16-Civ-A2 Elementary Structural Design · May 2013

Question 5 of 7: B2 — Moment and Shear Resistances of a Triple-T Floor Section

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Question 5: 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 1 600 mm wide × 200 mm thick; overall depth 800 mm; three stems each 200 mm wide with 500 mm clear between them, so $3(200)+2(500) = 1\,600$ mm confirms the flange width. Bottom reinforcement 6–25M ($A_s = 3\,000$ mm²), top reinforcement 6–15M, 15M stirrups, "65 typical" to the bottom steel. $f_c' = 35$ MPa, $f_y = 400$ MPa.

Find. The factored moment resistance $M_R$ and shear resistance $V_R$ of the section.

Approach. Treat the section as a T-beam with the flange in compression: compute the depth of the stress block from the tension steel and confirm it falls inside the flange, so the section behaves as a rectangular beam of width 1 600 mm; then compute the shear resistance using the combined web width of the three stems.

1. Effective depth and tension force. Taking the "65 typical" dimension as the distance from the tension face to the centroid of the bottom bars, $d = 800-65 = 735$ mm. The 6–25M bars give $$T_r=\phi_sA_sf_y=0.85(3\,000)(400)=1\,020\ \text{kN}$$
2. Depth of the compression block. Equating $T_r$ to the compressive resultant over the full flange width, $$a=\frac{T_r}{\alpha_1\phi_cf_c'b_f}=\frac{1\,020\,000}{0.7975(0.65)(35)(1\,600)}=\frac{1\,020\,000}{29\,029}=35.1\ \text{mm}$$ Since $a = 35.1$ mm is much less than the 200 mm flange thickness, the compression zone lies wholly within the flange and the section is analysed as a rectangular beam 1 600 mm wide — the stems play no part in flexure.
3. Confirm ductile, under-reinforced behaviour. The neutral axis depth is $c = a/\beta_1 = 35.1/0.8825 = 39.8$ mm, so $$\varepsilon_s=0.0035\,\frac{d-c}{c}=0.0035\,\frac{735-39.8}{39.8}=0.061$$ which is thirty times the yield strain of 0.002. The steel is deeply yielded, so the assumed stress $f_y$ is valid and the failure would be extremely ductile. Minimum reinforcement, $A_{s,min} = 0.2\sqrt{35}(600)(800)/400 = 1\,420$ mm², is also satisfied by the 3 000 mm² provided.
4. Moment resistance. The lever arm between the steel and the flange compression resultant is $d - a/2$: $$M_R=T_r\left(d-\frac{a}{2}\right)=1\,020\left(0.735-\frac{0.0351}{2}\right)=1\,020(0.7174)$$ $$\boxed{M_R=732\ \text{kN}\cdot\text{m}}$$
5. Effective shear depth and web width. All three stems resist shear together, so $b_w = 3(200) = 600$ mm, and $$d_v=\max(0.9d,\,0.72h)=\max(0.9\times735,\;0.72\times800)=\max(661.5,\;576)=661.5\ \text{mm}$$
6. Concrete contribution. The stems carry 15M stirrups, so at least minimum transverse reinforcement is present and the simplified method gives $\beta = 0.18$, $\theta = 35^\circ$: $$V_c=\phi_c\lambda\beta\sqrt{f_c'}\,b_wd_v=0.65(1.0)(0.18)\sqrt{35}(600)(661.5)=275\ \text{kN}$$
7. Stirrup contribution. With one 15M double-leg stirrup in each of the three stems, $A_v = 3(2)(200) = 1\,200$ mm², and taking the spacing as 300 mm on centre, $$V_s=\frac{\phi_sA_vf_yd_v\cot\theta}{s}=\frac{0.85(1\,200)(400)(661.5)(1.428)}{300}=1\,285\ \text{kN}$$
8. Shear resistance and upper bound. Adding the two contributions, $$\boxed{V_R=V_c+V_s=275+1\,285=1\,560\ \text{kN}}$$ This must not exceed the crushing limit of the diagonal compression field, $V_{r,max} = 0.25\phi_cf_c'b_wd_v = 0.25(0.65)(35)(600)(661.5) = 2\,258$ kN, and it does not, so the calculated value stands. Note how lopsided the section is: it can resist 1 560 kN of shear but only 732 kN·m of moment, because the six 25M bars are the true bottleneck while the 600 mm of combined web is generous.