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.
