Lintel Beam Design Example

 

In Lintel beams there are three type of loading.

1) Masonry load (Triangular Loading  concept is used as an angle of 60 degrees develops due to brick masonry bonding but in case of block masonry total load of block wall over the lintel will be considered).

2) Concentrated Load may be point load or may be the uniformly distributed load in case of slab load (Slab loads should be considered in case of bigger sizes of window lintels).

3) Self Weight of Lintel .

Example : Design RCC Lintel beam for window opening size of 6 Ft and brick wall over the Lintel is 6 Ft high. The width of wall is 9 inches and the Ratio of concrete is (1:2:4) having 3000 Psi  compressive strength.

Here in this case the triangular loading concept will be used as  three sided equilateral triangle may be assumed but when brick work is well bonded and extends to each side of the opening a distance equal to at least half the span of the lintel beam. The usual practice is to take a load Uniformly distributed equivalent to the weight of the brick work contained in an equilateral triangle. This takes account of the fact that if the lintel were to fail, only a triangular block (roughly) of brick work would drop out because of the arch effect exerted by the bonding. The area of the equilateral triangle is 0.433 x (Length of side )^2 . It will be assumed that the lintel may appear to be fixed at its supports, the amount of actual fixing is not great ,so that the bending moment may , therefore, be assumed to be Wl/8 .(The actual bending moment for triangular load is Wl/6 but as mentioned above , in case of lintels, it is quite usual to consider the load as being uniformly distributed.

Coming to the example the wall bearing will be considered as width of wall so the side length of equilateral triangle will be 6.75 Ft

Weight of brick work = 0.433 (6.75X6.75) X0.75X 120 = 1775 Pounds .

Assuming size of beam ( width must be equal to width of beam and depth must be equal to number of brick courses).

So Width = 9" and depth may be assumed 1 inch for each ft of span = 6.75 x1         = 6.75 Inches so the size is increasing from two brick courses of 6Inches therefore we must exceed size up to 9".

Now calculating weight of beam = 6.75 x0.75x0.75x150 = 570 Pounds

Total Load = 1775+ 570 = 2345 Pounds.

M= Wl/8

M= Bending moment in pound-inch

l = effective span in ft.

M= 2345 x  6.75 x12 /8 = 23743 Pounds-Inch

Moment of Resistance = 250 . b.d1.d1

we know that Moment of resistance = Bending moment

Putting values in above equation

23743 = 250.(9).(d1.d1)

d1= 3.25"

Adding 1" cover on top and bottom

Overall size = 3.25 + 1+1= 5.25

now here we must consider the depth of brick courses which is 6" for two courses so the overall depth will be 6".

and d1(Effective depth will be 4" instead of 3.25 inches.

Area of Steel in tension zone = Bending moment / Fst.Lever Arm

= 23743/20,000x0.75x4  =0.396 Sq.Inch

No of 1/2 Inch Dia bars = 0.396 / 0.196 =2 Nos

Spacing of Stirrups  = AW.FW.Lever Arm/ Shear Force

Aw= Area of Legs of stirrups

Fw. Permissble tensile strength of steel bars (20000 Psi ) considering factor of safety of 3 times.

Lever arm  in load factor method = 0.75 .d1

Shear Force =  W/2 = 2345 /2 = 1173 Pounds

so trying stirrups of 3/8" dia bars

Spacing = (0.11x2)x 20000x( 0.75x4) x 1/ 1173 = 11.25 Inch  Center / Center

So the results are as under

Width of beam = 9 Inches

Depth of beam  = 6"

2 Nos of 1/2 " dia bars at top and 2 No of supporting bars at top and 3/8" dia bar stirrups @ 11.25 Inch Center to Center.