16-Civ-A2 Elementary Structural Design · Undated paper

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

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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. A cast-in-place interior Tee formed by a 100 mm slab and a 300 mm wide stem, reinforced with 2–25M in the bottom of the stem, 2–10M in the flange and 10M closed stirrups at 300 mm.

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

Approach. Establish the effective flange width from A23.3 Cl. 10.3.3, solve force equilibrium by strain compatibility to locate the neutral axis, take moments to get $M_r$, then apply the general shear method of Cl. 11.3 with the minimum-stirrup value of $\beta$.

1. Establish the effective flange width. For a symmetrical interior Tee, A23.3 Cl. 10.3.3 limits the overhang each side to the least of one tenth of the span, half the clear distance to the next web, and twelve slab thicknesses: $$b_{oh}=\min\left(\frac{5000}{10},\ \frac{2000-300}{2},\ 12(100)\right)=\min(500,\ 850,\ 1200)=500\ \text{mm}$$ $$b_{eff}=300+2(500)=\boxed{1300\ \text{mm}}$$ The span controls, as it usually does on short-span floor framing.
2. Set up the section geometry. With 40 mm cover, 10M stirrups and 25M bars, $$d=500-40-11.3-\frac{25.2}{2}=436\ \text{mm},\qquad d'=40+11.3+\frac{11.3}{2}=57\ \text{mm}$$ The stress-block parameters at 30 MPa are $\alpha_1=0.85-0.0015(30)=0.805$ and $\beta_1=0.97-0.0025(30)=0.895$.
3. Locate the neutral axis by force equilibrium. Assume the block lies within the flange, so the compression width is $b_{eff}$. Ignoring the flange bars for a first pass, $$T=\phi_sA_sf_y=0.85(1000)(400)=340\ \text{kN}$$ $$a=\frac{T}{\alpha_1\phi_cf_c'b_{eff}}=\frac{340\times10^{3}}{0.805(0.65)(30)(1300)}=16.7\ \text{mm}\ \ll\ h_f=100\ \text{mm}$$ The assumption holds — the voids of the Tee are irrelevant and the section behaves as a 1300 mm wide rectangle.
4. Include the flange bars, which are in tension. The neutral axis at $c=a/\beta_1=18.6$ mm sits well above the 2–10M at $d'=57$ mm, so those bars are on the tension side and add to $T$ rather than to $C$. Re-solving equilibrium with both layers gives $$c=22.4\ \text{mm},\qquad a=\beta_1c=20.0\ \text{mm}$$ with the 10M bars at a strain of $0.0035(57-22.4)/22.4=0.0054$, comfortably yielded in tension.
5. Take moments for the resistance. Summing about the centroid of the compression block, $$M_r=\phi_sA_sf_y\left(d-\frac{a}{2}\right)+\phi_sA_s'f_y\left(d'-\frac{a}{2}\right)$$ $$M_r=340\times10^{3}(436-10.0)+68\times10^{3}(57-10.0)=\boxed{148.1\ \text{kN}\cdot\text{m}}$$ Ignoring the flange bars would give 145.4 kN·m, so they are worth 1.8 % — small, but the sign of their contribution is the instructive part.
6. Confirm minimum steel and ductility. For a flanged section in sagging, A23.3 Cl. 10.5.1.2 uses $b_t=b_w$: $$A_{s,\min}=\frac{0.2\sqrt{f_c'}}{f_y}b_wh=\frac{0.2\sqrt{30}}{400}(300)(500)=411\ \text{mm}^2\ <\ 1000\ \text{mm}^2\quad\checkmark$$ With $c/d=22.4/436=0.051$, the section is emphatically tension-controlled: it will yield and deflect visibly long before the concrete crushes.
7. Compute the concrete contribution to shear. The effective shear depth is $$d_v=\max(0.9d,\ 0.72h)=\max(392.5,\ 360)=392.5\ \text{mm}$$ Minimum transverse steel is satisfied, $A_{v,\min}=0.06\sqrt{f_c'}b_ws/f_y=73.9$ mm² against 200 mm² provided, so the simplified method permits $\beta=0.18$ and $\theta=35^{\circ}$: $$V_c=\phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v=0.65(1.0)(0.18)\sqrt{30}(300)(392.5)=75.5\ \text{kN}$$
8. Compute the stirrup contribution and total. $$V_s=\frac{\phi_sA_vf_yd_v\cot\theta}{s}=\frac{0.85(200)(400)(392.5)\cot 35^{\circ}}{300}=127.1\ \text{kN}$$ $$V_r=V_c+V_s=75.5+127.1=\boxed{202.6\ \text{kN}}$$
9. Check the web-crushing cap. $$V_{r,\max}=0.25\phi_cf_c'b_wd_v=0.25(0.65)(30)(300)(392.5)=574\ \text{kN}\ >\ 202.6\ \text{kN}\quad\checkmark$$ Diagonal crushing does not govern, so the stirrups are fully effective.

The section resists 148 kN·m in bending and 203 kN in shear. For context, a 5 m simply supported span developing 148 kN·m corresponds to a uniform factored load of about 47 kN/m, at which the end shear would be 118 kN — comfortably inside the shear capacity. The Tee is therefore flexure-governed, which is the normal and desirable balance for a one-way floor system.