07-Str-A2 · December 2014

Question 3 of 7: A3 — Welded rigid splice in a W360×79 beam

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

Notes on this paper

Paper format. National Examinations, December 2014 — 07-Str-A2 Elementary Structural Design, 3-hour duration, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel, CSA S16), Part B (B1–B3, reinforced concrete, CSA A23.3) and Part C (C1, timber, CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here so the set works as a complete study resource.

Reference texts. CSA S16, Design of Steel Structures, with the CISC Handbook of Steel Construction; CSA A23.3, Design of Concrete Structures, with the CAC Concrete Design Handbook; CSA O86, Engineering Design in Wood, with the CWC Wood Design Manual; National Building Code of Canada (load combinations); Kulak & Grondin, Limit States Design in Structural Steel; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design.

Check — load factors. Note 6 on page 1 states “All loads shown are unfactored, unless otherwise specified”, but the paper never splits the figure loads into dead and live components. Throughout Parts A and B the applied figure loads are therefore treated as live load, factored by 1.5, and self-weight of concrete members as dead load, factored by 1.25 (NBCC 4.1.3.2, combination 1.25D + 1.5L). In Part C the paper does name the components, so 1.25D + 1.5S is used directly. If an examiner intended a different split the method is unchanged — only the numerical factor moves.

Check — figure dimensions. Every figure on this paper is a hand sketch on page 3 whose OCR text is unusable; all geometry below was read from the printed figure of the original PDF. Figure A1 is not drawn to scale horizontally (its printed 600 and 300 dimensions are authoritative, not the drawn proportions). In Figure B1 the dimension “1 m × 1 m” runs from the section centreline to the outer face, so each cell is 1 m square and the box is 2000 mm wide × 1000 mm deep; the “600 × 600” dimension line brackets the void, giving 200 mm walls and a 400 mm central web. That reading is confirmed independently by the “65 typical” cover note: 65 + 65 cover plus two 15M tie legs plus one 30M bar needs exactly 200 mm of wall.

Question 3: A3 — Welded rigid splice in a W360×79 beam (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 W360×79 of G40.21 350W spanning 6 m overall with a 1 m overhang beyond the left support. The two halves AB and BC are each 3 m long and are joined by a shop splice at B.

Item	Position from A	Value
Point load at the free tip	0 m	50 kN
Roller support	1.0 m	—
Point load	2.0 m	180 kN
Splice B	3.0 m	—
Point load	4.5 m	180 kN
Pin support C	6.0 m	—
W360×79: d = 354, b = 205, t = 16.8, w = 9.4 mm		Z_x = 1430 × 10³ mm³

Find. The shear and moment that the splice at B must transmit, and a welded detail — flange and web welds — capable of carrying them.

[Figure not reproduced: Figure A3 — beam elevation. Distances in metres; loads as printed (unfactored). See the official exam paper.]

Approach. Solve the determinate beam for its reactions, extract the shear and moment at the splice section, factor them, then resolve the moment into a flange force couple and detail welds for the flanges (moment) and the web (shear).

Reactions. Taking moments about the roller at 1.0 m, with the tip load acting 1.0 m to its left: $$\sum M = -50(1.0) + 180(1.0) + 180(3.5) - R_C(5.0) = 0$$ $$R_C = \frac{760}{5.0} = \boxed{152\ \text{kN}} \qquad R_1 = 410 - 152 = 258\ \text{kN}$$
Shear and moment at the splice. Working from the left of section B, the forces are the 50 kN tip load, the 258 kN reaction and the 180 kN load at 2.0 m: $$V_B = -50 + 258 - 180 = 28\ \text{kN}$$ $$M_B = -50(3.0) + 258(2.0) - 180(1.0) = -150 + 516 - 180 = 186\ \text{kN}\cdot\text{m}$$ Checking from the right-hand free body confirms the value: $152(3.0) - 180(1.5) = 186$ kN·m.
Factor the actions. Applying the 1.5 live-load factor, $$V_f = 1.5(28) = \boxed{42\ \text{kN}} \qquad M_f = 1.5(186) = \boxed{279\ \text{kN}\cdot\text{m}}$$ The beam itself is adequate: $M_r = \phi Z_x F_y = 0.90(1430\times10^3)(350) = 450.5$ kN·m, comfortably above 279 kN·m.
Resolve the moment into a flange couple. A rigid splice is detailed by letting the flanges carry the bending as an equal-and-opposite pair of forces separated by the flange centroidal distance $d - t$: $$T_f = C_f = \frac{M_f}{d-t}=\frac{279\times10^{6}}{354-16.8}=\boxed{827\ \text{kN}}$$
Option 1 — complete-joint-penetration groove welds in the flanges. A CJP weld made with a matching E49xx electrode develops the base metal, so the check is on the flange itself: $$T_r = \phi\,b\,t\,F_y = 0.90(205)(16.8)(350) = 1085\ \text{kN} \; > \; 827\ \text{kN} \quad\checkmark$$ This is the cleanest shop detail: butt the two halves, bevel both flanges and the web, and specify CJP throughout. No further flange calculation is required.
Option 2 — bolted-free fillet-welded splice plates. Where a groove weld is impractical, cover the top and bottom flanges with 200 × 16 plates: $$T_r = \phi\,b_p t_p F_y = 0.90(200)(16)(350) = 1008\ \text{kN} \; > \; 827\ \text{kN} \quad\checkmark$$ Using 10 mm fillet welds of E49xx electrode along both longitudinal edges, the resistance per millimetre of a single line (S16 13.13.2.2, with $\theta = 0$) is $$v_r = 0.67\,\phi_w A_w X_u = 0.67(0.67)(0.707\times10)(490) = 1.555\ \text{N/mm per mm}\times10^{3}$$ so that each millimetre of weld carries 1.555 kN. Two lines are available on each side of the joint: $$L = \frac{827}{2(1.555)} = 266\ \text{mm} \;\Rightarrow\; \text{use } \boxed{270\ \text{mm each side}}$$ The plate is therefore 200 × 16 × 550 long, centred on the splice.
Web welds for the shear. The factored shear is only 42 kN. Two lines of 6 mm fillet weld give $$v_r = 2\left[0.67(0.67)(0.707\times6)(490)\right] = 1.866\ \text{kN/mm}$$ $$L_{\text{req}} = \frac{42}{1.866} = 22.5\ \text{mm}$$ This is far below the practical minimum, so the web splice is governed by detailing rather than strength: provide a pair of 250 × 10 web plates welded with continuous 6 mm fillets over the full available web depth (about 300 mm each side). That also gives the joint the stiffness it needs to behave as the rigid connection the question demands.

The two options are not equivalent in cost. The CJP butt splice needs edge preparation, backing and ultrasonic inspection but leaves a flush member; the plated splice needs no preparation and only visual inspection, but adds 60 kg of plate and a local stiffness step. For a shop-made joint in a 79 kg/m beam the CJP detail is normally the economical choice, and it is the one recommended here.

Quantity	Value
Reactions R_roller / R_C	258 kN / 152 kN
Shear at B (unfactored / factored)	28 kN / 42 kN
Moment at B (unfactored / factored)	186 / 279 kN·m
Beam resistance M_r	450.5 kN·m > 279 ✓
Flange force T_f	827 kN
Recommended detail	CJP groove welds, E49xx, both flanges and web
Alternative flange plates	200 × 16, 10 mm fillets, 270 mm each side
Web plates	2 × (250 × 10), continuous 6 mm fillets

Topic: Rigid (moment-resisting) welded beam splices — flange-couple idealisation and fillet-weld resistance.

Key relations: $T_f = M_f/(d-t)$; CJP resistance equals the base metal, $T_r = \phi A_g F_y$; fillet weld $V_r = 0.67\phi_w A_w X_u(1+0.50\sin^{1.5}\theta)$ with $\phi_w = 0.67$, $X_u = 490$ MPa for E49xx and throat $A_w = 0.707DL$.

Why this works: A splice must reproduce, across a cut, whatever internal actions the uncut member was carrying at that point. Bending is stored almost entirely in the flanges of an I-section, so replacing the moment by a force couple $M_f/(d-t)$ acting at the flange centroids is both accurate and conservative — it ignores the web’s share of $Z$, which for a W360×79 is about a fifth. Shear, by contrast, is carried almost entirely by the web, so the two load paths can be detailed independently: flange welds sized for $T_f$, web welds sized for $V_f$. Because the fillet-weld equation is written per unit throat area, the required length falls straight out of dividing the force by the resistance per millimetre.

Common pitfall: Using the full depth d instead of $d - t$ in the couple — a small error here, but on shallow sections it can flatter the design by 5–10 %. A more serious mistake is designing the flange welds for $M_f/d$ and then forgetting the web entirely: the joint then has ample moment capacity but no reliable shear path, and it is not the rigid connection the question asks for.

Source: CSA S16, Design of Steel Structures, Clauses 13.13.1 (groove welds) and 13.13.2.2 (fillet welds); CISC Handbook of Steel Construction, Part 3.

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