16-Civ-B2 Advanced Structural Design · December 2014

Question 4 of 7: Plastic design of the rigid frame and its welded corner

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2014 — 98-Civ-B2 Advanced Structural Design, 3 hours, closed book (handbooks and textbooks permitted). Seven design questions of equal value; any five constitute a complete paper.

Reference texts. CSA S16:19 Design of Steel Structures (Clause 13 members, Clause 14 plate girders, Clause 17 composite, Clause 21 plastic design); CISC Handbook of Steel Construction; CSA A23.3:19 Design of Concrete Structures (Clause 10 flexure and columns, Clause 11 shear, Clause 18 prestressed concrete); National Building Code of Canada 2020 for load combinations; Hibbeler, Structural Analysis, for the frame and continuous-beam analysis.

Design data printed on page 1. Concrete $f_c' = 30\ \text{MPa}$; structural steel $F_y = 350\ \text{MPa}$; reinforcing steel $f_y = 400\ \text{MPa}$. Prestressed concrete: $f_{ci}' = 35\ \text{MPa}$ at transfer, $f_c' = 50\ \text{MPa}$, modular ratio $n = 6$, $f_{ult} = 1750\ \text{MPa}$, $f_y = 1450\ \text{MPa}$, $f_{initial} = 1200\ \text{MPa}$, losses $= 240\ \text{MPa}$. Design in SI. All loads shown are unfactored.

Check — load factors. Page 1 states only that the loads shown are unfactored; it gives no dead/live split. Throughout this paper the figure loads are treated as a single variable action and factored by 1.5, while self weight introduced by the solver (the reinforced-concrete frame, the prestressed girder, the bridge deck) is factored by 1.25, per NBCC 2020 Case 2. Mechanisms, section classifications and interaction ratios are unaffected by this choice — only the magnitudes are. Serviceability checks (the no-tension prestress design of Q2 and the soil bearing of Q5) use unfactored loads, as they must.

Question 4: Plastic design of the rigid 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.

Quantity	Value
Beam AC / column CE	16.0 m / 10.0 m
Supports	roller at A (vertical only); fixed base at E
Vertical loads on AC	200 kN at 4 m, 200 kN at 12 m, 400 kN at C
Horizontal load at D	80 kN at mid-height of the column
Factored values (×1.5)	300, 300, 600 kN vertical; 120 kN horizontal
Degree of static indeterminacy	$(1 + 3) - 3 = 1$, so collapse needs two hinges

Find. The required plastic moment $M_p$ and a rolled section that supplies it, then the plate work needed at the knee joint C so that the corner can develop the members it joins.

Figure 4.1 — The frame, its factored loads, and the two-hinge collapse mechanism. A is a roller, so the frame is a sway frame with a single redundant and only two hinges are needed.

Approach. Because the frame is only once redundant, the whole plastic analysis reduces to a one-parameter minimax: express every critical-section moment in terms of the single unknown reaction at A, find the reaction that minimises the largest of them, and confirm the result with virtual work on the mechanism it implies.

Part (a) — write every critical moment in terms of one redundant. Take the vertical reaction $V_A$ at the roller as the redundant. Sagging positive in the beam, the moments at the four critical sections are $$M_{4} = 4V_A, \qquad M_{12} = 12V_A - 300(8), \qquad M_C = 16V_A - 300(12) - 300(4)$$ $$M_E = M_C - 120(5) = 16V_A - 5400$$ The 600 kN at C sits over the column and enters only the axial force. The column moment is constant between C and D, because no horizontal force acts above D.
Minimise the largest moment. The required plastic moment is $$M_p = \min_{V_A}\ \max\left(|M_4|,\ |M_{12}|,\ |M_C|,\ |M_E|\right)$$ The minimax is reached where the sagging moment under the first load balances the hogging moment at the base, $4V_A = 5400 - 16V_A$, giving $$V_A = 270\ \text{kN} \quad\Rightarrow\quad \boxed{M_p = 1080\ \text{kN}\cdot\text{m}}$$ with the other two sections comfortably inside: $M_{12} = 840$ and $M_C = -480\ \text{kN}\cdot\text{m}$. Because a statically admissible moment field with $|M| \le M_p$ everywhere has been exhibited, this is a lower bound as well as a mechanism, so it is the exact collapse load, not an estimate.
Confirm by virtual work on the implied mechanism. Hinges form at $x = 4\ \text{m}$ and at E. Rotating the rigid body from the hinge through C to E about E by $\varphi$ moves the point at 4 m down by $12\varphi$ and horizontally by $10\varphi$; compatibility at A gives the outer segment rotation $\theta = 3\varphi$. Then $$W_{ext} = 300(12\varphi) + 300(4\varphi) + 120(5\varphi) = 5400\varphi, \qquad W_{int} = M_p(4\varphi + \varphi) = 5M_p\varphi$$ $$M_p = \frac{5400}{5} = 1080\ \text{kN}\cdot\text{m} \quad\checkmark$$ Note that the 600 kN load at C does no external work: C moves horizontally only. The 120 kN horizontal load does work, so a pure beam mechanism would have under-designed the frame.
Select a section. $$Z_{req} = \frac{M_p}{\phi F_y} = \frac{1080\times10^{6}}{0.9(350)} = 3429\times10^{3}\ \text{mm}^3$$ Take a W610 × 125. Modelling it from its nominal plate dimensions ($d = 612$, $b = 229$, $t = 19.6$, $w = 11.9$ mm, fillets ignored, which is conservative): $$Z_x = 2A_f\frac{d-t}{2} + \frac{wh^2}{4} = 3635\times10^{3}\ \text{mm}^3, \qquad A = 15\,793\ \text{mm}^2$$ $$M_r = \phi Z_xF_y = 0.9(3635\times10^{3})(350) = 1145\ \text{kN}\cdot\text{m} \;>\; 1080\ \text{kN}\cdot\text{m}$$
Verify the section is Class 1, as plastic design demands. Flange: $b/2t = 5.84 \le 145/\sqrt{F_y} = 7.75$. Web: $h/w = 48.1 \le 1100/\sqrt{F_y} = 58.8$. Both pass, so the section can sustain the rotation the mechanism requires without local buckling. A Class 2 or 3 section, however strong, would be inadmissible here.
Check the bracing next to the plastic hinge. Clause 13.7 limits the unbraced length adjacent to a hinge that must rotate. With $r_y = 49.9\ \text{mm}$ and the moment falling to zero at A ($\kappa = 0$), $$L_{cr} = \frac{25\,000 + 15\,000\kappa}{F_y}\,r_y = \frac{25\,000}{350}(49.9) = 3564\ \text{mm}$$ The segment from A to the hinge is 4000 mm, so one additional brace is required in that bay — place it 3.5 m from A. This is a real design action, not a formality: without it the hinge buckles laterally before it can rotate and the mechanism never forms.
Part (b) — find the force the corner must carry. A rigid corner in plastic design must develop the members it joins, so design for the full flange yield force rather than the 480 kN·m that happens to act at C at collapse: $$T_f = A_fF_y = 229(19.6)(350) = 1571\ \text{kN}$$ delivered by the beam flange and turned through the corner into the column flange.
Check the knee panel in shear. The unstiffened web panel bounded by the two member depths resists $$V_r = 0.55\phi F_y w d_c = 0.55(0.9)(350)(11.9)(612) = 1262\ \text{kN} \;<\; 1571\ \text{kN}$$ so the panel is 309 kN short and must be reinforced. A knee joint always fails this check before the members do, because the panel is asked to carry the whole flange force in shear over a depth no greater than the member itself.
Size the diagonal stiffener. A stiffener on the compression diagonal at $\theta = 45^{\circ}$ (the panel is square, $d_b = d_c = 612$ mm) carries the deficit: $$A_{st} \ge \frac{(1571 - 1262)\times10^{3}}{\phi F_y\cos 45^{\circ}} = \frac{309\times10^{3}}{0.9(350)(0.707)} = 1388\ \text{mm}^2$$ $$\text{Provide } \boxed{\text{two plates } 100 \times 10\ \text{mm}} = 2000\ \text{mm}^2, \quad b/t = 10 \le 200/\sqrt{F_y} = 10.7$$
Detail the welds and the continuity plates. The beam and column flanges meet on complete-joint-penetration groove welds with matching E49xx electrodes, which develop the flange without calculation. Continuity plates 100 × 12 mm are fitted in line with each beam flange to carry the concentrated flange force into the column web. The diagonal and continuity plates are attached with 8 mm fillet welds, whose resistance is $$v_r = 0.67\phi_wA_wX_u = 0.67(0.67)(0.707 \times 8)(490) = 1244\ \text{N/mm}$$ so a pair of 8 mm fillets along the 573 mm panel depth develops 1425 kN — ample for the 630 kN the stiffener delivers. Backing bars are removed and the CJP root gouged and back-welded at the tension flange.

Result	Value
Redundant at collapse	$V_A = 270$ kN
Hinge locations	$x = 4$ m in the beam and at base E
Required plastic moment	$M_p = 1080$ kN·m (exact — upper and lower bounds coincide)
Required plastic modulus	$3429\times10^{3}$ mm³
Section	W610 × 125, $Z_x = 3635\times10^{3}$ mm³, $M_r = 1145$ kN·m
Classification	Class 1 (flange 5.84, web 48.1)
Extra brace required	at 3.5 m from A ($L_{cr} = 3564 < 4000$ mm)
Corner flange force	1571 kN
Panel shear resistance	1262 kN — deficit 309 kN
Corner detail	CJP flange welds; diagonal stiffener 2 × 100 × 10; continuity plates 2 × 100 × 12; 8 mm fillets

Figure 4.2 — Welded rigid corner at C: complete-joint-penetration groove welds at the flanges, a 45 degree diagonal stiffener carrying the 309 kN panel shear deficit, and continuity plates in line with the beam flanges.

Topic. Plastic (limit) design of a once-redundant sway frame, and the rigid knee connection that lets the mechanism form.

Key relations. Number of hinges for collapse = degree of indeterminacy + 1; upper bound $\sum M_p\theta = \sum P\delta$; lower bound: any statically admissible field with $|M| \le M_p$; $M_p = ZF_y$, $M_r = \phi ZF_y$; Clause 13.7 hinge bracing $L_{cr} = (25\,000 + 15\,000\kappa)r_y/F_y$; knee panel $V_r = 0.55\phi F_ywd_c$; diagonal stiffener $A_{st} = (V_f - V_r)/(\phi F_y\cos\theta)$.

Why this works. With one redundant the whole plastic analysis collapses to a one-variable minimax, and that is a far safer route than enumerating mechanisms by hand: the minimising value of the redundant automatically identifies which sections hinge, and because the resulting moment field is statically admissible everywhere, the upper and lower bound theorems close on the same number. The roller at A is the feature that makes this frame simple and also the feature most often misread — it provides no horizontal restraint, so the 80 kN load is carried entirely by the fixed base and does work in every mechanism that sways.

Common pitfall. Assuming a pin at A instead of reading the roller symbol. A pin would make the frame twice redundant, require three hinges, and give a different answer entirely. The second pitfall is designing the corner for the moment that acts there at collapse (480 kN·m) rather than for the member capacity: moments redistribute as hinges form, and a corner that yields before the members have finished redistributing turns a ductile mechanism into a brittle joint failure.

Source. CSA S16:19 Clause 13.5 (plastic moment resistance), 13.7 (bracing at hinges) and Clause 21 (plastic design); Neal, The Plastic Methods of Structural Analysis, Ch. 3–4; Kulak and Grondin, Limit States Design in Structural Steel, Ch. 5 and 12.

← Question 3 Next: Question 5 →