6 step procedure for determining stresses in steel and concrete | Doubly reinforced sections





Numerical example for determining stresses in steel and concrete

In our article series for “Design of Doubly reinforced sections”, we 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

In our previous article, we discussed a detailed 6 step procedure for determining stresses in steel and concrete. Now we shall move on with a numerical example in which we will use the 6 step procedure to solve the problem.

Problem Type two: Determining stresses in steel and concrete using the 6 step procedure

A rectangular beam is 200mm wide and 480mm deep. It has to resist a bending moment of 100 kN-m. The reinforcedment consists of four 25mm ⏀ bars on tension side and three 22mm⏀bars on compression side. The centres of bars being 30mm from the top and bottom edges of the beam. Find the stresses set up in steel and concrete. m=18.66

Given data is as follows:

Breadth of the beam = b = 200mm

Effective depth of the beam = d = 480 – 30 = 450mm

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

Bending moment = M = 100kN-m

Modular ratio = m = 18.66

Area of tensile steel = Ast = 4 π/4 x 25 x 25 = 1964 mm2

Area of compressive steel = Asc = 4 π/4 x 22 x 22 = 1140 mm2

Step one:

Find x:

bx.x/2 + (1.5m – 1)Asc (x – d’) = mAst(d-x)

200x2/2 + (1.5×18.66 – 1) 1140 (x – 30)

= 18.66 x 1964 x (450 – x)

Therefore, x2 + 674.17x – 174147 = 0

Solving for x, we get;

x = 199.36mm = 199mm

Step two:

Let σcbc  be the compressive stress in concrete at the top of the beam,

Then, σ’cbc = σcbc (x – d’)/x

= σcbc (199 – 30)/199

= 0.849 σcbc = 0.85 σcbc N/mm2

Step three:

To find Moment of resistance Mr:

Taking moments about tensile steel;

Mr = bx σcbc/2 (d – x/3) + (1.5m – 1) Asc σcbc (d – d’)

Substitute the value of σ’cbc in terms of σcbc

= 200 x 199 x σcbc/2 (450 – 199/3) + (1.5 x 18.66 – 1)1140 x 0.85 σcbc x (450 – 30)

= 18618356 σcbc N-mm

= 18.618 σcbc kN-m

Step four:

Equate Mr to M and find σcbc

18.618 σcbc = 100

σcbc = 5.37 N/mm2

Step five:

To find compressive stress in steel

σcbc/ (σst/m) = x/(d – x)

5.37/ (σst/18.66) = 199/(450 – 199)

Therefore, σst= 126.388 = 126.39 N/mm2

Step Six:

To find compressive stresses in concrete

σsc = 1.5m σcbc

= 1.5 x 18.66 x (0.85 x 5.37)

= 127.76 N/mm2



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  1. #1 by Afroz on September 12, 2012 - 6:48 pm

    Good job!
    Keep it up.

  2. #2 by PARIKSHIT on September 26, 2012 - 4:17 am

    THANK YOU VERY^1/0 MUCH……..

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