Solved numericals for Singly reinforced Sections | Design Method 1



Design of Singly reinforced Sections | Method 1

In our article series for Singly reinforced sections, we have covered the following:

 

Numerical Problem

An RC beam 200mm wide has an effective depth of 350mm. The permissible stresses in concrete and steel are 5N/mm2 and 140 N/mm2 respectively. Find the depth of neutral axis, area of steel and percentage of steel. (modular ratio (m) = 18.66)

Step One:

Given that:

b = breadth of a rectangular beam = 200mm

d = effective depth of a beam = 350mm

x = depth of neutral axis below the compression edge = ?

Ast = cross-sectional area of steel in tension = ?

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

σst = permissible stress in steel = 140 N/mm2

m = modular ratio = 18.66

From the concrete stress diagram, the formula is given as,

σcbc/(σst/m) = x/(d – x)

5/(140/18.66) = x/(350-x)

Therefore, x = 139.97mm

Step two:

To find area of steel

Equating total compressive force (C) to total tensile force (T)

C = T

C = area x average compressive stress

= (b.x) X (σcbc + 0)/2

= bx (σcbc/2)

T = area x tensile stress

= Ast x σst

Therefore, bx (σcbc/2) = Ast x σst

200 x 139.97 x 5/2 = Ast x 140

Therefore, Ast = 499.89 mm2

Calculating area of Steel (pt)

Area of steel is expressed as a percentage. The formula for percentage of steel is as follows;

pt = Ast x 100/ bd

= 499.89 x 100/(200×350)

= 0.714

 



8 thoughts on “Solved numericals for Singly reinforced Sections | Design Method 1”

    • Modular ratio is the ratio of modulus of elasticity of steel and concrete. Thus, m = Es/Ec. where Es is the modulus of elasticity of steel which is 200000 N/mm2, which is a constant. so, the value of m depends on the modulus of elasticity of concrete, which can change. In the above example, we know the strength of concrete used. So, we know the value of modular ratio.

Leave a Comment