16-Civ-A2 Elementary Structural Design · May 2013

Question 2 of 7: A2 — Welded Rigid Splice in an Overhanging W610×241

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

Question 2: A2 — Welded Rigid Splice in an Overhanging W610×241 (4 + 16 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. W610×241 of G40.21 350W ($F_y = 350$ MPa). Pin at A, roller at B, free end at C. Span $AB = 5.0$ m (dimensioned 1.5 + 2.0 + 1.5 m), overhang $BC = 2.0$ m. Unfactored live loads: 60 kN at 1.5 m from A, 60 kN at 3.5 m from A, and 50 kN at the free end C. Section properties: $d = 635$ mm, $b_f = 329$ mm, $t_f = 31.0$ mm, $w = 17.9$ mm, $Z_x = 7\,670\times10^3$ mm³. Electrodes E49xx ($X_u = 490$ MPa), $\phi_w = 0.67$.

Find. (a) the factored flexure and shear that the splice at B must carry; (b) a welded rigid connection detail, with all welds and splice plates sized and checked.

Approach. Find the reactions by statics, read the moment and the two shears at the splice section, factor them, then proportion a flange-plate and web-plate splice: the flange plates and their fillet welds carry the moment as a force couple, and the web plates and their welds carry the shear.

1. Reactions on the beam. Taking moments about A with $x$ measured from A, $$R_B=\frac{60(1.5)+60(3.5)+50(7.0)}{5.0}=\frac{650}{5.0}=130\ \text{kN}\uparrow$$ and vertical equilibrium gives $R_A = 60+60+50-130 = 40$ kN upward.
2. Internal actions at the splice section B. The cleanest route is the free body to the right of B, which carries only the 50 kN tip load on a 2.0 m lever: $$M_B=-50(2.0)=-100\ \text{kN}\cdot\text{m}\qquad \text{(hogging)}$$ Checking from the left, $40(5.0)-60(3.5)-60(1.5) = 200-210-90 = -100$ kN·m — the two agree. The shear is discontinuous across the support: just left of B, $V = 40-60-60 = -80$ kN; just right of B, $V = -80+130 = +50$ kN. The splice must be able to sit on either face of the reaction point, so the larger value, 80 kN, is used.
3. Factored design actions (part a). The figure loads are specified live loads, so $$M_f=1.5(100)=150\ \text{kN}\cdot\text{m}\qquad V_f=1.5(80)=120\ \text{kN}$$ $$\boxed{M_f=150\ \text{kN}\cdot\text{m},\quad V_f=120\ \text{kN}}$$ For reference the member itself is Class 1 with $M_r = 0.90(7\,670\times10^3)(350) = 2\,416$ kN·m, so the splice is nowhere near the member capacity. A splice designed for the actual forces is what the question asks for; a designer wishing to make the joint truly "full strength" would instead develop the flanges with complete-joint-penetration groove welds.
4. Convert the moment into a flange force couple. In a rigid I-beam splice the flanges carry essentially all of the moment, acting at the centroidal distance between them: $$T_f=C_f=\frac{M_f}{d-t_f}=\frac{150\times10^6}{635-31.0}=\frac{150\times10^6}{604}=248\ \text{kN}$$ The top flange is in tension (the moment is hogging) and the bottom flange in compression by the same amount.
5. Size the flange splice plates. Try one plate 250 mm × 16 mm on each flange, G40.21 350W. Gross-section yielding governs because the joint is welded and there are no bolt holes: $$T_r=\phi A_g F_y=0.90(250\times16)(350)=1\,260\ \text{kN} \;>\; 248\ \text{kN}\quad\checkmark$$ The plate is deliberately narrower than the 329 mm flange so that fillet welds can be laid along both of its longitudinal edges.
6. Design the flange welds. For a fillet weld loaded parallel to its axis ($\theta = 0$), CSA S16 Clause 13.13.2.2 gives the factored resistance per millimetre of an 8 mm leg as $$v_r=0.67\,\phi_w\,(0.707D)\,X_u=0.67(0.67)(0.707\times8)(490)=1.24\ \text{kN/mm}$$ The length required on each side of the splice is $L = 248/1.24 = 200$ mm. Providing two 150 mm longitudinal fillets (one along each edge of the plate) gives $2(150)(1.24) = 373$ kN, so $$\boxed{\text{8 mm fillets, 150 mm long each side of the splice, on both edges of each flange plate}}$$ The 8 mm leg also satisfies the minimum size for a 31 mm thick flange and stays below the maximum permitted on a 16 mm plate edge.
7. Design the web splice for shear. Provide two plates, 300 mm deep × 10 mm, one on each face of the web. Their shear resistance is $$V_r=0.66\,\phi\,A_w F_y=0.66(0.90)(2\times300\times10)(350)=1\,247\ \text{kN}\;\gg\;120\ \text{kN}\quad\checkmark$$ Welding them with 6 mm fillets loaded transversely ($\theta = 90^\circ$, so the $1+0.5\sin^{1.5}\theta$ term becomes 1.5) gives $v_r = 1.40$ kN/mm, so only $120/1.40 = 86$ mm of weld is arithmetically needed. Four vertical 250 mm fillets (two per plate, one each side of the splice line) are used, since the practical requirement is to weld the full depth of the plate rather than a calculated stub length.
8. Check the joint as a whole. Combining, the flange plates supply a moment resistance of $2(1\,260)(0.604)/2 = 761$ kN·m about the section centroid, comfortably above $M_f = 150$ kN·m, and the web plates supply $1\,247$ kN in shear against $V_f = 120$ kN. Because the flange plates are continuous across the joint and welded on both sides, the connection transmits moment without appreciable rotation, satisfying the "rigid" requirement.

The governing check is therefore the flange weld length, not any plate, and the detail is dominated by minimum-size and practical-length rules rather than by the calculated demand.