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.

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