Steel Beam Design
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India Pavilion at World Expo 2010, Shangai, China
Design of STEEL BEAM :--
Steel Beam Design
(for uniaxial bending)
Data given : Span of the beam :--
7.5 .m
u.d.l. On the beam :-
36.6 .kN/m
Point load on the beam:-
25 .kN
Yield stress of steel :-Beam type
( udl from slab + L.L + Landscapping load)
308 N/mm2
:-
1 (enter 1 – laterally supported & 2 – laterally unsupported)
End condition of the Column.
2 (enter 1 – hinged condition & 2- for fixed condition) 36.60 kN/m 25.00 kN
0.00 kN
a1 = 0.00 a2= 0.00 L = 7.50 m
figure no . 1 0
figure no . 2
(B.M. Dia. Due to POINT LOAD )
257.34375 ( B.M.Dia. Due to UDL )
Design for Bending Moment :-I) Bending moment due to UDL :-M = W * l 2/ =
=
8 36.6
257.34375
x 7.50 8
2
kN.m
ii) Bending moement due to POINT LOADS :M =
P*l/4 =
46.875
1
India Pavilion at World Expo 2010, Shangai, China
Steel Beam Design
Total Moment :-=
257.34 +
=
304.22 kN.m
46.875
Assume permissible bending comp. Stress in steel =
175 N/mm2
Z required = M / Stress Z required =
1738392.857 .mm3
Go for higher Z . Approximately 20 % of required area. Select section having Z = Select
2086071.429 .mm3
=
2086.1 cm3
ISMB 550
11.2 ISMB 550
All dimensions are in mm 550
19.3
Properties
ISMB 550 :--
Area Ab =
132.11 cm2
Ixxb =
64893.6 cm4
Iyyb =
1833.8 cm4
b
=
19.00 cm
twb
=
1.12 cm
=
1.93 cm
.rxxb =
22.16 cm
tfb
.ryyb =
3.73 cm
D
=
55.00 cm
Maximum compresive and tensile stresses at top and bottom fibers of the beam are :-Actual Bending Stress :-σ c = M yt / Ix
=
304.22 x 64893.60 x =
1000.00 x 10000.0
1000.00 x
275.00000
128.919 N/mm2 2
σ t = M yb / Ix
=
304.22 x 64893.60 x =
1000.00 x 10000.0
1000.00 x
275.00000
128.919 N/mm2
The allowable bending compressive stress has to be computed from the following parameters. dw twb
=
( 55 -
=
45.66
1.93 1.12
x 2 )
2
India Pavilion at World Expo 2010, Shangai, China
T t
=
Steel Beam Design
mean thickness of compression flange web thikness =
tfb twb
=
T t
=
D T
=
=
1.930 1.12
1.930 1.12 1.723
depth of beam = mean th. Of comp. Flange
=
55 1.9
28.497
ry = sqrt ( Iyb / Ab ) =
1833.8 132.11 3.73 cm
= Ly ry
=
750.00 3.73
=
( enter span when no secodary beam else distance of point load)
201.304
The allowable bending compression from IS 800 1984 Table 6.1 A to Table 6.1 f σ bc
=
160 N/mm2
O.K
Check for Shear :-shear force :-- W * l / 2 +
P
=
36.6
x 7.50 2
=
137.25 kN
+
0.00
Shear stress Tv = Vu / D tw =
137.25 550
=
22.28
Maximum Permissible Sheat stress :--
x 1000.00 x 11.20 N/mm2 ( as per IS 800 1984 # 6.4.1 )
Tvm = 0.45 * fy =
0.45 x
308 = 138.60 N/mm2
o.k
3
India Pavilion at World Expo 2010, Shangai, China
Steel Beam Design
Check for Deflection :-a) Due to POINT LOAD ONLY :-Maximum Deflection from Conjugate beam method. Max.def =
B.M at centre of B.M Diagram.
EI * v =
[ 0.00
x 0.00
x 0.50
[0.00
x 0.00
x 0.50 ]
x 1000.00 200000
x 1000.00 x 64893.60
EI * v =
0.000 kN.m3
v
=
0.000
=
0.000 mm
x 1.00 ]
x 1.00
( 0.00
x 0.33)
x 1000.00 x 10000.00
x 1000.00
x
x 3.75
b) Due to UDL :-Maximum Deflection :--
5 * W * L4 384 * EI
=
5 384
=
x 36.60 x 200000
x 7.50 x 64893.60
4
x
x 10000.00
11.618 mm
Total Deflection :--
0.000
Permissible Deflection :--
+
11.618
=
L / 325 =
7.500
=
23.08 mm
x 1000.00 325 O.K
4
11.62 mm
1E+12
Design of STEEL BEAM :--
Steel Beam Design
(for uniaxial bending)
Data given : Span of the beam :--
7.5 .m
u.d.l. On the beam :-
36.6 .kN/m
Point load on the beam:-
25 .kN
Yield stress of steel :-Beam type
( udl from slab + L.L + Landscapping load)
308 N/mm2
:-
1 (enter 1 – laterally supported & 2 – laterally unsupported)
End condition of the Column.
2 (enter 1 – hinged condition & 2- for fixed condition) 36.60 kN/m 25.00 kN
0.00 kN
a1 = 0.00 a2= 0.00 L = 7.50 m
figure no . 1 0
figure no . 2
(B.M. Dia. Due to POINT LOAD )
257.34375 ( B.M.Dia. Due to UDL )
Design for Bending Moment :-I) Bending moment due to UDL :-M = W * l 2/ =
=
8 36.6
257.34375
x 7.50 8
2
kN.m
ii) Bending moement due to POINT LOADS :M =
P*l/4 =
46.875
1
India Pavilion at World Expo 2010, Shangai, China
Steel Beam Design
Total Moment :-=
257.34 +
=
304.22 kN.m
46.875
Assume permissible bending comp. Stress in steel =
175 N/mm2
Z required = M / Stress Z required =
1738392.857 .mm3
Go for higher Z . Approximately 20 % of required area. Select section having Z = Select
2086071.429 .mm3
=
2086.1 cm3
ISMB 550
11.2 ISMB 550
All dimensions are in mm 550
19.3
Properties
ISMB 550 :--
Area Ab =
132.11 cm2
Ixxb =
64893.6 cm4
Iyyb =
1833.8 cm4
b
=
19.00 cm
twb
=
1.12 cm
=
1.93 cm
.rxxb =
22.16 cm
tfb
.ryyb =
3.73 cm
D
=
55.00 cm
Maximum compresive and tensile stresses at top and bottom fibers of the beam are :-Actual Bending Stress :-σ c = M yt / Ix
=
304.22 x 64893.60 x =
1000.00 x 10000.0
1000.00 x
275.00000
128.919 N/mm2 2
σ t = M yb / Ix
=
304.22 x 64893.60 x =
1000.00 x 10000.0
1000.00 x
275.00000
128.919 N/mm2
The allowable bending compressive stress has to be computed from the following parameters. dw twb
=
( 55 -
=
45.66
1.93 1.12
x 2 )
2
India Pavilion at World Expo 2010, Shangai, China
T t
=
Steel Beam Design
mean thickness of compression flange web thikness =
tfb twb
=
T t
=
D T
=
=
1.930 1.12
1.930 1.12 1.723
depth of beam = mean th. Of comp. Flange
=
55 1.9
28.497
ry = sqrt ( Iyb / Ab ) =
1833.8 132.11 3.73 cm
= Ly ry
=
750.00 3.73
=
( enter span when no secodary beam else distance of point load)
201.304
The allowable bending compression from IS 800 1984 Table 6.1 A to Table 6.1 f σ bc
=
160 N/mm2
O.K
Check for Shear :-shear force :-- W * l / 2 +
P
=
36.6
x 7.50 2
=
137.25 kN
+
0.00
Shear stress Tv = Vu / D tw =
137.25 550
=
22.28
Maximum Permissible Sheat stress :--
x 1000.00 x 11.20 N/mm2 ( as per IS 800 1984 # 6.4.1 )
Tvm = 0.45 * fy =
0.45 x
308 = 138.60 N/mm2
o.k
3
India Pavilion at World Expo 2010, Shangai, China
Steel Beam Design
Check for Deflection :-a) Due to POINT LOAD ONLY :-Maximum Deflection from Conjugate beam method. Max.def =
B.M at centre of B.M Diagram.
EI * v =
[ 0.00
x 0.00
x 0.50
[0.00
x 0.00
x 0.50 ]
x 1000.00 200000
x 1000.00 x 64893.60
EI * v =
0.000 kN.m3
v
=
0.000
=
0.000 mm
x 1.00 ]
x 1.00
( 0.00
x 0.33)
x 1000.00 x 10000.00
x 1000.00
x
x 3.75
b) Due to UDL :-Maximum Deflection :--
5 * W * L4 384 * EI
=
5 384
=
x 36.60 x 200000
x 7.50 x 64893.60
4
x
x 10000.00
11.618 mm
Total Deflection :--
0.000
Permissible Deflection :--
+
11.618
=
L / 325 =
7.500
=
23.08 mm
x 1000.00 325 O.K
4
11.62 mm
1E+12
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