## Numerical example 2 | Singly reinforced Sections

#### Guide to design of Singly reinforced Sections | Civil Engineering

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

Now we will move on with our next solved example in which we will make use of formulas derived earlier. That is why it is necessary that you go through the entire step by step guide in order to gain complete understanding.

#### Numerical Problem

Calculate the moment of resistance of an RC beam 250mm wide, the depth of the centre of reinforcement being 500mm. Assume σcbc = 5N/mm2, σst = 140 N/mm2, modular ratio = 18.66

#### Given that,

b = width of the beam = 250mm

d = depth of the beam = 500mm

σcbc = permissible compressive stress in concrete in bending = 5N/mm2

σst = permissible stress in steel = 140 N/mm2

m =  modular ratio = 18.66

#### To find Neutral Axis (NA)

σcbc /(σst/m) = xc/(d – xc)

5/(140/18.66) = xc/(500 – xc)

Xc = 199.95mm = 200mm

Read moreNumerical example 2 | Singly reinforced Sections

## Solved numerical examples | Design of Singly reinforced sections

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

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

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 σcbc = 5 N/mm2, σst = 140 N/mm2, m = 18.66

#### To find Neutral Axis (NA)

σcbc /(σst/m) = xc/(d – xc)

5/(140/18.66) = xc/(d – xc)

Therefore, xc  = 0.399d mm= 0.4dmm

Read moreSolved numerical examples | Design of Singly reinforced sections