#### Singly reinforced beams | ______1_7_0_3_3_f_c_b_7_b_7_e_b_2_1_a_8_2_5_4_2_d_2_5_5_3_7_5_e_f_d_a______

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

- Basic definitions and formulas
- Understanding stresses and modular ratios
- Assumptions for singly reinforced sections
- Design procedure for Singly reinforced section – I
- Solved Numericals for Singly reinforced beam | Method I
- Design of Singly reinforced sections | Design Method 2
- Solved Numericals for Singly reinforced beam | Method 2
- Moment of Resistance for Singly reinforced sections
- Solved numerical example | Moment of resistance
- Solved numerical example 2 | Guide to singly reinforced sections

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.**

**(σ _{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 = ?

A_{st} = cross-sectional area of steel in tension = 1531mm2

σ_{cbc} = permissible compressive stress in concrete in bending = 7N/mm^{2}

σ_{st} = permissible stress in steel = 140 N/mm^{2}

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):

σ_{cbc}/(σ_{st}/m) = x_{c}/(d – x_{c})

7/(140/13.33) = x_{c}/(500 – x_{c})

**x _{c} = 209.969mm = 210mm**

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

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

bxx/2 = mA_{st}(d – x)

250x^{2}/2 = 13.33 x 1521 (525 – x)

x^{2} + 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 > x_{c}

Since x is greater than x_{c}, 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:

M_{r}= C x z

= bx_{c}(σ_{cbc}/2)z

= 250 x 222 x (7/2) x (525 – 222/3)

= 87606750 N-mm

**= 87.6 kN-m**

You can leave a response, or trackback from your own site.