Numerical Examples | Moment of Resistance Calculations

Moment of Resistance calculations | Doubly reinforced sections

In our article series for “Doubly reinforced sections”, we have covered the following:

What are doubly reinforced sections?

Methods for determining Neutral Axis?

Solved numerical examples for determining Neutral Axis

Numerical examples for practice (Find Neutral axis)

Methods for calculating Moment of Resistance

Numerical example for calculating Moment of resistance

Types of problems in Doubly reinforced sections

Determining stresses in steel and concrete 

Numerical example | Stresses in steel and concrete

Now we shall move on with a solved example. This will help you understand the methods in a better way. I suggest that you do them yourselves too. Practice will help you make your concepts more concrete and clear.

Numerical Example:

An reinforced concrete 300mm x 600mm effective dimensions is provided with tensile and compressive reinforcement of 1256mm2 each. The compressive steel is placed 30mm from the top edge of the beam. If σ_cbc = 7N/mm², σ_st = 190N/mm² and m = 13.33, find the moment of resistance of beam by following two methods:

1)     Elastic theory method

2)     Steel beam theory method

Given that:

Width of the beam = b = 300mm

Effective depth of the beam = d = 600mm

Ast = Asc = 1256 mm2

Effective depth of the beam = 400 – 30 = 370mm

Distance of compressive steel from the top edge of the beam to the centre of the steel = d’ = 30mm

Permissible stress in concrete = σ_cbc = 7N/mm²

Permissible stress in steel = σ_st = 190N/mm²

Modular ratio = m = 13.33

σ_sc = 130N/mm²

To find Mr by Elastic Theory Method

To find x_c

σ_cbc/( σ_st/m) = x_c/(d – x_c)

7/(190/13.33) = x_c/(d – x_c)

x_c = 197.61mm

To find x

bxx/2 + (1.5m – 1)A_sc(x – d’) = mA_st (d – x)

300×2/2 + (1.5 x 13.33 -1)1256 (x – 30)

= 13.33 x 1256 (600 – x)

x² + 270.6x – 71741 = 0

Solving the above equation, we get,

x = 164.778 = 165mm

therefore, x < x_c

Hence, the beam is under-reinforced.

Therefore, σ_st = 190N/mm²

To find σ_cbc

σ_cbc/( σ_st/m) = x/(d – x)

σ_cbc/( 190/13.33) = 165/(600 – 165)

σ_cbc = 5.4 N/mm²

To find σ_cbc

σ_cbc = σ_cbc [(x – d’)/x)]

= 5.4 [(165-30)/165]

= 4.418 = 4.42 N/mm²

To find σ_sc

σ_sc = 1.5m x σ_cbc

= 1.5 x 13.33 x 4.42

= 88.377 = 88.38 N/mm²

<130 N/mm²

Hence, the design is all right.

To find Mr

Taking moments about tensile steel, we get,

M_r = bx. σ_cbc/2 (d – x/3)+(1.5m – 1)A_sc. σ_cbc (d – d’)

= 300 x 165 x (5.4/2)(600-165/3) + (1.5 x 13.33 – 1)1256 x 4.42 (600 – 30)

= 132946390 N-mm

= 132.946 kN-m

To find Moment of resistance using Steel beam theory Method

Mr = Ast.σst (d – d’)

= 1256 x 190 (600 – 30)

= 136024000 N-mm

Mr= 136.024 kN-m