#### Guide to design of Singly reinforced Sections | Building Construction

**For the “design of Singly reinforced sections” article series, we have covered the following:**

- Basic definitions and formulas
- Understanding stresses and modular ratios
- Assumptions for singly reinforced sections
- Design procedure for Singly reinforced section – I
- Solved Numericals for Singly reinforced beam | Method I
- Design of Singly reinforced sections | Design Method 2
- Solved Numericals for Singly reinforced beam | Method 2
- Moment of Resistance for Singly reinforced sections
- Solved numerical example | Moment of resistance
- Solved numerical example 2 | Guide to singly reinforced sections

**Now we will move on with our next solved numerical example in which we will make use of the formulas that we have derived in our earlier articles.**

#### Numerical Problem

**Determine the following:**

a) The position of the neutral axis

b) Lever arm

c) Moment of resistance

d) Percentage of steel

(For a rectangular beam section of width b mm and effective depth d mm. Take σ_{cbc} = 5 N/mm^{2}, σ_{st} = 140 N/mm^{2}, m = 18.66

#### To find Neutral Axis (NA)

σ_{cbc}/(σ_{st}/m) = x_{c}/(d – x_{c})

5/(140/18.66) = x_{c}/(d – x_{c})

Therefore, x_{c } = 0.399d mm= 0.4dmm

#### To find lever arm

z = d – x_{c}/3

= d – 0.4d/3

= 0.867dmm

= 0.87d mm

#### To find Moment of resistance

M_{r}= C x z

= bx_{c}(σ_{cbc}/2)z

= b (0.4d)(5/2)(0.87d)

= 0.87 bd^{2} N-mm

#### To find the percentage of steel

Equating, C = T

bx_{c}.σ_{cbc}/2 = A_{st}. σ_{st }

Therefore, Ast = bx_{c}.σ_{cbc}/2 σ_{st}

= [b (0.4d)(5/2)]/140

= bd/140 mm^{2}

P_{t}= Ast. 100/bd

= (bd/140) x (100/bd)

= 0.71

hi

plz can you provide me one 3 storey industrial RCC design with loading details.

thank you

Thanx. That time i’m doing diploma in civil engineering

Thanks Benjujk for your effort. Please provide sample of one complete multistoried building design using working stress method.