Example 03: Finding the Number of 32-mm Steel Bars for Doubly-Reinforced Concrete Propped Beam

The most critical section in bending is at the fixed support
 

 

$M_{max} = \dfrac{w_oL^2}{8} = \dfrac{28.8(8^2)}{8}$

$M_{max} = 230.4 ~ \text{kN}\cdot\text{m}$
 

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

$f_s = 140 ~ \text{MPa}$
 

 

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

$\dfrac{x_{bal}}{9.45} = \dfrac{540}{9.45 + 140/9}$

$x_{bal} = 204 ~ \text{mm}$
 

$C_{bal} = \frac{1}{2}f_c \, bx_{bal} = \frac{1}{2}(9.45)(400)(204)$

$C_{bal} = 385.56 ~ \text{kN}$
 

$M_{bal} = C_{bal}(d - \frac{1}{3}x_{bal}) = 385.56[ \, 0.540 - \frac{1}{3}(0.204) \, ]$

$M_{bal} = 181.98 ~ \text{kN}\cdot\text{m}$
 

M_max > M_bal, thus, the beam is doubly-reinforced
 

For Doubly-Reinforced Beam Section
Solve for A_s1 from Balanced Condition
$M_{bal} = T_{bal}(d - \frac{1}{3}x_{bal})$

$M_{bal} = f_s A_{s1}(d - \frac{1}{3}x_{bal})$

$181.98(1000^2) = 140A_{s1} [ \, 540 - \frac{1}{3}(204) \, ]$

$A_{s1} = 2753.93 ~ \text{mm}^2$
 

Solve A_s2 from the excess of M_max and M_bal
$M_{excess} = M_{max} - M_{bal} = 230.4 - 181.98$

$M_{excess} = 48.42 ~ \text{kN}\cdot\text{m}$
 

$M_{excess} = T_2(d - d')$

$M_{excess} = f_s A_{s2}(d - d')$

$48.42(1000^2) = 140A_{s2}(540 - 60)$

$A_{s2} = 720.54 ~ \text{mm}^2$
 

Total steel area in tension
$A_s = A_{s1} + A_{s2} = 2753.93 + 720.54$

$A_s = 3474.47 ~ \text{mm}^2$
 

 

Number of 32-mm tension bars
$N = \dfrac{A_s}{A_b} = \dfrac{3474.47}{\frac{1}{4}\pi (28^2)}$

$N = 4.32$

Use 5 - 32 mm ø bars           answer
 

Solve A_s' from the balanced condition using M_excess
(Note: if f_s' > f_s, use f_s' = f_s)
$\dfrac{f_s' / 2n}{x_{bal} - d'} = \dfrac{f_c}{x_{bal}}$

$f_s' = \dfrac{2nf_c(x_{bal} - d')}{x_{bal}}$

$f_s' = \dfrac{2(9)(9.45)(204 - 60)}{204}$

$f_s' = 120.07 ~ \text{MPa}$
 

You may also use the tension steel in the proportion:

 

f_s' < (f_s = 140 MPa), use f_s' = 120.07 MPa
$M_{excess} = T_2' (d - d')$

$M_{excess} = A_s'(2n - 1)(f_s' / 2n)(d - d')$

$48.42(1000^2) = A_s'[ \, 2(9) - 1 \, ][ \, 120.07 / (2 \times 9) \, ](540 - 60)$

$A_s' = 889.55 ~ \text{mm}^2$
 

Number of 32-mm compression bars
$N = \dfrac{A_s'}{A_b} = \dfrac{889.55}{\frac{1}{4}\pi (32^2)}$

$N = 1.1$

Use 2 - 32 mm ø bars           answer