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 σ_cbc  be the compressive stress in concrete at the top of the beam,

Then, σ’_cbc = σ_cbc (x – d’)/x

= σ_cbc (199 – 30)/199

= 0.849 σ_cbc = 0.85 σ_cbc N/mm²

Step three:

To find Moment of resistance Mr:

Taking moments about tensile steel;

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

Substitute the value of σ’_cbc in terms of σ_cbc

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

= 18618356 σ_cbc N-mm

= 18.618 σ_cbc kN-m

Step four:

Equate Mr to M and find σ_cbc

18.618 σ_cbc = 100

σ_cbc = 5.37 N/mm²

Step five:

To find compressive stress in steel

σ_cbc/ (σ_st/m) = x/(d – x)

5.37/ (σ_st/18.66) = 199/(450 – 199)

Therefore, σ_st= 126.388 = 126.39 N/mm²

Step Six:

To find compressive stresses in concrete

σ_sc = 1.5m σ_cbc

= 1.5 x 18.66 x (0.85 x 5.37)

= 127.76 N/mm²