07-Str-A5 · Undated paper

Question 3 of 7: A3: Two sections of unequal depth for Figure 1, with a continuity splice

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

Notes on this paper

Paper format. National Exams — 07-Str-A5 Advanced Structural Design, printed footer 07-Str-A5/May 2019. Three hours, "closed book" with handbooks and textbooks permitted. Seven questions of equal value (20 marks each) in three parts: A1–A3 (do two), B1–B3 (do two), C1 (do one) — five solutions make a complete paper. All seven are solved here, because the set is a study resource rather than an exam attempt.

Design data given on page 1. Solutions to CAN/CSA S16 (steel), CAN/CSA A23.3 (concrete) and CAN/CSA O86 (timber). All loads shown on the figures are unfactored. All structural steel is G40.21 300W — $F_y = 300$ MPa, $F_u = 450$ MPa, $E = 200\,000$ MPa, $G = 77\,000$ MPa. All reinforcement is 400W — $f_y = 400$ MPa. Concrete strength is stated per question. Steel sections are W-shapes unless noted otherwise.

Reference texts.

CSA S16:19, Design of Steel Structures — Cl. 11 (section classification), 13.3 (compression), 13.4 (shear), 13.5–13.6 (bending and lateral–torsional buckling), 13.8 (beam-columns), 13.13 (welds), 14.3.2 (web crippling), 8.4.3 (stability effects).
CSA A23.3:19, Design of Concrete Structures — Cl. 8.6 (material properties), 10 (flexure and axial load, slenderness), 11 (shear), 15 (footings), 18 (prestressed concrete).
National Building Code of Canada 2020, Part 4 — Table 4.1.3.2 load combinations, 4.1.6 snow and drift loads.
CISC, Handbook of Steel Construction, 11th ed. — W-shape dimensions and properties.
Salmon, Johnson & Malhas, Steel Structures: Design and Behavior, 5th ed. — Ch. 9 (LTB), 12 (beam-columns), 13 (connections).
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Canadian ed. — Ch. 4–6 (flexure), 11 (columns), 12 (slender columns), 15 (footings).
Collins & Mitchell, Prestressed Concrete Structures — Ch. 4 (service stresses), 6 (flexural strength); PCI Design Handbook, 8th ed. — multiplier method for long-term camber.

Check — load combinations. The figures split the loads into DEAD, LIVE, SNOW and WIND, so NBCC 2020 Table 4.1.3.2 is applied literally. Throughout Part A the governing case is $1.25D + 1.5L$ (case 2 with no companion snow or wind on the figure). For Figures 3 the governing case is $1.25D + 1.5S + 0.4W$ (case 3), with $1.25D + 1.4W + 0.5S$ and $1.4D$ checked and shown not to govern. The "LIVE/SNOW" label on W1 of Figure 3 is read as a single 3 kN/m specified roof load that may be either occupancy live or balanced snow, and the triangular W2 is read as the additional snow-drift surcharge that its shape implies.

Question 3 — A3: Two sections of unequal depth for Figure 1, with a continuity splice (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. The beam and loading of Figure 1 (see Question 1), now built from two W-shapes spliced over support B, with span A–B required to be about two-thirds the depth of span B–C. Continuity at B must be preserved by a welded connection.

Find. The two W-shapes, and a welded splice detail at B that transmits the hogging moment and shear without a hinge forming.

Approach. Because the redundant moment at B depends on the ratio of the two flexural stiffnesses, the analysis must be redone with $I_1 \ne I_2$; the shallower span attracts less hogging moment, which pushes more sagging moment into span B–C.

Choose the deeper section first. Span B–C carries the point load, so retain W530×72 ($d = 524$ mm, $I_x = 399\times10^6$ mm$^4$) from Question 1. Two-thirds of 524 mm is 349 mm, which points at the W360 series; try W360×45 ($d = 352$ mm, $I_x = 121\times10^6$ mm$^4$), giving a depth ratio of $352/524 = 0.672$.
Re-solve with unequal stiffness. The three-moment equation written in $L/I$ form gives $$2M_B\left(\frac{L_1}{I_1} + \frac{L_2}{I_2}\right) = -\left[\frac{w_fL_1^3}{4I_1} + \frac{3P_fL_2^2}{8I_2}\right]$$ and substituting $I_1 = 121\times10^6$, $I_2 = 399\times10^6$ mm$^4$ yields $$\boxed{M_B = -157.3\ \text{kN}\cdot\text{m}}$$ compared with 216.8 kN$\cdot$m for a prismatic beam — a 27% reduction.
Recover the new actions. $R_A = 52.92$ kN, $R_C = 143.54$ kN and $R_B = 322.29$ kN, so the sagging moment under the point load rises to $$M_{BC} = 143.54(2.5) = \boxed{358.9\ \text{kN}\cdot\text{m}}$$ while the peak sagging moment in span A–B rises to 41.5 kN$\cdot$m at 1.568 m from A. The shears at B become 115.8 kN on the A–B side and 206.5 kN on the B–C side.
Re-check the deeper span. The flatter moment diagram gives $\omega_2 = 1.311$ for segment B–C, so W530×72 now offers $M_r = 379.2$ kN$\cdot$m against a demand of 358.9 kN$\cdot$m — 95% utilised, still adequate. Nothing lighter is available; the section is now working essentially at capacity.
Check the shallow span. Segment A–B reverses curvature, so $\omega_2$ computes as 3.07 and is capped at 2.5. For W360×45, $M_p = Z_xF_y = 233.1$ kN$\cdot$m, $M_u = 347.1$ kN$\cdot$m and $$M_r = 1.15(0.9)(233.1)\left[1 - \frac{0.28(233.1)}{347.1}\right] = 195.9\ \text{kN}\cdot\text{m} \ \ge\ 157.3\ \text{kN}\cdot\text{m}$$ at 80% utilisation. Shear: $V_r = 432.8$ kN against 115.8 kN. The section is Class 2 ($b/2t = 8.72$, $h/w = 48.2$).
Size the splice from the transferred actions. The joint at B must carry the full hogging moment and the larger of the two adjacent shears, i.e. 157.3 kN$\cdot$m and 206.5 kN. The controlling member is the smaller one; its elastic flange stress there is $$f = \frac{M_B}{S_x} = \frac{157.3\times10^6}{688\times10^3} = 228.6\ \text{MPa} \ \lt\ \phi F_y = 270\ \text{MPa}$$ so the splice is required to develop the W360×45, not more.
Specify the welded detail. Align the top flanges (a common floor level) and let the depth step occur on the soffit. Then, at the face of a full-depth transition piece over the support:
- Both flanges: complete-joint-penetration (CJP) groove welds, single-V with backing, matching electrode E49XX. Under CSA S16 Cl. 13.13.1 a CJP groove weld with matching filler develops the full strength of the thinner part joined, so the splice moment resistance equals $M_r$ of the W360×45 — 195.9 kN$\cdot$m against the 157.3 kN$\cdot$m demand.
- Width and thickness transition: the flanges are 171 mm and 207 mm wide, and 9.8 mm and 10.9 mm thick. CSA W59 requires the transition to be tapered no steeper than 1 in 2.5, so the wider flange is flame-cut back over a 90 mm run each side.
- Web: CJP groove weld over the common depth, or a double-fillet-welded splice plate carrying $V_f = 206.5$ kN. The step in depth is closed by a full-depth web transition plate welded to both webs.
- Stiffeners and bracing: a pair of transverse stiffeners at B on the deeper section, aligned with the splice; lateral bracing to the bottom flange at B, since the analysis assumes restraint there.
Confirm the support does not need a bearing stiffener for the reaction. With a 150 mm bearing length, the W530×72 web gives an interior web-yielding resistance $B_r = 0.80\,w(N + 10t)F_y = 559.4$ kN and a web-crippling resistance $B_r = 1.45(0.80)w^2\sqrt{F_yE} = 727.8$ kN, both above $R_B = 322.3$ kN. The stiffeners specified above are for the splice geometry, not for bearing.

Question 3 — results
Quantity	Value
Section, span A–B	W360×45 ($d = 352$ mm)
Section, span B–C	W530×72 ($d = 524$ mm)
Depth ratio achieved	0.672 (target 2/3)
Hogging moment at B	157.3 kN$\cdot$m (was 216.8 prismatic)
Sagging moment, span B–C	358.9 kN$\cdot$m
Reactions $R_A$ / $R_B$ / $R_C$	52.9 / 322.3 / 143.5 kN
$M_r$ available: W360×45 / W530×72	195.9 / 379.2 kN$\cdot$m
Splice	CJP groove welds, E49XX, 1:2.5 flange taper, web transition plate
Bearing at B	$B_r = 559.4$ kN (yielding), 727.8 kN (crippling)

Topic. Stiffness-dependent moment distribution and welded continuity splices in a non-prismatic continuous beam.

Key relations. Three-moment equation in $L/I$ form; CSA S16 Cl. 13.13.1 (a CJP groove weld with matching electrode develops the base metal); CSA W59 taper limit of 1:2.5 at a change of section; interior bearing $B_r = 0.80w(N+10t)F_y$ and crippling $B_r = 1.45(0.80)w^2\sqrt{F_yE}$.

Why this works. Moment migrates towards stiffness in an indeterminate structure, so shallowing span A–B by a third does not merely make that span weaker — it actively sheds hogging moment onto span B–C, whose sagging moment rises from 329 to 359 kN$\cdot$m. That is the whole point of the question, and it is why the deeper section cannot simply be carried over unchanged from A1 without re-checking. The welded splice must reproduce the flexural continuity assumed in that analysis; a bolted shear-only connection would create a hinge at B and turn the structure into two simple spans, roughly doubling the sagging moment in B–C.

Common pitfall. Designing the two spans independently from the prismatic-beam moments of Question 1, which understates the B–C sagging moment by 9% and would leave W530×72 apparently at 86% when it is really at 95%. The second is detailing the splice for the moment only: the joint also carries 206 kN of shear across a change in web depth, which is why the web transition plate is not optional.

Source. CSA S16:19 Cl. 13.6, 13.13.1 and 14.3.2; CSA W59:18 Cl. 4.4 (transitions); MacGregor's companion treatment of stiffness-proportioned continuity in Salmon & Johnson, Ch. 13.

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