16-Civ-A2 Elementary Structural Design · December 2015

Question 4 of 7: B1 — Reinforced concrete box girder — $M_r$ and $V_r$

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

Notes on this paper

Paper format. National Examinations, December 2015 — 98-Civ-A2 Elementary Structural Design, 3-hour closed-book paper (textbooks and design handbooks 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, all of equal value. All seven are solved here, because the set is a study resource rather than an exam script.

Reference texts.

CSA S16, Design of Steel Structures — Clauses 11 (classification), 13.3 (compression), 13.5–13.6 (flexure), 13.8 (axial + bending), 13.11–13.12 (bolts, block shear).
CISC, Handbook of Steel Construction — section property tables for W-shapes and hollow structural sections.
CSA A23.3, Design of Concrete Structures — Clauses 10 (flexure and axial), 11 (shear), 10.15 (slender columns).
CSA O86, Engineering Design in Wood — Clause 6 (modification factors), Clause 7 (sawn lumber, bending and shear).
Canadian Wood Council, Wood Design Manual — specified strengths for Beam and Stringer grades.

Check — load factoring. Note 6 on page 1 states “all loads shown are unfactored”, so every load taken from a figure is a specified load and is factored here as 1.25D + 1.5L (NBCC Table 4.1.3.2, Case 2), treating the drawn point loads as live and member self-weight as dead. Where a question states a load directly in its text without a dead/live split (B3), that load is taken as already factored; this is stated again at that question.

Question 4: B1 — Reinforced concrete box girder — $M_r$ and $V_r$ (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.

Quantity	Value
Overall width / overall depth	2500 mm / 2400 mm
Wall thickness (constant)	200 mm
Top slab	200 mm thick, its top surface 1600 mm above the soffit
Web upstands above the top slab	800 mm
Bottom (tension) steel	7 – 30M
Top-slab steel / upstand steel	4 – 30M / 2 – 20M each side
Transverse steel, cover	15M closed ties at 200 mm, one per web; 70 mm typical
Materials	$f_c' = 35$ MPa, $f_y = 400$ MPa

Find. The factored moment resistance and factored shear resistance of the section.

[Figure not reproduced: Figure B1 (redrawn): the heavy 7–30M layer at the soffit identifies the bottom as the tension face, so the section is analysed in sagging with the upstands in compression. See the official exam paper.]

Approach. Establish the orientation from the reinforcement layout, then solve force equilibrium for the neutral-axis depth by strain compatibility — the compression zone lies inside the two 200 mm upstands, so its width is 400 mm, not 2500 mm — and take moments. For shear, use the A23.3 simplified method with the web widths summed.

Fix the orientation and the bar depths. Seven 30M bars sit at the soffit and only four at the top slab, so the soffit is the tension face and the section sags. With 70 mm cover, 15M ties (16 mm) and 30M bars (29.9 mm), $$d_{7\text{-}30M}=2400-\left(70+16+\tfrac{29.9}{2}\right)=2299\ \text{mm}$$ $$d_{4\text{-}30M}=2400-\left(1400+70+16+\tfrac{29.9}{2}\right)=899\ \text{mm},\qquad d'_{2\text{-}20M}=95.8\ \text{mm}$$ The 4–30M layer lies at the underside of the top slab and the 2–20M pairs at the head of each upstand.
Recognise the compression zone. Measured down from the top fibre, the section is only $2\times200 = 400$ mm wide for the first 800 mm, because the compression begins in the two upstands. Only below that does the full 2500 mm slab appear. The stress block therefore bears on a narrow 400 mm strip, and the neutral axis must be found before assuming any width.
Concrete parameters. With $f_c' = 35$ MPa, $$\alpha_1=0.85-0.0015f_c'=0.7975,\qquad \beta_1=0.97-0.0025f_c'=0.8825$$
Force equilibrium by strain compatibility. Setting the sum of the concrete compression, the compression steel in the upstands and the tension steel to zero, $$\alpha_1\phi_cf_c'(400)a+\phi_sA_s'f_s'-\phi_sA_{s}f_y=0$$ Solving iteratively (with $\varepsilon_{cu} = 0.0035$) gives $$c=348.5\ \text{mm},\qquad a=\beta_1c=307.5\ \text{mm}$$ and the strain check confirms every layer has yielded: the 2–20M pairs at $+0.00254$ in compression, the 4–30M at $-0.0055$ and the 7–30M at $-0.0196$, all beyond $\varepsilon_y = 0.002$. Since $c/d = 0.152$, the section is comfortably under-reinforced and will fail in a ductile manner.
Moment resistance. Taking moments of all internal forces about the top fibre and assembling the couple, $$M_r=\boxed{4306\ \text{kN}\cdot\text{m}}$$ Note that the 4–30M layer sits 899 mm below the top fibre, well below the neutral axis at 348.5 mm, so it acts in tension and adds roughly 670 kN·m; omitting it understates $M_r$ by about 15 %.
Shear — effective web width and shear depth. Only the webs resist shear, so $$b_w=2(200)=400\ \text{mm},\qquad d_v=\max\left(0.9d,\;0.72h\right)=\max(2069,\;1728)=2069\ \text{mm}$$
Concrete and steel contributions. One closed 15M tie per web puts four legs across a shear plane, so $A_v = 4(200) = 800\ \text{mm}^{2}$ at $s = 200$ mm — far above the minimum $A_{v,\min} = 71\ \text{mm}^{2}$. With minimum transverse reinforcement present, the A23.3 simplified method takes $\beta = 0.18$ and $\theta = 35^\circ$: $$V_c=\phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v=0.65(1.0)(0.18)\sqrt{35}(400)(2069)=573\ \text{kN}$$ $$V_s=\frac{\phi_sA_vf_yd_v\cot\theta}{s}=\frac{0.85(800)(400)(2069)(1.428)}{200}=4019\ \text{kN}$$
Factored shear resistance. Summing and checking the web-crushing cap, $$V_r=V_c+V_s=\boxed{4592\ \text{kN}}\qquad\text{against}\qquad V_{r,\max}=0.25\phi_cf_c'b_wd_v=4707\ \text{kN}$$ The section is just inside the crushing limit, with 98 % of the cap taken up — adding any more tie steel would buy almost nothing.

Result	Value
Effective depths $d$ / $d'$	2299 mm / 95.8 mm
Neutral axis $c$ / stress block $a$	348.5 mm / 307.5 mm
Ductility check $c/d$	0.152 (under-reinforced)
Moment resistance $M_r$	4306 kN·m
$b_w$ / $d_v$	400 mm / 2069 mm
$V_c$ / $V_s$	573 kN / 4019 kN
Shear resistance $V_r$	4592 kN (cap 4707 kN)

Topic. Flexural and shear strength of a reinforced concrete box section with a narrow compression zone.

Key relations. $\alpha_1 = 0.85-0.0015f_c'$, $\beta_1 = 0.97-0.0025f_c'$; equilibrium $C = T$ with $\varepsilon_{cu} = 0.0035$; $d_v = \max(0.9d,\,0.72h)$; $V_c=\phi_c\lambda\beta\sqrt{f_c'}b_wd_v$, $V_s=\phi_sA_vf_yd_v\cot\theta/s$, capped by $0.25\phi_cf_c'b_wd_v$.

Why this works. In flexure the concrete tension zone is ignored and the whole tensile force is carried by steel, so the analysis reduces to finding the one neutral-axis depth at which compression balances tension. What makes a box section different from a rectangular beam is that the width of the compression zone is a function of depth, so $C$ must be integrated over the varying width rather than written as $\alpha_1\phi_cf_c'ba$. In shear the reverse simplification applies: the flanges contribute essentially nothing to shear transfer, so only the sum of the web widths enters $b_w$. The very large $V_s$ here is a direct consequence of the enormous $d_v$, which is why the crushing cap rather than the tie steel becomes the real limit.

Common pitfall. Using the full 2500 mm width in the compression block. Here the stress block is only 307 mm deep and lies entirely within the 800 mm upstands, so the correct width is 400 mm; using 2500 mm would place the neutral axis far too high and overstate the lever arm. The mirror-image error in shear is using $b_w = 2500$ mm instead of the summed web widths, which overstates $V_c$ more than six-fold.

Source. CSA A23.3, Design of Concrete Structures, Clauses 10.1 (flexural assumptions), 11.3.4–11.3.6 (shear, simplified method) and 11.3.3 (maximum shear).

← Question 3 Next: Question 5 →