Beam 3:
Given:
Span = 4.0 m.
f c ' = 21 MPa
f y = 275 MPa
M EL = 51.53kN.m
β 1 = 0.85
E s = 200000 MPa
Ec = 18500 MPa
Dead Load:
Slab= (depth)(thickness)(w) = (3.25)(0.10)(23.6) = 7.67kN/m
Beam= (area)(unit weight)= (0.3)(0.4)(23.6) = 2.83kN/m
Partition = (3)(0.15)(19.6) = 8.82kN/m
Total Dead Load = 19.32kN/m
2
M DL = 19.32(4) = 38.64 kN.m
7
Live Load = (load per unit area)(tributary width)
Live Load = 1.92(3.25) = 6.24 kN/m
6.24( 4)2
M¿ = = 12.48 kN.m
8
Earthquake Load: Considering 5 percent of total earthquake moment at level two.
M EL = 0.05(51.53) = 2.58 kN.m
For Cross Section dimension, size and number of steel bars:
M u= 0.75[1.4( M DL )+1.7( M ¿)+1.87( M EL )]
M u= 0.75[1.4(38.64)+1.7(2.48)+1.87(2.58)] = 60.10 kN.m
M u= 0.9(DL)+1.43( M EL )
M u= 0.9(38.64)+1.43(2.58)= 38.46 kN.m
Use M u = = 60.10 kN.m
0.85 f c ' β1 600 0.85 ( 21 )( 0.85 ) (600)
Pmax = 0.75 Pb = 0.75( ) = 0.75( ) = 0.0284
(600+ f y ) f y (600+ 275)275
P = 0.5 Pmax = 0.5(0.0284) = 0.0142
fy 275
m = 0.85 f ' = 0.85(21) = 15.406
c
Pm ( 0.0142 ) (15.406)
Ru = P f y ( 1 - ) = 0.0142(275)( 1 – ) = 3.48
2 2
Try b = 250 mm
Mu Mu 179.61(10)6
2
bd = ∅ R , d =
u √ ∅ Rub
=
√ 0.9 ( 3.48 ) (250)
=277.243mm say 300 mm
Mu 60.10 (10)6
Ru = 2 = 2
¿ = 2.97
∅b d 0.9 ( 250 ) 300 ¿
1 2 m Ru 1
P = m (1 – 1−
fy √ ) = 15.406 (1 – 1− 2 15.406 (2.97) ) = 0.0119
√ (
275
)
Pmax = 0.75 Pb= 0.0284
P < Pmax (Compression reinforcement not necessary)
A s = Pbd = 0.0119(250)(300) = 890.96mm2
Try 16 mm dia. Bars
890.96
N = (16)2 = 5pcs.
π
4
2
A s = 5( π (16) ) = 1005.31 mm2
4
As f y 1005.31(275)
a = 0.85 f ' b = = 61.95 mm
c 0.85 ( 21 )( 250)
61.95
−a 0.90 (1005.31 ) ( 275 ) (300− )
M u=∅ A s f y (d ) = 2 =66.94kN.m>51.53kN.m
2
106
SAFE
For Size and Spacing of Stirrups:
1 1
Vc = √ f c ' bd = √ 21(250)(300) = 57282.196 N
6 6
W u = 1.4(19.32) + 1.7(6.24) = 37.656 kN/m
R = 37.656(3.25) = 122.382 kN
Shear Force at Critical Section
V u = 75.312– 0.5(37.656) = 56.484 kN = 56484 N
V c 0.85(57282.196)
∅ = = 24344.933 N
2 2
Vc
V u>∅ provide stirrups
2
Vu 56484
Vs = - V c = 0.85 – 24344.933 =9169.569 N
∅
2
A v = 2 π (10) = 157.08mm2
4
A v f y d ( 157.08 )( 275 )( 300)
S= Vs
= = 1413.273 mm
9169.569
1 1
√ f c ' bd = √ 21(250)(300) = 114564.392 N
3 3
1
V s< √ f ' bd
3 c
d 300
S = 2 = 2 = 150 mm
Use S = 150 mm
Minimum Required Area
bS 250(150)
Av = =¿45.455 mm2 < 157.080 mm2
3 f y = 3(275)
Use 10 mm U Shaped Stirrups Spaced at 225 mm on centers
For Transformed Section:
Gross Moment of Inertia
3 3
I g = b h = 250(300) =893229166.7 mm 4
12 12
f r = 0.7√ f c ' = 0.7√ 21 = 3.208
total depth 350
yt = = 2 = 175 mm
2
f r I g 3.208(893229166.7)
M cr = = =16373161.08N.mm =16373.16kN.m
yt 175
2
A s = 5 π (16) = 1005.310 mm2
4
Es 200000
n = E = 18500 = 10.81
c
b x2
= n A s(d – x)
2
(250)x 2
= (10.81)( 1005.31)(300 – x)
2
125 x 2 = 3260220.33 – 10867.40x
x = 123.78 mm
3
V cr = b x + n A s (d – x)2
3
3
V cr = 250(123.78) + (10.81)(1005.31)(300 – 123.78)2
3
V cr =533879930.5 mm 4
W T L2
Ma = = ¿ ¿ = 51.120 kN.m = 51120000 N.mm
8
Effective Moment of Inertia
3
M cr
IC =
[ ]
Ma g
I + [1 – ¿] I cr
3
I C = 16373161.08 (893229166.7) + [1 – ¿]( 533879930.5 )
[ 51120000 ]
I C = 545686991.1 mm 4
Instantaneous deflection
5W T L4 5 ( 25.56 ) (4000)4
δ= = = 0.781 mm
384 Ec I c 384 ( 200000 ) (545686991.1)
Allowable Deflection
L 4000
δ all = = = 11.111 mm>2.014mm (Safe for Deflection)
360 360
Development Length
π 162
Ldb =
0.02 A b f y
√f c'
=
0.02 ( )
4
(275)
= 241.31mm
√ 21
Ldb = 0.06d b f y = 0.06(16)(275) = 264 mm
Use Ldb = 264 mm
