each column (3.33 m). This is a real design consequence of choosing plastic design, and the paper's bracing note does not remove it.
8. (b) Design the welded knee at B. The corner must develop the full $M_p$ of the members. The horizontal shear in the connection panel is $$V_{pz} = \frac{M_p}{d - t_f} = \frac{1739.8 \times 10^6}{744 - 22} = 2410\ \text{kN},$$ whereas the panel web alone resists $V_r = \phi(0.66F_y)d_cw = 0.90(231)(744)(14) = 2165$ kN. The shortfall of 245 kN must be carried by a diagonal stiffener at $45^\circ$: $$A_{st} \ge \frac{V_{pz} - V_r}{\phi F_y\cos\alpha} = \frac{245 \times 10^3}{0.90(350)(0.707)} = \boxed{1096\ \text{mm}^2}$$ Provide a pair of $10 \times 80$ mm diagonal plates (1600 mm², $b/t = 8 \le 200/\sqrt{F_y} = 10.7$).
9. Complete the corner detail. The beam flange delivers $340(22)(350) = 2618$ kN into the column, so provide 22 mm horizontal continuity plates opposite both beam flanges, fitted and welded to the column web and flanges. Connect the beam flanges to the column with complete-joint-penetration groove welds using matching electrodes and run-off tabs, and the beam web with a double fillet weld sized for the web shear plus its share of the moment. Because a plastic hinge may form immediately adjacent to the joint, the welds must develop the member rather than merely the applied force, and the joint region must be laterally braced.