16-Civ-A2 Elementary Structural Design · December 2019

Question 3 of 7: A3 — Ultimate moment of a shored composite beam

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

Notes on this paper

Paper format. National Examinations, December 2019 — 16-Civ-A2 Elementary Structural Design. Three hours, CLOSED BOOK. Seven questions in three parts: Part A (A1–A3, steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). A candidate submits five solutions — two from Part A, two from Part B and the single Part C question — all of equal value. All seven are solved below, because this set is a study resource rather than an exam script. Note 6 on page 1 states that all loads shown are unfactored unless otherwise stated, so every load pattern is factored here before it meets a resistance; Note 7 fixes G40.21 300W steel and 400W reinforcement unless a question says otherwise.

Reference texts.

CSA S16:19, Design of Steel Structures, and CISC Handbook of Steel Construction, 11th ed. — Clauses 13.3 (compression), 13.13 (welds), 17 (composite beams), and the angle/W-shape section tables.
CSA A23.3:19, Design of Concrete Structures, and the Cement Association of Canada Concrete Design Handbook, 4th ed. — Clauses 10 (flexure and columns) and 11 (shear).
CSA O86:19, Engineering Design in Wood, and the Canadian Wood Council Wood Design Manual — Clause 6 (sawn lumber) and the service-condition and size-factor tables.
Kulak & Grondin, Limit States Design in Structural Steel, 11th ed., Chs. 4, 5, 8, 9.
MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design (Canadian edition), Chs. 4, 6, 11.
National Building Code of Canada, Table 4.1.3.2 (load combinations).

Check — assumptions declared under Note 1 of the paper. (i) Figure A1 shows a solid vertical between F and C as well as the load arrow at C; that member is required, because without it the truss counts m + r = 12 + 3 = 15 against 2n = 16 and is a mechanism. With it, m + r = 16 = 2n and the truss is determinate, which is how it is solved here. (ii) In A3 the number of shear studs is reported as a bonus; the question asks only for the ultimate moment. (iii) In B1 the two stacked dimensions read as a sum, so the tee is 100 mm flange plus 400 mm of stem, i.e. 500 mm overall, and the beam spacing of 1250 mm governs the effective flange width because no span is given. (iv) In C1 the 1.5 kPa / 1.0 kPa area loads are converted to line loads on the 0.9 m tributary width, and the deflection limits adopted are L/240 on the back span and 2L/180 on the cantilever.

Question 3: A3 — Ultimate moment of a shored composite beam (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 design starts from a W360×72 at 1000 mm centres spanning 8.0 m under a 100 mm slab of 25 MPa concrete, with 16 mm studs of 415 MPa ultimate stress, 100 % connection and shored construction.

Given data
Item	Value
Steel section	W360×72: A = 9100 mm², d = 350, b = 204, t = 15.1 mm
Span / spacing	8.0 m / 1000 mm on centre
Slab	100 mm thick, f′_c = 25 MPa
Studs	16 mm diameter, F_u = 415 MPa, 100 % connection
Construction	shored during casting
Steel grade	300W, F_y = 300 MPa

Find. The ultimate (factored) moment of resistance of the composite section, with the plastic neutral axis located explicitly.

Composite section: 100 mm slab on a W360x72 at 1000 mm centres, with the plastic neutral axis just inside the top flange.

Approach. Compare the compressive capacity of the effective slab with the tensile capacity of the steel section. Whichever is smaller tells you where the plastic neutral axis lies; then take moments of the resulting force blocks about any convenient point.

Effective slab width. S16 Clause 17.4.1 limits the effective width to the lesser of a quarter of the span each side and the beam spacing: $$b_{\text{eff}} = \min\left(\frac{8000}{4},\ 1000\right) = \boxed{1000\ \text{mm}}$$ The spacing governs, as it usually does on closely spaced floor beams.
Slab compressive capacity. $$C_r = 0.85\,\phi_c f′_c\, b_{\text{eff}}\, t_c = 0.85(0.65)(25)(1000)(100)/10^{3} = 1381\ \text{kN}$$
Steel tensile capacity. $$T_r = \phi A_s F_y = 0.90(9100)(300)/10^{3} = 2457\ \text{kN}$$
Locate the plastic neutral axis. Because $C_r = 1381 < T_r = 2457$ kN, the slab alone cannot balance the steel, so part of the steel must also be in compression and the neutral axis falls inside the steel section. Equilibrium requires the steel compression block $C_s$ to satisfy $T_r - C_s = C_r + C_s$, hence $$C_s = \frac{T_r - C_r}{2} = \frac{2457 - 1381}{2} = 538\ \text{kN}$$ The corresponding steel area is $$A_{\text{comp}} = \frac{C_s}{\phi F_y} = \frac{538\times10^{3}}{0.90(300)} = 1992\ \text{mm}^{2}$$ The top flange alone offers $204(15.1) = 3080$ mm², so the axis lies within the flange at a depth $$\bar{y} = \frac{1992}{204} = \boxed{9.77\ \text{mm below the top of the steel}}$$
Assemble the force blocks. Working downward from the top of the steel flange (the slab therefore sits at negative ordinates): the slab compression 1381 kN acts 50 mm above that datum; the flange compression 538 kN acts 4.89 mm below it; the remaining steel area $9100 - 1992 = 7108$ mm² carries $$T = 0.90(7108)(300)/10^{3} = 1919\ \text{kN}$$ at the centroid of the tension part, $$\bar{y}_T = \frac{9100(175) - 1992(4.89)}{7108} = 222.7\ \text{mm}$$ Equilibrium checks: $1381 + 538 = 1919$ kN.
Moment of resistance. Taking moments about the plastic neutral axis, $$M_r = 1381(9.77 + 50) + 538(9.77 - 4.89) + 1919(222.7 - 9.77)$$ $$M_r = 82\,540 + 2\,628 + 408\,617 = 493\,785\ \text{kN}\cdot\text{mm} = \boxed{494\ \text{kN}\cdot\text{m}}$$ Summing about the slab centroid instead gives $1919(272.7) - 538(54.9) = 494$ kN·m, confirming the arithmetic.
What composite action bought. The bare steel section would give $M_r = \phi Z_x F_y = 0.90(1280\times10^{3})(300) = 346$ kN·m, so the slab has added 43 % for the cost of the studs. Since the system is shored, the whole moment is carried by the composite section — nothing is locked into the bare beam during casting.
Shear studs (bonus check on the stated 100 % connection). With $A_{sc} = \pi(16)^{2}/4 = 201$ mm² and $E_c = 4500\sqrt{25} = 22\,500$ MPa, $$q_r = \min\left[0.5\phi_{sc}A_{sc}\sqrt{f′_c E_c},\ \phi_{sc}A_{sc}F_u\right] = \min(60.3,\ 66.8) = 60.3\ \text{kN per stud}$$ Full interaction needs $C_r/q_r = 1381/60.3 = 22.9$, i.e. 23 studs between the support and midspan, 46 over the span.

Final results
Quantity	Value
Effective slab width	1000 mm (spacing governs)
Slab compression C_r	1381 kN
Steel tension T_r	2457 kN
Plastic neutral axis	9.77 mm into the top flange
Steel compression / tension blocks	538 kN / 1919 kN
Ultimate moment M_r	494 kN·m
Bare-steel M_r for comparison	346 kN·m (+43 % from composite action)
Stud resistance q_r	60.3 kN each
Studs for 100 % connection	23 per half span (46 total)

Topic. Plastic moment of a fully composite steel-concrete beam.

Key relations. $b_{\text{eff}} = \min(L/4,\ s)$; $C_r = 0.85\phi_c f′_c b_{\text{eff}} t_c$; $T_r = \phi A_s F_y$; steel compression $C_s = (T_r - C_r)/2$; stud resistance $q_r = \min\left[0.5\phi_{sc}A_{sc}\sqrt{f′_c E_c},\ \phi_{sc}A_{sc}F_u\right]$.

Why this works. The plastic method assumes every fibre reaches its factored strength, so the section reduces to a small number of rectangular force blocks and the only real question is where the neutral axis sits. The $(T_r - C_r)/2$ expression comes straight from that: whatever the slab cannot balance must be shared equally between a compression block and the reduction it causes in the tension block, because moving the axis down by one unit adds area to compression and removes the same area from tension. Because the beam is shored, no stress is locked into the bare steel during casting and the entire service history can be referred to the composite section.

Common pitfall. Two errors dominate. Taking the neutral axis to be at the slab soffit whenever the slab is the weaker element ignores the steel compression block entirely and understates $M_r$; and using $A_s F_y$ moment arms measured to the slab top rather than to the block centroids inflates the answer. Always close the loop by summing forces (here 1381 + 538 = 1919 kN) before taking moments.

Source. CSA S16:19 Clause 17 (17.4.1 effective width, 17.9 flexural resistance, 17.7.2 stud resistance); Kulak & Grondin, Limit States Design in Structural Steel, Ch. 10.

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