07-Str-A2 · May 2016

Question 4 of 7: B1 — Moments of resistance of an L-shaped concrete girder

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

Notes on this paper

Paper format. National Examinations — May 2016, 07-Str-A2 Elementary Structural Design. Three-hour closed-book examination (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, structural steel to CSA S16), Part B (B1–B3, reinforced concrete to CSA A23.3), Part C (C1, timber to CSA O86). A candidate answers two from Part A, two from Part B and the one question in Part C — five solutions in total, all of equal value. All seven are solved here, because the set is a study resource rather than an examination script. Page 1 states that all loads shown are unfactored, so the load factors are applied within each solution.

Reference texts.

CSA S16:19, Design of Steel Structures — Clauses 11 (class of section), 13.3 (compressive resistance), 13.5–13.6 (bending), 13.8 (axial compression and bending), 13.13 (welds).
CISC, Handbook of Steel Construction, 12th ed. — section property tables, weld tables.
CSA A23.3:19, Design of Concrete Structures — Clauses 10 (flexure and axial load), 11 (shear), 7.6 (ties).
CSA O86:19, Engineering Design in Wood, with the Canadian Wood Council Wood Design Manual 2020 — Clause 7 (glued-laminated timber).
NBCC 2020 — Table 4.1.3.2 load combinations.

Check — assumptions common to the whole paper. (i) Every load drawn in Figures A1, A2, A3, B3 and quoted in C1 is a specified (unfactored) load; unless a question names the load type, the point loads are treated as live (factor 1.5) and concrete/steel self-weight as dead (factor 1.25), per NBCC Table 4.1.3.2 case 2. (ii) Steel is CSA G40.21 350W, so F_y = 350 MPa, E = 200 000 MPa. (iii) Question B3 states f'_c = 35 MPa and f_y = 400 MPa; because B1 and B2 quote no material strengths, the same pair is adopted for them and the assumption is stated in each solution. (iv) Section properties are recomputed from the nominal plate dimensions of each rolled shape rather than read off a table, so every number below is reproducible; they agree with the CISC Handbook to better than 1 %.

Question 4: B1 — Moments of resistance of an L-shaped concrete girder (12 + 8)

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 dimensions	1500 mm wide × 1500 mm deep
Top flange	1500 mm wide × 500 mm deep
Stem (right-hand side)	750 mm wide × 1000 mm deep
Flange reinforcement	8–25M: 5 at the flange top, 3 at the flange soffit
Stem reinforcement	2–20M at mid-depth of the stem, 4–30M at the bottom
Ties	20M
Bar centroids from the concrete face	65 mm (the “65 typical” cover convention of this paper series)
Materials (assumed — see callout)	f'_c = 35 MPa, f_y = 400 MPa

Find. The factored moment of resistance about each centroidal axis. Because the section is doubly asymmetric, four values are required: sagging and hogging about the horizontal axis a-a, and compression-on-the-left and compression-on-the-right about the vertical axis b-b.

[Figure not reproduced: Figure B1 (redrawn). The centroid of the gross section lies 625 mm below the top face and 937.5 mm from the left face; a-a and b-b are the horizontal and vertical axes through that point. See the official exam paper.]

Check — material strengths. Question B1 quotes no concrete or steel grade. The bracketed note in B3 on the same paper gives f'_c = 35 MPa and f_y = 400 MPa, and those values are adopted here so the three Part B answers are mutually consistent. If an examiner intended different grades, every moment below scales roughly with f_y and only weakly with f'_c, because all four cases are strongly tension-controlled.

Approach. Locate the gross centroid, then for each of the four bending senses find the neutral-axis depth that balances the rectangular stress block against the bar forces from strain compatibility, and sum moments about the centroidal axis.

Locate the centroid of the gross section. The flange and the stem happen to have equal areas of 750×10³ mm² each, so the centroid is simply the mid-point of their two centroids: $$\begin{aligned} \bar{y} &= \frac{750(250)+750(1000)}{1500} = \boxed{625\ \text{mm from the top}} \\ \bar{x} &= \frac{750(750)+750(1125)}{1500} = \boxed{937.5\ \text{mm from the left}} \end{aligned}$$ Total gross area A_g = 1.50×10⁶ mm².
Set up the stress-block constants. For f'_c = 35 MPa, A23.3 Clause 10.1.7 gives $$\begin{aligned} \alpha_{1} &= 0.85 - 0.0015f'_{c} = 0.7975 \\ \beta_{1} &= 0.97 - 0.0025f'_{c} = 0.8825 \end{aligned}$$ with φ_c = 0.65 and φ_s = 0.85. Every bar force follows from strain compatibility on a linear strain profile with ε_cu = 0.0035, capped at ±f_y.
Bending about a-a, sagging (top in compression). The tension zone contains the 3–25M at the flange soffit, the 2–20M in the stem and the 4–30M at the bottom — 4900 mm² in all — while the 5–25M top bars sit almost exactly on the neutral axis and contribute nothing. Balancing forces over the 1500 mm-wide flange gives a = 59.4 mm and c = 67.3 mm, well inside the flange, so the section behaves as a wide slab in compression. Summing moments about a-a: $$M_{r,a\text{-}a}^{+} = \boxed{1741\ \text{kN}\cdot\text{m}}$$
Bending about a-a, hogging (bottom in compression). Now the compression block is confined to the 750 mm-wide stem, so it must run deeper (a = 81.9 mm, c = 92.9 mm), but the tension steel — the eight 25M bars in the flange — sits much further from the centroid. The larger lever arm more than compensates for the narrower block: $$M_{r,a\text{-}a}^{-} = \boxed{1790\ \text{kN}\cdot\text{m}}$$ That the hogging capacity slightly exceeds the sagging capacity is a genuine feature of this section, not an arithmetic slip: the flange steel is both plentiful and remote from the centroid.
Bending about b-b, compression on the left. Rotating the problem through 90°, the “width” available to the compression block on the left-hand side is only the 500 mm flange depth, so the block runs a long way in (c = 232 mm), while every bar in the stem acts in tension at lever arms up to 500 mm: $$M_{r,b\text{-}b}^{\text{left}} = \boxed{2088\ \text{kN}\cdot\text{m}}$$
Bending about b-b, compression on the right. Reversing the sense puts the full 1500 mm depth of the stem into compression, so the block is shallow (c = 71.8 mm), but the tension steel is now only the flange bars at the far left, at smaller lever arms: $$M_{r,b\text{-}b}^{\text{right}} = \boxed{1347\ \text{kN}\cdot\text{m}}$$ This is the smallest of the four values and therefore governs any design that could see horizontal bending in that direction.
Check the minimum flexural reinforcement. A23.3 Clause 10.5.1.2 requires, on the tension face of the stem, $$A_{s,\min} = \frac{0.2\sqrt{f'_{c}}}{f_{y}}b_{t}h = \frac{0.2\sqrt{35}}{400}(750)(1500) = 3330\ \text{mm}^{2}$$ The 4–30M alone supply 2800 mm², but together with the 2–20M stem bars and the 3–25M at the flange soffit the tension zone carries 4900 mm² — comfortably above the minimum, so all four results are valid moments of resistance rather than cracking-controlled values.

Every one of the four cases is strongly tension-controlled: the deepest neutral axis found is 232 mm on a 1500 mm section, so the extreme tension bars reach strains an order of magnitude beyond yield and all four resistances are governed by the steel, not the concrete. That is why the answers are relatively insensitive to the assumed f'_c.

Result	Value
Gross area A_g	1.50×10⁶ mm²
Centroid from the top face / from the left face	625 mm / 937.5 mm
Stress-block constants α₁, β₁	0.7975, 0.8825
M_r about a-a, sagging (c = 67.3 mm)	1741 kN·m
M_r about a-a, hogging (c = 92.9 mm)	1790 kN·m
M_r about b-b, compression on the left (c = 232 mm)	2088 kN·m
M_r about b-b, compression on the right (c = 71.8 mm)	1347 kN·m (governs)
A_s,min required / provided in the stem tension zone	3330 mm² / 4900 mm²

Topic. Flexural strength of an asymmetric reinforced concrete section about both principal centroidal axes by strain compatibility.

Key relations. A23.3 Clause 10.1.7 rectangular stress block, C = α₁φ_cf'_cA_block with a = β₁c; linear strain profile ε_s = 0.0035(c − d_i)/c; bar force F_i = φ_sf_siA_bi with |f_si| ≤ f_y; force equilibrium ΣC = ΣT for pure flexure; Clause 10.5.1.2 minimum flexural steel.

Why this works. A moment of resistance is not a property of a section alone but of a section and a direction of bending, and this girder has neither a horizontal nor a vertical axis of symmetry. Four answers therefore exist, and none can be inferred from another. The general method — guess the neutral-axis depth, compute the concrete block area on whichever side is in compression, add the bar forces from the strain profile, and iterate until the net axial force vanishes — handles all four cases with the same code, which is why it is worth setting up once rather than deriving four closed-form expressions. Because the concrete block area changes discontinuously where the flange ends, closed-form algebra would in any case need a separate branch for each case.

Common pitfall. Computing only the sagging capacity and calling it “the” moment of resistance, which answers about a quarter of the question. A subtler error is placing the block by depth alone without tracking the varying width: for a-a hogging the compression zone is 750 mm wide, not 1500, and using the flange width overstates the capacity. A third is forgetting the compression-side bars — here they matter little because the neutral axis is so shallow, but on a more heavily loaded section they carry a real share.

Source. CSA A23.3:19 Clauses 10.1 (assumptions and stress block), 10.5.1 (minimum reinforcement) and 10.5.2; MacGregor and Bartlett, Reinforced Concrete: Mechanics and Design, Canadian edition, Chapter 4.

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