Example 02: Finding the Number of 28-mm Steel Bars of Singly-Reinforced Concrete Cantilever Beam

Problem
A reinforced concrete cantilever beam 4 m long has a cross-sectional dimensions of 400 mm by 750 mm. It is to carry a superimposed load of 29.05 kN/m including its own weight. The steel reinforcement has an effective depth of 685 mm. Use f’_c = 21 MPa, f_s = 165 MPa, and n = 9. Determine the required number of 28 mm ø reinforcing bars using Working Stress Design method.

Solution

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Maximum Moment
$M_{max} = 29.05(4)(2) = 232.4 ~ \text{kN}\cdot\text{m}$

For Balanced Capacity
$f_c = 0.45f_c' = 0.45(21) = 9.45 ~ \text{MPa}$

$f_s = 165 ~ \text{MPa}$

$\dfrac{x_{bal}}{f_c} = \dfrac{d}{f_c + f_s/n}$

$\dfrac{x_{bal}}{9.45} = \dfrac{685}{9.45 + 165/9}$

$x_{bal} = 233 ~ \text{mm}$

$C_{bal} = \frac{1}{2}f_c \, b x_{bal} = \frac{1}{2}(9.45)(400)(233)$

$C_{bal} = 440.37 ~ \text{kN}$

$M_{bal} = C_{bal}(d - \frac{1}{3}x_{bal}) = 440.37[ \, 0.685 – \frac{1}{3}(0.233) \, ]$

$M_{bal} = 267.45 ~ \text{kN}\cdot\text{m}$

M_max < M_bal, thus, the beam is singly-reinforced.

For Singly Reinforced Section
$f_s = 165 ~ \text{MPa}$

$f_c = ~ ?$

Accurate Solution for finding A_s (Not recommended)

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Locate the neutral axis

$Q_{\text{below NA}} = Q_{\text{above NA}}$

$400x(\frac{1}{2}x) = nA_s(d - x)$

$200x^2 = nA_s(685 - x)$

$nA_s = \dfrac{200x^2}{685 - x}$ ← Equation (1)

Moment of inertia
$I_{NA} = \dfrac{bx^3}{3}+ nA_s(d –x)^2$

$I_{NA} = \dfrac{400x^3}{3}+ \dfrac{200x^2}{685 - x}(685 –x)^2$

$I_{NA} = \frac{400}{3}x^3 + 200x^2(685 –x)$

Stress in steel
$\dfrac{f_s}{n} = \dfrac{M(d - x)}{I_{NA}}$

$\dfrac{165}{9} = \dfrac{232.4(685 - x)(1000^2)}{I_{NA}}$

$I_{NA} = 12\,676\,363.64(685 - x)$

Equate the I_NA
$I_{NA} = I_{NA}$

$\frac{400}{3}x^3 + 200x^2(685 –x) = 12\,676\,363.64(685 - x)$

$x = 219.6 ~ \text{mm}$

From Equation (1)
$9A_s = \dfrac{200(219.6^2)}{685 - 219.6}$

$A_s = 2302.17 ~ \text{mm}^2$

Number of bars
$N = \dfrac{A_s}{A_b} = \dfrac{2302.17}{\frac{1}{4}\pi (28^2)}$

$N = 3.74$

Use 4 - 28 mm ø bars answer

Approximate Solution for finding A_s (Recommended)
$M_{max} = T(d - \frac{1}{3}x_{bal})$

$M_{max} = A_sf_s(d - \frac{1}{3}x_{bal})$

$A_s = \dfrac{M_{max}}{f_s(d - \frac{1}{3}x_{bal})}$

$A_s = \dfrac{232.4(1000^2)}{165[ \, 685 - \frac{1}{3}(233) \, ]}$

$A_s = 2319.13 ~ \text{mm}^2$

Number of bars
$N = \dfrac{A_s}{A_b} = \dfrac{2319.13}{\frac{1}{4}\pi (28^2)}$

$N = 3.76$

Use 4 - 28 mm ø bars answer

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