07-Str-A2 · Undated paper

Question 2 of 7: A2 — Moment connection and adequacy of a cantilever beam

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

Question 2: A2 — Moment connection and adequacy of a cantilever beam (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. A W360x39 cantilever projects 2.5 m from the face of a continuous W200x71 column and carries a single point load at its free tip. Lateral restraint exists at the column and at the tip only, so the full 2.5 m is the unbraced length.

Find. (a) A bolted-web, welded-flange moment connection able to transfer the factored moment and shear into the column, including whether the column needs stiffening. (b) Whether the W360x39 itself is adequate in flexure, shear and deflection.

Approach. Factor the loads to NBCC Case 2, add the beam self-weight, obtain $M_f$ and $V_f$ at the support; check the member for lateral-torsional buckling over the 2.5 m unbraced length, then for shear and service deflection; finally resolve $M_f$ into a flange couple, test whether the bare beam flange can deliver it, and detail plates, welds, bolts and column stiffening accordingly.

1. Factor the loads and find the support actions. The critical combination is $1.25D+1.5L$: $$P_f=1.25(30)+1.5(20)=67.5\ \text{kN}$$ The beam weighs $39\ \text{kg/m}=0.383$ kN/m, factored $1.25(0.383)=0.478$ kN/m. At the face of the column, $$M_f=P_fL+\frac{w_fL^{2}}{2}=67.5(2.5)+\frac{0.478(2.5)^{2}}{2}=\boxed{170.2\ \text{kN}\cdot\text{m}}$$ $$V_f=67.5+0.478(2.5)=68.7\ \text{kN}$$
2. (b) Classify the section. At $F_y=300$ MPa, $$\frac{b}{2t}=\frac{64}{10.7}=5.98\ <\ \frac{145}{\sqrt{300}}=8.37,\qquad \frac{h}{w}=\frac{331.6}{6.48}=51.2\ <\ \frac{1100}{\sqrt{300}}=63.5$$ Both flange and web are Class 1, so the plastic moment is available and $Z_x$ may be used.
3. (b) Compute the elastic critical moment over the unbraced length. The moment falls linearly from $M_f$ at the column to zero at the tip, so $\kappa=0$ and $\omega_2=1.75+1.05\kappa+0.3\kappa^{2}=1.75$. Because the free end is laterally restrained, the "unbraced cantilever" default of $\omega_2=1.0$ does not apply. With $L=2500$ mm, $$M_u=\frac{\omega_2\pi}{L}\sqrt{EI_yGJ+\left(\frac{\pi E}{L}\right)^{2}I_yC_w}$$ $$M_u=\frac{1.75\pi}{2500}\sqrt{(2\times10^{5})(3.75\times10^{6})(77000)(1.58\times10^{5})+\left(\frac{\pi(2\times10^{5})}{2500}\right)^{2}(3.75\times10^{6})(1.09\times10^{11})}$$ $$M_u=409.3\ \text{kN}\cdot\text{m}$$
4. (b) Evaluate the flexural resistance. $M_p=Z_xF_y=647\times10^{3}(300)=194.1$ kN·m, and $M_u=409.3 > 0.67M_p=130.0$, so the inelastic branch of Cl. 13.6(a) applies: $$M_r=1.15\phi M_p\left(1-\frac{0.28M_p}{M_u}\right)=1.15(0.9)(194.1)\left(1-\frac{0.28(194.1)}{409.3}\right)=174.3\ \text{kN}\cdot\text{m}$$ which is just below the cap $\phi M_p=174.7$ kN·m. Hence $$\frac{M_f}{M_r}=\frac{170.2}{174.3}=\boxed{0.98\ \le\ 1.0\quad\checkmark}$$ The beam passes in flexure, but with only 2 % in hand — a result typical of how these papers are calibrated.
5. (b) Check shear and deflection. With $h/w=51.2 < 1014/\sqrt{300}=58.5$, the web is stocky and $F_s=0.66F_y=198$ MPa: $$V_r=\phi\,d\,w\,F_s=0.9(353)(6.48)(198)=407\ \text{kN}\ \gg\ V_f=68.7\ \text{kN}\quad\checkmark$$ Under specified load the tip deflects $$\Delta=\frac{PL^{3}}{3EI}+\frac{wL^{4}}{8EI}=\frac{50\times10^{3}(2500)^{3}}{3(2\times10^{5})(102\times10^{6})}+0.09=12.9\ \text{mm}$$ against a cantilever allowance of $2L/180=27.8$ mm. The beam is therefore adequate in all three respects, governed by lateral-torsional buckling.
6. (a) Resolve the moment into a flange couple. With the web bolted for shear only, the moment is delivered by equal and opposite flange forces separated by $d-t$: $$T_f=\frac{M_f}{d-t}=\frac{170.2\times10^{6}}{353-10.7}=497\ \text{kN}$$ Now test whether the beam flange alone can deliver that force: $$\phi btF_y=0.9(128)(10.7)(300)=370\ \text{kN}\ <\ 497\ \text{kN}$$ It cannot. No weld detail of any kind can transfer more than the flange itself can carry, so the moment must be routed through flange plates lapped over the beam flanges and carried past the joint. This is the engineering decision the question is really asking for.
7. (a) Size the flange plates. Treating each plate as a tension member at gross yield, $$A_g\ \ge\ \frac{T_f}{\phi F_y}=\frac{497\times10^{3}}{0.9(300)}=1841\ \text{mm}^2$$ Adopt a 140 × 16 mm plate ($A_g=2240$ mm²) on each flange, complete-joint-penetration welded to the column flange and fillet-welded back onto the beam flange: $$T_r=\phi A_gF_y=0.9(2240)(300)=605\ \text{kN}\ >\ 497\ \text{kN}\quad\checkmark$$
8. (a) Size the fillet welds returning the plate onto the beam. For an 8 mm fillet with $X_u=490$ MPa loaded longitudinally ($\theta=0$), $$v_r=0.67\phi_wA_wX_u=0.67(0.67)(0.707\times8)(490)=1.24\ \text{kN/mm}$$ $$L_w=\frac{T_f}{v_r}=\frac{497}{1.24}=400\ \text{mm total}$$ Provide 8 mm fillet welds, 210 mm long each side of each flange plate (420 mm total), which also satisfies the S16 Table 7 minimum leg for a 16 mm thicker part.
9. (a) Detail the bolted web shear connection. Using M20 A325 bolts in single shear with threads intercepted, through a 10 mm shear tab welded to the column flange: $$V_r=0.70(0.60)\phi_bmA_bF_u=0.70(0.60)(0.80)(1)(314)(830)=87.6\ \text{kN per bolt}$$ Two bolts give 175 kN against $V_f=68.7$ kN. Bearing on the 6.48 mm beam web is $$B_r=3\phi_{br}tdF_u=3(0.80)(6.48)(20)(450)=140\ \text{kN per bolt}\ >\ 34.4\ \text{kN}\quad\checkmark$$ Adopt 2–M20 A325 bolts in a 10 mm plate.
10. (a) Check the column for the concentrated flange force. S16 Cl. 14.3.2 with $\phi_{bi}=0.80$ and a bearing length $N$ equal to the plate thickness: $$B_r(\text{web yielding})=\phi_{bi}w(N+10t)F_y=0.80(10.2)(16+174)(300)=465\ \text{kN}$$ $$B_r(\text{flange bending})=\phi_{bi}(7t^{2})F_y=0.80(7)(17.4)^{2}(300)=509\ \text{kN}$$ Web yielding at 465 kN falls short of $T_f=497$ kN, so transverse web stiffeners are required opposite both beam flanges. A pair of 100 × 10 mm stiffeners fitted to the column flanges and fillet-welded to the web restores the load path with a large margin.

The connection therefore comprises a pair of 140 × 16 mm flange plates CJP-welded to the column and fillet-welded to the beam, a 10 mm shear tab with two M20 A325 bolts, and 100 × 10 mm transverse stiffeners in the column. The beam itself is adequate, but only just, and its governing limit state is lateral-torsional buckling rather than yielding — a reminder that the "lateral restraint at the column and end of the cantilever" clause in the question is load-bearing information, not decoration.