Ultimate Moment and Number of Bars of Simply Supported Concrete Beam

$w_u = 1.4(3.17 + 16) + 1.7(14)$

$w_u = 50.638 ~ \text{kN/m}$
 

Ultimate moment
$M_u = \dfrac{w_u L^2}{8}$

$M_u = \dfrac{50.638(5^2)}{8}$

$M_u = 158.24 \, \text{kN}\cdot\text{m}$   ←   answer

$R_n = \dfrac{200(1000^2)}{0.9(300)(380^2)}$

$R_n = 5.1298$
 

$\rho = \dfrac{0.85 f_c'}{f_y} \left[ 1 - \sqrt{1 - \dfrac{2R_n}{0.85f_c'}} \right]$

$\rho = \dfrac{0.85(30)}{415} \left[ 1 - \sqrt{1 - \dfrac{2(5.1298)}{0.85(30)}} \right]$

$\rho = 0.01394$
 

$\rho_{min} = \dfrac{1.4}{f_y} = \dfrac{1.4}{415}$

$\rho_{min} = 0.0034$
 

Use $\rho = 0.01394$
 

Number of 16-mm bars:
$N = \dfrac{\rho bd}{A_b} = \dfrac{0.01394(300)(380)}{\frac{1}{4}\pi 16^2}$

$N = 7.9 \cong 8 \, \text{bars}$   ←   answer

$M_u = \dfrac{1.4(3.17)(5^2)}{8} + \dfrac{50(5)}{4}$

$M_u = 76.369 \, \text{kN} \cdot \text{m}$
 

$R_n = \dfrac{M_u}{\phi bd^2}$

$R_n = \dfrac{76.369(1000^2)}{0.9(300)(380^2)}$

$R_n = 1.9588$
 

$\rho = \dfrac{0.85 f_c'}{f_y} \left[ 1 - \sqrt{1 - \dfrac{2R_n}{0.85f_c'}} \right]$

$\rho = \dfrac{0.85(30)}{415} \left[ 1 - \sqrt{1 - \dfrac{2(1.9588)}{0.85(30)}} \right]$

$\rho = 0.00492 \gt \rho_{min}$
 

$N = \dfrac{\rho bd}{A_b} = \dfrac{0.00492(300)(380)}{\frac{1}{4}\pi 16^2}$

$N = 2.79 \cong 3 \, \text{bars}$   ←   answer