Question 6 of 7: B3 — Moment and shear resistance of a triple-stem section
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2016 — 07-Str-A2 Elementary Structural Design, 3 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). The candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in all, of equal value. All seven are solved here, because the set is a study resource. Page 1 states that all loads shown are unfactored, so the load combination is applied by the candidate: the point and distributed loads drawn on the figures are treated as live (1.5) and self-weight as dead (1.25), per NBCC 2020 Table 4.1.3.2 case 2 (1.25D + 1.5L).
Reference texts.
CSA S16:19, Design of Steel Structures (Clauses 13.3, 13.8, 13.13) with the CISC Handbook of Steel Construction, 11th ed., Part 6 section tables.
CSA A23.3:19, Design of Concrete Structures (Clauses 10 and 11), with Brzev & Pao, Reinforced Concrete Design: A Practical Approach, 3rd ed.
CSA O86:19, Engineering Design in Wood, with the Canadian Wood Council Wood Design Manual, 2020.
Kassimali, Structural Analysis, 6th ed. (SI), Chapters 3, 5 and 12 for the determinate/indeterminate analysis feeding each design.
Check — figure page reconstruction. Page 3 of 3 of this paper carries all six figures as a single hand-drawn composite whose OCR text is unusable. One genuine inconsistency was found and is flagged in Question 5: Figure B2 carries two conflicting height dimensions (7 m against the column, 9 m at the right-hand margin) for the same distance between A and the beam.
Question 6: B3 — Moment and shear resistance of a triple-stem section (12 + 8 marks)
Given. A 2000 mm wide by 200 mm thick top slab on three 200 mm stems, overall depth 700 mm. Reading the dimension string across the top: 200 + 200 + 500 + 200 + 500 + 200 + 200 = 2000 mm, so the stems are 200 mm wide with 500 mm clear between them and 200 mm slab overhangs. Reinforcement: 2–30M in the bottom of each stem (6–30M = 4200 mm²), 6–25M near the top (3000 mm²), 20M @ 200 transverse slab bars, 10M ties typical. fc' = 35 MPa, fy = 400 MPa.
Find. The factored moment of resistance Mr and shear resistance Vr.
Figure B3 — triple-stem section. The 6–25M sit within the slab depth at the head of each stem.
Check — tie spacing assumed. The figure labels the stirrups “10 typical” but gives no spacing. A spacing of 200 mm is adopted here, matching the 20M @ 200 slab bars shown on the same figure and satisfying the maximum-spacing rules; Vr scales inversely with that spacing, so the steel contribution should be re-checked against the real detail before use.
Approach. Positive bending puts the wide slab in compression, so first test whether the stress block stays inside the 200 mm flange; then locate the neutral axis by axial equilibrium, check which side of it the 25M bars lie on, and take moments. For shear the web width is the sum of the three stems.
Effective depths. With 40 mm cover and 10M ties,
$$\begin{aligned} &d=700-40-11.3-\tfrac{29.9}{2}=633.8\ \text{mm} \\ &d'=40+11.3+\tfrac{25.2}{2}=63.9\ \text{mm} \end{aligned}$$
Trial: neutral axis inside the flange. Ignoring the top bars for a first estimate, \(T=\phi_sA_sf_y=0.85(4200)(400)=1428\) kN and
$$a=\frac{T}{\alpha_1\phi_cf_c'b_f}=\frac{1428\times10^{3}}{0.7975(0.65)(35)(2000)}=39.4\ \text{mm}\ \ll\ 200\ \text{mm}$$
The stress block lies well within the slab, so the section behaves as a 2000 mm wide rectangle and the stems play no part in flexure at all.
Locate the neutral axis exactly. Solving \(\sum F=0\) including the 25M bars gives
$$\begin{aligned} &c=\boxed{54.4\ \text{mm}} \\ &a=\beta_1c=0.8825(54.4)=48.0\ \text{mm} \end{aligned}$$
Check the 25M bars. They sit at d' = 63.9 mm, which is below the neutral axis at 54.4 mm, so they are in tension, not compression:
$$\varepsilon_s=0.0035\frac{63.9-54.4}{54.4}=614\times10^{-6}\ \Rightarrow\ f_s=614\times10^{-6}(200\,000)=123\ \text{MPa}$$
Moment of resistance. Taking moments about the compression resultant, with the 30M bars at yield and the 25M bars at 123 MPa,
$$M_r=\boxed{883\ \text{kN}\cdot\text{m}}$$
The 30M bars alone would give 877 kN·m, so the top bars add only 0.7 % — small, but of the correct sign; treating them as compression steel would push the answer the wrong way.
Shear: the web width is the sum of the stems.
$$\begin{aligned} &b_w=3(200)=600\ \text{mm} \\ &d_v=\max(0.9d,\,0.72h)=\max(570.4,\,504)=570.4\ \text{mm} \end{aligned}$$
Each stem carries one closed 10M tie, so six legs (\(A_v=600\) mm²) cross any shear plane. With \(\beta=0.18\) and \(\theta=35^\circ\):
$$V_c=0.65(0.18)\sqrt{35}(600)(570.4)=236.9\ \text{kN}$$
$$V_s=\frac{0.85(600)(400)(570.4)\cot35^\circ}{200}=830.9\ \text{kN}$$
$$V_r=236.9+830.9=\boxed{1068\ \text{kN}}$$
Crushing check. Clause 11.3.3 limits the total to
$$V_{r,max}=0.25\phi_cf_c'b_wd_v=0.25(0.65)(35)(600)(570.4)=1946\ \text{kN}$$
so the 1068 kN is a genuine resistance and not a diagonal-compression failure.
The result is characteristic of a multi-stem deck: enormous flexural lever arm from the 700 mm depth but very little concrete actually in compression (48 mm of a 2000 mm wide slab), and a shear capacity dominated by the steel term because three narrow webs give a small \(V_c\) but three ties give a large \(A_v\).
