Suppose we have a beam whose span is 30 Ft , and a Point Load "F"=10000 Pounds, which includes weight of beam and imposed load.
In this case we are going to design a
RCC Beam using Load Factor Theory in following steps.
1-Find out Support or Reactions
Left and Right Support = F/2= 5000 Pounds
In given symmetrical loading case
both reactions are equal
2-Find out Max shear force = Load /2
= 10000/2 =5000 Pounds
Shear force at support = +5000 Pounds
Shear force at Mid = +5000 + (-10000) = -5000 Pounds
3-Max Bending Moment at Center of Beam = Reaction
x L/2
Max MB= 5000x (30/2)x12= 900000 Pound-Inch--------1
4-Moment of Resistance = K.b.d1^2----------------EQ.2
5-b (width of beam ) = (Span x12)/30
=12"
Value of "K" = 250 in Load Factor Method using (1000-20000)
Concrete
Where fc =1000 PSI and Fs = 20000 PSI
We know that Moment of Resistance
=Bending Moment so comparing both
equations EQ----1 AND EQ-----2
6- 900000 = 250x12xd1^2
d1 (Effective Depth) = 17.33 inch Say
17.5"
7-
Area of Steel in Tension zone = BM /(FstxLever Arm)
Ast =900000/(20000x0.75/17.5)= 3.42 Sq.in
No of bars according to 3/4" dia bars = Ast/area of one bar =8 nos.
8-Area of steel in compression zone =(B.M - M.r)/fst.Lever
Arm
=((900000-(250x12x17.5^2))/20000x
0.75x17.5
=(900000-918750)/20000x0.75x17.5=
- (0.07)
Use (4+4) nos
of 3/4" dia bars and 2 no of 1/2" dia bar at top sides.
Use 3/8" dia bar stirrups 2-Leg @ 10 "
c/c
Minus sign shows that the M.r value is greater than the applied value of B.m so no need of Asc (Area of steel in compression zone ), Only two steel bars of 1/2" dia bars are enough to hold stirrups.
Use (4+4) nos of 3/4" dia bars
and 2 no of 1/2" dia bar at top sides.
9- Spacing of stirrups = Aw.Fw.Lever
arm/(Shear Force)
=(0.22x18000x0.75x17.5)
/5000 = 10" C/c
Here Aw = Area of 2 leg stirrup bars
Fw = Permissible value of Tensile stress of stirrups
Overall depth of beam = Effective
depth+ 2 concrete covers
concrete cover = 1"
Concrete ratio =1:2:4
Factor of safety in Fc = 3000/1000=3
Times (considering compressive strength of concrete)
The factor of safety in Fst= 60000/20000=3 Times
(considering Tensile strength of Steel).
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