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F P C L
A 3 1 0.5 1.5
B 0 0 0 0
C 0 0 0 0
D 0 0 0 0
E 0 0 0 0
F 0 0 0 0
G 0 0 0 0
F P C L
A 3 2 0.5 1.5
B 0 0 0 0
C 0 0 0 0
D 0 0 0 0
E 0 0 0 0
F 0 0 0 0
G 0 0 0 0
F P C L
A 3 3 0.5 1.5
B 0 0 0 0
C 0 0 0 0
D 0 0 0 0
E 0 0 0 0
F 0 0 0 0
G 0 0 0 0
F P C L
A 3 4 0.5 1.5
B 0 0 0 0
C 0 0 0 0
D 0 0 0 0
E 0 0 0 0
F 0 0 0 0
G 0 0 0 0
F P C L
LEV - 1 LOADING
LEV - 2 LOADING
LEV - 3 LOADING
LEV - 4 LOADING
LEV - 5 LOADING
A 3 5 0.5 1.5
B 0 0 0 0
C 0 0 0 0
D 0 0 0 0
E 0 0 0 0
F 0 0 0 0
G 0 0 0 0
F P C L
A 3 6 0.5 1.5
B 0 0 0 0
C 0 0 0 0
D 0 0 0 0
E 0 0 0 0
F 0 0 0 0
G 0 0 0 0
LEV - 6 LOADING
Page of
AUMS
f'c = N/mm2
Fy = N/mm2
Es = N/mm2
L = m
D = mm
b = mm
c = mm
d = mm
t = mm
h = m
ts = mm
Lc = mm
WD = kN/m
WL = kN/m
= kN/m
= kN/m
= kN/m 25 kN/m3
= kN/m
= kN/m
Pu1 Pu2 Pu3 Pu4 Pu5
0 0 0 0 0
a = 0 0 0 0 0
b = 6.9 6.9 6.9 6.9 6.9
Mu = Kn-m
r =
= mm2
= mm2
1.58 f 25 bars
3 f 25 bars = 2
3 f 20 bars
= mm2 > = mm
2
OK
Required
Use
Provided Ast 2414 Required Ast 776
# of Layers of steel
190.81
0.0034
Ast 776
Ast min 705.375
Wall load 0
Total ult Load Per meter Run 32.06
Point load
Moment :
Load Per meter Run :
From Slab 24.86
S/W of beam 7.2 gconcrete =
Total Dead Load 17.47
Live load 2.44
Total ult (DL+LL) 24.86
Wall Height 0.0
Slab Thickness 250
Clear distance to next beam 2000
Cover 40
Effective Depth 712.5
Block Wall over Beam 250
Span 6.9
Depth 800
Width 300
Concrete 32
Steel 420
Elasticity of Conc. 26587.2
Subject : Design of Simply Supported Beam Date :
Beam Identity : TRB-1 b/w Grid D2.1 & D3
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
P
a b
L
w
2*'**85.0*9.0
*211*
'*85.0
dbf
M
F
f
c
ult
y
c
reqr
Page of
AUMS
Subject : Design of Simply Supported Beam Date :
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Vu = kN
Vc = kN
Vus = kN
= 78.5 mm2 mm
2 f 10 @ S = 200
= 875 mm
= 1800 mm
= 1150 mm
= 875 mm
Neutral axis from bottom of beam Yb = 531.1 mm
Neutral axis from top of beam Yt = 268.9 mm
Ie =
Ie =
and Mcr = , fr = 3.51 Mpa
Ig = bd3/12 = mm
4= 54282 in4
Mcr = N-mm
Ma = N-mm 3
Icr =
b * x2
/ 2 +bf*ts*(x-ts/2)- n Ast(d-x) = 0,
n = = 7.52
b / 2 X2 + X - n * Ast * d+ bf*ts
2/2 = 0
150 X2 + X - = 0
X = mm
Icr = mm4
Ie = mm4
= in4
dcr = mm
L/240 = mm > mmAllowable Deflection 28.75 1.27
[ACI 318M-05 TABLE 9.5(b)] OK
154.85
6.0E+09
Effective Moment of inertia 2.3E+10 54282.19
Total immediate deflection 1.27
From top of the beam
where Es/Ec
Quadratic equation n * Ast + bf *ts
236908.16 40281441.59
2.95E+08
Mom due to service Loads 127207969
Cracked Moment of inertia b x3/3 + n Ast (d-x)
2
Location of the Nutral Axis
where Effective moment of Inertia Given By
fr * Ig / Yt 0.62 * √fc' =
Gross Moment of Inertia 2.26E+10
L- Action of Beams for Effective Flange Width
L /12 + bw
6 * ts + bw
( 1/2 * Lc ) + bw
Effective flange width bf
(Mcr/Ma)3*Ig+[1-(Mcr/Ma)
3]*Icr
Area of one leg SRequired = min reinf
No of legs .
Deflection :
Shear :
110.62
205.56 [ACI 318M-05 11.3.1.1]
-58
Page of
AUMS
Subject : Design of Simply Supported Beam Date :
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
dsus = mm
l∆ =
r =
= Reinforcement in middle top of the Beam
2 f 16 = Ast' = mm2
r =
x = 2.0
l = 1.83
= mm
= = mm
= 3.49 mm < = L/240 = 28.75 mm
2 f 16
1 f 10 / 200
3 f 25
300
Summary :
80
0
Total Deflection d+ldsus 3.49
d+ldsus Allowable
OK
Provided 402
0.00188
For five Years
Therefore Additional Long Term Deflection = l*dsus 2.22
Deflection due to sustain loads
1.21
For Additional Long Term deflection , Immediate deflection is Multiplied by a factor l given by
x / ( 1+ 50 * r' ) [ACI 318M-05 9.5.2.5]
Ast'/bd
Long Term Deflection
Page of
AUMS
f'c = N/mm2
Fy = N/mm2
Es = N/mm2
L = m
D = mm
b = mm
c = mm
d = mm
t = mm
h = m
ts = mm
Lc = mm
WD = kN/m
WL = kN/m
= kN/m
= kN/m
= kN/m 25 kN/m3
= kN/m
= kN/m
Pu1 Pu2 Pu3 Pu4 Pu5
95 125 0 0 0
a = 2.2 4.8 0 0 0
b = 6.8 4.2 9.0 9.0 9.0
Mu = Kn-m
r =
= mm2
= mm2
7.08 f 25 bars
8 f 25 bars = 2
8 f 25 bars
= mm2 > = mm
2
OK
Required
Use
Provided Ast 7850 Required Ast 3473
# of Layers of steel
896.65
0.0104
Ast 3473
Ast min 1039.5
Wall load 0
Total ult Load Per meter Run 51.50
Point load
Moment :
Load Per meter Run :
From Slab 40.70
S/W of beam 10.8 gconcrete =
Total Dead Load 28.98
Live load 3.70
Total ult (DL+LL) 40.70
Wall Height 0.0
Slab Thickness 300
Clear distance to next beam 2400
Cover 65
Effective Depth 787.5
Block Wall over Beam 250
Span 9.0
Depth 900
Width 400
Concrete 32
Steel 420
Elasticity of Conc. 26587.2
Subject : Design of Simply Supported Beam Date :
Beam Identity : TRB-3 on Grid B5
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
P
a b
L
w
2*'**85.0*9.0
*211*
'*85.0
dbf
M
F
f
c
ult
y
c
reqr
Page of
AUMS
Subject : Design of Simply Supported Beam Date :
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Vu = kN
Vc = kN
Vus = kN
= 78.5 mm2 mm
2 f 10 @ S = 200
= 1150 mm
= 2200 mm
= 1400 mm
= 1150 mm
Neutral axis from bottom of beam Yb = 596.8 mm
Neutral axis from top of beam Yt = 303.2 mm
Ie =
Ie =
and Mcr = , fr = 3.51 Mpa
Ig = bd3/12 = mm
4= 1E+05 in4
Mcr = N-mm
Ma = N-mm 3
Icr =
b * x2
/ 2 +bf*ts*(x-ts/2)- n Ast(d-x) = 0,
n = = 7.52
b / 2 X2 + X - n * Ast * d+ bf*ts
2/2 = 0
200 X2 + X - = 0
X = mm
Icr = mm4
Ie = mm4
= in4
dcr = mm
L/240 = mm > mmAllowable Deflection 37.50 6.37
[ACI 318M-05 TABLE 9.5(b)] OK
219.35
2.0E+10
Effective Moment of inertia 3.3E+10 79459.30
Total immediate deflection 6.37
From top of the beam
where Es/Ec
Quadratic equation n * Ast + bf *ts
404050.94 98252614.18
4.94E+08
Mom due to service Loads 597768889
Cracked Moment of inertia b x3/3 + n Ast (d-x)
2
Location of the Nutral Axis
where Effective moment of Inertia Given By
fr * Ig / Yt 0.62 * √fc' =
Gross Moment of Inertia 4.27E+10
L- Action of Beams for Effective Flange Width
L /12 + bw
6 * ts + bw
( 1/2 * Lc ) + bw
Effective flange width bf
(Mcr/Ma)3*Ig+[1-(Mcr/Ma)
3]*Icr
Area of one leg SRequired = 289
No of legs .
Deflection :
Shear :
361.86
302.92 [ACI 318M-05 11.3.1.1]
180
Page of
AUMS
Subject : Design of Simply Supported Beam Date :
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
dsus = mm
l∆ =
r =
= Reinforcement in middle top of the Beam
4 f 20 = Ast' = mm2
r =
x = 2.0
l = 1.67
= mm
= = mm
= 16.70 mm < = L/240 = 37.50 mm
4 f 20
1 f 10 / 200
8 f 25
400
Summary :
90
0
Total Deflection d+ldsus 16.70
d+ldsus Allowable
OK
Provided 1257
0.00399
For five Years
Therefore Additional Long Term Deflection = l*dsus 10.33
Deflection due to sustain loads
6.19
For Additional Long Term deflection , Immediate deflection is Multiplied by a factor l given by
x / ( 1+ 50 * r' ) [ACI 318M-05 9.5.2.5]
Ast'/bd
Long Term Deflection
Page of
AUMS
f'c = N/mm2
Fy = N/mm2
Es = N/mm2
L = m
D = mm
b = mm
c = mm
d = mm
t = mm
h = m
ts = mm
Lc = mm
WD = kN/m
WL = kN/m
= kN/m
= kN/m
= kN/m 25 kN/m3
= kN/m
= kN/m
Pu1 Pu2 Pu3 Pu4 Pu5
0 0 0 0 0
a = 0 0 0 0 0
b = 10.0 10.0 10.0 10.0 10.0
Mu = Kn-m
r =
= mm2
= mm2
7.41 f 25 bars
8 f 25 bars = 2
8 f 25 bars
= mm2 > = mm
2
OK
Required
Use
Provided Ast 7850 Required Ast 3636
# of Layers of steel
935.00
0.0109
Ast 3636
Ast min 1039.5
Wall load 0
Total ult Load Per meter Run 74.80
Point load
Moment :
Load Per meter Run :
From Slab 64.00
S/W of beam 10.8 gconcrete =
Total Dead Load 46.67
Live load 5.00
Total ult (DL+LL) 64.00
Wall Height 0.0
Slab Thickness 300
Clear distance to next beam 9350
Cover 65
Effective Depth 787.5
Block Wall over Beam 250
Span 10.0
Depth 900
Width 400
Concrete 32
Steel 420
Elasticity of Conc. 26587.2
Subject : Design of Simply Supported Beam Date :
Beam Identity : TRB-3 on Grid B5
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
P
a b
L
w
2*'**85.0*9.0
*211*
'*85.0
dbf
M
F
f
c
ult
y
c
reqr
Page of
AUMS
Subject : Design of Simply Supported Beam Date :
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Vu = kN
Vc = kN
Vus = kN
= 78.5 mm2 mm
2 f 10 @ S = 200
= 1233 mm
= 2200 mm
= 4875 mm
= 1233 mm
Neutral axis from bottom of beam Yb = 602.1 mm
Neutral axis from top of beam Yt = 297.9 mm
Ie =
Ie =
and Mcr = , fr = 3.51 Mpa
Ig = bd3/12 = mm
4= 1E+05 in4
Mcr = N-mm
Ma = N-mm 3
Icr =
b * x2
/ 2 +bf*ts*(x-ts/2)- n Ast(d-x) = 0,
n = = 7.52
b / 2 X2 + X - n * Ast * d+ bf*ts
2/2 = 0
200 X2 + X - = 0
X = mm
Icr = mm4
Ie = mm4
= in4
dcr = mm
L/240 = mm > mmAllowable Deflection 41.67 8.92
[ACI 318M-05 TABLE 9.5(b)] OK
215.99
2.1E+10
Effective Moment of inertia 3.3E+10 80013.97
Total immediate deflection 8.92
From top of the beam
where Es/Ec
Quadratic equation n * Ast + bf *ts
429050.94 102002614.2
5.12E+08
Mom due to service Loads 623333333
Cracked Moment of inertia b x3/3 + n Ast (d-x)
2
Location of the Nutral Axis
where Effective moment of Inertia Given By
fr * Ig / Yt 0.62 * √fc' =
Gross Moment of Inertia 4.35E+10
L- Action of Beams for Effective Flange Width
L /12 + bw
6 * ts + bw
( 1/2 * Lc ) + bw
Effective flange width bf
(Mcr/Ma)3*Ig+[1-(Mcr/Ma)
3]*Icr
Area of one leg SRequired = 265
No of legs .
Deflection :
Shear :
374.00
302.92 [ACI 318M-05 11.3.1.1]
196
Page of
AUMS
Subject : Design of Simply Supported Beam Date :
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
dsus = mm
l∆ =
r =
= Reinforcement in middle top of the Beam
4 f 20 = Ast' = mm2
r =
x = 2.0
l = 1.67
= mm
= = mm
= 23.18 mm < = L/240 = 41.67 mm
4 f 20
1 f 10 / 200
8 f 25
400
Summary :
90
0
Total Deflection d+ldsus 23.18
d+ldsus Allowable
OK
Provided 1257
0.00399
For five Years
Therefore Additional Long Term Deflection = l*dsus 14.26
Deflection due to sustain loads
8.55
For Additional Long Term deflection , Immediate deflection is Multiplied by a factor l given by
x / ( 1+ 50 * r' ) [ACI 318M-05 9.5.2.5]
Ast'/bd
Long Term Deflection
Page of
AUMS
f'c = N/mm2
Fy = N/mm2
Es = N/mm2
L = m
D = mm
b = mm
c = mm
d = mm
t = mm
h = m
ts = mm
Lc = mm
WD = kN/m
WL = kN/m
= kN/m
= kN/m
= kN/m 25 kN/m3
= kN/m
= kN/m
Pu1 Pu2 Pu3 Pu4 Pu5
0 0 0 0 0
a = 0 0 0 0 0
b = 11.2 11.2 11.2 11.2 11.2
Mu = Kn-m
r =
= mm2
= mm2
9.40 f 25 bars
8 f 25 bars = 2
8 f 25 bars
= mm2 > = mm
2
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Subject : Design of Simply Supported Beam Date :
Beam Identity : TRB-3 b/w Grid C2 & C3
Concrete 32
Steel 420
Elasticity of Conc. 26587.2
Span 11.2
Depth 900
Width 400
Cover 65
Effective Depth 787.5
Block Wall over Beam 250
Wall Height 0.0
Slab Thickness 250
Clear distance to next beam 3000
Total Dead Load 45.83
Live load 5.00
Total ult (DL+LL) 63.00
Load Per meter Run :
From Slab 63.00
S/W of beam 10.8 gconcrete =
Wall load 0
Total ult Load Per meter Run 73.80
Point load
Moment :
1157.18
0.0138
Ast 4614
Ast min 1039.5
Required
Use
Provided Ast 7850 Required Ast 4614
# of Layers of steel
OK
P
a b
L
w
2*'**85.0*9.0
*211*
'*85.0
dbf
M
F
f
c
ult
y
c
reqr
Page of
AUMS
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Subject : Design of Simply Supported Beam Date :
Vu = kN
Vc = kN
Vus = kN
= 78.5 mm2 mm
2 f 10 @ S = 200
= 1333 mm
= 1900 mm
= 1700 mm
= 1333 mm
Neutral axis from bottom of beam Yb = 606.3 mm
Neutral axis from top of beam Yt = 293.8 mm
Ie =
Ie =
and Mcr = , fr = 3.51 Mpa
Ig = bd3/12 = mm
4= 1E+05 in4
Mcr = N-mm
Ma = N-mm 3
Icr =
b * x2
/ 2 +bf*ts*(x-ts/2)- n Ast(d-x) = 0,
n = = 7.52
b / 2 X2 + X - n * Ast * d+ bf*ts
2/2 = 0
200 X2 + X - = 0
X = mm
Icr = mm4
Ie = mm4
= in4
dcr = mm
L/240 = mm > mm
Shear :
413.28
302.92 [ACI 318M-05 11.3.1.1]
248
Area of one leg SRequired = 209
No of legs .
Deflection :
L- Action of Beams for Effective Flange Width
L /12 + bw
6 * ts + bw
( 1/2 * Lc ) + bw
Effective flange width bf
(Mcr/Ma)3*Ig+[1-(Mcr/Ma)
3]*Icr
where Effective moment of Inertia Given By
fr * Ig / Yt 0.62 * √fc' =
Gross Moment of Inertia 4.43E+10
5.29E+08
Mom due to service Loads 771456000
Cracked Moment of inertia b x3/3 + n Ast (d-x)
2
Location of the Nutral Axis From top of the beam
where Es/Ec
Quadratic equation n * Ast + bf *ts
392384.27 88169280.85
203.58
2.1E+10
Effective Moment of inertia 2.9E+10 68950.49
Total immediate deflection 16.07
Allowable Deflection 46.67 16.07
[ACI 318M-05 TABLE 9.5(b)] OK
Page of
AUMS
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Subject : Design of Simply Supported Beam Date :
dsus = mm
l∆ =
r =
= Reinforcement in middle top of the Beam
4 f 20 = Ast' = mm2
r =
x = 2.0
l = 1.67
= mm
= = mm
= 41.74 mm < = L/240 = 46.67 mm
4 f 20
1 f 10 / 200
8 f 25
400
Long Term Deflection
Deflection due to sustain loads
15.39
For Additional Long Term deflection , Immediate deflection is Multiplied by a factor l given by
x / ( 1+ 50 * r' ) [ACI 318M-05 9.5.2.5]
Ast'/bd
Provided 1257
0.00399
For five Years
Therefore Additional Long Term Deflection = l*dsus 25.67
Summary :
90
0
Total Deflection d+ldsus 41.74
d+ldsus Allowable
OK
Page of
AUMS
f'c = N/mm2
Fy = N/mm2
Es = N/mm2
L = m
D = mm
b = mm
c = mm
d = mm
t = mm
h = m
ts = mm
Lc = mm
WD = kN/m
WL = kN/m
= kN/m
= kN/m
= kN/m 25 kN/m3
= kN/m
= kN/m
Pu1 Pu2 Pu3 Pu4 Pu5
137 0 0 0 0
a = 2.3 0 0 0 0
b = 6.7 9.0 9.0 9.0 9.0
Mu = Kn-m
r =
= mm2
= mm2
10.34 f 25 bars
8 f 25 bars = 2
8 f 25 bars
= mm2 > = mm
2
OK
Required
Use
Provided Ast 7850 Required Ast 5074
# of Layers of steel
1257.33
0.0152
Ast 5074
Ast min 1039.5
Wall load 0
Total ult Load Per meter Run 108.62
Point load
Moment :
Load Per meter Run :
From Slab 97.82
S/W of beam 10.8 gconcrete =
Total Dead Load 71.78
Live load 7.30
Total ult (DL+LL) 97.82
Wall Height 0.0
Slab Thickness 250
Clear distance to next beam 6000
Cover 65
Effective Depth 787.5
Block Wall over Beam 250
Span 9.0
Depth 900
Width 400
Concrete 32
Steel 420
Elasticity of Conc. 26587.2
Subject : Design of Simply Supported Beam Date :
Beam Identity : TRB-3 on Grid D4
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
P
a b
L
w
2*'**85.0*9.0
*211*
'*85.0
dbf
M
F
f
c
ult
y
c
reqr
Page of
AUMS
Subject : Design of Simply Supported Beam Date :
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Vu = kN
Vc = kN
Vus = kN
= 78.5 mm2 mm
2 f 10 @ S = 200
= 1150 mm
= 1900 mm
= 3200 mm
= 1150 mm
Neutral axis from bottom of beam Yb = 594.3 mm
Neutral axis from top of beam Yt = 305.7 mm
Ie =
Ie =
and Mcr = , fr = 3.51 Mpa
Ig = bd3/12 = mm
4= 1E+05 in4
Mcr = N-mm
Ma = N-mm 3
Icr =
b * x2
/ 2 +bf*ts*(x-ts/2)- n Ast(d-x) = 0,
n = = 7.52
b / 2 X2 + X - n * Ast * d+ bf*ts
2/2 = 0
200 X2 + X - = 0
X = mm
Icr = mm4
Ie = mm4
= in4
dcr = mm
L/240 = mm > mmAllowable Deflection 37.50 12.79
[ACI 318M-05 TABLE 9.5(b)] OK
211.96
2.1E+10
Effective Moment of inertia 2.5E+10 60511.18
Total immediate deflection 12.79
From top of the beam
where Es/Ec
Quadratic equation n * Ast + bf *ts
346550.94 82440114.18
4.90E+08
Mom due to service Loads 838218333
Cracked Moment of inertia b x3/3 + n Ast (d-x)
2
Location of the Nutral Axis
where Effective moment of Inertia Given By
fr * Ig / Yt 0.62 * √fc' =
Gross Moment of Inertia 4.27E+10
L- Action of Beams for Effective Flange Width
L /12 + bw
6 * ts + bw
( 1/2 * Lc ) + bw
Effective flange width bf
(Mcr/Ma)3*Ig+[1-(Mcr/Ma)
3]*Icr
Area of one leg SRequired = 107
No of legs Check
Deflection :
Shear :
590.78
302.92 [ACI 318M-05 11.3.1.1]
485
Page of
AUMS
Subject : Design of Simply Supported Beam Date :
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
dsus = mm
l∆ =
r =
= Reinforcement in middle top of the Beam
4 f 20 = Ast' = mm2
r =
x = 2.0
l = 1.67
= mm
= = mm
= 33.34 mm < = L/240 = 37.50 mm
4 f 20
1 f 10 / 200
8 f 25
400
Summary :
90
0
Total Deflection d+ldsus 33.34
d+ldsus Allowable
OK
Provided 1257
0.00399
For five Years
Therefore Additional Long Term Deflection = l*dsus 20.55
Deflection due to sustain loads
12.32
For Additional Long Term deflection , Immediate deflection is Multiplied by a factor l given by
x / ( 1+ 50 * r' ) [ACI 318M-05 9.5.2.5]
Ast'/bd
Long Term Deflection
Page of
AUMS
f'c = N/mm2
Fy = N/mm2
Es = N/mm2
L = m
D = mm
b = mm
c = mm
d = mm
t = mm
h = m
ts = mm
Lc = mm
WD = kN/m
WL = kN/m
= kN/m
= kN/m
= kN/m 25 kN/m3
= kN/m
= kN/m
Pu1 Pu2 Pu3 Pu4 Pu5
0 0 0 0 0
a = 0 0 0 0 0
b = 12.0 12.0 12.0 12.0 12.0
Mu = Kn-m
r =
= mm2
= mm2
12.86 f 32 bars
10 f 32 bars = 2
10 f 32 bars
= mm2 > = mm
2
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Subject : Design of Simply Supported Beam Date :
Beam Identity : TRB-4 on Grid B2 & B3
Concrete 32
Steel 420
Elasticity of Conc. 26587.2
Span 12.0
Depth 1200
Width 600
Cover 65
Effective Depth 1078.5
Block Wall over Beam 250
Wall Height 0.0
Slab Thickness 250
Total Dead Load 130.33
8000Clear distance to next beam
Live load 11.50
Total ult (DL+LL) 174.80
Load Per meter Run :
From Slab 174.80
S/W of beam 21.6 gconcrete =
Wall load 0
Total ult Load Per meter Run 196.40
Point load
Moment :
3535.20
0.0152
Ast 10337
Ast min 2135.43
Required
Use
Provided Ast 16077 Required Ast 10337
# of Layers of steel
OK
P
a b
L
w
2*'**85.0*9.0
*211*
'*85.0
dbf
M
F
f
c
ult
y
c
reqr
Page of
AUMS
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Subject : Design of Simply Supported Beam Date :
Vu = kN
Vc = kN
Vus = kN
= 113.1 mm2 mm
4 f 12 @ S = 200
= 1600 mm
= 2100 mm
= 4300 mm
= 1600 mm
Neutral axis from bottom of beam Yb = 769.6 mm
Neutral axis from top of beam Yt = 430.4 mm
Ie =
Ie =
and Mcr = , fr = 3.51 Mpa
Ig = bd3/12 = mm
4= 4E+05 in4
Mcr = N-mm
Ma = N-mm 3
Icr =
b * x2
/ 2 +bf*ts*(x-ts/2)- n Ast(d-x) = 0,
n = = 7.52
b / 2 X2 + X - n * Ast * d+ bf*ts
2/2 = 0
300 X2 + X - = 0
X = mm
Icr = mm4
Ie = mm4
= in4
dcr = mm
L/240 = mm > mm
Shear :
1178.40
622.29 [ACI 318M-05 11.3.1.1]
949
Area of one leg SRequired = 216
No of legs .
Deflection :
L- Action of Beams for Effective Flange Width
L /12 + bw
6 * ts + bw
( 1/2 * Lc ) + bw
Effective flange width bf
(Mcr/Ma)3*Ig+[1-(Mcr/Ma)
3]*Icr
Cracked Moment of inertia b x3/3 + n Ast (d-x)
2
Location of the Nutral Axis
where Effective moment of Inertia Given By
fr * Ig / Yt 0.62 * √fc' =
Gross Moment of Inertia 1.47E+11
From top of the beam
where Es/Ec
Quadratic equation n * Ast + bf *ts
520936.32 180429823.6
18.45
[ACI 318M-05 TABLE 9.5(b)] OK
295.93
7.9E+10
Effective Moment of inertia 8.8E+10 211398.42
Total immediate deflection 18.45
1.19E+09
Allowable Deflection 50.00
Mom due to service Loads 2356800000
Page of
AUMS
GULF CONSULTDesign by : MUDDASSIR
Date : 24-May-2014
Project : Checked by : AZEEM
Subject : Design of Simply Supported Beam Date :
dsus = mm
l∆ =
r =
= Reinforcement in middle top of the Beam
6 f 20 = Ast' = mm2
r =
x = 2.0
l = 1.75
= mm
= = mm
= 49.49 mm < = L/240 = 50.00 mm
6 f 20
2 f 12 / 200
10 f 32
600
Long Term Deflection
31.04
Deflection due to sustain loads
17.78
For Additional Long Term deflection , Immediate deflection is Multiplied by a factor l given by
x / ( 1+ 50 * r' ) [ACI 318M-05 9.5.2.5]
Ast'/bd
49.49
d+ldsus Allowable
OK
Provided 1885
0.00291
For five Years
Therefore Additional Long Term Deflection = l*dsus
Summary :
12
00
Total Deflection d+ldsus
