# Guide to design of Singly reinforced beams | Solved examples

#### For design of “Singly reinforced beam” article series, we have covered the following:

Now we will move on with another solved example where we will calculate the Moment of resistance and determine the position of the Neutral axis. For this we will have to use the formulas that we have derived earlier in our previous articles in the list above.

#### Numerical Problem

Calculate the moment of resistance of an RC beam 250x550mm overall. Reinforcement is 1521mm2 and is placed at a distance of 25mm from the bottom.

cbc = 7N/mm2, σst = 140N/mm2, m = 13.33)

#### Given that:

b = breadth of a rectangular beam = 250mm

d = effective depth of a beam = 550 – 25 = 525mm

x = depth of neutral axis below the compression edge = ?

Ast = cross-sectional area of steel in tension = 1531mm2

σcbc = permissible compressive stress in concrete in bending = 7N/mm2

σst = permissible stress in steel = 140 N/mm2

m =  modular ratio = 13.33

We have to find the value for Moment of resistance. To calculate Mr, we have to first calculate NA(critical) and NA (actual).

#### To find Neutral axis (NA) (critical):

σcbc /(σst/m) = xc/(d – xc)

7/(140/13.33) = xc/(500 – xc)

xc = 209.969mm = 210mm

#### To find N.A (actual):

Taking moment of area of compression and tension side about N.A., we get

bxx/2 = mAst (d – x)

250x2/2 = 13.33 x 1521 (525 – x)

x2 + 162.19x – 85154 = 0

On solving the above equation, we get two values out of which one is positive and the other negative.

x = 221.77mm = 222mm

Therefore, x > xc

Since x is greater than  xc, it is clear that the actual N.A. is below the critical N.A. Hence the beam is over-reinforced.

#### To find Moment of resistance:

Mr = C x z

= bxccbc/2)z

= 250 x 222 x (7/2) x (525 – 222/3)

= 87606750 N-mm

= 87.6 kN-m

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