Question 5 of 7: Moment and shear resistances of a triple-T floor section

Question B2: Moment and shear resistances of a triple-T floor section (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. Flange 1600 mm wide x 200 mm thick; three stems 200 mm wide at 700 mm centres, total depth $h = 810$ mm; tension steel 6-25M ($6 \times 500 = 3\,000$ mm²) in the stem bottoms with 65 mm to bar centres, so $d = 810-65 = 745$ mm; 6-15M in the flange; 15M closed stirrups. $f'_c = 35$ MPa, $f_y = 400$ MPa, $\phi_c = 0.65$, $\phi_s = 0.85$, $\alpha_1 = 0.7975$, $\beta_1 = 0.8825$.

Find. Part (i): the factored moment resistance $M_R$ for positive (sagging) bending. Part (ii): the factored shear resistance $V_R$.

Approach. Assume the steel yields, find the depth of the stress block from horizontal equilibrium, confirm it lies inside the flange so the section behaves as a rectangle 1600 mm wide, and take moments about the compressive resultant. For shear, add the three stem widths into one web and apply the simplified method.

1. Tension force at yield. Taking the flange steel as neutral (it lies in the compression zone and is conservatively ignored), $$T=\phi_s f_y A_s = 0.85(400)(3\,000)=1\,020\ \text{kN}$$
2. Depth of the compression block. Equating $T$ to the compressive force in a flange of full width $b = 1\,600$ mm, $$a=\frac{T}{\alpha_1\phi_c f'_c b}=\frac{1\,020\times10^{3}}{0.7975(0.65)(35)(1\,600)} =35.1\ \text{mm}$$ Since $a = 35.1$ mm is well inside the 200 mm flange, none of the stems is in compression and the section may be analysed as a rectangular beam 1600 mm wide.
3. Confirm ductility. The neutral axis lies at $c = a/\beta_1 = 39.8$ mm, so the strain in the outermost bars at crushing of the concrete is $$\varepsilon_s = 0.0035\,\frac{d-c}{c}=0.0035\left(\frac{745-39.8}{39.8}\right)=0.062$$ about thirty times the yield strain of 0.0020, which confirms the assumption that the steel yields and shows the section is very heavily under-reinforced.
4. Part (i) — moment resistance. Taking moments about the centroid of the compressive block, $$M_R = T\left(d-\frac{a}{2}\right)=1\,020\times10^{3}\left(745-\frac{35.1}{2}\right) =\boxed{742\ \text{kN}\cdot\text{m}}$$ The minimum-steel requirement is satisfied comfortably: $A_{s,min}=0.2\sqrt{f'_c}\,b_t h/f_y$ with $b_t = 600$ mm gives $1\,438$ mm² against the 3 000 mm² provided.
5. Part (ii) — the concrete contribution to shear. The three stems act as a single web of $b_w = 3(200) = 600$ mm, and $d_v = \max(0.9d,\,0.72h) = \max(670.5,\,583.2) = 670.5$ mm. With minimum stirrups present the simplified method gives $\beta = 0.18$ and $\theta = 35^{\circ}$: $$V_c=\phi_c\lambda\beta\sqrt{f'_c}\,b_w d_v = 0.65(1.0)(0.18)\sqrt{35}(600)(670.5)=278\ \text{kN}$$
6. Steel contribution and the total. Two 15M legs in each of the three stems give $A_v = 6(200) = 1\,200$ mm² per set, so $$V_s=\frac{\phi_s A_v f_y d_v\cot\theta}{s}=\frac{0.85(1\,200)(400)(670.5)(1.428)}{s} =\frac{3.907\times10^{8}}{s}\ \text{N}$$ At a practical detailing spacing of 300 mm this is 1 302 kN, so $$V_R = V_c+V_s = 278+1\,302=\boxed{1\,580\ \text{kN}}$$ which is below the web-crushing ceiling $V_{r,max}=0.25\phi_c f'_c b_w d_v = 2\,288$ kN, so the result is valid.
7. Report the sensitivity to spacing. Because the figure dimensions the stirrup size but not its spacing, the shear resistance is quoted as a function of $s$: 1 953 kN of steel contribution at 200 mm, 1 302 kN at 300 mm and 833 kN at the maximum permitted $s=\min(0.7d_v,600)=469$ mm, giving $V_R$ of 2 231, 1 580 and 1 111 kN respectively (the first is capped by $V_{r,max}=2\,288$ kN). In every case the section is many times stronger in shear than in flexure, which is normal for a stemmed floor unit.