WHP-East Piatu-Roark Calc & Stiffener-Rev0

1

SIDE WALL (1) DESIGN CALCULATION (@ Length = 3000mm ) TANK NO. :

T-980 / T-985 / T-987 63.0 in 110.24 in 118.11 in

Tank Height, H = Tank Width, W = Tank Length, L = Design Pressure = Design Temp. = Material =

1600 mm 2800 mm 3000 mm

FULL static head Deg C 131 A 240 316L

As per Table 11.4 Case No.1a Chapter 10 of Roark's Rectangular plate, all edges simply supported, with uniform loads over entire plate. g= ρ liq = a= b= a/b = b a g E

9.81 m/s2 1000 kg/m3 29.53 in 23.62 in 1.2500

S a

= =

750 mm 600 mm

b

S

S Loading q = ρ liq gH = 15696 N/m2 = 2.2759 psi = 2.2759 psi

= 0.3954 = 0.0655 = 0.4608 = 2.9E+07 psi

t= c.a = t (corr) =

S

0.2362 in 0.0000 in 0.2362 in

6.0 mm 0 mm 6.0 mm

At Center, Maximum Deflection, = = =

-(aqb4)/Et3 -0.12 0.12 in

t/2 =

0.118 in

Max Deflection < t/2

: O.K

Maximum Bending stress, s =(bqb2)/ t2 =

8,999 psi

<

σ allowable

16,700 psi.

Max Bending stress < σ allowable Material Yield Stress, sy = Stress Ratio, s/sy =

A 240 316L 25000 psi 0.360

At center of long side, Maximum reaction force per unit length normal to the plate surface, R

= = =

g qb 24.77 lb/in 2798.80 N/mm

: OK : O.K

2

SIDE WALL (1) HORIZONTAL STIFFENER CALCULATION TANK NO. :

T-980 / T-985 / T-987

Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam.

W 33.60 lb/in

L= 750 mm = ґ= 375 mm = Load q = 2.2759 psi unit load W = q x ґ psi = 33.60 lb/in

29.53 14.8

X 29.53 in Wb

Wa

Bending Moment As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 14.76 in Maximum moment, Mmax = WL2/8 = 3662 lb-in

Use FB 90 x 6 I/y Therefore,

s

M/I

=

s/y

(I/y)required

= =

M/s 0.146

in3

=

0.494

in3

>

(I/y)required

=

7409

psi

<

σallowable

O.K O.K 16700 psi

Deflection As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x =L/2=

14.76 in

δmax = (5WL4) 384EI =

0.013

< L/360 = 0.0820 in

The stiffener size used is adequate.

O.K

in in

3

SIDE WALL (2) DESIGN CALCULATION (@ Length = 2800mm ) TANK NO. :

T-980 / T-985 / T-987 63.0 in 110.24 in 118.11 in

Tank Height, H = Tank Width, W = Tank Length, L = Design Pressure = Design Temp. = Material =

1600 mm 2800 mm 3000 mm

FULL static head 131 Deg C A 240 316L

As per Table 11.4 Case No.1a Chapter 10 of Roark's Rectangular plate, all edges simply supported, with uniform loads over entire plate. g= ρ liq = a= b= a/b = b a g E

9.81 m/s2 1000 kg/m3 27.56 in 23.62 in 1.1667

S a

= =

700 mm 600 mm

b

S

S Loading q = ρ liq gH = 15696 N/m2 = 2.2759 psi = 2.2759 psi

= 0.3614 = 0.0587 = 0.4492 = 2.9E+07 psi

t= c.a = t (corr) =

S

0.2362 in 0.0000 in 0.2362 in

6.0 mm 0 mm 6.0 mm

At Center, Maximum Deflection, = = =

-(aqb4)/Et3 -0.11 0.11 in

t/2 =

0.118 in

Max Deflection < t/2

: O.K

Maximum Bending stress, s = (bqb2)/ t2 =

8,225 psi

<

σ allowable

16,700 psi

Max Bending stress < σ allowable Material Yield Stress, sy = Stress Ratio, s/sy =

A 240 316L 25000 psi 0.329

At center of long side, Maximum reaction force per unit length normal to the plate surface, R

= = =

g qb 24.15 lb/in 2728.44 N/mm

: OK : O.K

4

SIDE WALL (2) HORIZONTAL STIFFENER CALCULATION TANK NO. :

T-980 / T-985 / T-987

Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam.

W 31.36 lb/in

L= 700 mm = ґ= 350 mm = Load q = 2.2759 psi unit load W = q x ґ psi = 31.36 lb/in

27.56 13.8

X 27.56 in Wb

Wa

Bending Moment As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 13.78 in Maximum moment, Mmax = WL2/8 = 2977 lb-in

Use FB 90 x 6 I/y Therefore,

s

M/I

=

s/y

(I/y)required

= =

M/s 0.119

in3

=

0.494

in3

>

(I/y)required

=

6023

psi

<

σallowable

Deflection As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x =L/2=

13.78 in

δmax = (5WL4) 384EI =

0.009

< L/360 = 0.0766 in

The stiffener size used is adequate.

O.K O.K 16700 psi

O.K

in in

SIDE WALL (1&2) VERTICAL STIFFENER CALCULATION (Top Section) TANK NO. :

5

T-980 / T-985 / T-987 L= ґ=

W

600 300

mm mm

= =

23.62 11.8

in in

26.88 lb/in Load q = unit load W = =

2.2759 qxґ 26.88

psi psi lb/in

X

23.62 in Wb

Wa

Bending Moment As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.548L = 12.94 in Maximum moment, Mmax = 0.0215WL2 = 322 lb-in M/I = s/y (I/y)required = M/s = 0.013 in3 1. Checking Section Modulus (Z) of stiffener : Stiffener size = FB 90 x 6 Section Modulus of stiffener is OK Z

= =

Z stiffener

I/y 0.494

in3

>

2. Checking stiffener Bending stress (s ) : = M/Z s s stiffener

=

Therefore, s stiffener

=

0.013

dmax

=

=

M max / Z stiffener 652

psi

<

12.40 in 0.001309 x WL4 EI

0.0052 in

Z required

Max bending stress of stiffener is OK

16700 psi

Deflection As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.525L =

in3

< L/360)

0.0656

Therefore the size used is adequate.

in

σallowable

6

SIDE WALL (1&2) VERTICAL STIFFENER CALCULATION (Middle & Bottom Section) TANK NO. :

T-980 / T-985 / T-987 L= ґ=

W

500 250

mm mm

= =

19.69 9.8

22.40 lb/in Load q = unit load W = =

2.2759 qxґ 22.40

psi psi lb/in

X

19.69 in Wb

Wa

Bending Moment As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) 10.79 in At x = 0.548L = Maximum moment, Mmax = 0.0215WL2 = 187 lb-in M/I = s/y (I/y)required = M/s = 0.007 in3 1. Checking Section Modulus (Z) of stiffener : Stiffener size = FB 90 x 6 Section Modulus of stiffener is OK Z

= =

Z stiffener

I/y 0.494

in3

>

2. Checking stiffener Bending stress (s ) : = M/Z s s stiffener

=

Therefore, s stiffener

=

0.007

dmax

=

=

M max / Z stiffener 378

psi

<

10.33 in 0.001309 x WL4 EI

0.0014 in

Z required

Max bending stress of stiffener is OK

16700 psi

Deflection As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.525L =

in3

< L/360)

0.0547

Therefore the size used is adequate.

in

σallowable

in in

7

BOTTOM WALL DESIGN CALCULATION TANK NO. :

T-980 / T-985 / T-987 63.0 in 110.24 in 118.11 in

Tank Height, H = Tank Width, W = Tank Length, L = Design Pressure = Design Temp. = Material =

1600 mm 2800 mm 3000 mm

FULL static head Deg C 131 A 240 316L

As per Table 11.4 Case No.1a Chapter 10 of Roark's Rectangular plate, all edges simply supported, with uniform loads over entire plate. g= ρ liq = a= b= a/b = b a g E

9.81 m/s2 1000 kg/m3 29.53 in 27.56 in 1.0714

S a

= =

750 mm 700 mm

b

S

S Loading q = ρ liq gH = 15696 N/m2 = 2.2759 psi = 2.2759 psi

= 0.3191 = 0.0505 = 0.4325 = 2.9E+07 psi

t= c.a = t (corr) =

S

0.2362 in 0.0000 in 0.2362 in

6.0 mm 0 mm 6.0 mm

At Center, Maximum Deflection, = = =

-(aqb4)/Et3 -0.17 0.17 in

t/2 =

0.118 in

Max Deflection < t/2

: O.K

Maximum Bending stress, s =(bqb2)/ t2 =

9,885 psi

<

σ allowable

16,700 psi

: OK

Max Bending stress < σ allowable : O.K Material Yield Stress, sy = Stress Ratio, s/sy =

A 240 316L 25000 psi 0.395

At center of long side, Maximum reaction force per unit length normal to the plate surface, R

= = =

g qb 27.13 lb/in 3065.07 N/mm

8

BOTTOM WALL STIFFENER CALCULATION (1) TANK NO. :

T-980 / T-985 / T-987

Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam.

W 33.60 lb/in

L= 750 mm = ґ= 375 mm = Load q = 2.2759 psi unit load W = q x ґ psi = 33.60 lb/in

29.53 14.8

X 29.53 in Wb

Wa

Bending Moment As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 14.76 in Maximum moment, Mmax = WL2/8 = 3662 lb-in

Use FB 90 x 6 I/y Therefore,

s

M/I

=

s/y

(I/y)required

= =

M/s 0.146

in3

=

0.494

in3

>

(I/y)required

=

7409

psi

<

σallowable

Deflection As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x =L/2=

14.76 in

δmax = (5WL4) 384EI =

0.013

< L/360 = 0.0820 in

The stiffener size used is adequate.

O.K O.K 16700 psi

O.K

in in

BOTTOM WALL STIFFENER CALCULATION (2) TANK NO. :

9

T-980 / T-985 / T-987 L= ґ=

W

700 350

mm mm

= =

27.56 13.8

in in

31.36 lb/in Load q = unit load W = =

2.2759 qxґ 31.36

psi psi lb/in

X

27.56 in Wb

Wa

Bending Moment As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.548L = 15.10 in Maximum moment, Mmax = 0.0215WL2 = 512 lb-in M/I = s/y (I/y)required = M/s = 0.020 in3 1. Checking Section Modulus (Z) of stiffener : Stiffener size = FB 90 x 6 Section Modulus of stiffener is OK Z

= =

Z stiffener

I/y 0.494

in3

>

2. Checking stiffener Bending stress (s ) : = M/Z s s stiffener

=

Therefore, s stiffener

=

0.020

dmax

=

=

M max / Z stiffener 1036

psi

<

14.47 in 0.001309 x WL4 EI

0.0152 in

Z required

Max bending stress of stiffener is OK

16700 psi

Deflection As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.525L =

in3

< L/360)

0.0766

in

Therefore the size used is adequate.

σallowable

10

ROOF WALL DESIGN CALCULATION TANK NO. :

T-980 / T-985 / T-987

Tank Height, H

63.0 in

1600 mm

Tank Width, W

110.24 in

2800 mm

Tank Length, L

118.11 in

3000 mm

Design Pressure Design Temp. Material =

= =

Roof weight =

FULL static head 131 Deg C A 240 316L

872.13

lb

Misc. weight = 2662.17 Live load,LL = 0.00 Total dead load,TDL = 0.27 Total conc. load, CL = 0.00

lb psi psi psi

As per Table 11.4 Case No.1a Chapter 10 of Roark's Rectangular plate, all edges simply supported, with uniform loads over entire plate. S g= ρ liq = a= b= a/b = b a g E

9.81 m/s2 1000 kg/m3

a S

55.12 in 39.37 in 1.4000

1400 mm 1000 mm

S

S

Loading q

= 0.4530 = 0.0770 = 0.4780 = 2.90E+07 psi

t= c.a = t (corr) =

b

= =

Live load + Conc.Load + TotalDeadLoad 0.271 psi

=

0.2362 in 0.0000 in 0.2362 in

6.0 mm 0 mm 6.0 mm

At Center, Maximum Deflection,= = =

-(aqb4)/Et3 -0.13 0.13 in

t/2 =

0.118 in

Max Deflection < t/2

: O.K

Maximum Bending stress, s = (bqb2)/ t2 =

3,416 psi

σallowable < 16,700 psi Max Bending stress < σ allowable

Material A 240 316L 25000 psi Yield Stress, sy = 0.137 Stress Ratio, s/sy = At center of long side, Maximum reaction force per unit length normal to the plate surface, R

= = =

g qb 5.11 lb/in 577.19 N/mm

: OK : O.K

11

ROOF WALL STIFFENER CALCULATION (1) TANK NO. :

T-980 / T-985 / T-987

Maximum bending moment occurs at the point where dM/dx = 0 and shear force is zero, that is, at the middle of the beam.

W 7.48 lb/in

L= ґ= Load q = unit load W = =

1400 700 0.2715 qxґ 7.48

mm = mm = psi psi lb/in

55.12 27.6

X 55.12 in Wb

Wa

Bending Moment As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x = L/2 = 27.56 in Maximum moment, Mmax = WL2/8 = 2841 lb-in

Use FB 90 x 6 I/y Therefore,

s

M/I

=

s/y

(I/y)required

= =

M/s 0.114

in3

=

0.494

in3

>

(I/y)required

=

5747

psi

<

σallowable

Deflection As per Table 8.1 Case 2e of Roark's (Uniform load on entire span) At x =L/2=

27.56 in

δmax = (5WL4) 384EI =

0.035

< L/360 = 0.1531 in

The stiffener size used is adequate.

O.K O.K 16700 psi

O.K

in in

ROOF WALL STIFFENER CALCULATION (2) TANK NO. :

12

T-980 / T-985 / T-987 L= ґ=

W

1000 500

mm mm

= =

39.37 19.7

in in

5.34 lb/in Load q = unit load W = =

0.2715 qxґ 5.34

psi psi lb/in

X

39.37 in Wb

Wa

Bending Moment As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.548L = 21.57 in Maximum moment, Mmax = 0.0215WL2 = 178 lb-in M/I = s/y (I/y)required = M/s = 0.007 in3 1. Checking Section Modulus (Z) of stiffener : Stiffener size = FB 90 x 6 Section Modulus of stiffener is OK Z

= =

Z stiffener

I/y 0.494

in3

>

2. Checking stiffener Bending stress (s ) : = M/Z s s stiffener

=

Therefore, s stiffener

=

0.007

dmax

=

=

M max / Z stiffener 360

psi

<

20.67 in 0.001309 x WL4 EI

0.0221 in

Z required

Max bending stress of stiffener is OK

16700 psi

Deflection As per Table 8.1 Case 2d of Roark's (Uniformly increasing load) At x = 0.525L =

in3

< L/360)

0.1094

The stiffener size used is adequate.

in

σallowable

SECTIONAL STIFFENER PROPERTIES CALCULATION TANK NO NO. :

T 980 / T T-980 T-985 985 / T T-987 987

Stiffener Size M t i l, Material, Yield Stress Stress,

FB 90 x 6 A 240 316L σy

25000 psi

Allowable o ab e Stress, St ess,

σ allowable

16700 6 00 ps psi

Stiffener

13

b1

h

Where :

1

d1

d1 =

90 mm

b1 =

6 mm

y1 1 C

PART

Area (a)

y mm 45 45

1 TOTAL

2

h mm2 0 0

Second Moment of Inertia of Stiffener I = 364500.0 364500 0 mm4

=

0 8757 0.8757

iin4

Section Modulus of Stiffener Z = 8100 mm3

=

0 4943 0.4943

in3

mm 540 540

axy

2

h mm 0 00 0.00 0 00 0.00

2

3

mm 24300 24300

axh mm4 0 0

Calculating Sectional Properties of stiffener : C =

Ay A

C =

45.00

=

24300 540

mm

3

bd /12 mm4 364500 364500

I section mm4 364500 0 364500.0 364500 0 364500.0