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

Question 4: Plastic design of the rigid steel frame (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.

Find. (a) The required plastic moment and a Class 1 section that provides it; (b) the welds and stiffening of the rigid corner at \(C\).

Approach. The frame has four reaction components and three equations of equilibrium, so it is indeterminate to the first degree and needs two hinges to become a mechanism. Enumerate the independent mechanisms, take the smallest collapse load (upper bound), and confirm it with a statically admissible moment field (lower bound) so the answer is the exact plastic collapse value. The corner is then detailed to develop the member's full plastic moment.

1. Degree of indeterminacy and possible hinge locations. The roller at \(A\) supplies one reaction, the fixed base at \(E\) three: \(4 - 3 = 1\) redundant. Hinges can form under each 200 kN load, at joint \(C\) and at base \(E\). No hinge is possible at \(A\) (a roller carries no moment), and the column between \(C\) and \(D\) carries no shear, so its moment there is constant and \(D\) can never be more critical than \(C\).
2. Beam mechanisms. With the hinge under the 4 m load and a second at \(C\), rotating segment \(A\)–4 m by \(\theta\) drops that load 4\(\theta\) and the 12 m load \(4\theta/3\):
   $$200(4\theta) + 200\left(\tfrac{4\theta}{3}\right) = M_p\left(\tfrac{4\theta}{3}\right) + M_p\left(\tfrac{\theta}{3}\right) \;\Rightarrow\; M_p = 640\ \text{kN}\cdot\text{m}$$
   The companion mechanism, hinges at 12 m and \(C\), gives \(M_p = 3200/7 = 457.1\ \text{kN}\cdot\text{m}\). The 400 kN at \(C\) sits directly over the column and does no work in either.
3. Sway mechanism. Because \(A\) is a roller the beam is free to translate horizontally, so the column can sway with hinges at \(C\) and \(E\). Rotating the column by \(\theta\) moves \(D\) through \(5\theta\):
   $$80(5\theta) = 2M_p\theta \;\Rightarrow\; M_p = 200\ \text{kN}\cdot\text{m}$$
4. Combined mechanism — the one that governs. Adding the first beam mechanism to the sway mechanism in the proportion that cancels the hinge at \(C\) (sway rotation \(\theta_2 = \theta_1/3\)) leaves hinges at \(x = 4\ \text{m}\) and at \(E\) only. Equivalently, rotate the rigid body 4 m–\(C\)–\(E\) about \(E\) through \(\varphi\): the 200 kN at 4 m falls \(12\varphi\), the 200 kN at 12 m falls \(4\varphi\), the 400 kN at \(C\) moves horizontally only, and the 80 kN moves \(5\varphi\) in its own direction. The hinge at 4 m rotates \(4\varphi\) and the base hinge \(\varphi\):
   $$200(12\varphi) + 200(4\varphi) + 80(5\varphi) = M_p(4\varphi) + M_p(\varphi)$$
   $$3600\varphi = 5M_p\varphi \;\Rightarrow\; \boxed{M_p = 720\ \text{kN}\cdot\text{m}\ \text{(for the service loads)}}$$
   This exceeds every other mechanism, so it is the critical one.
5. Lower-bound confirmation. Writing the four critical moments in terms of the single redundant \(R_A\) and a load factor \(\lambda\),
   $$M_4 = 4R_A, \quad M_{12} = 12R_A - 1600\lambda, \quad M_C = 16R_A - 3200\lambda, \quad M_E = 16R_A - 3600\lambda$$
   Maximising \(\lambda\) subject to \(|M| \le M_p\) at all four sections gives \(R_A = M_p/4\) and \(M_p/\lambda = 720\ \text{kN}\cdot\text{m}\), identical to the kinematic result — the collapse load is exact, with \(M_C = -0.44M_p\) and \(M_{12} = 0.50M_p\) both safely inside the yield surface.
6. Required factored plastic moment. Applying the load factor to the printed (live) loads,
   $$M_{p,req} = 1.5(720) = \boxed{1080\ \text{kN}\cdot\text{m}}, \qquad Z_{req} = \frac{1080\times10^{6}}{0.90(350)} = 3.43\times10^{6}\ \text{mm}^3$$
7. Section selection. Plastic design demands Class 1 sections (Cl. 13.5). Adopt a welded I-section with flanges \(260 \times 22\ \text{mm}\) and web \(520 \times 10\ \text{mm}\), \(d = 564\ \text{mm}\):
   $$\frac{b}{t} = 5.68 \le \frac{145}{\sqrt{350}} = 7.75, \qquad \frac{h}{w} = 52.0 \le \frac{1100}{\sqrt{350}} = 58.8 \quad\text{(Class 1)}$$
   $$Z = 3.776\times10^{6}\ \text{mm}^3, \qquad \phi Z F_y = 1189.5\ \text{kN}\cdot\text{m} \;\ge\; 1080\ \text{kN}\cdot\text{m}$$
   The column axial force at collapse is \(C_f = 1.5(800) - M_{p,req}/4 = 930\ \text{kN}\), so \(C/C_y = 0.160\) and the reduced plastic moment is
   $$\phi M_{pc} = 0.90(1.18)Z F_y\left(1 - \frac{C}{C_y}\right) = 1179.5\ \text{kN}\cdot\text{m} \;\ge\; 1080\ \text{kN}\cdot\text{m}$$
   Adopt a uniform welded I-section, 564 mm deep, 130.6 kg/m.
8. Bracing consequence. The paper's note that lateral support exists "at all joints and load points" is not sufficient near a plastic hinge. CSA S16 Cl. 13.7 limits the unbraced length adjacent to a hinge to \(L_{cr} = (25\,000 + 15\,000\kappa)r_y/F_y\); with \(r_y = 62.3\ \text{mm}\) and \(\kappa \approx 0\) this gives \(L_{cr} \approx 4.45\ \text{m}\), against the 4 m and 8 m intervals the note implies. Additional bracing must be specified at the quarter points of the 8 m interior beam segment and at the column mid-height, otherwise the assumed mechanism cannot form.
9. (b) The welded corner at C. A rigid corner in a plastically designed frame must develop the members' plastic moment so the hinge forms in the member, not the joint. Splitting \(\phi Z F_y\) between the plates,
   $$M_{flanges} = \phi b_f t_f (d - t_f)F_y = 976.6\ \text{kN}\cdot\text{m}, \qquad M_{web} = \phi\frac{t_w h_w^2}{4}F_y = 212.9\ \text{kN}\cdot\text{m}$$
   their sum is exactly 1189.5 kN·m. The flange force to be transferred is
   $$T_f = \phi b_f t_f F_y = 0.90(260)(22)(350) = \boxed{1801.8\ \text{kN}}$$
   Use complete-joint-penetration groove welds on both flanges with matching E49xx electrodes — a CJP weld develops the full plate strength, so no weld sizing is needed — and a 10 mm double fillet on the web.
10. Knee panel-zone shear. The corner panel must carry the flange couple as shear:
    $$V_{pz} = \frac{\phi Z F_y}{d - t_f} - V_{col} = \frac{1189.5\times10^{6}}{542} \times 10^{-3} - 120 = 2074.7\ \text{kN}$$
    $$V_{r,pz} = 0.90(0.55F_y)d\,w = 0.90(0.55)(350)(564)(10) = 977.1\ \text{kN} \;<\; V_{pz}$$
    The unreinforced panel carries less than half the demand, so a diagonal stiffener across the square knee (\(542 \times 542\ \text{mm}\), \(45^{\circ}\)) takes the balance:
    $$F_{diag} = \frac{2074.7 - 977.1}{\cos 45^{\circ}} = 1552.2\ \text{kN}, \qquad A_{req} = \frac{1552.2\times10^{3}}{0.90(350)} = 4928\ \text{mm}^2$$
    Provide two plates \(210 \times 12\ \text{mm}\) (5040 mm²), one each side of the web.
11. Continuity stiffeners. Opposite each beam flange the column web must receive 1801.8 kN. Provide horizontal continuity stiffeners of
    $$A_{req} = \frac{1801.8\times10^{3}}{0.90(350)} = 5720\ \text{mm}^2 \;\Rightarrow\; \text{two plates } 145 \times 20\ \text{mm (5800 mm}^2\text{)}$$
    fillet-welded to the column web and flanges, with the ends against the beam flanges ground to bear.