Exactely, but they not especify if the shears are for the wind case o earthquake case. I need to know these cases for trasmit the loads to civil department.
An example:
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 * ( 4392.0 /2 + 0 ) * 1590.0000 / 2443.9238
= 4286.1 Kg
Saddle Reaction Force due to Wind Fl or Friction [Fwl]:
= Max( Fl, Friction Load, Sum of X Forces) * B / Ls
= Max( 793.70 , 5969.35 , 0 ) * 1590.0000 / 5350.0005
= 1774.1 Kg
Saddle Reaction Force due to Earthquake Fl or Friction [Fsl]:
= Max( Fl, Friction Force, Sum of X Forces ) * B / Ls
= Max( 26212.75 , 5969.35 , 0 ) * 1590.0000 / 5350.0005
= 7790.3 Kg
Saddle Reaction Force due to Earthquake Ft [Fst]:
= Ftr * ( Ft/Num of Saddles + Z Force Load ) * B / E
= 3.00 * ( 0 /2 + 0 ) * 1590.0000 / 2443.9238
= 0.0 Kg
Load Combination Results for Q + Wind or Seismic [Q]:
= Saddle Load + Max( Fwl, Fwt, Fsl, Fst )
= 19897 + Max( 1774 , 4286 , 7790 , 0 )
= 27688.2 Kg
Formulas and Substitutions for Horizontal Vessel Analysis:
Note: Wear Plates are assumed to be Welded 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 = 0.0410 K8 = 0.3405
K9 = 0.2711 K10 = 0.0581 K1* = 0.1923
Note: Dimension a is greater than or equal to Rm / 2.
Moment per Equation 4.15.3 [M1]:
= -Q*a [1 - (1- a/L + (R²-h2²)/(2a*L))/(1+(4h2)/3L)]
= -27688*1200.00[1-(1-1200.00/7850.00+(1411.000²-706.000²)/
(2*1200.00*7850.00))/(1+(4*706.00)/(3*7850.00))]
= -5744.4 Kg.m
Moment per Equation 4.15.4 [M2]:
= Q*L/4(1+2(R²-h2²)/(L²))/(1+(4h2)/( 3L))-4a/L
= 27688*7849/4(1+2(1411²-706²)/(7849²))/(1+(4*706)/
(3*7849))-4*1200/7849
= 17649.7 Kg.m
Longitudinal Stress at Top of Shell (4.15.6) [Sigma1]:
= P * Rm/(2t) - M2/(pi*Rm²t)
= 5.0 * 1411.000 /(2*10.00 ) - 17649.7 /(pi*1411.0²*8.00 )
= 317.49 Kg/cm²
Longitudinal Stress at Bottom of Shell (4.15.7) [Sigma2]:
= P * Rm/(2t) + M2/(pi * Rm² * t)
= 5.2 * 1411.000 /(2 * 10.00 ) + 17649.7 /(pi * 1411.0² * 8.00 )
= 399.92 Kg/cm²
Longitudinal Stress at Top of Shell at Support (4.15.8) [Sigma3]:
= P * Rm/(2t) - M1/(pi * Rm² * t)
= 5.0 * 1411.000 /(2 * 10.00 ) - -5744.4 /(pi * 1411.0² * 10.00 )
= 361.93 Kg/cm²
Longitudinal Stress at bottom of Shell at Support (4.15.9) [Sigma4]:
= P * Rm/(2t) + M1/(pi*Rm²t)
= 5.0 * 1411.000 /(2*10.00 ) + -5744.4 /(pi*1411.0²*10.00 )
= 343.57 Kg/cm²
Maximum Shear Force in the Saddle (4.15.5) [T]:
= Q(L-2a)/(L+(4*h2/3))
= 27688 ( 7850.00 - 2 * 1200.00 )/(7850.00 + ( 4 * 706.00 /3))
= 17164.7 Kg
Shear Stress in the shell with a single ring (4.15.13) [tau1]:
= T / ( pi * Rm * t )
= 17164.70 / ( pi * 1411.0000 * 10.0000 )
= 38.72 Kg/cm²
Decay Length (4.15.22) [x1,x2]:
= 0.78 * sqrt( Rm * t )
= 0.78 * sqrt( 1411.000 * 10.000 )
= 92.653 mm.
Circ. Stress in shell w/ring in Plane of Saddle (4.15.32) [sigma6*]:
= -K5 * Q * k/A
= -0.760 * 27688 * 0.1/55.63
= -37.84 Kg/cm²
In this example I can find the shears for Wind or Earthquake, but where is the moments indicates?
The maximum shear is in wind case or earthquake? It's not clear.
On the other hand, when I have a little vessel support with legs, I suppose the shears and moments indicates are in the base, not in the junction with the shell. It is correct?
Thanks in advance for your cooperation.
Regards
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