# Design of a Lintel

Posted in Design Sheets, Steelworks, Student Corner | Email This Post |**INTRODUCTION**

Before coming to the design part, it is important to know the importance of lintels in structure. A lintel is a horizontal structural member that is present on top of doors, windows, etc. to support the top load falling on these openings. These are used for load bearing purposes, to transfer its load to the side walls and sometimes for decorative purposes. These can be made of wood or concrete; however, concrete is widely used due to its strength and durability. These can also be prestressed for better load carrying capacity.

The breadth of the lintel should be equal to the thickness of openings& depth should be in the range of l/12 to l/8 of the span. The minimum breadth of the lintel must be 80 mm. Moreover, these must be supported sufficiently at each end. Also, the length of lintel in case of masonry wall is calculated by taking the measurement of the total width of the opening and adding 150 mm for end-bearings at each end. In a house, the level of lintel is generally the soffit level of a door or window. For residential purpose, its height is kept 2.1 meters or 7 feet. The height of the door jam, which is not fastened or secured, is the lintel level.

Here an example is shown to calculate the reinforcements and other design aspects of lintel. The design calculation of lintel is similar to beam design calculations. The difference is there in the diameter taken for reinforcements as in case of beam large diameters are used while in case of lintels small diameters are used.

A clear cover of 25 mm is taken for lintels which is same as the clear covers for beams. Design aid SP-16 is used for calculation. Grade of concrete used is M20 and for steel Fe 500 is used. Since partial fixity is there hence the bending moment formula wl^{2}/10 is used instead of wl^{2}/8. Other data if not mentioned are assumed in this example. The calculation is done in simple manner for easy understanding of the problem. The height of the building is taken 3 meters and in this height beam width of beam is also included for easy deduction in the example.

__DESIGN OF LINTEL__

Height of storey including beam= 3 m.

Let the lintel beam size= 125 mm × 201 mm

Height of the wall above lintel= 3-2.02-0.201-0.45

= 0.33 mm.

Using 12 φ steel reinforcement bar and a clear cover of 25 mm.

Therefore, d_{available = }201-25- [12/2] = 170 mm.

__Calculation of Load__

Load of wall above the lintel = 0.33 × 0.125 × 25 = 1.031 kN/m.

Self-Weight of lintel = 0.125 × 0.201 × 25 = 0.63 kN/m

Total Load = 1.031 + 0.63 = 1.661 kN/m

__Calculation of Design BM__

Let us use maximum span length = 4.42 m.

Considering partial fixity at ends,

Maximum Bending Moment (BM) = wL^{2}/10

= [1.661× 4.42^{2}]/10

= 3.24 kN-m.

Factored Bending Moment (BM), M_{u }= 1.5 × 3.24 = 4.86 kN-m.

Now, M_{u} / bd^{2 }

= [4.86 x 10^{6}] / [125 x 170^{2 }] = 1.24 N/mm^{2}

Hence single reinforcement is provided.

From table 2 of Design Aid of SP-16.

P_{t }= 0.382 %

Therefore, 100A_{st} / bd= 0.382

A_{st} = [0.382 x bd]/100

A_{st} = [0.382 x 125 x 170]/100

A_{st} = 81.17mm^{2}

= 0.81 cm^{2} approx.

No of rebar = 2 no. s

Let us provide 2- 12 mm dia rebar at tensile reinforcement (226 mm^{2}) and

2-10m dia bars as nominal reinforcement at top layer.

__Check for Shear__

Maximum Shear Force (SF), V= [wL]/2

=[1.661×4.42]/2

= 3.67 kN

Factored Shear Force, V_{u} = 1.5 × 3.67 = 5.505 kN

Again, P_{t }= [100A_{st}]/bd = [100 x 226] / [125 x 170] = 1.06%

Therefore, τ_{c = }0.64 N/mm^{2} (As per Design Aid Table 61 of SP-16)

Induced Shear Stress, τV_{u}/bd

= [5.505 x 1000 ] / [125 x 170] = 0.26 N/mm^{2}

Therefore, τ_{c}>τ_{v}

Hence, safe in shear.

Spacing of shear reinforcement should be least of the following

i. 300 mm

ii. 0.75 d = 0.75×170 = 127.5 mm.

Hence, we use 2-L 6φ @ 150 mm c/c.

Therefore,

We provide section (125 mm×201 mm)

2-12 φ as bottom reinforcement and 2-10 φ as top reinforcement.

Provide 2-L 6φ @ 150 mm c/c shear reinforcement.