07-Str-A2 · May 2016

Question 2 of 7: A2 — Welded connection and capacity of a stub beam

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

Notes on this paper

Paper format. National Examinations — May 2016, 07-Str-A2 Elementary Structural Design. Three-hour closed-book examination (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, structural steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here, because the set is a study resource rather than an examination script. Page 1 states that all loads shown are unfactored, so the load factors are applied within each solution.

Reference texts.

CSA S16:19, Design of Steel Structures — Clauses 11 (class of section), 13.3 (compressive resistance), 13.5–13.6 (bending), 13.8 (axial compression and bending), 13.13 (welds).
CISC, Handbook of Steel Construction, 12th ed. — section property tables, weld tables.
CSA A23.3:19, Design of Concrete Structures — Clauses 10 (flexure and axial load), 11 (shear), 7.6 (ties).
CSA O86:19, Engineering Design in Wood, with the Canadian Wood Council Wood Design Manual 2020 — Clause 7 (glued-laminated timber).
NBCC 2020 — Table 4.1.3.2 load combinations.

Check — assumptions common to the whole paper. (i) Every load drawn in Figures A1, A2, A3, B3 and quoted in C1 is a specified (unfactored) load; unless a question names the load type, the point loads are treated as live (factor 1.5) and concrete/steel self-weight as dead (factor 1.25), per NBCC Table 4.1.3.2 case 2. (ii) Steel is CSA G40.21 350W, so F_y = 350 MPa, E = 200 000 MPa. (iii) Question B3 states f'_c = 35 MPa and f_y = 400 MPa; because B1 and B2 quote no material strengths, the same pair is adopted for them and the assumption is stated in each solution. (iv) Section properties are recomputed from the nominal plate dimensions of each rolled shape rather than read off a table, so every number below is reproducible; they agree with the CISC Handbook to better than 1 %.

Question 2: A2 — Welded connection and capacity of a stub beam (8 + 12)

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
Stub beam	W360 × 79: d = 354 mm, b = 205 mm, t = 16.8 mm, w = 9.4 mm
Column	W460 × 106: d = 469 mm, b = 194 mm, t = 20.6 mm, w = 12.6 mm
Steel	G40.21M 350W: F_y = 350 MPa
Cantilever projection A to the column face	2.0 m
Tip load at A (unfactored)	80 kN
Weld metal	E49xx electrode, X_u = 490 MPa, φ_w = 0.67

Find. The fillet weld sizes required at B and C to transfer the end moment and shear into the column flange, and a verdict on whether the W360 × 79 stub itself is adequate in flexure, shear and lateral-torsional buckling.

[Figure not reproduced: Figure A2 (redrawn). The stub beam cantilevers 2 m from the column face; the fillet welds at B and C form the moment couple, while the web weld carries the vertical shear. See the official exam paper.]

Approach. Factor the tip load and the beam self-weight, compute the moment and shear at the column face, check the stub against φZF_y, shear and lateral-torsional buckling, then resolve the end moment into a pair of equal and opposite flange forces and size transverse fillet welds for them.

Factor the loads. The 80 kN tip load is live and the beam’s own 79 kg/m is dead: $$\begin{aligned} P_{f} &= 1.5(80) = 120\ \text{kN} \\ w_{f} &= 1.25\left(\frac{79 \times 9.81}{1000}\right) = 0.969\ \text{kN/m} \end{aligned}$$ $$\begin{aligned} M_{f} &= 120(2.0) + \frac{0.969(2.0)^{2}}{2} = \boxed{241.9\ \text{kN}\cdot\text{m}} \\ V_{f} &= 120 + 0.969(2.0) = 121.9\ \text{kN} \end{aligned}$$
Classify the stub section. For a Class 1 flange the limit is 145/√F_y = 7.75 and for the web 1100/√F_y = 58.8: $$\begin{aligned} \frac{b}{2t} &= \frac{102.5}{16.8} = 6.10 \;<\; 7.75 \\ \frac{h}{w} &= \frac{320.4}{9.4} = 34.1 \;<\; 58.8 \end{aligned}$$ Both satisfied, so the section is Class 1 and the plastic moment governs.
Flexural resistance of the stub. From the plate geometry, $$Z_{x} = 2bt\left(\frac{d-t}{2}\right) + \frac{w(d-2t)^{2}}{4} = 1.403\times10^{6}\ \text{mm}^{3}$$ $$M_{r} = \phi Z_{x}F_{y} = 0.9\left(1.403\times10^{6}\right)(350) = \boxed{442\ \text{kN}\cdot\text{m}}$$ The utilisation is 241.9/442 = 0.55, so the stub has substantial reserve in bending.
Confirm that lateral-torsional buckling does not reduce that resistance. The unbraced length is the 2 m projection. With I_y = 24.1×10⁶ mm⁴, J = 0.737×10⁶ mm⁴, C_w = 6.85×10¹¹ mm⁶ and a conservative ω₂ = 1.0, $$M_{u} = \frac{\omega_{2}\pi}{L}\sqrt{EI_{y}GJ + \left(\frac{\pi E}{L}\right)^{2}I_{y}C_{w}} = 2170\ \text{kN}\cdot\text{m}$$ Since M_u = 2170 far exceeds 0.67M_p = 329 kN·m, S16 Clause 13.6(a) caps M_r at φM_p — the value already used. A short, deep cantilever of this kind is simply too stocky laterally to buckle.
Shear. With h/w = 34.1 well below 1014/√F_y = 54.2 the web develops its full shear yield: $$V_{r} = \phi\left(0.66F_{y}\right)dw = 0.9(0.66)(350)(354)(9.4) = 691\ \text{kN} \;>\; V_{f} = 121.9\ \text{kN}$$ The W360 × 79 stub beam is adequate — flexure governs at 55 % utilisation.
Resolve the end moment into flange forces. The moment is transferred as a couple between the two flange welds, whose lever arm is the distance between flange centroids: $$T_{f} = C_{f} = \frac{M_{f}}{d-t} = \frac{241.9\times10^{6}}{354-16.8} = \boxed{717\ \text{kN}}$$ The top flange at C pulls away from the column and the bottom flange at B pushes against it.
Size the flange welds. Fillet welds loaded transversely (θ = 90°) gain the full 50 % strength bonus of S16 Clause 13.13.2.2: $$V_{r} = 0.67\,\phi_{w}A_{w}X_{u}\left(1.00+0.50\sin^{1.5}\theta\right)$$ Per millimetre of weld length and per millimetre of leg size D, the throat is 0.707D, so $$v_{r} = 0.67(0.67)(0.707)(490)(1.50) = 233\ \text{N/mm per mm of leg}$$ Welding both faces of each flange gives 2(205) = 410 mm of weld, hence $$D_{\text{req}} = \frac{717\times10^{3}}{233(410)} = 7.5\ \text{mm} \quad\Longrightarrow\quad \boxed{\text{8 mm fillet welds at B and C}}$$ This also satisfies the S16 Table 8 minimum of 8 mm for a thickest connected part exceeding 20 mm (the 20.6 mm column flange).
Size the web weld. The web weld carries the shear alone and is loaded parallel to its axis (θ = 0), so v_r = 155.5 N/mm per mm of leg over 2(320.4) = 641 mm: $$D_{\text{req}} = \frac{121.9\times10^{3}}{155.5(641)} = 1.2\ \text{mm}$$ Strength is nowhere near critical; the minimum size rules. Specify an 8 mm fillet all round for a single, unambiguous weld symbol.
Check the column for the concentrated flange forces. The 717 kN couple must be delivered into a W460 × 106 without stiffeners. Web local yielding opposite the compression flange (S16 Clause 14.3.2, bearing length N ≈ 25 mm) and flange bending opposite the tension flange give $$B_{r} = \phi_{bi}w\left(N+10t\right)F_{y} = 0.80(12.6)\left(25+206\right)(350) = 815\ \text{kN}$$ $$B_{r} = \phi_{bi}\left(7t^{2}\right)F_{y} = 0.80(7)\left(20.6\right)^{2}(350) = 832\ \text{kN}$$ Both exceed 717 kN, so no transverse stiffeners are strictly required — but with only 14 % reserve the detail is tight, and a pair of horizontal stiffeners aligned with the stub flanges is cheap insurance if the tip load is ever increased.

Result	Value
Factored moment / shear at the column face	241.9 kN·m / 121.9 kN
Class of stub section	Class 1
M_r of W360 × 79 (LTB does not govern, M_u = 2170 kN·m)	442 kN·m — utilisation 0.55
V_r of W360 × 79	691 kN — utilisation 0.18
Flange force couple T_f = C_f	717 kN
Flange fillet weld required / specified	7.5 mm / 8 mm E49xx, both faces of each flange
Web fillet weld required / specified	1.2 mm / 8 mm (minimum size governs)
Column W460 × 106, web yielding / flange bending	815 kN / 832 kN — no stiffeners required
Verdict	Stub beam AB is adequate; connection is 8 mm fillet weld all round

Topic. Moment-resisting welded connection of a cantilever stub to a column flange, plus a member check of the cantilever.

Key relations. Flange force couple T_f = M_f/(d − t); S16 Clause 13.13.2.2 fillet weld resistance V_r = 0.67φ_wA_wX_u(1.00 + 0.50 sin^1.5θ); throat = 0.707D for an equal-leg fillet; Clause 13.5 M_r = φZF_y; Clause 13.6 M_u for lateral-torsional buckling; Clause 14.3.2 web yielding and flange bending resistances of the supporting member.

Why this works. A welded moment connection is statically determinate at the level of the flanges: the web is close to the neutral axis and contributes little to the moment, so the couple is carried almost entirely by the two flanges and the required weld follows directly from M_f divided by the flange lever arm. Splitting the transfer this way also explains why the two welds are sized so differently — the flange welds resist a 717 kN couple while the web weld resists only the 122 kN shear, a six-fold difference. The 50 % strength bonus for transverse loading is real, not a fudge: a fillet loaded across its axis fails on a plane inclined to the throat, mobilising more metal than a longitudinally-loaded fillet.

Common pitfall. Dividing the moment by the full beam depth d rather than the flange-centroid distance (d − t), which understates the flange force by about 5 %; and forgetting to check the supporting column, since a heavy stub can punch a thin column flange long before the weld itself fails. A third trap is treating the whole weld group as one and applying the moment as a stress distribution over its polar modulus — correct for a bolt group in shear, but needlessly conservative for a flange-and-web weld where the load path is obvious.

Source. CSA S16:19 Clauses 13.5, 13.6, 13.13.2.2, 13.13.4 and 14.3.2; CISC Handbook of Steel Construction, 12th ed., Part 3 (welded connections) and Table 8 (minimum fillet weld sizes).

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