# Design of Steel Reinforcement of Concrete Beams by WSD Method

Steps is for finding the required steel reinforcements of beam with known *M _{max}* and other beam properties using Working Stress Design method.

Given the following, direct or indirect:

*b*

Effective depth =

*d*

Allowable stress for concrete =

*f*

_{c}Allowable stress for steel =

*f*

_{s}Modular ratio =

*n*

Maximum moment carried by the beam =

*M*

_{max}

**Step 1: Solve for the balanced moment capacity**

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

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

- If
*M*≤_{max}*M*, design the beam as singly reinforced (go to Step 2)_{bal} - If
*M*>_{max}*M*, design the beam as doubly reinforced (go to Step 3)_{bal}

**Step 2: Singly Reinforced Beam ( M_{max} ≤ M_{bal})**

*x*from the equation below

$\dfrac{bx^3}{3} + \dfrac{bx^2(d - x)}{2} = \dfrac{nM_{max}(d - x)}{f_s}$

$A_s = \dfrac{bx^2}{2n(d - x)}$

You can also use the approximate formula for the amount of *A _{s}*

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

**Step 3: Doubly Reinforced Beam ( M_{max} > M_{bal})**

Additional given: Embedment depth of compresion steel =

*d'*

*A*

_{s1}from Balanced Condition

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

Solve *A*_{s2} from the excess of *M _{max}* and

*M*

_{bal}$M_{excess} = M_{max} - M_{bal}$

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

Total steel area in tension

$A_s = A_{s1} + A_{s2}$

*A*from the balanced condition and use

_{s}'*M*

_{excess}(Note: if

*f*>

_{s}'*f*, use

_{s}*f*=

_{s}'*f*)

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

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