Question 1 of 7: Welded moment connection at B

Question A1: Welded moment connection at B (20 marks: 5 + 10 + 5)

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. The welds — type, size and length — that carry the factored end moment and end shear at B from the beam into the column flange, together with whatever the column itself needs to receive that pair of flange forces.

[Figure not reproduced: Figure A1 as drawn on the exam paper, with the fixed-end bending-moment diagram for a central point load. The end and midspan ordinates are both $PL/8$ and the moment crosses zero at the quarter points. See the official exam paper.]

Approach. Read the fixed-fixed beam case out of the Handbook to get the end actions, factor them, replace the moment by a couple of equal and opposite flange forces acting at the flange centroids, size the flange welds for that force and the web welds for the shear, then check that the column can accept the concentrated flange forces.

1. Factor the applied load. With the single specified load treated as live, $$P_f = \alpha_L P = 1.5 \times 300 = 450\ \text{kN}$$
2. Take the end moment and end shear from the Handbook beam diagrams. For a prismatic beam fixed at both ends carrying a concentrated load at midspan the CISC Handbook of Steel Construction beam-diagram table gives $M = PL/8$ at each end and at midspan, and $R = P/2$: $$M_f = \frac{P_f L}{8} = \frac{450 \times 6.0}{8} = 337.5\ \text{kN}\cdot\text{m} \qquad V_f = \frac{P_f}{2} = 225\ \text{kN}$$ Note that the diagram is antisymmetric about the quarter points: the moment is hogging over the outer quarters and sagging over the middle half, with the same peak magnitude everywhere. These are the actions the connection at B must deliver.
3. Confirm the beam itself is adequate, so the connection is the real question. The W530 × 92 is a Class 1 section in bending, so $$M_r = \phi Z_x F_y = 0.90 \times 2360 \times 10^3 \times 350 \times 10^{-6} = 743.4\ \text{kN}\cdot\text{m} \;>\; 337.5\ \text{kN}\cdot\text{m}$$ The member is loaded to 45 per cent of its flexural resistance; nothing about the beam constrains the connection design.
4. Replace the moment by a flange couple. In a directly welded moment connection the flanges carry essentially all of the moment as an equal and opposite pair acting at the flange centroids, a lever arm of $d - t$: $$T_f = C_f = \frac{M_f}{d - t} = \frac{337.5 \times 10^6}{533 - 15.6} = \frac{337.5 \times 10^6}{517.4} = 652\,300\ \text{N}$$ $$\boxed{T_f = C_f = 652.3\ \text{kN per flange}}$$ The bottom flange is in tension and the top flange in compression at the hogging end; the connection must be detailed identically at both because the moment reverses under pattern loading.
5. Size the flange welds — complete-joint-penetration groove welds. With a matching electrode, a CJP groove weld develops the base metal, so the check is on the gross flange: $$T_r = \phi b t F_y = 0.90 \times 209 \times 15.6 \times 350 \times 10^{-3} = 1027\ \text{kN}$$ $$\boxed{\frac{T_f}{T_r} = \frac{652.3}{1027} = 0.64 \;<\; 1.0\ \ \text{(CJP groove weld at each flange)}}$$ This is the detail to adopt: it is the standard field-welded moment connection, it needs no calculation of weld length, and it leaves the flange — not the weld — as the governing element.
6. Fillet-weld alternative, for comparison. Had fillets been used instead, S16 Cl.13.13.2.2 gives the factored resistance of a fillet weld as $$V_r = 0.67\,\phi_w A_w X_u \left(1.00 + 0.50 \sin^{1.5}\theta\right)$$ with $\phi_w = 0.67$ and $A_w = 0.707 D \ell$. For a transverse weld, $\theta = 90^{\circ}$, the resistance per millimetre of length and per millimetre of leg size is $0.67(0.67)(0.707)(490)(1.5) = 233\ \text{N/mm}^2$. Welding across both faces of the flange gives an effective length of $2b = 418$ mm, so $$D = \frac{652\,300}{418 \times 233} = 6.7\ \text{mm} \;\rightarrow\; 8\ \text{mm fillets}$$ The groove weld is preferred here because 8 mm fillets on both faces of a 15.6 mm flange put more weld metal in than the CJP and are harder to inspect.
7. Size the web welds for the shear. The web welds carry $V_f = 225$ kN as longitudinal welds ($\theta = 0$), for which the unit resistance is $0.67(0.67)(0.707)(490) = 155.5\ \text{N/mm}^2$. Allowing a 10 mm cope at each flange, the available length is $\ell = 533 - 2(15.6) - 20 = 481.8$ mm on each side. Trying the minimum permitted size for the thicker part joined (the 19.6 mm column flange), S16 Table 5.8 gives $D_{\min} = 6$ mm: $$V_r = 2 \ell (155.5) D = 2(481.8)(155.5)(6) \times 10^{-3} = 899\ \text{kN}$$ $$\boxed{\frac{V_f}{V_r} = \frac{225}{899} = 0.25 \;\rightarrow\; \text{6 mm fillet, both sides of the web, 480 mm long}}$$ Minimum size, not strength, governs the web weld — which is the usual outcome for a deep beam whose moment is taken entirely by the flanges.
8. Check the column for the concentrated flange forces. A 652 kN force delivered into the column flange over the 15.6 mm thickness of the beam flange must be spread into the column web. Local yielding of the column web (S16 Cl.14.3.2, dispersion on a 2.5:1 slope through the column flange and fillet) gives $$B_r = \phi w_c (t + 5k) F_y = 0.90 \times 11.9 \times \left(15.6 + 5 \times 30\right) \times 350 \times 10^{-3} = 619\ \text{kN}$$ $$\boxed{B_r = 619\ \text{kN} \;<\; T_f = 652\ \text{kN} \;\rightarrow\; \text{continuity stiffeners required}}$$ The shortfall is 31.5 kN, but the check is not satisfied and the column flange would also be asked to bend locally on the tension side, so a pair of transverse stiffeners is provided opposite each beam flange.
9. Proportion the continuity stiffeners. Try a pair of $16 \times 100$ plates, one each side of the column web, welded to the web and to both column flanges: $$A_r = \phi (2 b_s t_s) F_y = 0.90 \times 2 \times 100 \times 16 \times 350 \times 10^{-3} = 1008\ \text{kN} \;>\; 31.5\ \text{kN}$$ with the proportioning rules all satisfied: $t_s = 16 \geq t/2 = 7.8$ mm; $b_s/t_s = 6.25 \leq 200/\sqrt{350} = 10.7$ (Class 3 outstand); and $2b_s + w_c = 212 \leq b_c = 229$ mm, so the plates fit inside the column flange. Fillet welds of 6 mm all round attach them.