Design of Steel Reinforcement of Concrete Beams by WSD Method

Steps is for finding the required steel reinforcements of beam with known Mmax and other beam properties using Working Stress Design method.

Given the following, direct or indirect:

Width or breadth = b
Effective depth = d
Allowable stress for concrete = fc
Allowable stress for steel = fs
Modular ratio = n
Maximum moment carried by the beam = Mmax


Step 1: Solve for the balanced moment capacity

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

$C_{bal} = \frac{1}{2}f_c \, bx_{bal}$

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

  • If MmaxMbal, design the beam as singly reinforced (go to Step 2)
  • If Mmax > Mbal, design the beam as doubly reinforced (go to Step 3)


Step 2: Singly Reinforced Beam (MmaxMbal)

Locate the neutral axis by solving x from the equation below
$\dfrac{bx^3}{3} + \dfrac{bx^2(d - x)}{2} = \dfrac{nM_{max}(d - x)}{f_s}$


WSD Singly Reinforced Beam


Solve for the required steel area
$A_s = \dfrac{bx^2}{2n(d - x)}$

You can also use the approximate formula for the amount of As
$A_s = \dfrac{M_{max}}{f_s(d - \frac{1}{3}x_{bal})}$


Step 3: Doubly Reinforced Beam (Mmax > Mbal)
Additional given: Embedment depth of compresion steel = d'

Solve for As1 from Balanced Condition
$A_{s1} = \dfrac{M_{bal}}{f_s(d - \frac{1}{3}x_{bal})}$

Solve As2 from the excess of Mmax and Mbal
$M_{excess} = M_{max} - M_{bal}$

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

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




Solve As' from the balanced condition and use Mexcess
(Note: if fs' > fs, use fs' = fs)
$f_s' = \dfrac{2nf_c(x_{bal} - d')}{x_{bal}}$   or   $f_s' = \dfrac{2f_s(x_{bal} - d')}{d - x_{bal}}$

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