BEAM 13 & 29:
E. Compute the total load to be carried by the
Design Criteria:
beam.
Live Load:
W u =11.4584+ 2.491+51.83
Residential – 4.8 kPa
W u =65.7794 kN /m
f’c = 20.7 Mpa
F. Moment at support
fy = 276 Mpa
γc = 23.56 kN/m3
β=0.85 M u=−W u ¿ ¿
CC =40 mm M u=−65.7794 ¿¿
Transferred load=51.83 kN/m M u=−65.7794 kN ∙ m
Slab thickness: 100 mm Compute for steel ratio:
Diameter of main reinforcement: 16 mm 0.85 f ' cβ 600
ρbal =
b=250 mm f y (600+ f y )
d=1.5b=1.5 ( 250 )=375 0.85(20.7)(0.85) 600
h=375+8+ 40=423 ρbal =
276(600+276)
ρbal =0.03711
A. Weight of slab. (Assuming the beam is
carrying the same Dead load and Live load) ρmax =0.75 ρbal
ρmax =0.75( 0.03711)
DL=( 0.10 ) (23.56)
DL=2.356 kPa ρmax =0.02784
¿=4.8 kPa 1.4
ρmin =
fy
B. Weight of beam. 1.4
ρmin =
W b =γ c bh 276
W b =23.56 ( 0.25 ) (0.423) ρmin =0.005072
W b =2.491 kN /m
ρmax > ρmin *use ρmax
C. Weight of wall
W w =γ c bh G. Computing for the strength ratio ( w)
W w =23.56 ( 3 ) (0.2)
ρmax f y
W w =14.136 kN /m w=
f 'c
D. Compute the factored load. 0.02784(276)
w=
U =1.4 DL+1.7≪¿ 20.7
U =1.4 ( 2.356 ) +1.7( 4.8) w=0.3712
U =11.4584 kN /m2
H. Compute for the value of M umax
M umax=∅ f ' c wb d 2 (1−0.59 w)
M umax=0.9(20.7)(0.3712)( 250)¿ M. Compute for steel ratio ( ρ)
M umax=189.876 kN ∙ m
M u=∅ Rnb d 2
M u < M umax (singly reinforced )
75.176 ×106 =0.9 Rn( 250)(375)2
Rn=2.376 Mpa
I. Compute for steel ratio ( ρ)
M u=∅ Rnb d 2
65.7794 × 106=0.9 Rn(250)(375)2 0.85 f ' c 2 Rn
Rn=2.079 Mpa ρ=
fy ( √1− 1−
0.85 f ' c )
0.85 (20.7) 2(2.376)
ρ=
0.85 f ' c
fy ( √1− 1−
2 Rn
0.85 f ' c ) ρ=
276 ( √
1− 1−
0.85(20.7) )
ρ=0.00928
0.85 (20.7) 2(2.079)
ρ=
276 ( √
1− 1−
0.85(20.7) )
ρ=0.00804 N. Computing for required steel area ( A s)
J. Computing for required steel area ( A s)
A s=ρbd
A s=0.00928 ( 250 ) ( 375 )
A s=ρbd
A s=0.00804 ( 250 ) ( 375 ) A s=870.454 mm 2
A s=753.707 mm2 O. Computing for required number of
reinforcing steel bars
K. Computing for required number of
reinforcing steel bars As
N=
A¯¿ ¿
As
N= 870.454
A¯¿ ¿ N=
π (162)
753.707
N= 2 4
π (16 )
N=4.329 pcs .*say 5 pcs.
4
N=3.749 pcs .*say 4 pcs. Use 5-16 mm∅ RSB for tension bars
Use 4-16 mm∅ RSB for tension bars P. Moment at end support
M u=−U ¿ ¿
L. Moment at midspan
M u=−65.7794 ¿¿
M u=U ¿ ¿
M u=−105.247 kN ∙ m
M u=65.7794 ¿ ¿
M u=75.176 kN ∙ m
M u < M umax (singly reinforced )
M u < M umax (singly reinforced )
Q. Compute for steel ratio ( ρ)
M u=∅ Rnb d 2 U. For allowable shearing stress.
105.247 ×106 =0.9 Rn( 250)(375)2
Rn=3.326 Mpa V a =0.09 √ f ' c
V a =0.09 √20.7
0.85 f ' c 2 Rn V a =0.409 Mpa <υ *not ok need for
ρ=
fy ( √1− 1−
0.85 f ' c ) stirrups
0.85 (20.7) 2(3.326)
ρ=
276 ( √
1− 1−
0.85(20.7) ) v' =υ−V a
v' =1.4−0.409
ρ=0.0135 v' =0.991 Mpa
R. Computing for required steel area ( A s)
A s=ρbd
V. Spacing of stirrups *use 10 mm∅ RSB for
A s=0.0135 ( 250 ) ( 375 ) stirrups
A s=1263.263 mm2
π D2
S. Computing for required number of Ab =
4
reinforcing steel bars
π 102
Ab =
4
As
N= Ab =78.54 mm2
A¯¿ ¿
1263.263
N= 2
A v =2 A b
π (16 )
A v =2(78.54)
4
N=6.283 pcs .*say 7 pcs. A v =157.08 mm2
Use 7-16 mm∅ RSB for tension bars Av f y
S=
v' b
T. Check for shear. (157.08)(276)
S=
0.991(250)
V S=174.991mm∗say say 170 mm
υ=
bd
W u La d 375
V= −Wd Smax = = =187.5 mm∗say 180 mm
2 2 2
65.7794 (4.0)
V= −65.7794 (0.375)
2 *adopt 170 mm O.C. spacing of bars
V =106.891 kN
106,891
υ=
(250)(375)
υ=1.4 Mpa
