stiffener calculation.xls

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1

DESIGN OF STIFFENERS

Prepared by : Mahesh .B .J

Checked by :

Spacing of horizontal stiffener = 700 mm

Max Internal pressure P = 725 0.00712

Max. Eff. Span L = 3 m

Fy = 250 Mpa

Used Section for Horizontal = ISMC 150

Section modulus = Zxx = 103.9

UDL acting on beam (Channel) W = 725 x 9.81 x 0.7

= 4978.58 N-m

L = 3000 mm

700 mm

P = 725 kg/m2

Max. Bending moment =

8

M = 4978.575 x 3 x 3 / 8

M = 5600.896875 N m

= 5600896.875 N mm

permissible bending stress = 0.66 Fy = 165 N/mm2

Z req =

BM/σ

Z req = 33944.8295455 mm3

Provided Section ISMC 150 , Zxx = 103.9 cm3 Zxx = 103900.00 mm3

Z provided > Z required Hence OK

1. HORIZONTAL STIFFENER

kg/m2 _N/mm2

cm3

(2)

2 Section usign as vertical stiffener = ISA 90 x 90 x 10 Propoerties of section P = 9 0 m m P = 90 mm Q = 90 mm Thickness = t = 12 mm Cxx = 2.66 cm 26.6 mm Cxx Moment of inertia = Is = 147.9 1.48E+06 Q = 90 mm C/s Area of stiffener = As = 20.19 2019 ISA 90 x 90 x 10 w = 700 mm d = 66.4 h = 96

Distance between stiffeners = W = 700 mm

Thickness of panel = t = 6 mm

Cross Sectional Area of plate width = Ap = 4200

Dist.between C.G of panel & Stiffener = d = 66.4 mm Total dist between plate & stiffener = h = 96 mm The resultant moment of Inertia is calculated usign below formula

therefore,

I = 1479000 + ( ( 4200 x 6^2 ) / 12 ) + ( ( 2019 x 4200 x 66.4^2)/( 2019 + 4200) )

I = 7.50E+06

I = 7.50E-06

Distance from neutral axis of whole section to outer fiber of plate is,

= 2019 x 66.4 + 6

2019 + 4200 2

Cp = 24.5568 mm

Cp = 0.0246 m

= ( 96 - 24.5568 ) 2. VERTICAL STIFFENER cm4 mm4 cm2 mm2 mm2 mm4 m4 NA of whole section Cs Cp CG of stiffener CG of plate

I =I

_s

+

A

p

×

t

2

12 +

A

_s

×

A

_p

×

d

2 (

A

_s

+

A

_p

₎

Cp =

[

(

A

s

×

d

)

(

A

_s

+

A

_p

₎

+

t

2 ]

C

_s

=

₍

h − C

_p

₎

(3)

3 Cs = 71.4432 mm Cs = 0.0714 m M max = 8

W uniform load Vmax = W b L

2

w = 725 kg /m2 Spacing of vertical stiffener= 700 mm 0.7 m M max = 725 x 0.7 x 0.7^2 8 = 31.085 kg-m M max = 304.95 N-m Stress in panel = 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 = 2.90E+06 N/m2

= 2.90 Mpa

Maximum Stress = 2.90 Mpa

Allowable stress = 0.66 Fy = 165 Mpa Hence OK Vmax = W b L

2

W b L2

(4)

4 = 725 x 0.7 x 0.7 2 Vmax = 177.63 Kg Vmax = 1742.50 N

Allowable stress for Fillet Weld = 108

Assume size of the fillet weld = 6 mm Strength of weld = 0.7 x 6 x 108 = 453.6 N/mm

Length of weld = 1742.50 453.6

= 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,

Length of plate = L = 700 mm

Breadth of plate = B = 700 mm

Thickness of plate = t = 6 mm

Yeild stress of plate = Fy = 250

k = 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 W = 4281.35

This calculated pressure should be greater than given pressure

<

725.0 kg/m2

<

4281.4 kg/m2 Hence Ok

N/mm2

Uniformly distrubuted load on plate / pressure on the plate = W (N/mm2₎

N/mm2 N/mm2 Kg/m2

W

_Required

W

_Provided

W =

4 × F

y

×

t

2

3 × k × B

2

[

_{1 +}

14

75 (

1 − k

)

+

20

57 (

1 − k

)

2

]

k =

(

L

4

L

4

₊

_B

4

)

(5)

5

Calculation of Maximum deflection occuring at serviceability

where,

d = Maximum deflection occuring at serviceability (mm)

Poisson's ratio = m = 3 Load Factors

k = 0.5 Loading

Breadth of plate = B = 700 mm Dead

Uniformly distributed load on plate = W = 0.0071 Vertical loading

Thckness of plate = t = 6 mm Horizontal loading

Type of Loading = Hzt + Vert loading Hzt + Vert loading

1.4 Young's modulus = E = 2.05E+05

d = 3 ^2 - 1

X

( 5 x 0.5 x 0.00712 x 700^4)

x (

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

3^2 ( 32 x 1.4 x 205000 x 6^3 ) 175 201

d = 2.31 mm

Minimum spanning dimension of plate = B = 700 mm therefore deflection , d = B 100 d = 700 100 d = 7.00 mm

>

7.0 mm

>

2.3 mm

Hence Ok N/mm2 Load factor = γ = N/mm2

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

d

_allowable

d

_actual

d =

m

2

−

1 m

2

×

5 × k × W × B

4

32 × γ × E × t

3

×

[

1 +

37

175 (

1 − k

)

+

79

201 (

1 − k

)

2

]

(6)

(7)

7 DESIGN OF STIFFENERS

(8)

8 ISA 90 x 90 x 10

t = 6 mm CG of plate

(9)

(10)

10 CHECK FOR SPACING OF STIFFENER

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11 Calculation of Maximum deflection occuring at serviceability

Load Factors Factors 1.4 1.6 1.6 1.4 1 + 37 x ( 1 - 0.5 ) + 79 x ( 1 -0.5)^2

)

201

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(13)

13

DESIGN OF STIFFENERS USING FLAT

Prepared by : Checked by :

Spacing of horizontal stiffener = 700 mm

Max Internal pressure P = 400 0.00393

Max. Eff. Span L = 3 m

Fy = 250 Mpa

Used Section for Horizontal = ISMC 150

Section modulus = Zxx = 103.9 UDL acting on beam (Channel) W = 400 x 9.81 x 0.7

= 2746.80 N-m

L = 3000 mm

700 mm

P = 400 kg/m2

Max. Bending moment =

8

M = 2746.8 x 3 x 3 / 8

M = 3090.15 N m

= 3090150 N mm

permissible bending stress = 0.66 Fy = 165 Z req =

BM/σ

Z req = 18728.181818 mm3

Provided Section ISMC 150 , Zxx = 103.9 cm3 Zxx = 103900.00 mm3

Z provided > Z required Hence OK 1. HORIZONTAL STIFFENER

kg/m2 _N/mm2

cm3

(14)

14 W = 700 mm d = 53 h =106 FLAT 100 x 10

Distance between stiffeners = W = 700 mm

Thickness of panel = t = 6 mm

Cross Sectional Area of plate width = Ap = 4200

Section used for vertical stiffener = FLAT 100 x 10 width =

Cross Sectional Area of stiffener = As = 1000 Thickness

Dist.between C.G of panel & Stiffener = d = 53 mm

Total dist between plate & stiffener = h = 106 mm

Moment of Inertia of Stiffener = Is = 8.33E+05

The resultant moment of Inertia is calculated usign below formula

therefore,

I = 833333.333333333 + ( ( 4200 x 6^2 ) / 12 ) + ( ( 1000 x 4200 x 53^2)/( 1000 + 4200) )

I = 3.11E+06

I = 3.11E-06

= 1000 x 53 + 6 1000 + 4200 2

Cp = 13.1924 mm

Cp = 0.0132 m

= ( 106 - 13.1924 ) Cs = 92.8076 mm Cs = 0.0928 m 2. VERTICAL STIFFENER mm2 mm2 mm4 mm4 m4 Cs Cp NA of whole section CG of stiffener

I=I

_s

+

A

p

×

t

2

12 +

A

_s

×

A

_p

×

d

2 (

A

_s

+

A

_p

₎

Cp =

[

(

A

s

×

d

)

(

A

_s

+

A

_p

)

+

t

2

]

C

_s

=

₍

h − C

_p

₎

(15)

15

M max =

8 W uniform load Vmax = W b L

2 w = 400 kg /m2 Spacing = 700 mm = 0.7 m M max = 400 x 0.7 x 0.7^2 8 = 17.15 kg-m M max = 168.24 N-m 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 N/m2 = 5.01 Mpa

Maximum Stress = 5.01 Mpa

Allowable stress = 0.66 Fy = 165 Mpa Hence OK

Vmax = W b L 2

W b L2

(16)

16 = 400 x 0.7 x 0.7 2 Vmax = 98.00 Kg Vmax = 961.38 N

Allowable stress for Fillet Weld = 108

Assume size of the fillet weld = 6 mm

Strength of weld = 0.7 x 6 x 108 = 453.6 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,

Length of plate = L = 700 mm

Breadth of plate = B = 700 mm Thickness of plate = t = 6 mm Yeild stress of plate = Fy = 250

k = 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 W = 4281.35

This calculated pressure should be greater than given pressure

<

400.0 kg/m2

<

4281.4 kg/m2 Hence Ok N/mm2

Uniformly distrubuted load on plate / pressure on the plate = W (N/mm2₎

N/mm2 N/mm2 Kg/m2

W

_Required

W

_Provided

W =

4 × F

y

×

t

2

3 × k × B

2

[

1 +

14

75 (

1 − k

)

+

20

57 (

1 − k

)

2

]

k =

(

L

4

L

4

₊

_B

4

)

(17)

17

Calculation of Maximum deflection occuring at serviceability

where,

d = Maximum deflection occuring at serviceability (mm)

Poisson's ratio = m = 3 Load Factors

k = 0.5 Loading

Breadth of plate = B = 700 mm Dead

Uniformly distributed load on plate = W = 0.0039 Vertical loading

Thckness of plate = t = 6 mm Horizontal loading

Type of Loading = Hzt + Vert loading Hzt + Vert loading

1.4 Young's modulus = E = 2.05E+05

d = 3 ^2 - 1

X

( 5 x 0.5 x 0.00393 x 700^4)

x (

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

3^2 ( 32 x 1.4 x 205000 x 6^3 ) 175

d = 1.27 mm

Minimum spanning dimension of plate = B = 700 mm therefore deflection , d = B 100 d = 700 100 d = 7.00 mm

>

7.0 mm

_>

1.3 mm

_{Hence Ok} N/mm2 Load factor = γ

=

N/mm2

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

d

_allowable

d

_actual

d =

m

2

−

1 m

2

×

5 × k × W × B

4

32 × γ × E × t

3

×

[

1 +

37

175 (

1 − k

)

+

79

201 (

1 − k

)

2

]

(18)

(19)

19 DESIGN OF STIFFENERS USING FLAT

Mahesh .B .J

N/mm2

(20)

20 t = 6 h =106 10 100 FLAT 100 x 10 100 10 CG of plate

(21)

21 Hence OK

(22)

22 CHECK FOR SPACING OF STIFFENER

(23)

23 Calculation of Maximum deflection occuring at serviceability

Load Factors Loading Factors Dead 1.4 Vertical loading 1.6 Horizontal loading 1.6 Hzt + Vert loading 1.4 1 + 37 x ( 1 - 0.5 ) + 79 x ( 1 -0.5)^2

)

201

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