1. HORIZONTAL STIFFENER Spacing of horizontal stiffener = Max Internal pressure P = Max. Eff. Span L = Fy = Used Section for Horizontal =
700 725 3 250 ISMC 150 103.9
Section modulus = Zxx =
mm kg/m2 m Mpa
0.00712
N/mm2
cm3
725 x 9.81 x 0.7
UDL acting on beam (Channel) W =
=
4978.58 N-m L = 3000 mm
700 mm
P=
725 kg/m2
Max. Bending moment =
M M
WL2 8 =
4978.575 x 3 x 3 / 8
=
5600.896875 N m
=
5600896.875
N mm
permissible bending stress = Z req =
0.66 Fy
165 N/mm2
BM/σ
Z req = 33944.8295455 mm3 Provided Section ISMC 150 , Zxx = Zxx = Z provided
=
>
1
103.9 103900.00 Z required
cm3 mm3 Hence OK
2. VERTICAL STIFFENER ISA 90 x 90 x 10
P = 90 mm
Section usign as vertical stiffener = Propoerties of section 90 P= mm 90 Q= mm 12 Thickness = t = mm 2.66 cm Cxx = 26.6 mm 147.9 Moment of inertia = Is =
Cxx cm4
1.48E+06 mm4 C/s Area of stiffener = As =
20.19
cm2
2019
mm2
Q = 90 mm ISA 90 x 90 x 10 w = 700 mm
Cp
d = 66.4
CG of plate
NA of whole section
h = 96 Cs CG of stiffener
Distance between stiffeners = W = Thickness of panel = t = Cross Sectional Area of plate width = Ap = Dist.between C.G of panel & Stiffener = d = Total dist between plate & stiffener = h =
700 6
mm mm
4200 66.4
mm2 mm mm
96
The resultant moment of Inertia is calculated usign below formula
A p× t2 A s× A p × d 2 I =I s + + 12 ( A s + A p) therefore, I=
1479000 + ( ( 4200 x 6^2 ) / 12 ) + ( ( 2019 x 4200 x 66.4^2)/( 2019 + 4200) ) 7.50E+06
I=
mm4
7.50E-06 I= m4 Distance from neutral axis of whole section to outer fiber of plate is,
Cp =
[
=
( As × d ) t + ( A s + A p) 2 2019 x 66.4 2019 + 4200
Cp Cp
= =
+
] 6 2
24.5568 mm 0.0246 m
Distance from neutral axis of whole section to outer fiber of plate is,
Cs = =
( h − Cp) ( 96 - 24.5568 )
2
Cs = Cs =
71.4432 mm 0.0714 m
W uniform load
w=
725
kg /m2
M max =
W b L2 8
Vmax =
WbL 2
Spacing of vertical stiffener=
700 mm
0.7 m
M max =
725 x 0.7 x 0.7^2 8
= M max =
31.085
kg-m
304.95
N-m
Stress in panel =
M max x Cp I 304.95 x 0.0246 0.00000750335414182345
=
=
1.00E+06
N/m2
=
1.00
Mpa
Stress in stiffener =
M max x Cs I =
304.95 x 0.0714432 0.00000750335414182345 =
Maximum Stress =
2.90E+06
= 2.90
Mpa
2.90
Mpa
Allowable stress = 0.66 Fy =
Vmax =
N/m2
165 Mpa
WbL 2
3
Hence OK
725 x 0.7 x 0.7 2
=
Vmax = Vmax =
177.63
Kg
1742.50
N
N/mm2
Assume size of the fillet weld =
108 6
Strength of weld =
453.6
N/mm
Allowable stress for Fillet Weld = 0.7 x 6 x 108 =
mm
1742.50
Length of weld =
453.6 4
=
mm
Provide Intermwdiate weld of weld length 100 mm @ spacing 50 mm CHECK FOR SPACING OF STIFFENER The pressure coming on the plate is calculated by usign the formula,
4×F
W= 3×k×B
2
[
1+
y
×t
2
14 20 (1 − k ) + (1 − k ) 75 57
2
]
Uniformly distrubuted load on plate / pressure on the plate = W (N/mm 2) Length of plate = L = mm 700 Breadth of plate = B = mm 700 Thickness of plate = t = mm 6 Yeild stress of plate = Fy = 250 N/mm2
k = k=
(
L 4 L 4+ B
4
)
700^4 ( 700 ^4 + 700^4 )
k=
0.5
W=
( 4 x 250 x 6^ 2 ) 3 x 0.5 x 700 ^2 ( 1 + 14/ 75 x ( 1 - 0.5 ) + 20/57 x ( 1 - 0.5 )^2)
W=
0.0420
N/mm2
W=
4281.35
Kg/m2
This calculated pressure should be greater than given pressure
W
Required
725.0 kg/m2
< <
W
Provided
4281.4 kg/m2
Hence Ok
4
Calculation of Maximum deflection occuring at serviceability
m2−1 5 × k × W × B 4 37 79 d= × × 1+ ( 1− k ) + ( 1− k ) 2 2 3 175 201 m 32 × γ × E × t
[
]
where, d = Maximum deflection occuring at serviceability (mm) Poisson's ratio = m =
Load Factors Loading
3
k=
0.5 700
Breadth of plate = B =
Dead
mm
Uniformly distributed load on plate = W = mm 6 Thckness of plate = t = Hzt + Vert loading Type of Loading = Load factor = γ =
2.05E+05 N/mm2
3 ^2 - 1
X
3^2 d=
Vertical loading Horizontal loading Hzt + Vert loading
1.4
Young's modulus = E = d=
0.0071 N/mm2
2.31
( 5 x 0.5 x 0.00712 x 700^4) ( 32 x 1.4 x 205000 x 6^3 )
x(
1 + 37 x ( 1 - 0.5 ) + 79 x ( 1 -0.5)^2 175
201
mm
Note: Deflection of floor plating under the action of dead & imposed loads should not exceed B/100,where B is the minimum spanning dimension of the plate Minimum spanning dimension of plate = B = therefore deflection , d=
B 100
d=
700
d=
7.00
700 mm
100
d
allowable
7.0 mm
> >
d
mm
actual
2.3 mm
Hence Ok
5
6
Mahesh .B .J
7
A 90 x 90 x 10
t = 6 mm CG of plate
8
9
10
Load Factors Factors 1.4 1.6 1.6 1.4
5 ) + 79 x ( 1 -0.5)^2 201
)
uld not exceed
11
12
DESIGN OF STIFFENERS USING FLAT Prepared by : Checked by : 1. HORIZONTAL STIFFENER Spacing of horizontal stiffener = Max Internal pressure P = Max. Eff. Span L = Fy = Used Section for Horizontal =
700 400 3 250 ISMC 150 103.9
Section modulus = Zxx =
mm kg/m2 m Mpa
0.00393
N/mm2
cm3
400 x 9.81 x 0.7
UDL acting on beam (Channel) W =
=
2746.80 N-m L = 3000 mm
700 mm
P=
400 kg/m2
Max. Bending moment =
WL2 8
M
=
2746.8 x 3 x 3 / 8
M
=
3090.15 N m
=
3090150
permissible bending stress = Z req =
N mm 0.66 Fy
=
165
BM/σ
Z req = 18728.181818 mm3 Provided Section ISMC 150 , Zxx = Zxx = Z provided >
13
103.9 103900.00 Z required
cm3 mm3 Hence OK
2. VERTICAL STIFFENER W = 700 mm
Cp
d = 53
Cs
NA of whole section
h =106
CG of stiffener FLAT 100 x 10
Distance between stiffeners = W = Thickness of panel = t =
700 6
mm mm
4200
mm2
Cross Sectional Area of plate width = Ap = Section used for vertical stiffener =
FLAT 100 x 10
Cross Sectional Area of stiffener = As = Dist.between C.G of panel & Stiffener = d = Total dist between plate & stiffener = h =
1000 53 106
mm mm mm
8.33E+05
mm4
Moment of Inertia of Stiffener = Is =
width = 2
Thickness
The resultant moment of Inertia is calculated usign below formula
A p× t I =I s + 12
2
A s× A p × d
+
2
( A s + A p)
therefore, I=
833333.333333333 + ( ( 4200 x 6^2 ) / 12 ) + ( ( 1000 x 4200 x 53^2)/( 1000 + 4200) ) 3.11E+06
I=
mm4
3.11E-06 I= m4 Distance from neutral axis of whole section to outer fiber of plate is,
Cp =
[ =
Cp Cp
= =
( As ( As
× d) + A p)
+
1000 x 53 1000 + 4200
t 2
] 6 2
+
13.1924 mm 0.0132 m
Distance from neutral axis of whole section to outer fiber of plate is,
Cs =
( h − Cp)
=
( 106 - 13.1924 )
Cs = Cs =
92.8076 mm 0.0928 m 14
W uniform load
w=
400
kg /m2
Spacing = =
M max =
W b L2 8
Vmax =
WbL 2
700 0.7
mm m
400 x 0.7 x 0.7^2
M max =
8 = M max =
17.15
kg-m
168.24
N-m
M max x Cp
Stress in panel =
I 168.2415 x 0.0131924 0.00000311474102564103
=
=
7.13E+05
N/m2
=
0.71
Mpa
Stress in stiffener =
M max x Cs I =
168.2415 x 0.0928076 0.00000311474102564103
=
5.01E+06
= 5.01
Allowable stress = 0.66 Fy =
Vmax =
Mpa
5.01
Maximum Stress =
Mpa
165
WbL 2
15
N/m2
Mpa
Hence OK
=
400 x 0.7 x 0.7 2
Vmax = Vmax =
Kg N
98.00 961.38 108 6 453.6
Allowable stress for Fillet Weld = Assume size of the fillet weld = Strength of weld = 0.7 x 6 x 108 =
N/mm2 mm N/mm
Length of weld =
961.38 453.6 =
2
mm
Provide Intermwdiate weld of weld length 100 mm @ spacing 50 mm
CHECK FOR SPACING OF STIFFENER The pressure coming on the plate is calculated by usign the formula,
4×F
W= 3×k×B
2
[
1+
(
k=
L 4 L 4+ B
4
×t
2
14 20 (1 − k ) + (1 − k ) 75 57
Uniformly distrubuted load on plate / pressure on the Length of plate = L = 700 Breadth of plate = B = 700 Thickness of plate = t = 6 Yeild stress of plate = Fy = 250
k =
y
2
]
plate = W (N/mm 2) mm mm mm N/mm2
)
700^4 ( 700 ^4 + 700^4 )
k=
0.5
W=
( 4 x 250 x 6^ 2 ) 3 x 0.5 x 700 ^2 ( 1 + 14/ 75 x ( 1 - 0.5 ) + 20/57 x ( 1 - 0.5 )^2)
W=
0.0420
N/mm2
W=
4281.35
Kg/m2
This calculated pressure should be greater than given pressure
W
Required
400.0 kg/m2
< <
W
Provided
4281.4 kg/m2
Hence Ok
16
Calculation of Maximum deflection occuring at serviceability 2
4
[
m −1 5 × k × W × B 37 79 d= × × 1+ ( 1− k ) + ( 1− k ) 2 2 3 175 201 m 32 × γ × E × t
]
where, d = Maximum deflection occuring at serviceability (mm) Poisson's ratio = m = 3
k=
0.5 700
Breadth of plate = B =
mm
Uniformly distributed load on plate = W = mm 6 Thckness of plate = t = Hzt + Vert loading Type of Loading = Load factor = γ =
Load Factors Loading Dead
0.0039 N/mm2
Vertical loading Horizontal loading Hzt + Vert loading
1.4
Young's modulus = E = d=
3 ^2 - 1 3^2
d=
1.27
2.05E+05 N/mm2
X
( 5 x 0.5 x 0.00393 x 700^4) ( 32 x 1.4 x 205000 x 6^3 )
x(
1 + 37 x ( 1 - 0.5 ) + 79 x ( 1 -0.5)^2 175
mm
Note: Deflection of floor plating under the action of dead & imposed loads should not exceed B/100,where B is the minimum spanning dimension of the plate Minimum spanning dimension of plate = B = therefore deflection , d= B 100 d= 700
700 mm
100 d=
d
allowable
7.0 mm
> >
d
mm
7.00
actual
1.3 mm
Hence Ok
17
18
Mahesh .B .J
N/mm2
Hence OK
19
t=6
h =106
CG of plate
10
100 FLAT 100 x 10 100 10
20
Hence OK
21
22
Load Factors Loading Factors Dead 1.4 Vertical loading Horizontal loading Hzt + Vert loading
