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PV Elite 2014 SP1 Licensee: SPLM Licensed User FileName : AKL-new- Earthquake Load Calculation : Step: 10 8:56amApr 15,2014 Earthquake Analysis Results The UBC Zone Factor for the Vessel is ............. 0.0000 The Importance Factor as Specified by the User is . 1.000 The UBC Frequency and Soil Factor (C) is ......... 2.750 The UBC Force Factor as Specified by the User is .. 3.000 The UBC Total Weight (W) for the Vessel is ........ 11811.8 kgf The UBC Total Shear (V) for the Vessel is ......... 0.0 kgf The UBC Top Shear (Ft) for the Vessel is .......... 0.0 kgf Earthquake Load Calculation | | Earthquake | Earthquake | Element | From| To | Height | Weight | Ope Load | | | cm | kgf | kgf | ------------------------------------------------- 10| 20| 45.0000 | 787.453 | ... | 20| 30| 45.0000 | 787.453 | ... | 30| 40| 45.0000 | 787.453 | ... | 40| 50| 45.0000 | 787.453 | ... | 50| 60| 45.0000 | 787.453 | ... | 60| 70| 76.9415 | 787.453 | ... | 70|Sadl| 76.9415 | 787.453 | ... | Sadl| 80| 76.9415 | 787.453 | ... | 70| 80| 76.9415 | 787.453 | ... | 80| 90| 76.9415 | 787.453 | ... | 90| 100| 76.9415 | 787.453 | ... | 100| 110| 45.0000 | 787.453 | ... | 110| 120| 45.0000 | 787.453 | ... | 120| 130| 45.0000 | 787.453 | ... | 130| 140| 45.0000 | 787.453 | ... | PV Elite is a trademark of Intergraph CADWorx & Analysis Solutions, Inc. 2014 1

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PAGE 1PV Elite 2014 SP1 Licensee: SPLM Licensed User

FileName : AKL-new-

Earthquake Load Calculation : Step: 10 8:56amApr 15,2014

TC "Earthquake Load Calculation :" /f C Earthquake Analysis Results The UBC Zone Factor for the Vessel is ............. 0.0000

The Importance Factor as Specified by the User is . 1.000

The UBC Frequency and Soil Factor (C) is ......... 2.750

The UBC Force Factor as Specified by the User is .. 3.000

The UBC Total Weight (W) for the Vessel is ........ 11811.8 kgf

The UBC Total Shear (V) for the Vessel is ......... 0.0 kgf

The UBC Top Shear (Ft) for the Vessel is .......... 0.0 kgf

Earthquake Load Calculation | | Earthquake | Earthquake | Element |

From| To | Height | Weight | Ope Load |

| | cm | kgf | kgf |

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

10| 20| 45.0000 | 787.453 | ... |

20| 30| 45.0000 | 787.453 | ... |

30| 40| 45.0000 | 787.453 | ... |

40| 50| 45.0000 | 787.453 | ... |

50| 60| 45.0000 | 787.453 | ... |

60| 70| 76.9415 | 787.453 | ... |

70|Sadl| 76.9415 | 787.453 | ... |

Sadl| 80| 76.9415 | 787.453 | ... |

70| 80| 76.9415 | 787.453 | ... |

80| 90| 76.9415 | 787.453 | ... |

90| 100| 76.9415 | 787.453 | ... |

100| 110| 45.0000 | 787.453 | ... |

110| 120| 45.0000 | 787.453 | ... |

120| 130| 45.0000 | 787.453 | ... |

130| 140| 45.0000 | 787.453 | ... |

PV Elite is a trademark of Intergraph CADWorx & Analysis Solutions, Inc. 2014 TC "Center of Gravity Calculation :" /f C Shop/Field Installation Options :Note : The CG is computed from the first Element From Node Center of Gravity of Saddles 357.700 cm

Center of Gravity of Nozzles 250.613 cm

Center of Gravity of Tubesheet(s) 357.318 cm

Center of Gravity of Tubes 357.318 cm

Center of Gravity of Bare Shell New and Cold 355.857 cm

Center of Gravity of Bare Shell Corroded 356.026 cm

Vessel CG in the Operating Condition 353.558 cm

Vessel CG in the Fabricated (Shop/Empty) Condition 353.661 cm

Vessel CG in the Test Condition 354.614 cm

PV Elite is a trademark of Intergraph CADWorx & Analysis Solutions, Inc. 2014 TC "Horizontal Vessel Analysis (Ope.) :" /f C 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 CaseNote: Wear Pad Width (225.00) is less than 1.56*sqrt(rm*t) and less than 2a. The wear plate will be ignored.Minimum Wear Plate Width to be considered in analysis [b1]: = min( b + 1.56*sqrt( Rm * t ), 2a )

= min( 180.000 + 1.56*sqrt( 777.2650 * 9.7000 ), 2 * 436.000 )

= 315.4550 mm

Input and Calculated Values: Vessel Mean Radius Rm 777.27 mm

Stiffened Vessel Length per 4.15.6 L 705.40 cm

Distance from Saddle to Vessel tangent a 436.00 mm

Saddle Width b 180.00 mm

Saddle Bearing Angle theta 124.00 degrees

Shell Allowable Stress used in Calculation 1406.14 kgf/cm

Head Allowable Stress used in Calculation 1406.14 kgf/cm

Circumferential Efficiency in Plane of Saddle 1.00

Circumferential Efficiency at Mid-Span 1.00

Saddle Force Q, Operating Case 6692.29 kgf

Horizontal Vessel Analysis Results: Actual Allowable -------------------------------------------------------------------

Long. Stress at Top of Midspan 491.07 1406.14 kgf/cm

Long. Stress at Bottom of Midspan 590.69 1406.14 kgf/cm

Long. Stress at Top of Saddles 535.78 1406.14 kgf/cm

Long. Stress at Bottom of Saddles 543.72 1406.14 kgf/cm

Tangential Shear in Shell 86.16 1124.91 kgf/cm

Circ. Stress at Horn of Saddle 235.23 1757.68 kgf/cm

Circ. Compressive Stress in Shell 16.31 1406.14 kgf/cm

Intermediate Results: Saddle Reaction Q due to Wind or SeismicSaddle Reaction Force due to Wind Ft [Fwt]: = Ftr * ( Ft/Num of Saddles + Z Force Load ) * B / E

= 3.00 * ( 428.1/2 + 0 ) * 1100.0000/1372.5685

= 514.6 kgf

Saddle Reaction Force due to Wind Fl or Friction [Fwl]: = max( Fl, Friction Load, Sum of X Forces) * B / Ls

= max( 113.12 , 0.00 , 0 ) * 1100.0000/1800.0000

= 69.1 kgf

Load Combination Results for Q + Wind or Seismic [Q]: = Saddle Load + Max( Fwl, Fwt, Fsl, Fst )

= 6177 + Max( 69 , 514 , 0 , 0 )

= 6692.3 kgf

Summary of Loads at the base of this Saddle: Vertical Load (including saddle weight) 6865.58 kgf

Transverse Shear Load Saddle 214.04 kgf

Longitudinal Shear Load Saddle 113.12 kgf

Formulas and Substitutions for Horizontal Vessel Analysis:Note: Wear Plate is Welded to the Shell, k = 0.1The Computed K values from Table 4.15.1: K1 = 0.1130 K2 = 1.1076 K3 = 0.8127 K4 = 0.3848

K5 = 0.7457 K6 = 0.0496 K7 = 0.0169 K8 = 0.3364

K9 = 0.2650 K10 = 0.0552 K1* = 0.2030

Note: Dimension a is greater than or equal to Rm / 2.Moment per Equation 4.15.3 [M1]: = -Q*a [1 - (1- a/L + (R-h2)/(2a*L))/(1+(4h2)/3L)]

= -6692*43.60[1-(1-43.60/705.40+(77.726-0.000)/

(2*43.60*705.40))/(1+(4*0.00)/(3*705.40))]

= 106.2 kgf-m.

Moment per Equation 4.15.4 [M2]: = Q*L/4(1+2(R-h2)/(L))/(1+(4h2)/( 3L))-4a/L

= 6692*705/4(1+2(77-0)/(705))/(1+(4*0)/

(3*705))-4*43/705

= 9170.6 kgf-m.

Longitudinal Stress at Top of Shell (4.15.6) [Sigma1]: = P * Rm/(2t) - M2/(pi*Rmt)

= 13.500 * 777.265/(2*9.700 ) - 9170.6/(pi*777.3*9.700 )

= 491.07 kgf/cm

Longitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]: = P * Rm/(2t) + M2/(pi * Rm * t)

= 13.500 * 777.265/(2 * 9.700 ) + 9170.6/(pi * 777.3 * 9.700 )

= 590.69 kgf/cm

Longitudinal Stress at Top of Shell at Support (4.15.10) [Sigma*3]: = P * Rm/(2t) - M1/(K1*pi*Rmt)

= 13.500*777.265/(2*9.700)-106.2/(0.1130*pi*777.3*9.700)

= 535.78 kgf/cm

Longitudinal Stress at Bottom of Shell at Support (4.15.11) [Sigma*4]: = P * Rm/(2t) + M1/(K1* * pi * Rm * t)

= 13.500*777.265/(2*9.700)+106.2/(0.2030*pi*777.3*9.700)

= 543.72 kgf/cm

Maximum Shear Force in the Saddle (4.15.5) [T]: = Q(L-2a)/(L+(4*h2/3))

= 6692 ( 705.40 - 2 * 43.60 )/(705.40 + ( 4 * 0.00/3))

= 5865.0 kgf

Shear Stress in the shell no rings, not stiffened (4.15.14) [tau2]: = K2 * T / ( Rm * t )

= 1.1076 * 5865.00/( 777.2650 * 9.7000 )

= 86.16 kgf/cm

Decay Length (4.15.22) [x1,x2]: = 0.78 * sqrt( Rm * t )

= 0.78 * sqrt( 777.265 * 9.700 )

= 67.727 mm

Circumferential Stress in shell, no rings (4.15.23) [sigma6]: = -K5 * Q * k / ( t * ( b + X1 + X2 ) )

= -0.7457 * 6692 * 0.1/( 9.700 * ( 180.00 + 67.73 + 67.73 ) )

= -16.31 kgf/cm

Circ. Comp. Stress at Horn of Saddle, L>=8Rm (4.15.24) [sigma7]: = -Q/(4*t*(b+X1+X2)) - 3*K7*Q/(2*t)

= -6692/(4*9.700 *(180.000 +67.727 +67.727 )) -

3*0.0169 *6692/(2*9.700)

= -235.23 kgf/cm

Effective reinforcing plate width (4.15.1) [B1]: = min( b + 1.56 * sqrt( Rm * t ), 2a )

= min( 180.00 + 1.56 * sqrt( 777.265 * 9.700 ), 2 * 43.600 )

= 315.45 mm

Free Un-Restrained Thermal Expansion between the Saddles [Exp]: = Alpha * Ls * ( Design Temperature - Ambient Temperature )

= 0.122E-04 * 1800.000 * ( 120.0 - 21.1 )

= 2.177 mm

Results for Vessel Ribs, Web and Base: Baseplate Length Bplen 1380.0000 mm

Baseplate Thickness Bpthk 12.7000 mm

Baseplate Width Bpwid 152.4000 mm

Number of Ribs ( inc. outside ribs ) Nribs 4

Rib Thickness Ribtk 12.7000 mm

Web Thickness Webtk 12.7000 mm

Web Location Webloc Center

Moment of Inertia of Saddle - Lateral Direction Y A AY Io

Shell 5. 35. 16938. 11.

Wearplate 16. 29. 45863. 77.

Web 119. 25. 292805. 4256.

BasePlate 222. 19. 430160. 9563.

Totals 362. 107. 785766. 13907.

Value C1 = Sumof(Ay)/Sumof(A) = 73. mm

Value I = Sumof(Io) - C1*Sumof(Ay) = 8159. cm**4

Value As = Sumof(A) - Ashell = 73. cm

K1 = (1+Cos(beta)-.5*Sin(beta) )/(pi-beta+Sin(beta)*Cos(beta)) = 0.2108

Fh = K1 * Q = 0.2108 * 6692.286 = 1410.7662 kgf

Tension Stress, St = ( Fh/As ) = 19.4577 kgf/cm

Allowed Stress, Sa = 0.6 * Yield Str = 1468.0079 kgf/cm

d = B - R*Sin(theta) / theta = 370.7596 mm

Bending Moment, M = Fh * d = 523.0558 kgf-m.

Bending Stress, Sb = ( M * C1 / I ) = 46.8884 kgf/cm

Allowed Stress, Sa = 2/3 * Yield Str = 1631.1200 kgf/cm

Minimum Thickness of Baseplate per Moss : = ( 3 * ( Q + Saddle_Wt ) * BasePlateWidth / ( 4 * BasePlateLength *

AllStress ))

= ( 3 * (6692 + 173 ) * 152.40/( 4 * 1380.000 * 1631.120 ))

= 5.904 mm

Calculation of Axial Load, Intermediate Values and Compressive StressEffective Baseplate Length [e]: = ( Bplen - Clearance ) / ( Nribs - 1)

= ( 1380.0000 - 25.4 )/( 4 - 1 ) = 451.5333 mm

Baseplate Pressure Area [Ap]: = e * Bpwid / 2

= 451.5333 * 152.4000/2 = 344.0684 cm

Axial Load [P]: = Ap * Bp

= 344.1 * 3.18 = 1094.9 kgf

Area of the Rib and Web [Ar]: = ( Bpwid - Clearance - Webtk ) * Ribtk + e/2 * Webtk

= ( 152.400 - 25.4 - 12.700 ) * 12.700 + 451.5333/2 * 12.700

= 43.188 cm

Compressive Stress [Sc]: = P/Ar

= 1094.9/43.1885 = 25.3506 kgf/cm

Check of Outside Ribs:Inertia of Saddle, Outer Ribs - Longitudinal Direction Y A AY Ay Io

Rib 90.0 17.9 160934.4 0.0 382.8

Web 90.0 28.7 258051.3 0.0 7.7

Values 90.0 46.6 418985.7 0.0 390.5

Bending Moment [Rm]: = Fl /( 2 * Bplen ) * e * rl / 2

= 113.1/( 2 * 1380.00 ) * 451.533 * 633.83/2

= 5.865 kgf-m.

KL/R < Cc ( 21.0793 < 128.2550 ) per AISC E2-1

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

Sca = ( 1-( 21.08 )/(2 * 128.25 )) * 2446/

( 5/3+3*(21.08 )/(8* 128.25 )-( 21.08)/(8*128.25)

Sca = 1396.99 kgf/cm

AISC Unity Check on Outside Ribs ( must be No Uplift in Longitudinal direction)

Bolt Area due to Shear Load [Bltarears]: = Fl / (Stba * Nbolts)

= 113.12/(1757.68 * 8.00 )

= 0.0080 cm

Bolt Area due to Transverse LoadMoment on Baseplate Due to Transverse Load [Rmom]: = B * Ft + Sum of X Moments

= 1100.00 * 214.04 + 0.00

= 235.45 kgf-m.

Eccentricity (e): = Rmom / QO

= 235.45/6350.97

= 37.07 mm < Bplen/6 --> No Uplift in Transverse direction

Bolt Area due to Transverse Load [Bltareart]: = 0 (No Uplift)

Required of a Single Bolt [Bltarear] = max[Bltarearl, Bltarears, Bltareart]

= max[0.0000 , 0.0080 , 0.0000 ]

= 0.0080 cm

ASME Horizontal Vessel Analysis: Stresses for the Right Saddle (per ASME Sec. VIII Div. 2 based on the Zick method.)Note: Wear Pad Width (225.00) is less than 1.56*sqrt(rm*t) and less than 2a. The wear plate will be ignored.Minimum Wear Plate Width to be considered in analysis [b1]: = min( b + 1.56*sqrt( Rm * t ), 2a )

= min( 180.000 + 1.56*sqrt( 777.2650 * 9.7000 ), 2 * 436.000 )

= 315.4550 mm

Input and Calculated Values: Vessel Mean Radius Rm 777.27 mm

Stiffened Vessel Length per 4.15.6 L 705.40 cm

Distance from Saddle to Vessel tangent a 436.00 mm

Saddle Width b 180.00 mm

Saddle Bearing Angle theta 124.00 degrees

Shell Allowable Stress used in Calculation 1406.14 kgf/cm

Head Allowable Stress used in Calculation 1406.14 kgf/cm

Circumferential Efficiency in Plane of Saddle 1.00

Circumferential Efficiency at Mid-Span 1.00

Saddle Force Q, Operating Case 6148.73 kgf

Horizontal Vessel Analysis Results: Actual Allowable -------------------------------------------------------------------

Long. Stress at Top of Midspan 495.11 1406.14 kgf/cm

Long. Stress at Bottom of Midspan 586.65 1406.14 kgf/cm

Long. Stress at Top of Saddles 536.19 1406.14 kgf/cm

Long. Stress at Bottom of Saddles 543.49 1406.14 kgf/cm

Tangential Shear in Shell 79.17 1124.91 kgf/cm

Circ. Stress at Horn of Saddle 216.13 1757.68 kgf/cm

Circ. Compressive Stress in Shell 14.98 1406.14 kgf/cm

Intermediate Results: Saddle Reaction Q due to Wind or SeismicSaddle Reaction Force due to Wind Ft [Fwt]: = Ftr * ( Ft/Num of Saddles + Z Force Load ) * B / E

= 3.00 * ( 428.1/2 + 0 ) * 1100.0000/1372.5685

= 514.6 kgf

Saddle Reaction Force due to Wind Fl or Friction [Fwl]: = max( Fl, Friction Load, Sum of X Forces) * B / Ls

= max( 113.12 , 0.00 , 0 ) * 1100.0000/1800.0000

= 69.1 kgf

Load Combination Results for Q + Wind or Seismic [Q]: = Saddle Load + Max( Fwl, Fwt, Fsl, Fst )

= 5634 + Max( 69 , 514 , 0 , 0 )

= 6148.7 kgf

Summary of Loads at the base of this Saddle: Vertical Load (including saddle weight) 6322.02 kgf

Transverse Shear Load Saddle 214.04 kgf

Longitudinal Shear Load Saddle 113.12 kgf

Formulas and Substitutions for Horizontal Vessel Analysis:Note: Wear Plate is Welded to the Shell, k = 0.1The Computed K values from Table 4.15.1: K1 = 0.1130 K2 = 1.1076 K3 = 0.8127 K4 = 0.3848

K5 = 0.7457 K6 = 0.0496 K7 = 0.0169 K8 = 0.3364

K9 = 0.2650 K10 = 0.0552 K1* = 0.2030

Note: Dimension a is greater than or equal to Rm / 2.Moment per Equation 4.15.3 [M1]: = -Q*a [1 - (1- a/L + (R-h2)/(2a*L))/(1+(4h2)/3L)]

= -6148*43.60[1-(1-43.60/705.40+(77.726-0.000)/

(2*43.60*705.40))/(1+(4*0.00)/(3*705.40))]

= 97.6 kgf-m.

Moment per Equation 4.15.4 [M2]: = Q*L/4(1+2(R-h2)/(L))/(1+(4h2)/( 3L))-4a/L

= 6148*705/4(1+2(77-0)/(705))/(1+(4*0)/

(3*705))-4*43/705

= 8425.8 kgf-m.

Longitudinal Stress at Top of Shell (4.15.6) [Sigma1]: = P * Rm/(2t) - M2/(pi*Rmt)

= 13.500 * 777.265/(2*9.700 ) - 8425.8/(pi*777.3*9.700 )

= 495.11 kgf/cm

Longitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]: = P * Rm/(2t) + M2/(pi * Rm * t)

= 13.500 * 777.265/(2 * 9.700 ) + 8425.8/(pi * 777.3 * 9.700 )

= 586.65 kgf/cm

Longitudinal Stress at Top of Shell at Support (4.15.10) [Sigma*3]: = P * Rm/(2t) - M1/(K1*pi*Rmt)

= 13.500*777.265/(2*9.700)-97.6/(0.1130*pi*777.3*9.700)

= 536.19 kgf/cm

Longitudinal Stress at Bottom of Shell at Support (4.15.11) [Sigma*4]: = P * Rm/(2t) + M1/(K1* * pi * Rm * t)

= 13.500*777.265/(2*9.700)+97.6/(0.2030*pi*777.3*9.700)

= 543.49 kgf/cm

Maximum Shear Force in the Saddle (4.15.5) [T]: = Q(L-2a)/(L+(4*h2/3))

= 6148 ( 705.40 - 2 * 43.60 )/(705.40 + ( 4 * 0.00/3))

= 5388.6 kgf

Shear Stress in the shell no rings, not stiffened (4.15.14) [tau2]: = K2 * T / ( Rm * t )

= 1.1076 * 5388.64/( 777.2650 * 9.7000 )

= 79.17 kgf/cm

Decay Length (4.15.22) [x1,x2]: = 0.78 * sqrt( Rm * t )

= 0.78 * sqrt( 777.265 * 9.700 )

= 67.727 mm

Circumferential Stress in shell, no rings (4.15.23) [sigma6]: = -K5 * Q * k / ( t * ( b + X1 + X2 ) )

= -0.7457 * 6148 * 0.1/( 9.700 * ( 180.00 + 67.73 + 67.73 ) )

= -14.98 kgf/cm

Circ. Comp. Stress at Horn of Saddle, L>=8Rm (4.15.24) [sigma7]: = -Q/(4*t*(b+X1+X2)) - 3*K7*Q/(2*t)

= -6148/(4*9.700 *(180.000 +67.727 +67.727 )) -

3*0.0169 *6148/(2*9.700)

= -216.13 kgf/cm

Effective reinforcing plate width (4.15.1) [B1]: = min( b + 1.56 * sqrt( Rm * t ), 2a )

= min( 180.00 + 1.56 * sqrt( 777.265 * 9.700 ), 2 * 43.600 )

= 315.45 mm

Results for Vessel Ribs, Web and Base Baseplate Length Bplen 1380.0000 mm

Baseplate Thickness Bpthk 12.7000 mm

Baseplate Width Bpwid 152.4000 mm

Number of Ribs ( inc. outside ribs ) Nribs 4

Rib Thickness Ribtk 12.7000 mm

Web Thickness Webtk 12.7000 mm

Web Location Webloc Center

Moment of Inertia of Saddle - Lateral Direction Y A AY Io

Shell 5. 35. 16938. 11.

Wearplate 16. 29. 45863. 77.

Web 119. 25. 292805. 4256.

BasePlate 222. 19. 430160. 9563.

Totals 362. 107. 785766. 13907.

Value C1 = Sumof(Ay)/Sumof(A) = 73. mm

Value I = Sumof(Io) - C1*Sumof(Ay) = 8159. cm**4

Value As = Sumof(A) - Ashell = 73. cm

K1 = (1+Cos(beta)-.5*Sin(beta) )/(pi-beta+Sin(beta)*Cos(beta)) = 0.2108

Fh = K1 * Q = 0.2108 * 6148.729 = 1296.1818 kgf

Tension Stress, St = ( Fh/As ) = 17.8773 kgf/cm

Allowed Stress, Sa = 0.6 * Yield Str = 1468.0079 kgf/cm

d = B - R*Sin(theta) / theta = 370.7596 mm

Bending Moment, M = Fh * d = 480.5724 kgf-m.

Bending Stress, Sb = ( M * C1 / I ) = 43.0800 kgf/cm

Allowed Stress, Sa = 2/3 * Yield Str = 1631.1200 kgf/cm

Minimum Thickness of Baseplate per Moss : = ( 3 * ( Q + Saddle_Wt ) * BasePlateWidth / ( 4 * BasePlateLength *

AllStress ))

= ( 3 * (6148 + 173 ) * 152.40/( 4 * 1380.000 * 1631.120 ))

= 5.666 mm

Calculation of Axial Load, Intermediate Values and Compressive StressEffective Baseplate Length [e]: = ( Bplen - Clearance ) / ( Nribs - 1)

= ( 1380.0000 - 25.4 )/( 4 - 1 ) = 451.5333 mm

Baseplate Pressure Area [Ap]: = e * Bpwid / 2

= 451.5333 * 152.4000/2 = 344.0684 cm

Axial Load [P]: = Ap * Bp

= 344.1 * 2.92 = 1005.9 kgf

Area of the Rib and Web [Ar]: = ( Bpwid - Clearance - Webtk ) * Ribtk + e/2 * Webtk

= ( 152.400 - 25.4 - 12.700 ) * 12.700 + 451.5333/2 * 12.700

= 43.188 cm

Compressive Stress [Sc]: = P/Ar

= 1005.9/43.1885 = 23.2916 kgf/cm

Check of Outside Ribs:Inertia of Saddle, Outer Ribs - Longitudinal Direction Y A AY Ay Io

Rib 90.0 17.9 160934.4 0.0 382.8

Web 90.0 28.7 258051.3 0.0 7.7

Values 90.0 46.6 418985.7 0.0 390.5

Bending Moment [Rm]: = Fl /( 2 * Bplen ) * e * rl / 2

= 113.1/( 2 * 1380.00 ) * 451.533 * 633.83/2

= 5.865 kgf-m.

KL/R < Cc ( 21.0793 < 128.2550 ) per AISC E2-1

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

Sca = ( 1-( 21.08 )/(2 * 128.25 )) * 2446/

( 5/3+3*(21.08 )/(8* 128.25 )-( 21.08)/(8*128.25)

Sca = 1396.99 kgf/cm

AISC Unity Check on Outside Ribs ( must be No Uplift in Longitudinal direction)

Bolt Area due to Shear Load [Bltarears]: = Fl / (Stba * Nbolts)

= 113.12/(1757.68 * 8.00 )

= 0.0080 cm

Bolt Area due to Transverse LoadMoment on Baseplate Due to Transverse Load [Rmom]: = B * Ft + Sum of X Moments

= 1100.00 * 214.04 + 0.00

= 235.45 kgf-m.

Eccentricity (e): = Rmom / QO

= 235.45/5807.41

= 40.54 mm < Bplen/6 --> No Uplift in Transverse direction

Bolt Area due to Transverse Load [Bltareart]: = 0 (No Uplift)

Required of a Single Bolt [Bltarear] = max[Bltarearl, Bltarears, Bltareart]

= max[0.0000 , 0.0080 , 0.0000 ]

= 0.0080 cm

PV Elite is a trademark of Intergraph CADWorx & Analysis Solutions, Inc. 2014 TC "Horizontal Vessel Analysis (Test) :" /f C ASME Horizontal Vessel Analysis: Stresses for the Left Saddle (per ASME Sec. VIII Div. 2 based on the Zick method.)Horizontal Vessel Stress Calculations : Test CaseNote: Wear Pad Width (225.00) is less than 1.56*sqrt(rm*t) and less than 2a. The wear plate will be ignored.Minimum Wear Plate Width to be considered in analysis [b1]: = min( b + 1.56*sqrt( Rm * t ), 2a )

= min( 180.000 + 1.56*sqrt( 775.7650 * 12.7000 ), 2 * 436.000 )

= 334.8430 mm

Input and Calculated Values: Vessel Mean Radius Rm 775.77 mm

Stiffened Vessel Length per 4.15.6 L 705.40 cm

Distance from Saddle to Vessel tangent a 436.00 mm

Saddle Width b 180.00 mm

Saddle Bearing Angle theta 124.00 degrees

Shell Allowable Stress used in Calculation 2538.08 kgf/cm

Head Allowable Stress used in Calculation 2538.08 kgf/cm

Circumferential Efficiency in Plane of Saddle 1.00

Circumferential Efficiency at Mid-Span 1.00

Saddle Force Q, Test Case, no Ext. Forces 11387.87 kgf

Horizontal Vessel Analysis Results: Actual Allowable -------------------------------------------------------------------

Long. Stress at Top of Midspan 473.40 2538.08 kgf/cm

Long. Stress at Bottom of Midspan 603.36 2538.08 kgf/cm

Long. Stress at Top of Saddles 531.79 2538.08 kgf/cm

Long. Stress at Bottom of Saddles 542.05 2538.08 kgf/cm

Tangential Shear in Shell 112.20 2030.47 kgf/cm

Circ. Stress at Horn of Saddle 247.03 3807.12 kgf/cm

Circ. Compressive Stress in Shell 19.97 2538.08 kgf/cm

Intermediate Results: Saddle Reaction Q due to Wind or SeismicSaddle Reaction Force due to Wind Ft [Fwt]: = Ftr * ( Ft/Num of Saddles + Z Force Load ) * B / E

= 3.00 * ( 141.3/2 + 0 ) * 1100.0000/1369.9198

= 170.2 kgf

Saddle Reaction Force due to Wind Fl or Friction [Fwl]: = max( Fl, Friction Load, Sum of X Forces) * B / Ls

= max( 113.12 , 0.00 , 0 ) * 1100.0000/1800.0000

= 22.8 kgf

Load Combination Results for Q + Wind or Seismic [Q]: = Saddle Load + Max( Fwl, Fwt, Fsl, Fst )

= 11217 + Max( 22 , 170 , 0 , 0 )

= 11387.9 kgf

Summary of Loads at the base of this Saddle: Vertical Load (including saddle weight) 11561.16 kgf

Transverse Shear Load Saddle 70.63 kgf

Longitudinal Shear Load Saddle 37.33 kgf

Hydrostatic Test Pressure at center of Vessel: 17.628 kgf/cmFormulas and Substitutions for Horizontal Vessel Analysis:Note: Wear Plate is Welded to the Shell, k = 0.1The Computed K values from Table 4.15.1: K1 = 0.1130 K2 = 1.1076 K3 = 0.8127 K4 = 0.3848

K5 = 0.7457 K6 = 0.0496 K7 = 0.0170 K8 = 0.3364

K9 = 0.2650 K10 = 0.0552 K1* = 0.2030

Note: Dimension a is greater than or equal to Rm / 2.Moment per Equation 4.15.3 [M1]: = -Q*a [1 - (1- a/L + (R-h2)/(2a*L))/(1+(4h2)/3L)]

= -11387*43.60[1-(1-43.60/705.40+(77.576-0.000)/

(2*43.60*705.40))/(1+(4*0.00)/(3*705.40))]

= 178.9 kgf-m.

Moment per Equation 4.15.4 [M2]: = Q*L/4(1+2(R-h2)/(L))/(1+(4h2)/( 3L))-4a/L

= 11387*705/4(1+2(77-0)/(705))/(1+(4*0)/

(3*705))-4*43/705

= 15603.2 kgf-m.

Longitudinal Stress at Top of Shell (4.15.6) [Sigma1]: = P * Rm/(2t) - M2/(pi*Rmt)

= 17.628 * 775.765/(2*12.700 ) - 15603.2/(pi*775.8*12.700 )

= 473.40 kgf/cm

Longitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]: = P * Rm/(2t) + M2/(pi * Rm * t)

= 17.628 * 775.765/(2 * 12.700 ) + 15603.2/(pi * 775.8 * 12.700 )

= 603.36 kgf/cm

Longitudinal Stress at Top of Shell at Support (4.15.10) [Sigma*3]: = P * Rm/(2t) - M1/(K1*pi*Rmt)

= 17.628*775.765/(2*12.700)-178.9/(0.1130*pi*775.8*12.700)

= 531.79 kgf/cm

Longitudinal Stress at Bottom of Shell at Support (4.15.11) [Sigma*4]: = P * Rm/(2t) + M1/(K1* * pi * Rm * t)

= 17.628*775.765/(2*12.700)+178.9/(0.2030*pi*775.8*12.700)

= 542.05 kgf/cm

Maximum Shear Force in the Saddle (4.15.5) [T]: = Q(L-2a)/(L+(4*h2/3))

= 11387 ( 705.40 - 2 * 43.60 )/(705.40 + ( 4 * 0.00/3))

= 9980.1 kgf

Shear Stress in the shell no rings, not stiffened (4.15.14) [tau2]: = K2 * T / ( Rm * t )

= 1.1076 * 9980.13/( 775.7650 * 12.7000 )

= 112.20 kgf/cm

Decay Length (4.15.22) [x1,x2]: = 0.78 * sqrt( Rm * t )

= 0.78 * sqrt( 775.765 * 12.700 )

= 77.421 mm

Circumferential Stress in shell, no rings (4.15.23) [sigma6]: = -K5 * Q * k / ( t * ( b + X1 + X2 ) )

= -0.7457 * 11387 * 0.1/( 12.700 * ( 180.00 + 77.42 + 77.42 ) )

= -19.97 kgf/cm

Circ. Comp. Stress at Horn of Saddle, L>=8Rm (4.15.24) [sigma7]: = -Q/(4*t*(b+X1+X2)) - 3*K7*Q/(2*t)

= -11387/(4*12.700 *(180.000 +77.421 +77.421 )) -

3*0.0170 *11387/(2*12.700)

= -247.03 kgf/cm

Effective reinforcing plate width (4.15.1) [B1]: = min( b + 1.56 * sqrt( Rm * t ), 2a )

= min( 180.00 + 1.56 * sqrt( 775.765 * 12.700 ), 2 * 43.600 )

= 334.84 mm

Results for Vessel Ribs, Web and Base: Baseplate Length Bplen 1380.0000 mm

Baseplate Thickness Bpthk 12.7000 mm

Baseplate Width Bpwid 152.4000 mm

Number of Ribs ( inc. outside ribs ) Nribs 4

Rib Thickness Ribtk 12.7000 mm

Web Thickness Webtk 12.7000 mm

Web Location Webloc Center

Moment of Inertia of Saddle - Lateral Direction Y A AY Io

Shell 6. 48. 30581. 26.

Wearplate 19. 29. 54435. 108.

Web 121. 24. 291895. 4253.

BasePlate 222. 19. 430160. 9563.

Totals 368. 120. 807072. 13950.

Value C1 = Sumof(Ay)/Sumof(A) = 67. mm

Value I = Sumof(Io) - C1*Sumof(Ay) = 8534. cm**4

Value As = Sumof(A) - Ashell = 72. cm

K1 = (1+Cos(beta)-.5*Sin(beta) )/(pi-beta+Sin(beta)*Cos(beta)) = 0.2108

Fh = K1 * Q = 0.2108 * 11387.869 = 2400.6179 kgf

Tension Stress, St = ( Fh/As ) = 33.2850 kgf/cm

Allowed Stress, Sa = 0.6 * Yield Str = 1468.0079 kgf/cm

d = B - R*Sin(theta) / theta = 370.2075 mm

Bending Moment, M = Fh * d = 888.7278 kgf-m.

Bending Stress, Sb = ( M * C1 / I ) = 69.8720 kgf/cm

Allowed Stress, Sa = 2/3 * Yield Str = 1631.1200 kgf/cm

Minimum Thickness of Baseplate per Moss : = ( 3 * ( Q + Saddle_Wt ) * BasePlateWidth / ( 4 * BasePlateLength *

AllStress ))

= ( 3 * (11387 + 173 ) * 152.40/( 4 * 1380.000 * 1631.120 ))

= 7.662 mm

Calculation of Axial Load, Intermediate Values and Compressive StressEffective Baseplate Length [e]: = ( Bplen - Clearance ) / ( Nribs - 1)

= ( 1380.0000 - 25.4 )/( 4 - 1 ) = 451.5333 mm

Baseplate Pressure Area [Ap]: = e * Bpwid / 2

= 451.5333 * 152.4000/2 = 344.0684 cm

Axial Load [P]: = Ap * Bp

= 344.1 * 5.41 = 1863.0 kgf

Area of the Rib and Web [Ar]: = ( Bpwid - Clearance - Webtk ) * Ribtk + e/2 * Webtk

= ( 152.400 - 25.4 - 12.700 ) * 12.700 + 451.5333/2 * 12.700

= 43.188 cm

Compressive Stress [Sc]: = P/Ar

= 1863.0/43.1885 = 43.1376 kgf/cm

Check of Outside Ribs:Inertia of Saddle, Outer Ribs - Longitudinal Direction Y A AY Ay Io

Rib 90.0 17.9 160934.4 0.0 382.8

Web 90.0 28.7 258051.3 0.0 7.7

Values 90.0 46.6 418985.7 0.0 390.5

Bending Moment [Rm]: = Fl /( 2 * Bplen ) * e * rl / 2

= 37.3/( 2 * 1380.00 ) * 451.533 * 630.83/2

= 1.926 kgf-m.

KL/R < Cc ( 20.9796 < 128.2550 ) per AISC E2-1

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

Sca = ( 1-( 20.98 )/(2 * 128.25 )) * 2446/

( 5/3+3*(20.98 )/(8* 128.25 )-( 20.98)/(8*128.25)

Sca = 1397.40 kgf/cm

AISC Unity Check on Outside Ribs ( must be No Uplift in Longitudinal direction)

Bolt Area due to Shear Load [Bltarears]: = Fl / (Stba * Nbolts)

= 37.33/(1757.68 * 8.00 )

= 0.0027 cm

Bolt Area due to Transverse LoadMoment on Baseplate Due to Transverse Load [Rmom]: = B * Ft + Sum of X Moments

= 1100.00 * 70.63 + 0.00

= 77.70 kgf-m.

Eccentricity (e): = Rmom / QO

= 77.70/11391.01

= 6.82 mm < Bplen/6 --> No Uplift in Transverse direction

Bolt Area due to Transverse Load [Bltareart]: = 0 (No Uplift)

Required of a Single Bolt [Bltarear] = max[Bltarearl, Bltarears, Bltareart]

= max[0.0000 , 0.0027 , 0.0000 ]

= 0.0027 cm

ASME Horizontal Vessel Analysis: Stresses for the Right Saddle (per ASME Sec. VIII Div. 2 based on the Zick method.)Note: Wear Pad Width (225.00) is less than 1.56*sqrt(rm*t) and less than 2a. The wear plate will be ignored.Minimum Wear Plate Width to be considered in analysis [b1]: = min( b + 1.56*sqrt( Rm * t ), 2a )

= min( 180.000 + 1.56*sqrt( 775.7650 * 12.7000 ), 2 * 436.000 )

= 334.8430 mm

Input and Calculated Values: Vessel Mean Radius Rm 775.77 mm

Stiffened Vessel Length per 4.15.6 L 705.40 cm

Distance from Saddle to Vessel tangent a 436.00 mm

Saddle Width b 180.00 mm

Saddle Bearing Angle theta 124.00 degrees

Shell Allowable Stress used in Calculation 2538.08 kgf/cm

Head Allowable Stress used in Calculation 2538.08 kgf/cm

Circumferential Efficiency in Plane of Saddle 1.00

Circumferential Efficiency at Mid-Span 1.00

Saddle Force Q, Test Case, no Ext. Forces 10644.17 kgf

Horizontal Vessel Analysis Results: Actual Allowable -------------------------------------------------------------------

Long. Stress at Top of Midspan 477.64 2538.08 kgf/cm

Long. Stress at Bottom of Midspan 599.12 2538.08 kgf/cm

Long. Stress at Top of Saddles 532.22 2538.08 kgf/cm

Long. Stress at Bottom of Saddles 541.81 2538.08 kgf/cm

Tangential Shear in Shell 104.87 2030.47 kgf/cm

Circ. Stress at Horn of Saddle 230.90 3807.12 kgf/cm

Circ. Compressive Stress in Shell 18.67 2538.08 kgf/cm

Intermediate Results: Saddle Reaction Q due to Wind or SeismicSaddle Reaction Force due to Wind Ft [Fwt]: = Ftr * ( Ft/Num of Saddles + Z Force Load ) * B / E

= 3.00 * ( 141.3/2 + 0 ) * 1100.0000/1369.9198

= 170.2 kgf

Saddle Reaction Force due to Wind Fl or Friction [Fwl]: = max( Fl, Friction Load, Sum of X Forces) * B / Ls

= max( 113.12 , 0.00 , 0 ) * 1100.0000/1800.0000

= 22.8 kgf

Load Combination Results for Q + Wind or Seismic [Q]: = Saddle Load + Max( Fwl, Fwt, Fsl, Fst )

= 10474 + Max( 22 , 170 , 0 , 0 )

= 10644.2 kgf

Summary of Loads at the base of this Saddle: Vertical Load (including saddle weight) 10817.46 kgf

Transverse Shear Load Saddle 70.63 kgf

Longitudinal Shear Load Saddle 37.33 kgf

Hydrostatic Test Pressure at center of Vessel: 17.628 kgf/cmFormulas and Substitutions for Horizontal Vessel Analysis:Note: Wear Plate is Welded to the Shell, k = 0.1The Computed K values from Table 4.15.1: K1 = 0.1130 K2 = 1.1076 K3 = 0.8127 K4 = 0.3848

K5 = 0.7457 K6 = 0.0496 K7 = 0.0170 K8 = 0.3364

K9 = 0.2650 K10 = 0.0552 K1* = 0.2030

Note: Dimension a is greater than or equal to Rm / 2.Moment per Equation 4.15.3 [M1]: = -Q*a [1 - (1- a/L + (R-h2)/(2a*L))/(1+(4h2)/3L)]

= -10644*43.60[1-(1-43.60/705.40+(77.576-0.000)/

(2*43.60*705.40))/(1+(4*0.00)/(3*705.40))]

= 167.2 kgf-m.

Moment per Equation 4.15.4 [M2]: = Q*L/4(1+2(R-h2)/(L))/(1+(4h2)/( 3L))-4a/L

= 10644*705/4(1+2(77-0)/(705))/(1+(4*0)/

(3*705))-4*43/705

= 14584.2 kgf-m.

Longitudinal Stress at Top of Shell (4.15.6) [Sigma1]: = P * Rm/(2t) - M2/(pi*Rmt)

= 17.628 * 775.765/(2*12.700 ) - 14584.2/(pi*775.8*12.700 )

= 477.64 kgf/cm

Longitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]: = P * Rm/(2t) + M2/(pi * Rm * t)

= 17.628 * 775.765/(2 * 12.700 ) + 14584.2/(pi * 775.8 * 12.700 )

= 599.12 kgf/cm

Longitudinal Stress at Top of Shell at Support (4.15.10) [Sigma*3]: = P * Rm/(2t) - M1/(K1*pi*Rmt)

= 17.628*775.765/(2*12.700)-167.2/(0.1130*pi*775.8*12.700)

= 532.22 kgf/cm

Longitudinal Stress at Bottom of Shell at Support (4.15.11) [Sigma*4]: = P * Rm/(2t) + M1/(K1* * pi * Rm * t)

= 17.628*775.765/(2*12.700)+167.2/(0.2030*pi*775.8*12.700)

= 541.81 kgf/cm

Maximum Shear Force in the Saddle (4.15.5) [T]: = Q(L-2a)/(L+(4*h2/3))

= 10644 ( 705.40 - 2 * 43.60 )/(705.40 + ( 4 * 0.00/3))

= 9328.4 kgf

Shear Stress in the shell no rings, not stiffened (4.15.14) [tau2]: = K2 * T / ( Rm * t )

= 1.1076 * 9328.36/( 775.7650 * 12.7000 )

= 104.87 kgf/cm

Decay Length (4.15.22) [x1,x2]: = 0.78 * sqrt( Rm * t )

= 0.78 * sqrt( 775.765 * 12.700 )

= 77.421 mm

Circumferential Stress in shell, no rings (4.15.23) [sigma6]: = -K5 * Q * k / ( t * ( b + X1 + X2 ) )

= -0.7457 * 10644 * 0.1/( 12.700 * ( 180.00 + 77.42 + 77.42 ) )

= -18.67 kgf/cm

Circ. Comp. Stress at Horn of Saddle, L>=8Rm (4.15.24) [sigma7]: = -Q/(4*t*(b+X1+X2)) - 3*K7*Q/(2*t)

= -10644/(4*12.700 *(180.000 +77.421 +77.421 )) -

3*0.0170 *10644/(2*12.700)

= -230.90 kgf/cm

Effective reinforcing plate width (4.15.1) [B1]: = min( b + 1.56 * sqrt( Rm * t ), 2a )

= min( 180.00 + 1.56 * sqrt( 775.765 * 12.700 ), 2 * 43.600 )

= 334.84 mm

Results for Vessel Ribs, Web and Base Baseplate Length Bplen 1380.0000 mm

Baseplate Thickness Bpthk 12.7000 mm

Baseplate Width Bpwid 152.4000 mm

Number of Ribs ( inc. outside ribs ) Nribs 4

Rib Thickness Ribtk 12.7000 mm

Web Thickness Webtk 12.7000 mm

Web Location Webloc Center

Moment of Inertia of Saddle - Lateral Direction Y A AY Io

Shell 6. 48. 30581. 26.

Wearplate 19. 29. 54435. 108.

Web 121. 24. 291895. 4253.

BasePlate 222. 19. 430160. 9563.

Totals 368. 120. 807072. 13950.

Value C1 = Sumof(Ay)/Sumof(A) = 67. mm

Value I = Sumof(Io) - C1*Sumof(Ay) = 8534. cm**4

Value As = Sumof(A) - Ashell = 72. cm

K1 = (1+Cos(beta)-.5*Sin(beta) )/(pi-beta+Sin(beta)*Cos(beta)) = 0.2108

Fh = K1 * Q = 0.2108 * 10644.168 = 2243.8420 kgf

Tension Stress, St = ( Fh/As ) = 31.1112 kgf/cm

Allowed Stress, Sa = 0.6 * Yield Str = 1468.0079 kgf/cm

d = B - R*Sin(theta) / theta = 370.2075 mm

Bending Moment, M = Fh * d = 830.6882 kgf-m.

Bending Stress, Sb = ( M * C1 / I ) = 65.3089 kgf/cm

Allowed Stress, Sa = 2/3 * Yield Str = 1631.1200 kgf/cm

Minimum Thickness of Baseplate per Moss : = ( 3 * ( Q + Saddle_Wt ) * BasePlateWidth / ( 4 * BasePlateLength *

AllStress ))

= ( 3 * (10644 + 173 ) * 152.40/( 4 * 1380.000 * 1631.120 ))

= 7.411 mm

Calculation of Axial Load, Intermediate Values and Compressive StressEffective Baseplate Length [e]: = ( Bplen - Clearance ) / ( Nribs - 1)

= ( 1380.0000 - 25.4 )/( 4 - 1 ) = 451.5333 mm

Baseplate Pressure Area [Ap]: = e * Bpwid / 2

= 451.5333 * 152.4000/2 = 344.0684 cm

Axial Load [P]: = Ap * Bp

= 344.1 * 5.06 = 1741.4 kgf

Area of the Rib and Web [Ar]: = ( Bpwid - Clearance - Webtk ) * Ribtk + e/2 * Webtk

= ( 152.400 - 25.4 - 12.700 ) * 12.700 + 451.5333/2 * 12.700

= 43.188 cm

Compressive Stress [Sc]: = P/Ar

= 1741.4/43.1885 = 40.3204 kgf/cm

Check of Outside Ribs:Inertia of Saddle, Outer Ribs - Longitudinal Direction Y A AY Ay Io

Rib 90.0 17.9 160934.4 0.0 382.8

Web 90.0 28.7 258051.3 0.0 7.7

Values 90.0 46.6 418985.7 0.0 390.5

Bending Moment [Rm]: = Fl /( 2 * Bplen ) * e * rl / 2

= 37.3/( 2 * 1380.00 ) * 451.533 * 630.83/2

= 1.926 kgf-m.

KL/R < Cc ( 20.9796 < 128.2550 ) per AISC E2-1

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

Sca = ( 1-( 20.98 )/(2 * 128.25 )) * 2446/

( 5/3+3*(20.98 )/(8* 128.25 )-( 20.98)/(8*128.25)

Sca = 1397.40 kgf/cm

AISC Unity Check on Outside Ribs ( must be No Uplift in Longitudinal direction)

Bolt Area due to Shear Load [Bltarears]: = Fl / (Stba * Nbolts)

= 37.33/(1757.68 * 8.00 )

= 0.0027 cm

Bolt Area due to Transverse LoadMoment on Baseplate Due to Transverse Load [Rmom]: = B * Ft + Sum of X Moments

= 1100.00 * 70.63 + 0.00

= 77.70 kgf-m.

Eccentricity (e): = Rmom / QO

= 77.70/10647.31

= 7.30 mm < Bplen/6 --> No Uplift in Transverse direction

Bolt Area due to Transverse Load [Bltareart]: = 0 (No Uplift)

Required of a Single Bolt [Bltarear] = max[Bltarearl, Bltarears, Bltareart]

= max[0.0000 , 0.0027 , 0.0000 ]

= 0.0027 cm

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