Question 3 of 7: Two sections for the Figure 1 beam, with a welded moment splice at B

Question A3: Two sections for the Figure 1 beam, with a welded moment splice at B (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.

Find. The lightest pair of W sections satisfying the two-thirds depth relation, and a welded splice detail at B that transfers the full hogging moment and shear.

Approach. Repeat the three-moment solution with unequal second moments of area, recognising that a softer span A–B sheds hogging moment onto the sagging region of B–C, then design the splice for the hogging moment and shear that the analysis returns at B.

1. Restate the compatibility equation for unequal stiffness. With $EI$ no longer constant, the three-moment equation carries the flexibility of each span explicitly: $$2M_B\!\left(\frac{L_1}{I_1} + \frac{L_2}{I_2}\right) = -\frac{1}{I_1}\frac{w L_1^{3}}{4} - \frac{1}{I_2}\frac{3PL_2^{2}}{8}$$ Reducing $I_1$ therefore reduces $|M_B|$ — the shallow span simply cannot attract as much hogging — and everything shed at B reappears as sagging in span B–C.
2. Choose the pair. Since span B–C carries the heavier moment it sets the depth. Trialling the W460, W530 and W610 families against a 2/3-depth partner, the lightest pair that satisfies every limit state is W360x51 for A–B and W530x82 for B–C: $$\frac{d_{AB}}{d_{BC}} = \frac{355}{528} = 0.672 \approx \tfrac{2}{3}\ \checkmark$$ $$I_{AB} = 138.9\times10^{6}\ \text{mm}^4, \quad I_{BC} = 468.1\times10^{6}\ \text{mm}^4$$ Both are Class 1.
3. Re-run the load cases with the new stiffness ratio. Including each member's own self weight (0.490 and 0.795 kN/m), the envelope becomes $$M_B = -158.6\ \text{kN}\!\cdot\!\text{m}\ \text{(full factored load)}$$ $$M^{+}_{BC} = 385.1\ \text{kN}\!\cdot\!\text{m}\ (0.9D\ \text{on A--B}), \quad M^{+}_{AB} = 53.5\ \text{kN}\!\cdot\!\text{m}\ (\text{live on A--B only})$$ $$V_f = 209.2\ \text{kN}$$ Compare A1: the hogging at B has dropped from 219.5 to 158.6 kN·m and the sagging in B–C has risen from 345.9 to 385.1 kN·m — exactly the redistribution the stiffness change predicts.
4. Check span B–C. The quarter-point formula gives $\omega_2 = 1.30$ over the 5 m unbraced segment, and for W530x82 ($Z_x = 2.028\times10^{6}\ \text{mm}^3$): $$\phi M_p = 547.5\ \text{kN}\!\cdot\!\text{m}, \quad M_r = 456.3\ \text{kN}\!\cdot\!\text{m}$$ $$\boxed{\frac{385.1}{456.3} = 0.844 \le 1.0}$$
5. Check span A–B, which must carry the hogging moment at the splice. The shallower member is the one present at B, so it takes the full $M_B$. With $\omega_2 = 2.5$ over the 5 m segment and $Z_x = 0.879\times10^{6}\ \text{mm}^3$: $$\phi M_p = 237.4\ \text{kN}\!\cdot\!\text{m}, \quad M_r = 226.5\ \text{kN}\!\cdot\!\text{m}$$ $$\boxed{\frac{158.6}{226.5} = 0.700; \quad \text{sagging: } \frac{53.5}{183.9} = 0.291}$$ Shear in A–B is 209.2 kN against $V_r = 455.5\ \text{kN}$, a ratio of 0.46.
6. Confirm the pair is worth the extra fabrication. $$\text{two-section beam} = (50.6 + 80.4)(5) = 654.6\ \text{kg}$$ $$\text{prismatic W530x72 from A1} = 70.9(10) = 709.0\ \text{kg}$$ The saving is 7.7 per cent of the steel, or about 54 kg on this beam — real but modest, which is worth saying plainly: the two-section solution is justified here by the depth constraint the question imposes rather than by economy.
7. Size the splice for the forces it must carry. The connection is a full-strength moment splice at B, designed for $M_B = 158.6\ \text{kN}\!\cdot\!\text{m}$ and $V = 209.2\ \text{kN}$. The flange force is the couple: $$F_f = \frac{M_B}{d_{AB} - t_f} = \frac{158.6\times10^{6}}{355 - 11.6} = 461.9\ \text{kN}$$ $$T_r = \phi\,b_f t_f F_y = 0.9(171)(11.6)(300) = 535.6\ \text{kN} \ \Rightarrow\ \frac{461.9}{535.6} = 0.86\ \checkmark$$ The shallower flange, not the deeper one, is the limiting element, so the splice is full strength if it develops the W360x51 flange.
8. Define the welded detail. Align the two members on a common top of steel, which puts the top flanges in one plane and confines the depth change to the soffit. Then: (i) top flange joined by a complete-joint-penetration groove weld with matching E49xx electrode, backing bar and run-off tabs, ground flush — a CJP weld with matching filler develops the base metal, so no calculation beyond the flange check above is required; (ii) the bottom flange of the deeper W530x82 tapered up to meet the W360x51 soffit at a slope not steeper than 1 in 2.5, as CSA W59 requires for a transition between parts of unequal thickness or width, and terminated in a CJP weld; (iii) the web spliced with a CJP groove weld over the full depth of the shallower member, which alone carries 455 kN in shear against the 209 kN demand; (iv) a pair of transverse stiffeners in the W530x82 web, aligned with the underside of the W360x51 bottom flange, to carry the 462 kN flange force into the deeper web instead of letting it punch into an unstiffened web. Sizing them for the flange force at $\phi = 0.9$ needs $A_{st} \ge 462\times10^{3}/(0.9 \times 300) = 1711\ \text{mm}^2$, met by two 90 x 12 mm plates.
9. Detail the splice for fatigue and erection. Locate the splice on the centre-line of support B where the deeper member is already stiffened by the bearing detail, specify all groove welds as CJP category B, and require ultrasonic inspection of the flange welds. If field welding is unattractive, the identical forces can be transferred by a bolted flange-plate splice; the question asks specifically for a welded connection, so the welded form is developed here.