07-Str-A2 · May 2018

Question 4 of 7: B1 — Moment and shear resistances of a double-T concrete section

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

Notes on this paper

Paper format. National Examinations, May 2018 — 07-Str-A2 Elementary Structural Design. Three hours, CLOSED BOOK (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete), Part C (C1, timber). A candidate submits five solutions — two from Part A, two from Part B and the one question in Part C — all of equal value. All seven questions are solved below, because this set is a study resource rather than an exam script. Page 1 states that all loads shown are unfactored, so every load case is factored here before any resistance is compared against it.

Reference texts.

CSA S16:19, Design of Steel Structures — and CISC, Handbook of Steel Construction, 11th ed. (section tables, Table 2 class limits, Table 7 minimum fillet sizes).
CSA A23.3:19, Design of Concrete Structures — and Cement Association of Canada, Concrete Design Handbook, 4th ed.
CSA O86:19, Engineering Design in Wood — and Canadian Wood Council, Wood Design Manual.
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 2015, Table 4.1.3.2 (load combinations).

Check — one dimension is missing from the source. Question A2 describes a stub cantilever welded to a column but the paper contains no Figure A2 and never states the cantilever projection. The projection is therefore taken as L = 1.5 m from the column face throughout question A2; every result below is also given in the general form so any other projection can be substituted directly. All other data are read from the printed text and from Figure A3, B1, B2 and B3 on page 3.

Question 4: B1 — Moment and shear resistances of a double-T concrete section (10 + 10 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.

Quantity	Value
Flange (slab)	3000 mm wide (2 × 1.5 m about y-y) × 200 mm thick
Stems	two, each 200 mm wide, clear spacing 700 mm
Overall depth, h	1200 mm (200 slab + 1000 stem)
Tension reinforcement	4–25M (2 per stem), A_s = 2000 mm²
Transverse reinforcement	closed 15M @ 200 mm in each stem (4 legs, A_v = 800 mm²)
Materials	f_c′ = 35 MPa, f_y = 400 MPa, φ_c = 0.65, φ_s = 0.85

Find. The factored moment resistance M_r (sagging, flange in compression) and the factored shear resistance V_r.

[Figure not reproduced: Figure 4.1 — Double-T section. The two stacked dimension lines on the source drawing share a datum at the slab soffit, so the overall depth is 200 + 1000 = 1200 mm. See the official exam paper.]

Approach. Locate the compression block, confirm it lies within the flange so the section behaves as a wide rectangular beam, take moments about the steel for M_r, then apply the A23.3 simplified sectional method for shear with b_w equal to the sum of the two webs.

Effective depth and material constants. With 40 mm clear cover to a 15M stirrup and 25M bars, $$d=1200-40-16-\tfrac{25.4}{2}=1131\ \text{mm}$$ $$\alpha_{1}=0.85-0.0015f_{c}'=0.7975,\qquad \beta_{1}=0.97-0.0025f_{c}'=0.8825$$
Depth of the compression block. Equating the steel tension to the concrete compression over the full 3000 mm flange width, $$T=\phi_{s}f_{y}A_{s}=0.85(400)(2000)=680\ \text{kN}$$ $$a=\frac{T}{\alpha_{1}\phi_{c}f_{c}'b}=\frac{680\times10^{3}}{0.7975(0.65)(35)(3000)}=12.5\ \text{mm}\ \lt\ 200\ \text{mm}$$ The block is only 12.5 mm deep, so it stays comfortably inside the flange and the section computes as a 3000 mm wide rectangular beam — the voids between the stems are irrelevant in flexure.
Moment resistance. Taking moments of the steel force about the centroid of the block, $$M_{r}=T\left(d-\frac{a}{2}\right)=680\times10^{3}\left(1131-6.2\right)=\boxed{765\ \text{kN}\cdot\text{m}}$$ The neutral axis sits at c = a/β₁ = 14.2 mm, giving c/d = 0.013 — the section is extremely tension-controlled, so the steel is fully yielded and failure would be ductile.
Confirm minimum flexural steel (A23.3 Cl. 10.5.1.2). For a T-beam with the flange in compression the width b_t is the web width, 2 × 200 = 400 mm: $$A_{s,\min}=\frac{0.2\sqrt{f_{c}'}}{f_{y}}b_{t}h=\frac{0.2\sqrt{35}}{400}(400)(1200)=1420\ \text{mm}^{2}\ \lt\ 2000\ \text{mm}^{2}$$ so the provided steel satisfies the minimum.
Shear — effective shear depth and web width. This is the step where the double-T differs from a solid section. In shear only the webs carry the diagonal field, so $$b_{w}=2(200)=400\ \text{mm},\qquad d_{v}=\max(0.9d,\,0.72h)=\max(1018,\,864)=1018\ \text{mm}$$
Concrete and steel contributions. The stirrups exceed the minimum area (A_v,min = 0.06√f_c′b_ws/f_y = 71 mm² against 800 mm² provided), so the simplified method of Cl. 11.3.6.3 applies with β = 0.18 and θ = 35°: $$V_{c}=\phi_{c}\lambda\beta\sqrt{f_{c}'}\,b_{w}d_{v}=0.65(1.0)(0.18)\sqrt{35}(400)(1018)=282\ \text{kN}$$ $$V_{s}=\frac{\phi_{s}A_{v}f_{y}d_{v}\cot\theta}{s}=\frac{0.85(800)(400)(1018)(1.428)}{200}=1978\ \text{kN}$$ Each stem contains one closed 15M stirrup, so four legs cross any shear plane — using two legs would halve V_s.
Apply the web-crushing cap. A23.3 Cl. 11.3.3 limits the total to $$V_{r,\max}=0.25\phi_{c}f_{c}'b_{w}d_{v}=0.25(0.65)(35)(400)(1018)=2316\ \text{kN}$$ Since V_c + V_s = 282 + 1978 = 2260 kN is just below that ceiling, $$\boxed{V_{r}=2260\ \text{kN}}$$ with only 2.4 % of reserve against crushing of the diagonal struts. Any closer stirrup spacing would buy nothing at all.

The contrast between the two answers is instructive: a section 1.2 m deep develops only 765 kN·m in flexure because it carries just 2000 mm² of tension steel, yet the same section is close to its absolute shear ceiling. This member is heavily under-reinforced in flexure and heavily reinforced in shear.

Quantity	Value
Effective depth, d	1131 mm
Compression block depth, a	12.5 mm (inside the flange)
c/d	0.013 — tension-controlled
Moment resistance, M_r	765 kN·m
b_w / d_v	400 mm / 1018 mm
V_c / V_s / crushing cap	282 / 1978 / 2316 kN
Shear resistance, V_r	2260 kN

Topic. Flexural and shear resistance of a multi-web (double-T) reinforced concrete section by the A23.3 simplified sectional method.

Key relations. a = φ_sf_yA_s/(α₁φ_cf_c′b); M_r = φ_sf_yA_s(d − a/2); d_v = max(0.9d, 0.72h); V_c = φ_cλβ√f_c′b_wd_v with β = 0.18; V_s = φ_sA_vf_yd_vcotθ/s with θ = 35°; V_r,max = 0.25φ_cf_c′b_wd_v.

Why this works. Flexure and shear see completely different parts of a voided section. Flexure only cares where the compression block lands: if it stays inside the flange, the concrete below the block is doing nothing anyway and the voids are invisible to the calculation. Shear, by contrast, is carried by a diagonal compression field running through the webs, so only the material that actually spans between flanges counts — hence b_w is the sum of the webs. The crushing cap exists because that diagonal field is itself a concrete strut with a finite capacity; once it is reached, adding stirrups changes nothing and the web must be thickened.

Common pitfall. Using the full 3000 mm flange width as b_w in the shear equations, which would overstate V_r by a factor of 7.5. The mirror-image mistake is using the web width in flexure, which understates M_r. A third trap is the source drawing itself: the 200 mm and 1 m dimensions are stacked about a shared datum at the slab soffit, so the overall depth is their sum.

Source. CSA A23.3:19 Cl. 10.1, 10.5.1.2, 11.3.3, 11.3.4, 11.3.6.3; MacGregor & Bartlett, Reinforced Concrete: Mechanics and Design, Chs. 4 and 6.

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