16-Civ-A2 Elementary Structural Design · May 2018

Question 2 of 7: A2 — Welded bracket connection and stub-beam check

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

Question 2: A2 — Welded bracket connection and stub-beam check (10 + 10 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) A welded connection that transfers the factored moment and shear from the stub into the column flange, including any column stiffening; (b) whether the W360 × 79 itself is adequate in bending, shear and deflection.

Approach. Factor the tip load, resolve the moment into a flange couple and the shear into web welds, size each weld group, then check the supporting column flange and web against the concentrated flange force before verifying the beam itself for flexure, lateral-torsional buckling, shear and deflection.

1. Factored actions at the column face. The 100 kN tip load is a specified live load (page 1: all loads shown are unfactored), so $$P_{f}=1.5(100)=150\ \text{kN},\qquad V_{f}=150\ \text{kN},\qquad M_{f}=P_{f}L=150(1.5)=\boxed{225\ \text{kN}\cdot\text{m}}$$ In general M_f = 150L with L in metres, so any other projection substitutes directly here.
2. Resolve the moment into a flange couple. Almost all of a W-shape's moment is carried by the flanges, so the connection is designed for equal and opposite flange forces separated by the distance between flange centroids, (d − t): $$T_{f}=C_{f}=\frac{M_{f}}{d-t}=\frac{225\times10^{6}}{354-16.8}=\boxed{667\ \text{kN}}$$
3. Can the flange itself deliver that force? The tensile capacity of the bare beam flange is $$B_{r}=\phi\,b\,t\,F_{y}=0.90(205)(16.8)(350)=1085\ \text{kN}\ \gt\ 667\ \text{kN}$$ Because the flange is not overstressed, no flange plates are needed: the flanges may be welded directly to the column. Use complete-joint-penetration (CJP) groove welds at both flanges with matching E49XX electrodes; a CJP weld of matching metal develops the full flange, so its resistance equals the 1085 kN just computed (utilisation 0.61).
4. Fillet welds for the web shear. Two lines of longitudinal fillet weld run down the web, each of length L_w = 354 − 2(16.8) − 20 = 300 mm to clear the fillets, so 600 mm of weld is available. The resistance of a longitudinal fillet weld per millimetre of length and per millimetre of leg D is $$v_{r}=0.67\,\phi_{w}(0.707D)X_{u}=0.67(0.67)(0.707)(490)D=155.5D\ \text{N/mm}$$ Setting 600(155.5D) = 150 × 10³ N gives D_required = 1.6 mm. The minimum fillet size of S16 Table 7 is set by the thicker part joined — the 22.2 mm column flange — which requires 8 mm. Specify 8 mm fillets, 300 mm long each side; the weld group is roughly five times stronger than it needs to be, which is normal when a minimum size governs.
5. Check the supporting column for the concentrated flange force (S16 Cl. 14.3.2). With N = t_beam = 16.8 mm bearing length, column flange t = 22.2 mm and web w = 13.1 mm:

   web local yielding   B_r = 0.80w(N + 10t)F_y = 0.80(13.1)(16.8 + 222)(350) = 876 kN;

   column flange bending   B_r = 0.80(7t²)F_y = 0.80(7)(22.2)²(350) = 966 kN;

   web crippling   B_r = 0.60(0.80)w²[1 + 3(N/d)(w/t)^1.5]√(F_yE) = 715 kN.

   All three exceed T_f = 667 kN, so no transverse stiffeners are required — although web crippling leaves only 7 % reserve, and a lighter column would fail it outright.
6. (b) Beam classification and moment resistance. Flange b/2t = 205/33.6 = 6.10 against the Class 1 limit 145/√350 = 7.75, and web (d − 2t)/w = 320.4/9.4 = 34.1 against 1100/√350 = 58.8, so the section is Class 1. The cantilever is unbraced along its 1.5 m length, so check lateral-torsional buckling with ω₂ = 1.0 (S16 Cl. 13.6 for cantilevers): $$M_{u}=\frac{\omega_{2}\pi}{L}\sqrt{EI_{y}GJ+\left(\frac{\pi E}{L}\right)^{2}I_{y}C_{w}}=3802\ \text{kN}\cdot\text{m}$$ Since M_u = 3802 > 0.67M_p = 335 kN·m the member is in the inelastic range, but the resulting value is capped by the plastic moment: $$M_{r}=1.15\phi M_{p}\left(1-\frac{0.28M_{p}}{M_{u}}\right)=499\ \text{kN}\cdot\text{m}\ \gt\ \phi M_{p}=\boxed{450\ \text{kN}\cdot\text{m}}$$ so M_r = 450 kN·m against M_f = 225 kN·m — a utilisation of 0.50. The short projection makes the tube-like torsional term dominate and lateral-torsional buckling is simply not a threat here.
7. Shear and deflection. The web is stocky, (d − 2t)/w = 34.1 < 1014/√350 = 54.2, so F_s = 0.66F_y = 231 MPa and $$V_{r}=\phi\,d\,w\,F_{s}=0.90(354)(9.4)(231)=692\ \text{kN}\ \gg\ V_{f}=150\ \text{kN}$$ At the specified (unfactored) load the tip deflects $$\Delta=\frac{PL^{3}}{3EI_{x}}=\frac{(100\times10^{3})(1500)^{3}}{3(200000)(227\times10^{6})}=2.5\ \text{mm}$$ which is L/610, far inside the usual L/180 cantilever guideline.

The W360 × 79 is therefore adequate on every count, with bending the closest check at half its capacity. The engineering interest in this question lies almost entirely on the connection side: the flange couple of 667 kN is the number that decides whether flange plates are needed, and it is also the number the column must be checked against.