AAPL Pipe Support/Span Calculation Sheet Cl Client:
Description:
Cu Customer No.:
Item No.:
Prepared By:
Approval:
Date:
Rev.: 0 1 2
Ow Owner No.:
Dwg. No.:
Service: Design Temperature:
F
3
SG =
1.00
C
100
FS =
Factor of Safety:
Imperial
Measurement Units: Deg. F
Corrosion, erosion, mech. allow:
4.0
Ice Thickness (external):
C= Tice =
Pi Pipe or Tubing Material:
Pipe or Tube Dimensions:
ASTM A 53-B Type S C arbon Steel Pipe
Nom. Size:
NPS 8 (DN 200)
E = 2. 2.93 934E 4E+0 +07 7 ps psii
Nom. Thk.:
Sch 40 per ASME B36.10M
Hot modulus of elasticity
2.93E+ 2.9 3E+07 07 psi
Hot allowable stress Design stress, Sh / FS
0
inch
psi
Sh =
20,000
psi
Actual OD:
8.62 8. 625 5 in inch ch
Do =
8.625
inch
5,000
psi
Sa =
5,000
psi
Nominal thickness:
0.32 0. 322 2 in inch ch
Tnom =
0.322
inch
lb/i lb /in^ n^3 3
Th Thi ck ckness tolerance:
Mil l =
and, h =
0.000
inch
2.83 833E 3E-0 -01 1 Pden = 2.
R el elative weight factor (vs. carbon steel)
Mat =
none
Liner on ID:
7.0
lb/ft^3
Thk, Ti =
2
inch
12.5%
1.000
Semi-rigid Fiberglass
Density, Ins =
Comments:
inch
20,000
Material density
Insulation:
0.0000
4.05 051E 1E-0 -03 3 Iden = 4.
lb/i lb /in^ n^3 3
Density:
0.000 0.0 00 lb/ lb/in^ in^3 3
Thickness:
0.00 0. 000 0 in inch ch
0.000E+00 E+00 lb/in lb/in^3 ^3 Lden = 0.000 0.000
Tr =
inch
Weight of concentrated load assumed at center of
x
span (valves + flanges + animal, etc.):
x
0.0 0. 0 lb lbs s
x Minimum pipe thickness,
t =
Tnom - (Tnom * Mill % / 100 ) - c - h
Outside radius of pipe, R = Do / 2 Pipe moment of inertia (based on t ),
I = π /
4
4
64 [ D o - (Do - 2 t ) ]
7.241E 7.2 41E+01 +01 in^ in^4 4
4 π / 32 (D o -
4
d ) / Do 2
1.681E 1.6 81E+01 +01 in^ in^3 3
2
0
lbs
t =
0.3216
inch
R=
4.313
inch
I =
d=
Pipe inside diameter, diameter, d = Do - 2 Tnom Pipe section modulus, Z =
Wc =
7.24 7. 241E 1E+0 +01 1 in in^4 ^4 7.981
inch
Z = 1. 1.68 681E 1E+0 +01 1 in in^3 ^3
Metal area of pipe cross-section, A m = ( Do - d ) * π / 4
Am =
8.40
i n^ 2
Conversion factor, water density, density, dw = 62.4 lb / cu ft
dw =
62.40
lb/ft^3
Conversion factor:
n = 8.3 8.3333 333E-0 E-02 2 ft / in
n = 1 ft / 12 in
Pipe weight per unit length, length, Wp = Pden * Am / n Pipe or liner or refractory refractory inside diameter, diameter, d L = d - 2 T r 2
2
Liner area cross-section, A r = ( d - dL ) * π / 4 Liner weight per unit length, length, Wr = Lden * Ar / n 2
Wp =
28.55
lb / ft
dL =
7.981
inch
Ar =
0.00
i n^ 2
Wr =
0.00
lb / ft
5.00 003E 3E+0 +01 1 in in^2 ^2 Af = 5.
Flow area, Af = d * L * π / 4 2
Fluid weight per per unit length, Wf = SG * A f * n * dw
Wf =
21.68
lb / ft
Insulation outside outside diameter, Dio = Do + 2 Ti
Dio =
12.63
inch
Ai =
66.76
i n^ 2
Wi =
3.63
lb / ft
Aice =
0.00
i n^ 2
Wice =
0.00
lb / ft
W=
53.87
lb / ft
2
2
Insulation area area cross-section, A i = ( Dio - Do ) * π / 4 Insulation weight per unit length (w/ factor factor 1.12 for lagging), 2
2
W i = 1.12 * A i * n * Ins
2
Ice area cross-section, A ice = [ (Dio + ( 2 * T ice )) - Dio ] * π / 4 2
Ice weight (external w/ S.G. = 0.9) 0.9) per unit length, W ice = 0.9 * A ice * n * dw Total weight per unit length, W = Wp + Wr + Wf + Wi + Wice
(excluding Wc)
53.87 53. 87 lb / ft
2 of 2
Pipe or Tubing Support Span Calculations
8/4/2010, Sup.xls
Method ( A ) -- Reference book Design Of Piping Systems, pp. 239 - 240 & 356 - 358, by M. W. Kellogg, Second Edition, 1956
Type of support span
Single span, free ends
Length of support span
32.00 feet
Factor for modifying stress value obtained by the basic formula for S 2
Maximum calculated bending stress due to loads, S = fs ( 1.2 W L / Z ) ( 2 W c / W L )
4922.3 psi
L=
32.00
fs =
1.250
S=
4,922.3
feet
psi
The calculated bending stress is within limits. Factor for modifying deflection value obtained by the basic formula for y 4
Deflection, y = fy (17.1 ( W L / E * I ) ) Natural frequency estimate, fn = 3.13 / y
0.598 inch 0.5
fy =
1.316
y=
0.598
inch
fn =
4.046
Hz
OK. Natural frequency above 4 Hz is recommended for most spans. Consider 8 Hz or more for pulsating line. 2
2
2
Slope of pipe between supports for drainage, h = (12 * L) * y / ( 0.25 (12 * L) - y )
2.393 inch
h=
2.393
inch
L1 =
32.00
feet
4,927.8
psi
0.598
inch
Method ( B ) -- Reference Book: Piping Handbook, pg 22-47, by Crocker and King, Fifth Edition, 1967, McGraw-Hill Wc L1
Length of support span (Single Span w/ Free Ends)
32.00 feet
2
Maximum bending stress, Sw1 = ( 0.75 W * (L 1) + 1.5 W c * L1 ) * Do / I 4
Sw1 =
3
Deflection, y1 = (22.5 W * (L 1) + 36 Wc * (L1) ) / ( E * I )
0.60 inch
The calculated bending stress is within limits.
y1 =
Deflection is within recommended limits.
0.5
Natural frequency estimate, fn1 = 3.13 / ( y 1)
fn1 =
4.047
Hz
OK. Natural frequency above 4 Hz is recommended for most spans. Consider 8 Hz or more for pulsating line. 2
2
2
Slope between supports for drainage, h1 = (12 * L 1 ) * y1 / ( 0.25 (12 * L 1 ) - (y1) )
2.393 inch
h1 =
2.393
inch
39.00 feet
L2 =
39.00
feet
4,879.7
psi
0.264
inch
Wc L 2
Length of support span (Continuous Straight Run) 2
Maximum bending stress, Sw2 = ( 0.5 W * L 2 + 0.75 W c * L2) * Do / I 4
3
Deflection, y2 = ( 4.5 W * (L 2) + 9 Wc * (L2) ) / ( E *
Sw2 =
I )
0.26 inch
The calculated bending stress is within limits.
y2 =
Deflection is within recommended limits.
0.5
Natural frequency estimate, fn2 = 3.13 / (y 2)
fn2 =
6.092
Hz
OK. Natural frequency above 4 Hz is recommended for most spans. Consider 8 Hz or more for pulsating line. 2
2
2
Slope between supports for drainage, h2 = (12 * L 2 ) * y2 / ( 0.25 (12 * L 2 ) - (y2) )
1.056 inch
h2 =
1.056
inch
