# Posts Tagged 6 step procedure for determining stresses in steel and concrete

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

Posted by Benzu JK in Building Construction on September 9, 2012

#### 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 mm^{2}

Area of compressive steel = Asc = 4 π/4 x 22 x 22 = 1140 mm^{2}

#### Step one:

Find x:

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

200x^{2}/2 + (1.5×18.66 – 1) 1140 (x – 30)

= 18.66 x 1964 x (450 – x)

Therefore, x^{2} + 674.17x – 174147 = 0