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DESIGN OF DEEP BEAM Beam Type
Dimensions & Parameters:
Width of Beam (bw) Depth of Beam (h)
8 78
Tot Total Spa Span of Beam Beam Clea Clearr Span Span of Beam Beam
4.50 4.500 0 4.50 4.5000 00
2 inches inches ft ft
Check whether the beam is Deep Beam or Simple Beam h/ln > 0.4 for continous span Clause 10.7 ACI 318-99
h/ln > 0.8 for simply supported span
h/ln
=
1 .4 4 4 4 4 4 4 4 4
>
0.4
It is a DEEP BEAM
Loadings:
1) Self weight of beam =
0.65
k/ft
a) Partition Load =
0
k/ft
b) Load from Slab =
0 .3 2 5
k/ft
0 .3 2 5
k/ft
2) Dead Load :
3) Live Load : Load from Slab =
Total Total load load (w ) = 1.4D 1.4D + 1.7L 1.7L =
2
Mu = w.l /8 =
1.91 1.9175 75
k/ft k/ft
4.8537
k.ft
Since As = Mu Ф.y.fy For y :
ln/h = For ln/h = 0.6 , For ln/h = 0.8 , For ln/h = 1.0 , For ln/h = 2.0 , For ln/h =3.0 ,
0.6923 y = 0.2d y = 0.4d y = 0.6d y = 0.8d y=d
#VALUE!
for ln/h = y = y =
0 .6 9 2 3 #VALUE! 2 2 .1 9 1
inches
fy f' c Ф
60 3 0.9
As(actual =
ksi ksi
2
0.0486050
in /ft
70.2
inches
For Min.Flexural Reinforcement :
Assume d = 0.9.h = For f'c <4500 psi: Asmin = (200/fy).bw.d Asmin =
1.872
2
in
Criteria for choosing As(min) or As : If As(actual) is less than Asmin then i ncrease As(actual) by 33 % i.e Multiply As(actual) by 1.33 and then compare As(actual) with As(min) and then Select the minimum As from both As(min) and As(actual) i.e As(Actual) x 1.33 =
0.06464
Therefore
4
Choose
5
As(actual)
bars
=
2
<
in
controls
1.2265625
2 bars on each face distributed within h/5 = of the tension zone of the beam
Spacing of bars =
15.6 inches 5 i.e.
2
3.12 ''
5 bars @
0 .0 6 4 6 4
2
in
15.6 inches
or
4 '' c/c
4 ''
on each face
fy f' c Ф
60 3 0.9
As(actual =
ksi ksi
2
0.0486050
in /ft
70.2
inches
For Min.Flexural Reinforcement :
Assume d = 0.9.h = For f'c <4500 psi: Asmin = (200/fy).bw.d Asmin =
1.872
2
in
Criteria for choosing As(min) or As : If As(actual) is less than Asmin then i ncrease As(actual) by 33 % i.e Multiply As(actual) by 1.33 and then compare As(actual) with As(min) and then Select the minimum As from both As(min) and As(actual) i.e As(Actual) x 1.33 =
0.06464
Therefore
4
Choose
5
As(actual)
bars
=
2
<
in
controls
1.2265625
2 bars on each face distributed within h/5 = of the tension zone of the beam
Spacing of bars =
15.6 inches 5 i.e.
2
3.12 ''
5 bars @
0 .0 6 4 6 4
2
in
15.6 inches
or
4 '' c/c
4 ''
on each face
Design for Shear : (1) Vu @ distance x = 0.15ln from the face of support : 0.15ln =
8.1 ''
or
0.68 ft
Vu = w. w.l/2 - w.(0.15ln) =
3.0201
Mu = w.l/2.(0.15ln) - w.(0.15ln).(0.15ln/2) Mu/ Vu.d =
<
d=
5.85 ft
kips
=
2.47537
kft
0 .1 4 0 1 1
For Vc : Check whether 3.5 - 2.5.(Mu/Vu.d) < 2.5 or > 2.5 i.e. (3.5 - 2.5.(Mu/Vu.d)
=
3.150
Use
Recalculate Mu/Vu.d:
2.5
2.5 = 3.5 -2.5.(Mu/ -2.5.(Mu/Vu.d) Vu.d) i.e. Mu/Vu.d =
