Question 6 of 7: B3 — Moment and shear resistances of a triple-T section
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, December 2017 — 07-Str-A2 Elementary Structural Design. Three hours, closed book (handbooks and textbooks permitted). Seven questions in three parts: Part A (A1–A3, steel), Part B (B1–B3, reinforced concrete) and Part C (C1, timber). The candidate answers two from Part A, two from Part B and the single Part C question — five solutions of equal value. All seven are solved here, because the set is a study resource. Page 1 states that all loads shown are unfactored, so every question below applies its own load combination.
Check: figure page. All five figures are hand-drawn on page 3 of the paper. Every dimension used below was read from the printed figures, cross-checked against the labelled dimension chains (for example 200 overhang + 3 × 200 stems + 2 × 600 voids + 200 overhang = 2200 in Figure B3). Where the exam leaves a quantity undimensioned (the tie length in Figure A3, member thicknesses in Figure B1) the assumption is stated in the question concerned, as NOTE 1 on page 1 invites.
Question 6: B3 — Moment and shear resistances of a triple-T section (12 + 8 = 20 marks)
Find. The factored moment resistance $M_r$ and shear resistance $V_r$ of the section as detailed.
Figure B3 - triple-T section: 2200 x 150 flange over three 200 mm stems (200 mm overhangs, 600 mm clear voids), 800 mm overall; 8-20M in the flange, 6-30M in the stems, 20M ties at 200 mm.
Approach. For sagging moment the wide flange is in compression, so find the neutral axis by equilibrium (checking whether the top bars fall above or below it) and take moments; for shear, use only the summed stem widths as $b_w$ and test the result against the web-crushing cap.
Geometry and effective depth. The dimension chain closes exactly: two 200 mm overhangs, three 200 mm stems and two 600 mm voids sum to 2200 mm. With 40 mm cover, 20M ties and 30M main bars, the tension steel sits at$$d = 800-40-19.5-15 = 725\ \text{mm}$$ and the flange bars sit 70 mm below the top face.
Locate the neutral axis. Assume the compression block is inside the flange, so its width is the full 2200 mm and $\alpha_1\phi_cf_c' = 18.14$ MPa. Balancing the 6-30M tension force alone would put $a \approx 36$ mm, that is $c \approx 41$ mm — above the flange bars at 70 mm. The 8-20M therefore lie below the neutral axis and act in tension, not compression, so they must be added to the tension side and the axis re-solved. Iterating on $c$ until the net force is zero:$$c = 53.1\ \text{mm},\qquad a = \beta_1c = 46.8\ \text{mm}\;<\;h_f = 150\ \text{mm}\;\checkmark$$ At that depth the flange bars carry a tensile stress of only 217 MPa (they are too close to the axis to yield).
Moment resistance. Taking moments of the concrete resultant and both bar groups about mid-depth:$$M_r = \boxed{1023\ \text{kN}\cdot\text{m}}$$ Ignoring the flange bars entirely would give 1010 kN·m, so their genuine contribution is $+1.2\%$ — small, but the sign matters: treating them as compression steel would move the answer the wrong way and, more importantly, would mask the fact that the entire section is a very shallow block on a deep lever arm.
Confirm the failure mode. With $c/d = 53.1/725 = 0.073$ the section is far into the tension-controlled range, so the 30M bars are yielding well before the concrete crushes and the section has ample ductility. The stress block occupies less than a third of the flange thickness, which is why the voids between the stems are irrelevant to flexure — in bending, this triple-T computes exactly as a 2200 mm wide slab.
Shear: the effective web width. The voids are not irrelevant to shear. A diagonal crack must cross concrete, and only the stems provide any, so$$b_w = 3(200) = 600\ \text{mm},\qquad d_v = \max(0.9d,\;0.72h) = 653\ \text{mm}$$ Using the full 2200 mm flange width here would overstate the shear capacity by nearly four times — this is the single most common error on this question type.
Concrete and steel contributions. With the simplified method ($\beta = 0.18$, $\theta = 35^\circ$, minimum stirrups present) and one closed 20M tie per stem giving six legs across a shear plane ($A_v = 1800$ mm2 at $s = 200$ mm):$$V_c = 0.65(0.18)\sqrt{35}(600)(653)/10^3 = 271\ \text{kN}$$$$V_s = \frac{\phi_sA_vf_yd_v\cot\theta}{s}= \frac{0.85(1800)(400)(653)(1.428)}{200(10^3)} = 2853\ \text{kN}$$ The sum is 3124 kN — an enormous figure that cannot actually be delivered.
Web crushing governs. A23.3 11.3.3 caps the shear at the load that crushes the diagonal concrete struts:$$V_{r,\max}=0.25\phi_cf_c'b_wd_v = 0.25(0.65)(35)(600)(653)/10^3= 2228\ \text{kN}$$ Since $V_c+V_s = 3124 > 2228$ kN, the answer is the cap:$$\boxed{V_r = 2228\ \text{kN}}$$ The practical reading is that the 20M ties at 200 mm are far heavier than the section can use; closer or larger ties would buy nothing, and any real increase in shear capacity would have to come from thicker stems or stronger concrete.
