Pipe or Tubing Support Span Calculations8/4/2010, Sup.xls

Client: Description:

Customer No.: Item No.:

Owner No.: Dwg. No.:

The user is responsible for verifying that method and results are correct.

Service: new process

Design Temperature: F

Factor of Safety: FS =

Pipe or Tubing Material:

ASTM A 53-B Type S Carbon Steel Pipe

Hot modulus of elasticity 2.93E+07

Hot allowable stress 20,000

Design stress, Sh / FS 5,000

Material density Pden =

Relative weight factor (vs. carbon steel) Mat =

Insulation: Semi-rigid Fiberglass

Density, Ins = 7

Thk, Ti = 2

Comments: x

x

x

Minimum pipe thickness, t = Tnom - (Tnom * Mill % / 100 ) - c - h

Outside radius of pipe, R = Do / 2

4

Pipe moment of inertia (based on t ), I = π / 64 [ Do

Pipe inside diameter, d = Do - 2 Tnom

4

Pipe section modulus, Z = π / 32 (Do

2

Metal area of pipe cross-section, Am = ( Do - d

Conversion factor, water density, dw

Conversion factor: n = 1 ft / 12 in

Pipe weight per unit length, Wp = Pden * Am / n

Pipe or liner or refractory inside diameter, dL = d - 2 Tr

2

Liner area cross-section, Ar = ( d

Liner weight per unit length, Wr = Lden * Ar / n

2

Flow area, Af = d * L * π / 4

2

Fluid weight per unit length, Wf = SG * Af * n

Insulation outside diameter, Dio = Do + 2 Ti

2

Insulation area cross-section, Ai = ( Dio

Insulation weight per unit length (w/ factor 1.12 for lagging),

2

Ice area cross-section, Aice = [ (Dio + ( 2 * Tice ))

Ice weight (external w/ S.G. = 0.9) per unit length, Wice =

Total weight per unit length, W = Wp + Wr + Wf + Wi + Wice

Pipe or Tubing Support Span Calculations 1 of 2

Prepared By: Approval: Date: Rev.:

0

1

2

3

4

SG = 1 Measurement Units: Imperial

C 100 Deg. F Corrosion, erosion, mech. allow: C = 0 inch

4 Ice Thickness (external): Tice = 0 inch

Pipe or Tube Dimensions:

Nom. Size: NPS 8 (DN 200)

psi E = 2.934E+07 psi Nom. Thk.: Sch 40 per ASME B36.10M

psi Sh = 20,000 psi Actual OD: 8.625 inch Do = 8.625

psi Sa = 5,000 psi Nominal thick 0.322 inch Tnom 0.322

2.833E-01 lb/in^3 Thickness tolerance: Mill = 12.5% and, h = 0 inch

1

Liner on ID: none

lb/ft^3 Iden = 4.051E-03 lb/in^3 Density: 0 lb/in^3 Lden0.000E

inch Thickness: 0 inch Tr = 0 inch

Weight of concentrated load assumed at center of

span (valves + flanges + animal, etc.):

0 lbs Wc = 0 lbs

t = 0.3216 inch

R = 4.313 inch

4

- (Do - 2 t ) ] 7.241E+01 in^4 I = 7.241E+in^4

d = 7.981 inch

4

- d ) / Do 1.681E+01 in^3 Z = 1.681E+in^3

2

) * π / 4 Am = 8.4 in^2

= 62.4 lb / cu ft dw = 62.4 lb/ft^3

n = 8.3333Eft / in

Wp = 28.55 lb / ft

dL = 7.981 inch

2

- dL ) * π / 4 Ar = 0 in^2

Wr = 0 lb / ft

Af = 5.003E+in^2

* dw Wf = 21.68 lb / ft

Dio = 12.63 inch

2

- Do ) * π / 4 A =

i 66.76 in^2

2

W = 1.12 * A* Ins W =

i i 3.63 lb / ft

2

- Dio ] * π / 4 Aice = 0 in^2

2

0.9 * Aice * * dw Wice = 0 lb / ft

(excluding Wc) 53.87 lb / ft W = 53.87 lb / ft

inch

inch

lb/in^3

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

Length of support span

Factor for modifying stress value obtained by the basic formula for

Maximum calculated bending stress due to loads,

Factor for modifying deflection value obtained by the basic formula for

4

Deflection, y = fy (17.1 ( W L

Natural frequency estimate, fn = 3.13 / y

Slope of pipe between supports for drainage,

Method ( B ) -- Reference Book:

Length of support span (Single Span w/ Free Ends)

Maximum bending stress, Sw1 = ( 0.75 W * (L1)

4

Deflection, y1 = (22.5 W * (L1)

The calculated bending stress is within limits.

Natural frequency estimate,

Slope between supports for drainage,

Length of support span (Continuous Straight Run)

Maximum bending stress, Sw2 = ( 0.5 W * L2

4

Deflection, y2 = ( 4.5 W * (L2)

The calculated bending stress is within limits.

Natural frequency estimate,

Slope between supports for drainage,

Pipe or Tubing Support Span Calculations

Reference book Design Of Piping Systems, pp. 239 - 240 & 356 - 358, by M. W. Kellogg, Second Edition, 1956

Single span, free ends

S

2

S = fs ( 1.2 W L / Z ) ( 2 Wc / W

y

/ E * I ) )

0.5

OK. Natural frequency above 4 Hz is recommended for most spans. Consider 8 Hz or more for pulsating line.

2

h = (12 * L) * y / ( 0.25 (12

Piping Handbook, pg 22-47, by Crocker and King, Fifth Edition, 1967, McGraw-Hill

Wc

L 1

2

+ 1.5 Wc * L1 ) * Do / I

3

+ 36 Wc * (L1) ) / ( E * I )

0.5

fn1 = 3.13 / ( y1)

OK. Natural frequency above 4 Hz is recommended for most spans. Consider 8 Hz or more for pulsating line.

2 2

h1 = (12 * L1 ) * y1 / ( 0.25 (1

Wc

L 2

2

+ 0.75 Wc * L2) * Do / I

3

+ 9 Wc * (L2) ) / ( E * I )

0.5

fn2 = 3.13 / (y2)

OK. Natural frequency above 4 Hz is recommended for most spans. Consider 8 Hz or more for pulsating line.

2 2

h2 = (12 * L2 ) * y2 / ( 0.25 (1

2 of 2

32 feet L = 32 feet

fs = 1.25

4922.3 psi S = ### psi

The calculated bending stress is within limits.

fy = 1.32

0.598 inch y = 0.6 inch

fn 4 Hz

2 2

- y ) 2 inch h 2 inch

32 feet L1 32 feet

Sw1 = ### psi

0.6 inch y1 0.6 inch

Deflection is within recommended limits.

fn1 4 Hz

2

- (y1) ) 2 inch h1 2 inch

39 feet L2 39 feet

Sw2 = ### psi

0.26 inch y2 0.26 inch

Deflection is within recommended limits.

fn2 6.09 Hz

2

- (y2) ) 1 inch h2 1 inch