Example 01: Total Compression Force at the Section of Concrete Beam

Area of reinforcing bars
$A_s = 4 \times \frac{1}{4}\pi(25^2) = 625\pi ~ \text{mm}^2$

$nA_s = 9(625\pi) = 5625\pi ~ \text{mm}^2$
 

 

Location of the neutral axis
$Q_{\text{above NA}} = Q_{\text{below NA}}$

$250x(\frac{1}{2}x) = nA_s(d - x)$

$125x^2 = 5625\pi(500 - x)$

$125x^2 + 5625\pi x - 2\,812\,500\pi = 0$

$x = 204.42 \, \text{ and } \, -345.79$
 

Use $x = 204.42 ~ \text{mm}$
 

Moment of inertia
$I_{NA} = \dfrac{250x^3}{3} + nA_s(d - x)^2$

$I_{NA} = \dfrac{250(204.42^3)}{3} + 5625\pi(500 - 204.42)^2$

$I_{NA} = 2\,255\,762\,492 ~ \text{mm}^4$
 

Bending stresses
$f_b = \dfrac{Mc}{I}$

Concrete
$f_c = \dfrac{Mx}{I_{NA}}$

$f_c = \dfrac{150(204.42)(1000^2)}{2\,255\,762\,492}$

$f_c = 13.59 ~ \text{MPa}$           answer
 

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

$\dfrac{f_s}{9} = \dfrac{150(500 - 204.42)(1000^2)}{2\,255\,762\,492}$

$f_s = 176.89 ~ \text{MPa}$           answer

 

Total compressive force in concrete
$C = \frac{1}{2}f_c bx = \frac{1}{2}(13.59)(250)(204.42)$

$C = 347.26 ~ \text{kN}$           answer