Diesel Tank Example

7/27/2019 Diesel Tank Example

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PV Elite 2013 SP2 Licensee: ZISHAN ENGINEERS (PVT.) LTD.

FileName : Diesel_tank_Example --------------------------- Page 2

Horizontal Vessel Analysis (Ope.) : Step: 10 11:44p Sep 8,2013

ASME Horizontal Vessel Analysis: Stresses for the Left Saddle

(per ASME Sec. VIII Div. 2 based on the Zick method.)

Horizontal Vessel Stress Calculations : Operating Case

Note: There are more than two saddles present in this model. The load on

the saddle supports has been taken as the total weight divided by the

number of saddles. Insure that this is an appropriate assumption for

this vessel model.

Input and Calculated Values:

Vessel Mean Radius Rm 1504.80 mm

Stiffened Vessel Length per 4.15.6 L 13610.00 mm

Distance from Saddle to Vessel tangent a 96.00 mm

Saddle Width b 350.00 mm

Saddle Bearing Angle theta 120.00 degrees

Wear Plate Width b1 464.00 mm

Wear Plate Bearing Angle theta1 133.00 degrees

Wear Plate Thickness tr 8.0 mm

Wear Plate Allowable Stress Sr 114.46 N/mm

Inside Depth of Head h2 751.60 mm


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Shell Allowable Stress used in Calculation 114.46 N/mm

Head Allowable Stress used in Calculation 114.46 N/mm

Circumferential Efficiency in Plane of Saddle 0.80

Circumferential Efficiency at Mid-Span 0.80

Saddle Force Q, Operating Case 108788.17 N

Horizontal Vessel Analysis Results: Actual Allowable

-------------------------------------------------------------------

Long. Stress at Top of Midspan 12.12 91.57 N/mm

Long. Stress at Bottom of Midspan 27.09 91.57 N/mm

Long. Stress at Top of Saddles 19.48 91.57 N/mm

Long. Stress at Bottom of Saddles 19.72 91.57 N/mm

Tangential Shear in Shell 9.94 91.57 N/mm

Tangential Shear in Head 7.57 91.57 N/mm

Circ. Stress at Horn of Saddle 20.24 143.07 N/mm

Addl. Stress in Head as Stiffener 33.23 143.07 N/mm

Circ. Compressive Stress in Shell 29.92 114.46 N/mm

Intermediate Results: Saddle Reaction Q due to Wind or Seismic

Saddle Reaction Force due to Wind Ft [Fwt]:

= Ftr * ( Ft/Num of Saddles + Z Force Load ) * B / E

= 3.00 * ( 37111.3/4 + 0 ) * 1976.0001/2700.0000

= 20370.0 N


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Saddle Reaction Force due to Wind Fl or Friction [Fwl]:

= max( Fl, Friction Load, Sum of X Forces) * B / Ls

= max( 8762.34 , 0.00 , 0 ) * 1976.0001/8670.0000

= 1997.0 N

Saddle Reaction Force due to Earthquake Fl or Friction [Fsl]:

= max( Fl, Friction Force, Sum of X Forces ) * B / Ls

= max( 76403.10 , 0.00 , 0 ) * 1976.0001/8670.0000

= 17413.2 N

Saddle Reaction Force due to Earthquake Ft [Fst]:


= 3.00 * ( 76403/4 + 0 ) * 1976.0001/2700.0000

= 41936.8 N

Load Combination Results for Q + Wind or Seismic [Q]:

= Saddle Load + Max( Fwl, Fwt, Fsl, Fst )

= 66851 + Max( 1997 , 20369 , 17413 , 41936 )

= 108788.2 N

Summary of Loads at the base of this Saddle:

Vertical Load (including saddle weight) 115062.24 N

Transverse Shear Load Saddle 19100.78 N

Longitudinal Shear Load Saddle 76403.10 N


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Formulas and Substitutions for Horizontal Vessel Analysis:

Note: Wear Plate is not Welded to the Shell, k = 1.0

The Computed K values from Table 4.15.1:

K1 = 0.1066 K2 = 1.1707 K3 = 0.8799 K4 = 0.4011

K5 = 0.7603 K6 = 0.0529 K7 = 0.0132 K8 = 0.3405

K9 = 0.2711 K10 = 0.0581 K1* = 0.1923 K6p = 0.0427

K7P = 0.0107

The suffix 'p' denotes the values for a wear plate if it exists.

Note: Dimension a is less than Rm/2.

Moment per Equation 4.15.3 [M1]:

= -Q*a [1 - (1- a/L + (R-h2)/(2a*L))/(1+(4h2)/3L)]

= -108788*96.00[1-(1-96.00/13610.00+(1504.800-751.600)/

(2*96.00*13610.00))/(1+(4*751.60)/(3*13610.00))]

= 5543.9 N-m


= Q*L/4(1+2(R-h2)/(L))/(1+(4h2)/( 3L))-4a/L

= 108788*13610/4(1+2(1504-751)/(13610))/(1+(4*751)/

(3*13610))-4*95/13610

= 340786.9 N-m

Longitudinal Stress at Top of Shell (4.15.6) [Sigma1]:


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= P * Rm/(2t) - M2/(pi*Rmt)

= 166.733 * 1504.800/(2*6.400 ) - 340786.9/(pi*1504.8*6.400 )

= 12.12 N/mm

Longitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]:

= P * Rm/(2t) + M2/(pi * Rm * t)

= 166.733 * 1504.800/(2 * 6.400 ) + 340786.9/(pi * 1504.8 * 6.400 )

= 27.09 N/mm

Longitudinal Stress at Top of Shell at Support (4.15.8) [Sigma3]:

= P * Rm/(2t) - M1/(pi * Rm * t)

= 166.733 * 1504.800/(2 * 6.400 ) - 5543.9/(pi * 1504.8 * 6.400 )

= 19.48 N/mm

Longitudinal Stress at bottom of Shell at Support (4.15.9) [Sigma4]:

= P * Rm/(2t) + M1/(pi*Rmt)

= 166.733 * 1504.800/(2*6.400 ) + 5543.9/(pi*1504.8*6.400 )

= 19.72 N/mm

Maximum Shear Force in the Saddle (4.15.5) [T]:

= Q(L-2a)/(L+(4*h2/3))

= 108788 ( 13610.00 - 2 * 96.00 )/(13610.00 + ( 4 * 751.60/3))

= 99897.8 N

Shear Stress in the shell no rings, stiffened (4.15.15) [tau3]:

= K3 * Q / ( Rm * t )


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= 0.8799 * 108788/( 1504.7999 * 6.4000 )

= 9.94 N/mm

Shear Stress in the head, shell stiffened (4.15.16) [tau3*]:

= K3 * Q / ( Rm * th )

= 0.8799 * 108788/( 1504.7999 * 8.4000 )

= 7.57 N/mm

Membrane stress in the head as a stiffener (4.15.18) [sigma5]:

= K4 * Q/(Rm*th) + ((P*Ri)/(2*th)) * (Ri/h2)

= 0.4011 * 108788/(1504.80 * 8.400 ) +

((166.73 * 1501.60 )/(2 * 8.400 ))*(1501.60/751.60 )

= 33.23 N/mm

Decay Length (4.15.22) [x1,x2]:

= 0.78 * sqrt( Rm * t )

= 0.78 * sqrt( 1504.800 * 6.400 )

= 76.546 mm

Circumferential Stress in shell, no rings (4.15.23) [sigma6]:

= -K5 * Q * k / ( t * ( b + X1 + X2 ) )

= -0.7603 * 108788 * 1.0/( 6.400 * ( 350.00 + 76.55 + 76.55 ) )

= -25.69 N/mm

Effective reinforcing plate width (4.15.1) [B1]:

= min( b + 1.56 * sqrt( Rm * t ), 2a )

= min( 350.00 + 1.56 * sqrt( 1504.800 * 6.400 ), 2 * 96.000 )


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= 192.00 mm

Wear Plate/Shell Stress ratio (4.15.29) [eta]:

= min( Sr/S, 1 )

= min( 114.457/114.457 , 1 )

= 1.0000

Circumferential Stress at wear plate (4.15.26) [sigma6,r]:

= -K5 * Q * k / ( B1( t + eta * tr ) )

= -0.7603 * 108788 * 1.0/( 192.000 ( 6.400 + 1.000 * 8.000 ) )

= -29.92 N/mm

Circ. Comp. Stress at Horn of Saddle, L>=8Rm (4.15.27) [sigma7,r]:

= -Q/(4(t+eta*tr)b1) - 3*K7*Q/(2(t+eta*tr))

= -108788/(4(6.400 + 1.000 * 8.000 )192.000 ) -

3 * 0.013 * 108788/(2(6.400 + 1.000 * 8.000 ))

= -20.24 N/mm

Results for Vessel Ribs, Web and Base:

Baseplate Length Bplen 2700.0000 mm

Baseplate Thickness Bpthk 24.0000 mm

Baseplate Width Bpwid 400.0000 mm

Number of Ribs ( inc. outside ribs ) Nribs 7

Rib Thickness Ribtk 12.0000 mm

Web Thickness Webtk 8.0000 mm

Web Location Webloc Side


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Moment of Inertia of Saddle - Lateral Direction

Y A AY Io

Shell 3. 3948. 12635. 53908.

Wearplate 10. 3712. 38605. 421287.

Web 235. 3533. 830915. 252841984.

BasePlate 468. 9600. 4492800. 2103089280.

Totals 717. 20793. 5374954. 2356406528.

Value C1 = Sumof(Ay)/Sumof(A) = 258. mm

Value I = Sumof(Io) - C1*Sumof(Ay) = 967001152. mm**4

Value As = Sumof(A) - Ashell = 16845. mm

K1 = (1+Cos(beta)-.5*Sin(beta) )/(pi-beta+Sin(beta)*Cos(beta)) = 0.2035

Fh = K1 * Q = 0.2035 * 108788.172 = 22140.7559 N

Tension Stress, St = ( Fh/As ) = 1.3145 N/mm

Allowed Stress, Sa = 0.6 * Yield Str = 148.9320 N/mm

d = B - R*Sin(theta) / theta = 739.7867 mm

Bending Moment, M = Fh * d = 16386.0762 N-m

Bending Stress, Sb = ( M * C1 / I ) = 4.3789 N/mm

Allowed Stress, Sa = 2/3 * Yield Str = 165.4800 N/mm


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Minimum Thickness of Baseplate per Moss :

= ( 3 * ( Q + Saddle_Wt ) * BasePlateWidth / ( 2 * BasePlateLength *

AllStress ))

= ( 3 * (108788 + 6274 ) * 400.00/( 2 * 2700.000 * 165.480 ))

= 12.431 mm

Calculation of Axial Load, Intermediate Values and Compressive Stress

Effective Baseplate Length [e]:

= ( Bplen - Clearance ) / ( Nribs - 1)

= ( 2700.0000 - 25.4 )/( 7 - 1 ) = 445.7667 mm

Baseplate Pressure Area [Ap]:

= e * Bpwid / 2

= 445.7667 * 400.0000/2 = 89153.3359 mm

Axial Load [P]:

= Ap * Bp

= 89153.3 * 0.10 = 8980.4 N

Area of the Rib and Web [Ar]:

= ( Bpwid - Clearance - Webtk ) * Ribtk + e/2 * Webtk

= ( 400.000 - 25.4 - 8.000 ) * 12.000 + 445.7667/2 * 8.000

= 6182.267 mm

Compressive Stress [Sc]:


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= P/Ar

= 8980.4/6182.2666 = 1.4527 N/mm

Check of Outside Ribs:

Inertia of Saddle, Outer Ribs - Longitudinal Direction

Y A AY Ay Io

Rib 191.3 4251.6 813331.1 13021346.0 47555920.0

Web 4.0 1783.1 7132.3 31048530.0 19019.4

Values 136.0 6034.7 820463.3 44069876.0 47574936.0

Bending Moment [Rm]:

= Fl /( 2 * Bplen ) * e * rl / 2

= 76403.1/( 2 * 2700.00 ) * 445.767 * 1227.60/2

= 3872.824 N-m

KL/R < Cc ( 10.0827 < 126.0993 ) per AISC E2-1

Sca = (1-(Klr)/(2*Cc))*Fy/(5/3+3*(Klr)/(8*Cc)-(Klr)/(8*Cc)

Sca = ( 1-( 10.08 )/(2 * 126.10 )) * 248/

( 5/3+3*(10.08 )/(8* 126.10 )-( 10.08)/(8*126.10)

Sca = 145.84 N/mm

AISC Unity Check on Outside Ribs ( must be


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Check of Inside Ribs

Inertia of Saddle, Inner Ribs - Axial Direction

Y A AY Ay Io

Rib 191.3 4399.2 841566.9 30933996.0 52565752.0

Web 4.0 3566.1 14264.5 38160340.0 19019.4

Values 107.4 7965.3 855831.4 69094336.0 52584772.0

KL/R < Cc ( 6.1123 < 126.0993 ) per AISC E2-1


Sca = ( 1-( 6.11 )/(2 * 126.10 )) * 248/

( 5/3+3*(6.11 )/(8* 126.10 )-( 6.11)/(8*126.10)

Sca = 147.15 N/mm

AISC Unity Check on Inside Ribs ( must be


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Distance from Saddle to Vessel tangent a 230.00 mm

Saddle Width b 350.00 mm

Saddle Bearing Angle theta 120.00 degrees

Wear Plate Width b1 464.00 mm

Wear Plate Bearing Angle theta1 133.00 degrees

Wear Plate Thickness tr 8.0 mm

Wear Plate Allowable Stress Sr 114.46 N/mm

Inside Depth of Head h2 751.60 mm

Shell Allowable Stress used in Calculation 114.46 N/mm

Head Allowable Stress used in Calculation 114.46 N/mm

Circumferential Efficiency in Plane of Saddle 0.80

Circumferential Efficiency at Mid-Span 0.80

Saddle Force Q, Operating Case 108788.17 N

Horizontal Vessel Analysis Results: Actual Allowable

-------------------------------------------------------------------

Long. Stress at Top of Midspan 10.71 91.57 N/mm

Long. Stress at Bottom of Midspan 25.04 91.57 N/mm

Long. Stress at Top of Saddles 17.78 91.57 N/mm

Long. Stress at Bottom of Saddles 17.97 91.57 N/mm

Tangential Shear in Shell 11.90 91.57 N/mm


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Circ. Compressive Stress in Shell 2.57 114.46 N/mm

Stiffener Circ. Stress at Shell 25.71 143.07 N/mm

Stiffener Circ. Stress at Tip 66.63 143.07 N/mm

Intermediate Results: Saddle Reaction Q due to Wind or Seismic

Saddle Reaction Force due to Wind Ft [Fwt]:


= 3.00 * ( 37111.3/4 + 0 ) * 1976.0001/2700.0000

= 20370.0 N

Saddle Reaction Force due to Wind Fl or Friction [Fwl]:

= max( Fl, Friction Load, Sum of X Forces) * B / Ls

= max( 8762.34 , 20055.41 , 0 ) * 1976.0001/8670.0000

= 4570.9 N

Saddle Reaction Force due to Earthquake Fl or Friction [Fsl]:

= max( Fl, Friction Force, Sum of X Forces ) * B / Ls

= max( 76403.10 , 20055.41 , 0 ) * 1976.0001/8670.0000

= 17413.2 N

Saddle Reaction Force due to Earthquake Ft [Fst]:


= 3.00 * ( 76403/4 + 0 ) * 1976.0001/2700.0000

= 41936.8 N


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Load Combination Results for Q + Wind or Seismic [Q]:

= Saddle Load + Max( Fwl, Fwt, Fsl, Fst )

= 66851 + Max( 4570 , 20369 , 17413 , 41936 )

= 108788.2 N

Summary of Loads at the base of this Saddle:

Vertical Load (including saddle weight) 115062.24 N

Transverse Shear Load Saddle 19100.78 N

Longitudinal Shear Load Saddle 76403.10 N

Formulas and Substitutions for Horizontal Vessel Analysis:

Note: Wear Plate is Welded to the Shell, k = 0.1

The Computed K values from Table 4.15.1:

K1 = 0.1066 K2 = 1.1707 K3 = 0.8799 K4 = 0.4011

K5 = 0.7603 K6 = 0.0529 K7 = 0.0132 K8 = 0.3405

K9 = 0.2711 K10 = 0.0581 K1* = 0.1923 K6p = 0.0427

K7P = 0.0107

The suffix 'p' denotes the values for a wear plate if it exists.

Note: Dimension a is less than Rm/2.


= -Q*a [1 - (1- a/L + (R-h2)/(2a*L))/(1+(4h2)/3L)]


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= -108788*230.00[1-(1-230.00/13610.00+(1504.800-751.600)/

(2*230.00*13610.00))/(1+(4*751.60)/(3*13610.00))]

= 4218.4 N-m


= Q*L/4(1+2(R-h2)/(L))/(1+(4h2)/( 3L))-4a/L

= 108788*13610/4(1+2(1504-751)/(13610))/(1+(4*751)/

(3*13610))-4*230/13610

= 326203.3 N-m

Longitudinal Stress at Top of Shell (4.15.6) [Sigma1]:

= P * Rm/(2t) - M2/(pi*Rmt)

= 152.030 * 1504.800/(2*6.400 ) - 326203.3/(pi*1504.8*6.400 )

= 10.71 N/mm

Longitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]:

= P * Rm/(2t) + M2/(pi * Rm * t)

= 152.030 * 1504.800/(2 * 6.400 ) + 326203.3/(pi * 1504.8 * 6.400 )

= 25.04 N/mm

Longitudinal Stress at Top of Shell at Support (4.15.8) [Sigma3]:

= P * Rm/(2t) - M1/(pi * Rm * t)

= 152.030 * 1504.800/(2 * 6.400 ) - 4218.4/(pi * 1504.8 * 6.400 )

= 17.78 N/mm

Longitudinal Stress at bottom of Shell at Support (4.15.9) [Sigma4]:


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= P * Rm/(2t) + M1/(pi*Rmt)

= 152.030 * 1504.800/(2*6.400 ) + 4218.4/(pi*1504.8*6.400 )

= 17.97 N/mm

Maximum Shear Force in the Saddle (4.15.5) [T]:

= Q(L-2a)/(L+(4*h2/3))

= 108788 ( 13610.00 - 2 * 230.00 )/(13610.00 + ( 4 * 751.60/3))

= 97902.5 N

Shear Stress in the shell with double rings (4.15.14) [tau2]:

= K2 * T / ( Rm * t )

= 1.1707 * 97902.52/( 1504.7999 * 6.4000 )

= 11.90 N/mm

Decay Length (4.15.22) [x1,x2]:

= 0.78 * sqrt( Rm * t )

= 0.78 * sqrt( 1504.800 * 6.400 )

= 76.546 mm

Circ. Stress in shell w/rings adjacent to Saddle (4.15.37) [sigma6]:

= -K5 * Q * k/(t(b + 2 * x2))

= -0.760 * 108788 * 0.1/(6.400 (350.000 + 2 * 76.546 ))

= -2.57 N/mm

Note: Double Rings adjacent to Saddle Inside the Shell.

Circ. + Bending Stress in shell, rings adjacent to Saddle (4.15.38) [sigma10]


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= -K9 * Q/A + K10 * Q * Rm * c1/I

= -0.2711*108788/5283.58+0.0581*108788*1504.80*44.53/13533118

= 25.71 N/mm

Circ. + Bending Stress in ring, rings adjacent to Saddle (4.15.39) [sigma11]:

= -K9 * Q/A - K10 * Q * Rm * c2/I

= -0.2711*108788/5283.58-0.0581*108788*1504.80*86.87/26397912

= -66.63 N/mm

Results for Vessel Ribs, Web and Base

Baseplate Length Bplen 2700.0000 mm

Baseplate Thickness Bpthk 24.0000 mm

Baseplate Width Bpwid 400.0000 mm

Number of Ribs ( inc. outside ribs ) Nribs 7

Rib Thickness Ribtk 12.0000 mm

Web Thickness Webtk 8.0000 mm

Web Location Webloc Side

Moment of Inertia of Saddle - Lateral Direction

Y A AY Io

Shell 3. 3948. 12635. 53908.

Wearplate 10. 3712. 38605. 421287.

Web 235. 3533. 830915. 252841984.

BasePlate 468. 9600. 4492800. 2103089280.


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Totals 717. 20793. 5374954. 2356406528.

Value C1 = Sumof(Ay)/Sumof(A) = 258. mm

Value I = Sumof(Io) - C1*Sumof(Ay) = 967001152. mm**4

Value As = Sumof(A) - Ashell = 16845. mm

K1 = (1+Cos(beta)-.5*Sin(beta) )/(pi-beta+Sin(beta)*Cos(beta)) = 0.2035

Fh = K1 * Q = 0.2035 * 108788.172 = 22140.7559 N

Tension Stress, St = ( Fh/As ) = 1.3145 N/mm

Allowed Stress, Sa = 0.6 * Yield Str = 148.9320 N/mm

d = B - R*Sin(theta) / theta = 739.7867 mm

Bending Moment, M = Fh * d = 16386.0762 N-m

Bending Stress, Sb = ( M * C1 / I ) = 4.3789 N/mm

Allowed Stress, Sa = 2/3 * Yield Str = 165.4800 N/mm

Minimum Thickness of Baseplate per Moss :

= ( 3 * ( Q + Saddle_Wt ) * BasePlateWidth / ( 2 * BasePlateLength *

AllStress ))

= ( 3 * (108788 + 6274 ) * 400.00/( 2 * 2700.000 * 165.480 ))

= 12.431 mm

Calculation of Axial Load, Intermediate Values and Compressive Stress


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Effective Baseplate Length [e]:

= ( Bplen - Clearance ) / ( Nribs - 1)

= ( 2700.0000 - 25.4 )/( 7 - 1 ) = 445.7667 mm

Baseplate Pressure Area [Ap]:

= e * Bpwid / 2

= 445.7667 * 400.0000/2 = 89153.3359 mm

Axial Load [P]:

= Ap * Bp

= 89153.3 * 0.10 = 8980.4 N

Area of the Rib and Web [Ar]:

= ( Bpwid - Clearance - Webtk ) * Ribtk + e/2 * Webtk

= ( 400.000 - 25.4 - 8.000 ) * 12.000 + 445.7667/2 * 8.000

= 6182.267 mm

Compressive Stress [Sc]:

= P/Ar

= 8980.4/6182.2666 = 1.4527 N/mm

Check of Outside Ribs:

Inertia of Saddle, Outer Ribs - Longitudinal Direction

Y A AY Ay Io

Rib 191.3 4251.6 813331.1 13021346.0 47555920.0

Web 4.0 1783.1 7132.3 31048530.0 19019.4


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Values 136.0 6034.7 820463.3 44069876.0 47574936.0

Bending Moment [Rm]:

= Fl /( 2 * Bplen ) * e * rl / 2

= 76403.1/( 2 * 2700.00 ) * 445.767 * 1227.60/2

= 3872.824 N-m

KL/R < Cc ( 10.0827 < 126.0993 ) per AISC E2-1


Sca = ( 1-( 10.08 )/(2 * 126.10 )) * 248/

( 5/3+3*(10.08 )/(8* 126.10 )-( 10.08)/(8*126.10)

Sca = 145.84 N/mm

AISC Unity Check on Outside Ribs ( must be


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Sca = ( 1-( 6.11 )/(2 * 126.10 )) * 248/

( 5/3+3*(6.11 )/(8* 126.10 )-( 6.11)/(8*126.10)

Sca = 147.15 N/mm

AISC Unity Check on Inside Ribs ( must be


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Diesel Tank Example

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