## 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

- Assume a suitable depth of lintel. The breadth of the lintel is usually equal to the thickness of wall above lintel.
- 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.
- Calculate the load as follows:

- 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.
- 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.
- 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;

- Moment capacity of section based on brickwork strength is: M
_{r}= 0.172bd²f_{wk }……….(1) - Moment capacity of section based on steel strength:

M_{r} = 0.696f_{y} A_{s} d ……….(2)

c. For balanced section, As is given by:

As = 0.248db f_{wk}/f_{y } ………..(3)

d. X_{u} = 2.03 (f_{y}/f_{wk}) (A_{s}/B) …………………………(4)

e. X_{u }= (E_{wu} x D)/(e_{wu}+ e_{su})

X_{u}= Depth of neutral axis at failure

where E_{wu} = Maximum compressive strain at failure

E_{su} = 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:

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

Effective depth, d = 80 mm,

b=220mm

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

OR,

= Clear span + Effective depth

= 1.8 +0.08

= 1.88 m

**Load on the lintel**

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*(1.88^{2})}/8 + (3.89*1.88)/6

= 0.20+1.22

= 1.42kN/m

Design moment, M_{2} = 1.5 x 1.42 = 2.13 kN/m

**5. Section design**

M_{ud} = 2.113kN-m = 2.13 x 10^{6} N-mm

F_{wk }= 10N/mm^{2}

F_{y }= 250 N/mm^{2} ( mild steel bar)

B = 220 mm

**a.** **Depth of section **

M_{ud }= 0.172 bd^{2}f_{wk }

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 **

M_{ud }= 0.696f_{y}A_{s}d

Or, A_{s }= (2.13×10^{6}) / (0.696x250x75)

.’. A_{s} = 163.2mm^{2}

** c. Balanced area of steel **

A_{s(balanced)} = 0.248db(f_{wk}/f_{y}) = 0.248x 220x 75x 10/250

= 163.7mm^{2}

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,

X_{u }= 2.03 x (f_{y}/f_{wk}) x (A_{s}/b) = 2.03 x (250/10) x (179/220)

= 41.3mm

X_{u }= {E_{w}/(E_{w}+E_{s})}/d

Hence,

E_{w }= {X_{u}/(d-X_{u})} t_{s}

_{ }Assuming E_{s} = 0.002;

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

**That is failure by initial yielding of steel.**

I hope this article on “**Window lintel design**” remains helpful for you.

Happy Learning – Civil Concept

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