Question 4 of 7: Reinforced concrete beam for the Figure 3 loading

Question B1: Reinforced concrete beam for the Figure 3 loading (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.

Find. A rectangular reinforced concrete cross-section, its flexural reinforcement and its shear reinforcement, adequate for the full NBCC load-combination envelope.

Approach. Assume a trial section so that self weight can be included, run every NBCC principal load combination over the combined uniform-plus-triangular loading to find the governing moment and shear, then design a singly reinforced rectangular section and check that stirrups are needed.

1. Assume a section and compute its self weight. A span of 4.5 m needs at least $L/16 = 281\ \text{mm}$ of depth to escape a deflection calculation under A23.3 Table 9.2. Take $b = 250\ \text{mm}$, $h = 450\ \text{mm}$: $$w_{sw} = 0.250(0.450)(24) = 2.70\ \text{kN/m}$$ so the total dead load is $4.5 + 2.7 = 7.2\ \text{kN/m}$.
2. Set out the statics of the triangular load. For a load of peak intensity $p$ at A falling to zero at $a = 2.5\ \text{m}$, the resultant and its position are $$W = \tfrac{1}{2}pa = 1.25p, \quad \bar{x} = \tfrac{a}{3} = 0.833\ \text{m from A}$$ and the moment it produces at a section $x \le a$ is $$M_{tri}(x) = p\left(\frac{x^{2}}{2} - \frac{x^{3}}{6a}\right)$$ Beyond $x = a$ the whole resultant acts, so $M_{tri}(x) = W(x - \bar{x})$.
3. Run the load combinations. Wind is treated as an additional vertical uniform load (see the callout below). Working through NBCC Table 4.1.3.2 with $D = 7.2$, $L = 3.0$, $W = 3.6\ \text{kN/m}$ uniform and $S = 6.0\ \text{kN/m}$ triangular:

   Combination	uniform (kN/m)	triangular peak	$M_{max}$ (kN·m)
   $1.4D$	10.08	—	25.5
   $1.25D + 1.5L + 0.5S$	13.50	3.0	35.8
   $1.25D + 1.5S + 1.0L$	12.00	9.0	35.2
   $1.25D + 1.4W + 0.5L$	15.54	—	39.3
   $1.25D + 1.4W + 0.5S$	14.04	3.0	37.1
   $0.9D + 1.4W$ (uplift)	1.44	—	3.7

   $$\boxed{M_f = 39.3\ \text{kN}\!\cdot\!\text{m}\ \text{at mid-span}, \quad V_f = 36.2\ \text{kN at A}}$$ The maximum shear comes from a different case, $1.25D + 1.5S + 1.0L$, because the triangular snow load piles its resultant near support A. Note also that even with the wind reversed the net load stays downward, so no uplift case governs.
4. Compute the effective depth. With 40 mm cover, 10M stirrups and 15M main bars: $$d = 450 - 40 - 11.3 - \tfrac{16.0}{2} = 390.7\ \text{mm}$$
5. Solve for the flexural steel. A23.3 Clause 10.1.7 gives the rectangular stress block parameters $$\alpha_1 = 0.85 - 0.0015f_c' = 0.8125, \quad \beta_1 = 0.97 - 0.0025f_c' = 0.9075$$ Equilibrium of the section with $a = \phi_s f_y A_s/\left(\alpha_1\phi_c f_c' b\right)$ substituted into $M_r = \phi_s f_y A_s\left(d - a/2\right)$ gives a quadratic in $A_s$; solving it for $M_r = M_f$: $$A_{s,req} = 309\ \text{mm}^2$$ Minimum reinforcement under Clause 10.5.1.2: $$A_{s,min} = \frac{0.2\sqrt{f_c'}\,b_t h}{f_y} = \frac{0.2(5)(250)(450)}{400} = 281\ \text{mm}^2$$ so the strength requirement governs, though only just.
6. Choose bars and confirm the resistance. Provide 2-15M ($A_s = 400\ \text{mm}^2$) in one layer: $$a = \frac{0.85(400)(400)}{0.8125(0.65)(25)(250)} = 41.2\ \text{mm}, \quad c = \frac{a}{\beta_1} = 45.4\ \text{mm}$$ $$\boxed{M_r = 0.85(400)(400)\left(390.7 - \tfrac{41.2}{2}\right) = 50.3\ \text{kN}\!\cdot\!\text{m} > 39.3\ \text{kN}\!\cdot\!\text{m}}$$ The utilisation is 0.78. Two 15M bars at 250 mm width leave 128 mm clear between them, well above the greater of $1.4d_b$ and 30 mm required by Clause 6.6.5.2.
7. Check that the section is under-reinforced. $$\frac{c}{d} = \frac{45.4}{390.7} = 0.116 \ll 0.5$$ The steel is far past yield at crushing, so the beam fails in a fully ductile manner with ample warning — which is what allows $\phi_s f_y$ to be used at all.
8. Check shear. The critical section is at $d_v$ from the support face. Using the simplified method of Clause 11.3.6.3 for a member without transverse reinforcement: $$d_v = \max\left(0.9d,\ 0.72h\right) = \max(351.6,\ 324) = 351.6\ \text{mm}$$ $$\beta = \frac{230}{1000 + d_v} = 0.170$$ $$V_c = \phi_c\lambda\beta\sqrt{f_c'}\,b_w d_v = 0.65(1.0)(0.170)(5)(250)(351.6) = 48.6\ \text{kN}$$ The shear at $d$ from the support is 28.9 kN, so $V_f < V_c$ and Clause 11.2.8.1 does not require minimum stirrups. 10M stirrups at 175 mm are nonetheless provided over the outer third of each end and at 300 mm elsewhere, to hold the cage, to confine the anchorage zone and to give the section a reserve the calculation does not credit.
9. Check deflection and detailing. $h = 450\ \text{mm}$ exceeds $L/16 = 281\ \text{mm}$, so Clause 9.8.2.1 waives the explicit deflection calculation for a simply supported beam not supporting brittle finishes. Extend both 15M bars the full span and anchor them past the support centre-line with a standard hook, since Clause 12.11.3 requires the development of $M_r/V_f$ at a simple support.