stiffener calculation.xls

DESIGN OF STIFFENERS Prepared by : Checked by :

Mahesh .B .J

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

1.6 1.6 1.4

37 x ( 1 - 0.5 ) + 79 x ( 1 -0.5)^2 201

)

ads should not exceed

23

24