16-Civ-A2 Elementary Structural Design · December 2019

Question 4 of 7: B1 — Moment and shear resistance of a reinforced concrete tee

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

Question 4: B1 — Moment and shear resistance of a reinforced concrete tee (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 the tee of Figure B1 — a 100 mm flange over a 400 mm stem 250 mm wide, five 25M bars in two layers, 2-20M in the flange and 10M stirrups at 200 mm — in 30 MPa concrete at 1250 mm centres.

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

Approach. Locate the bar centroids, assume the rectangular stress block lies inside the flange and verify it, solve horizontal equilibrium by strain compatibility (the 2-20M turn out to be below the neutral axis and therefore in tension), then apply the A23.3 simplified shear method.

1. Effective flange width. The question gives no span, so A23.3 Clause 10.3 reduces to the physical limit: the tee cannot claim more slab than the beam spacing, $$b_{\text{eff}} = 1250\ \text{mm}$$
2. Bar positions, measured from the top fibre. With 40 mm clear cover, a 10M stirrup (11.3 mm) and 25M bars (25.2 mm), the bottom layer sits at $500 - (40 + 11.3 + 12.6) = 436.1$ mm and the second layer 50.2 mm above it at 385.9 mm. The effective depth is the centroid of the five bars: $$d = \frac{3(500)(436.1) + 2(500)(385.9)}{2500} = \boxed{416.0\ \text{mm}}$$ The 2-20M in the flange sit at $40 + 11.3 + 9.75 = 61.0$ mm from the top.
3. Trial: is the stress block inside the flange? With $\alpha_1 = 0.85 - 0.0015(30) = 0.805$ and $\beta_1 = 0.97 - 0.0025(30) = 0.895$, the compression per unit depth over the full flange is $\alpha_1\phi_c f′_c b_{\text{eff}} = 0.805(0.65)(30)(1250) = 19\,622$ N/mm. The five 25M bars deliver $T = \phi_s A_s f_y = 0.85(2500)(400) = 850$ kN, which needs only $a = 850\,000/19\,622 = 43$ mm. That is well inside the 100 mm flange, so the tee behaves as a 1250 mm wide rectangular section.
4. Solve equilibrium by strain compatibility. Bisecting on the neutral-axis depth until the net axial force vanishes gives $$c = 52.0\ \text{mm},\qquad a = \beta_1 c = 46.5\ \text{mm}$$ confirming the rectangular assumption.
5. The 2-20M bars are in tension, not compression. At $c = 52.0$ mm and a bar level of 61.0 mm, the strain is $$\varepsilon = 0.0035\,\frac{c - y}{c} = 0.0035\,\frac{52.0 - 61.0}{52.0} = -0.00061$$ i.e. tensile. Those bars sit below the neutral axis even though they are near the top of the member, because the wide flange pulls the axis up so hard. They add a small tensile force on a short lever arm, worth about 0.3 % on $M_r$ — correct to include, but not decisive here.
6. Moment resistance. Taking moments of the concrete block and the steel forces about the tensile resultant, $$\boxed{M_r = 336\ \text{kN}\cdot\text{m}}$$ A quick hand check confirms the magnitude: $T(d - a/2) = 850(416.0 - 23.3)/10^{3} = 334$ kN·m.
7. Minimum steel. A23.3 Clause 10.5.1.2 requires $$A_{s,\min} = \frac{0.2\sqrt{f′_c}}{f_y}\,b_t h = \frac{0.2\sqrt{30}}{400}(250)(500) = 342\ \text{mm}^{2}$$ against 2500 mm² provided. Comfortably satisfied, and the section is strongly tension-controlled ($c/d = 0.125$), so it will warn before it fails.
8. Effective shear depth. $$d_v = \max(0.9d,\ 0.72h) = \max(374.4,\ 360) = 374.4\ \text{mm}$$
9. Concrete contribution. The stirrups exceed the minimum, so the A23.3 Clause 11.3.6.3 simplified method applies with $\beta = 0.18$ and $\theta = 35^\circ$: $$V_c = \phi_c \lambda \beta \sqrt{f′_c}\, b_w d_v = 0.65(1.0)(0.18)\sqrt{30}\,(250)(374.4)/10^{3} = 60.0\ \text{kN}$$
10. Stirrup contribution and total. $$V_s = \frac{\phi_s A_v f_y d_v \cot\theta}{s} = \frac{0.85(200)(400)(374.4)(1.428)}{200}\cdot\frac{1}{10^{3}} = 181.8\ \text{kN}$$ $$\boxed{V_r = V_c + V_s = 60.0 + 181.8 = 242\ \text{kN}}$$ The web-crushing cap is not approached: $0.25\phi_c f′_c b_w d_v = 456$ kN. Minimum stirrup area is also met, $A_{v,\min} = 0.06\sqrt{30}(250)(200)/400 = 41$ mm² against 200 mm² supplied.