16-Civ-A2 Elementary Structural Design · December 2019

Question 2 of 7: A2 — Shear connection and welded splice for a W200×27

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

Question 2: A2 — Shear connection and welded splice for a W200×27 (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 a laterally supported W200×27 beam spanning 5.0 m into the flange of a W200×59 column, M20 A325 bolts and E49XX electrodes, and splice actions of 40 kN·m moment with 50 kN shear.

Find. (a) the end reaction the shear connection must carry and a bolted double-angle connection to the column flange that develops it; (b) a flange-and-web fillet-welded splice for the stated moment and shear.

Approach. For (a), work backwards: the largest uniformly distributed load a laterally supported W200×27 can carry over 5 m follows from $M_r = \phi Z F_y$, and 60 % of that load fixes the end reaction. Then check bolts, bearing, the angles and the welds against it. For (b), split the actions the way the section carries them: the flanges take the moment as a couple, the web takes the shear.

1. Moment resistance of the beam. Laterally supported and Class 1, so $$M_r = \phi Z_x F_y = 0.90\,(277\times10^{3})(300) = \boxed{74.8\ \text{kN}\cdot\text{m}}$$
2. Maximum uniformly distributed load. For a simple span, $M = wL^{2}/8$, so $$w_{\max} = \frac{8M_r}{L^{2}} = \frac{8(74.8)}{5^{2}} = 23.93\ \text{kN/m}$$ Shear does not limit it: $V_r = \phi A_w (0.66F_y) = 0.90(207)(6.4)(198)/10^{3} = 236$ kN, which corresponds to about 94 kN/m. Flexure governs, as the phrase "laterally supported" implies.
3. Design reaction for the connection. At 60 % of that load, $$w = 0.60(23.93) = 14.36\ \text{kN/m},\qquad V_f = \frac{wL}{2} = \frac{14.36(5)}{2} = \boxed{35.9\ \text{kN}}$$
4. Bolt group. Try two M20 A325 bolts through the beam web in double shear (one angle each side), at 70 mm pitch with 35 mm end distances, so each angle is 140 mm long. With threads intercepted by the shear planes, $$V_r = 0.60\,\phi_b\, n\, m\, A_b\, F_u (0.70) = 0.60(0.80)(1)(2)(314)(830)(0.70)/10^{3} = 175\ \text{kN per bolt}$$ Two bolts give 350 kN against 35.9 kN required — the connection is governed by detailing rules, not by strength, which is normal for a shear tab on a light beam.
5. Bearing on the beam web. The 6.4 mm web is the thin part: $$B_r = 3\,\phi_{br}\, t\, d\, F_u = 3(0.80)(6.4)(20)(450)/10^{3} = 138\ \text{kN per bolt}$$ so 276 kN for the pair. Adequate.
6. Angles. Two L90×90×8, each 140 mm long. Gross shear of the pair is $$V_r = 2\,\phi(0.66F_y)A_g = 2(0.90)(198)(8)(140)/10^{3} = 399\ \text{kN}$$ and the block-shear and net-section paths are far above the 35.9 kN demand at this size.
7. Weld to the column flange. Each angle is fillet welded along its two vertical edges to the W200×59 flange. Take two 6 mm fillets, 140 mm long, loaded longitudinally ($\theta = 0$): $$V_r = 2\left[0.67\,\phi_w\,(0.707D L)\,X_u\right] = 2\left[0.67(0.67)(0.707)(6)(140)(490)\right]/10^{3} = 261\ \text{kN}$$ The 6 mm leg is the S16 Table 7 minimum for the 14.2 mm flange it is welded to, so it cannot be reduced.
8. Splice, part (b) — flange couple. Assign the whole moment to the flanges. The couple acts on the lever arm between flange centroids, $d - t$: $$T_f = \frac{M_f}{d - t} = \frac{40\times10^{6}}{207 - 8.4} = \boxed{201.4\ \text{kN}}$$
9. Flange splice plates. A 120×10 plate fits inside the 133 mm flange with room for the fillet on each side: $$T_r = \phi A_g F_y = 0.90(120)(10)(300)/10^{3} = 324\ \text{kN} \;>\; 201.4\ \text{kN}\quad(\text{utilisation }0.62)$$
10. Flange welds. Each end of the plate is welded with two 120 mm longitudinal fillets down the sides plus one 120 mm transverse fillet across the end. Using $V_r = 0.67\phi_w(0.707DL)X_u(1 + 0.50\sin^{1.5}\theta)$ with 6 mm legs, $$V_r = 2(133) + 126 = 392\ \text{kN} \;>\; 201.4\ \text{kN}$$ The transverse run earns the 1.50 directional multiplier; the side runs do not.
11. Web splice plates and welds. Give the web the shear alone. Two 6×150 plates, one each side: $$V_r = 2\,\phi(0.66F_y)A = 2(0.90)(198)(6)(150)/10^{3} = 321\ \text{kN} \;>\; 50\ \text{kN}$$ and four 5 mm longitudinal fillets 150 mm long give $V_r = 467$ kN. Both are far above the 50 kN demand, so the plate size is set by giving the welds room rather than by strength.