# Window lintel design Procedure with Numerical Example

## What is lintel?

The lintel is the beam above the opening parts of the wall to support the load above the opening. The opening may be a window, door, etc. It is the same as the beam but a little bit has small dimensions than the beam. Beam carry the load of the slab of the structure and transfer it to the column but lintel carries load above the opening and transfer it to the column.

## Steps for design of lintel

1. Assume a suitable depth of lintel. The breadth of the lintel is usually equal to the thickness of wall above lintel.
2. Calculate the effective span of lintel. This is usually equal to the clear span plus half the width of bearing on either ends of span.
3. Calculate the load as follows:
1. In cased the lintel is so located that the triangular loading is to be considered, calculated the load w of effective triangle of masonry. Also calculate the self-weight of the lintel per meter run inclusive of self-weight of the lintel.
2. In case the loading on the lintel is to be considered in the form the uniformly distributed load, calculate total load per meter run inclusive of self-weight of lintel.
3. For other cases of loading, like, concentrated load in the vicinity of the lintel, opening in the wall above lintel, etc., load coming. on lintel may be calculated by procedure discussed earlier.

4. Calculate the maximum bending moment in the lintel to find the design moment.

5. Assume the strength of brickwork and steel according to quality              of material used and then find the depth and corresponding area             of steel of lintel using equations as follows;

1. Moment capacity of section based on brickwork strength is:                                    Mr = 0.172bd²fwk              ……….(1)
2. Moment capacity of section based on steel strength:

Mr = 0.696fy As d              ……….(2)

c. For balanced section, As is given by:

As = 0.248db fwk/fy           ………..(3)

d. Xu = 2.03 (fy/fwk) (As/B)          …………………………(4)

e. Xu = (Ewu x D)/(ewu+ esu)

Xu= Depth of neutral axis at failure

where Ewu = Maximum compressive strain at failure

Esu =  Acceptable tensile strain at failure = 0.002

6) The depth is usually taken in multiple brick modular sizes. The external cover to the reinforcement should be at least equal to the diameter of the reinforcement bar.

7) Check the design for shear and bending stress.

EXAMPLE

Q) Design a linter for a window opening of span 1.8 m. The thickness of the wall is 220 mm and the height of the brickwork above the lintel is 1.1 m. The length of the wall on either side of the lintel is more than half the span of the lintel. Use brickwork having characteristics strength of 10 N/mm² and mild steel bars.

Solution:

1. Depth of the lintel is assumed as one brick on edge that is 105 mm total depth.

Effective depth, d = 80 mm,

b=220mm

1. Effective span = Clear span + Half width of bearing on either side

OR,

= Clear span + Effective depth

= 1.8 +0.08

= 1.88 m

Unit weight of plain brickwork = 20 kN/m³

Unit weight of reinforced brickwork = 21 kN/m³ Effective L = 1.88 m

Height of the apex of the triangle =1.88/2 = 0.94 m

Since the height of brickwork above the lintel is 1.1 m and there is a sufficient length wall on either side of the lintel, arch action is possible and so the lintel will carry a triangular wall load in addition to its own weight.

Self weight, w = 21 x 0.22 x 0.105 × 1 = 0.46 kN

Weight of triangular brickwork wall, (w’) = ½ x 20 × 0.22 × 0.94 x 1.88

= 3.89 KN

4. Maximum bending moment

M = w’2/8 + w’l/6

= {0.46*(1.882)}/8  +  (3.89*1.88)/6

= 0.20+1.22

= 1.42kN/m

Design moment, M2 = 1.5 x 1.42 = 2.13 kN/m

5.     Section design

Mud = 2.113kN-m = 2.13 x 106 N-mm

Fwk = 10N/mm2

Fy = 250 N/mm2 ( mild steel bar)

B = 220 mm

a. Depth of section

Mud = 0.172 bd2fwk

d= 75mm

so, provide a lintel made of one brick on edge having total depth = 105 mm and effective depth = 75 mm.

b. Area of steel

Mud = 0.696fyAsd

Or, As = (2.13×106) / (0.696x250x75)

.’. As = 163.2mm2

c. Balanced area of steel

As(balanced) = 0.248db(fwk/fy)    = 0.248x 220x 75x 10/250

= 163.7mm2

Provide three bars of 8 mm diameter and one bar of 6 mm dia – a total area of 179 mm².

Since the provided steel area is slightly more than the balanced area, the section is lightly over-reinforced, and hence the strain in brickwork should be evaluated.

d. Strain in brickwork assuming strain in steel = 0.002

Now,

Xu = 2.03 x (fy/fwk) x (As/b)   = 2.03 x (250/10) x (179/220)

= 41.3mm

Xu = {Ew/(Ew+Es)}/d

Hence,

Ew = {Xu/(d-Xu)} ts

Assuming Es = 0.002;

Ew = {41.3/(75-413)}    = 0.00245<0.003

That is failure by initial yielding of steel.

Happy Learning – Civil Concept