A1 — Welded rigid connection at B in a W610 × 241 column: 07-Str-A2, December 2016, Q1

Question 1: A1 — Welded rigid connection at B in a W610 × 241 column (12 + 8 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 W610 × 241 column of G40.21-300W steel, fixed at the base A and rigidly restrained at the head C, made up of two 4 m lengths AB and BC spliced at mid-height B. The head carries a vertical load of 300 kN on the column centreline together with an applied moment of 400 kN·m. Section properties are taken from the CISC Handbook.

Quantity	Symbol	Value
Overall depth	d	635 mm
Flange width	b	329 mm
Flange thickness	t	31.0 mm
Web thickness	w	17.9 mm
Gross area	A	30 800 mm²
Yield strength	F_y	300 MPa
Applied head moment / axial load	M / N	400 kN·m / 300 kN (unfactored)

Find. The factored axial force, moment and shear transmitted through the splice at B, and a welded splice detail (plate sizes, fillet weld sizes and lengths) that resists them.

Figure A1 — the 8 m column ABC, fixed at both ends, spliced at mid-height B.

Approach. Find the internal moment and shear at mid-height from the moment distribution in a member whose far end is fixed, factor the loads, then split the resultant into a flange couple carried by flange splice plates and a web force carried by web splice plates, and size E49xx fillet welds for each.

Distribute the applied head moment. The moment is applied to the column head, whose far end A is fixed. For a prismatic member the far-end carry-over factor is one half, so $$M_A=\tfrac{1}{2}M_C=\tfrac{1}{2}(400)=200\ \text{kN}\cdot\text{m}$$ acting in the opposite sense to the head moment.
Shear from end-moment equilibrium. With no transverse load along the column the shear is constant over the whole height: $$V=\frac{M_C+M_A}{L}=\frac{400+200}{8}=75.0\ \text{kN}$$
Internal moment at the splice. The moment therefore varies linearly from +400 kN·m at C to −200 kN·m at A, crossing zero 5.33 m below the head. At mid-height, 4 m below C, $$M_B=400-75.0(4)=\boxed{100\ \text{kN}\cdot\text{m}}$$ and the axial force is unchanged at 300 kN.
Factor the loads. The figure loads are unfactored live load, so 1.5 applies throughout: $$\begin{aligned} &C_f=1.5(300)=450\ \text{kN} \\ &M_f=1.5(100)=150\ \text{kN}\cdot\text{m} \\ &V_f=1.5(75.0)=112.5\ \text{kN} \end{aligned}$$

The splice must therefore transfer 450 kN of compression, 150 kN·m of moment and 112.5 kN of shear across the cut at B. The standard elementary treatment shares these between a flange couple and a web connection in proportion to the areas of the elements that carry them.

Flange forces. One flange has area $A_f=bt=329(31.0)=10\,199\ \text{mm}^2$, so the two flanges carry a fraction $2A_f/A = 20\,398/30\,800 = 0.662$ of the axial load. The couple arm is the flange-centroid separation $d-t$: $$\begin{aligned} &F_M=\frac{M_f}{d-t}=\frac{150\times10^{6}}{635-31.0}=248.3\ \text{kN} \\ &F_N=\frac{0.662\,C_f}{2}=149.0\ \text{kN} \end{aligned}$$ so the flange plates are designed for $$F_{f}=248.3+149.0=\boxed{397.4\ \text{kN}\ \text{(compression)}}$$ while the opposite flange carries only 248.3 − 149.0 = 99.3 kN of tension.
Size the flange splice plates. Adopt one 250 × 20 plate on the outside of each flange (narrower than the 329 mm flange so that longitudinal fillet welds can be run along both edges). Gross-section yield, S16 Clause 13.2: $$T_r=\phi A_g F_y=0.90(250\times20)(300)=1350\ \text{kN}\ \gg\ 397.4\ \text{kN}$$
Fillet welds to the flange plates. Use E49xx electrode, $X_u=490$ MPa. S16 Clause 13.13.2.2 gives, per millimetre of weld, $$v_r=0.67\,\phi_w(0.707D)X_u\bigl(1+0.50\sin^{1.5}\theta\bigr)$$ With $D=8$ mm (the S16 Table 7 minimum for a 31 mm thick part): $v_r=1244$ N/mm longitudinally and $1866$ N/mm transversely. Providing a 250 mm end weld plus 100 mm along each side, on each side of the splice, $$\begin{aligned} &V_r=250(1.866)+2(100)(1.244)=\boxed{715\ \text{kN}} \\ &\frac{397.4}{715}=0.56 \end{aligned}$$
Web splice. The web carries the balance of the axial load together with the whole shear: $$\begin{aligned} &N_w=(1-0.662)(450)=152.0\ \text{kN} \\ &R_w=\sqrt{152.0^{2}+112.5^{2}}=189.1\ \text{kN} \end{aligned}$$ Two 10 × 350 plates, one each side of the web, welded with 6 mm fillets (the Table 7 minimum for a 17.9 mm part) along four lines of 350 mm give $$V_r=4(350)(0.933)=1306\ \text{kN}\ \gg\ 189.1\ \text{kN}$$

The connection is comfortably governed by the flange plates; the web plates are set by minimum weld size and plate stiffness rather than by strength. Both plates are welded on both sides of the splice line, which is what makes the joint rigid: the flange couple restores full moment continuity and the web plates restore shear and web axial continuity, so the spliced column behaves as the single 8 m member assumed in Step 1.

Result	Value
Carry-over moment at the fixed base A	200 kN·m
Shear in the column	75.0 kN
Service moment at the splice, M_B	100 kN·m
Factored actions at B	C_f = 450 kN, M_f = 150 kN·m, V_f = 112.5 kN
Design flange force	397.4 kN (compression) / 99.3 kN (tension)
Flange splice plates	2 – 250 × 20 plates, T_r = 1350 kN
Flange welds	8 mm E49xx fillet, 250 mm end + 2 × 100 mm sides, V_r = 715 kN
Web splice plates	2 – 10 × 350 plates, 6 mm fillets, V_r = 1306 kN vs R_w = 189.1 kN

Topic. Rigid (moment-resisting) welded splices in a column, and the carry-over of an applied end moment.

Key relations. Carry-over factor $\tfrac12$ for a prismatic member with a fixed far end; $V=(M_C+M_A)/L$; flange couple $F_M=M_f/(d-t)$; axial load shared in proportion to element area; fillet weld $v_r=0.67\phi_w(0.707D)X_u(1+0.5\sin^{1.5}\theta)$.

Why this works. A splice is a cut through the member: it must carry exactly the internal forces that cross that cut, no more. Separating those forces into a flange couple and a web force reproduces the way the rolled section itself carries them, so the plates are stressed the way the parent material is and no artificial stress concentration is introduced.

Common pitfall. Designing the flange plates for the moment alone. The axial load adds roughly 60 % to the compression-flange force here (397 kN against 248 kN), and it subtracts from the tension flange — both flanges must be checked, and the governing one is not always obvious.

Source. CSA S16:19 Clauses 13.2, 13.13.2.2 and Table 7; CISC Handbook of Steel Construction, 11th ed., Part 3 (welded connections).

Question 1 of 7: A1 — Welded rigid connection at B in a W610 × 241 column

Question 1: A1 — Welded rigid connection at B in a W610 × 241 column (12 + 8 marks)