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Determination of the length between pipe supports by Requirements Pipe yield strengththe method of the "Maximum Ref. 2a. Table A-1 Bending Stress" [1] The ratio Lenght/deflection 25 is defined with a minimumData value of Pipe basic allowable stress atNominal pipe size (NPS) (L/y)min = 600 given temperature
300 mm Ref. 2a. Table A-1Material: A 312 TP316L Maximum bending stress t = 72Pipe pressure and temperature ratio 16.7
P = 20 bar 0.3 115.1t = 22 ºC
Pipe filled with water bending stress Maximum bending stress
It will be considered the weight in tension according B31.3 115.1of two men at the center of Table A-1 34.54the pipe. 3.5E+07Concentrated weight The lenght is designed for the
200 kg pipe working with the Steel elasticity module1961 N maximum allowed bending Sheet SS 316
Pipe insulation weigth E = 2E+110 N/m
Pipe shedule is to be determined
Water load Concentrated weight1961 N
1000 kg/m³V = #VALUE! m³/m Pipe length (eq. (h)
#VALUE! N/mb = 245
Pipe load (steel) a = #VALUE!a = w/16 c = #VALUE!
(from sheet SS 316) w = #VALUE! N/m #VALUE!8000 kg/m³ a = #VALUE! Let
#VALUE! m 10#VALUE! m b =#VALUE! N/m 1961 N
b = 245 hast to meet two conditions:
st =
dn =
sallow =sb/sallow = sallow =sb : Maximum allowed
sallow : Pipe allowed stress sb = sallow * 0.3sallow =sb =sb =
mc =wc =
stress sb.wi =
ww = rw * g * V wc =rw =
ww = Lcalc = (-b + (b^2 - 4*a*c)^0.5 ) / (2*a)
wp = rp*g*(p/4)*(de^2-di^2)
Lcalc =rp =de = Lselected =di = wc /8wp = wc = The selected length Lselected
w16
⋅L2+ wc8⋅L-
σb⋅Id
=0 (h )
Insulation load (no insulation)0 N/m c = -sb * I /de the condition for the maximum
3.5E+07 Pa bending stress.
Total load I = #VALUE!w = #VALUE! m the additional design condition.
#VALUE! N/m c = #VALUE!#VALUE! N/m
0 N/mw = #VALUE! N/m
Shedule Maximum bending stress Weld joint strenght reduction Material A 312 TP316L Material: A 312 TP316L factor WNominal pipe size (NPS) Ref. 2a. Table A-1 Ref. 2c. Table 302,3,5
dn = 300 mm W = 1Schedule, to be determined Yield strength Table A-1
Pressure in pipe 25 ksi Coefficient YP = 20 bar Table 304.1.1
Temperature Allowed stress Table A-1 Ref. 2 ASME B31.3t = 22 at the temperature Y = 0.4
16.7 ksiPipe thickness 115.1 MPa Check requirementEquation 3a, ASME B312.3-2008 P / S*E <= 0.385
t = P*D / ( 2*(S*E*W + P*Y) ) Maximum bending stress P = 2t: Pressure design thickness 34.54 MPa S = 34.54P: Internal design gauge pressure E = 1D: Outside pipe diameter Quality factor "E" P / S*E = 0.057899d: Inside pipe diameter Ref. 2b. Table A-1Bc: sum of mechanical, corrosion Seamless tube A 316L Eq. (3a) is applicable, since P/(S*Eand erosion allowances E = 1S: Material stress value. Table A-1
E: Quality factor from table A-1A or A-1B Exterior pipe diameterW: Weld joint strength reduction factor per dn = 300 mmpara. 302.3.5€ and Table 302.3.5 #VALUE! mmY: Coefficient from Table 304.1.1, valid for t <D/6
For t>= D/6, Y = (d + 2*c) / ( D + d + 2*c )
Presure design thickness (acc. Eq. 3a) For a SS pipe
1. Lselected <= Lcalculated to meet
wi =sb =
2. ycalc <= Lselected / 600 to meet
ww + wp + wi de =ww =wp =wi =
syield =
sallow =sallow =
sb =
de =
#VALUE! mm dn = 300Corrosion allowance TC = 1.59 mm sch = 40S
Thread depth TD = 0 mm s = #VALUE!
Over thickness OT = TC + TDOT = 1.59 mm A schedule of 40S
Required thickness #VALUE! #VALUE!#VALUE! mm #VALUE!
Mill tolerance MT = 12.5 % Selecting the next scheduleMinimum thickness For a SS pipe
#VALUE! mm dn = 300MT = 12.50 % sch = 80S
#VALUE! mm s = #VALUE!
A schedule of 80S
For more information on these correction factors, see #VALUE! #VALUE!#VALUE!
Pipes. Wall thickness calculation according ASME B31.3
Selected shedule
P * dext / ( 2* (sallow * E*W + P* Y) )P = 2
de = #VALUE!Sb = 34.53E = 1W = 1y = 0.4
#VALUE!#VALUE!
The reference has a calculation mistake
tdis =
treq = tdis + OTtreq =
tmin = treq * (100 /(100 - MT))treq =
tmin =
www.piping-tools.net
t =
t =
t =
t = 9.2 mm
ASME B31.3-2008
t=P⋅dext2⋅[σ⋅E⋅W +P⋅Y ]
(3a )
SS pipedn Schin -
1/8 5S 1/4 10S 3/8 40S 1/2 80S 3/41 1 1/41 1/22 2 1/23 3 1/2
4568
101214161820222430
Corrosion allowance Pipe dimensionsTC = 1.59 mm The schedule is to be determined
ksi Thread depthTD = 0 mm Schedule initially assumed
Pipe basic allowable stress at Mill tolerance Sch = 40SMT = 12.5 %
Exterior diameter (of SS pipe)ºF Note. 300 mmksi The above values are not considered Pipe_SS_Dext_dnMPa in the example of the reference [1]. #VALUE! mm
#VALUE! m Maximum bending stress
Interior diameterMPa dn = 300 mmMPa Pipe_SS_Dint_dn_schPa Sch = 40S
#VALUE! mmSteel elasticity module #VALUE! m
PaPipes. Maximum span between pipe supports for a given maximum bending stress.xlsm
Deflection Result Deflection at ther center of a simple Forsupported beam with a uniform load 10 mw and a concentrated load "wc". there is a deflextion(from sheet equations) #VALUE! mm
Requirement
m 10 mDeflection value for the 0.0167 m
m selected length 16.67 mm
w = #VALUE! N/m Check10.0 m #VALUE! mm
dn =de =de =de =
di =
di =di =
www.piping-tools.net
cjcruz@piping-tools.net
Lselected =
= (-b + (b^2 - 4*a*c)^0.5 ) / (2*a) ycalc =
ycalc <= Lselected / 600Lselected =
Lselected / 600 =
Lselected / 600 =
ycalc = (5*w*L^4 + 8*wc*L^3) / (384*E*I)
Lselected = ycalc =
y=5⋅w⋅L4+8⋅wc⋅L
3
384⋅E⋅I (q )
1961 16.67 mmthe condition for the maximum E = 2.00E+11 Pa thus
I = #VALUE! #VALUE!#VALUE! m #VALUE!
the additional design condition. #VALUE! mm
Weld joint strenght reduction Design thicknessRef. 2 ASME B31.3 ASME B31.3-2008ASME B31.3 paragraph 304.1.2, Ec. (3a)
P = 2.0 Mpa#VALUE! mm
34.54 Mpa
E = 1.0 -
Mpa W = 1.00 -
MPa Y = 0.4 -
t = #VALUE! mm
Design thickness requirementEq. (3a) is applicable, since P/(S*E t = #VALUE! mm
#VALUE! mm#VALUE! mm#VALUE!
wc = Lselected / 600 =
m4 ycalc Lselected / 600
ycalc =ycalc =
t = P * dext / ( 2* (sallow * E*W + P* Y) )
dext =sb =
dext =dext/6 =
ASME B31.3-2008
t=P⋅dext2⋅[σ⋅E⋅W+P⋅Y ]
(3a )
mm For the given data, a span L = 10 m will ensure that the maximum
mm 34.5 MpaIt was considered L = 10
gives a thickness Additionally it holds that L/y>= 600 -#VALUE! The model used is a simple supported beam y = #VALUE!
#VALUE!
#VALUE!Selecting the next schedule The selected shedule is sch = 80S
#VALUE!mm
mm
gives a thickness#VALUE!
#VALUE!
80S
g =
P * dext / ( 2* (sallow * E*W + P* Y) )MPam MPa
mmm
The reference has a calculation mistake
bending stress is less than sb =sb = sallow * 0.3
spipe =
srequired =
ASME B31.3-2008
t=P⋅dext2⋅[σ⋅E⋅W +P⋅Y ]
(3a )
Rev. cjc. 01.06.2017
Page 1 of 5
Equatorial inertia momentI =
#VALUE! m#VALUE! m
I = #VALUE!
Area of pipe sectionA =
#VALUE! mA = #VALUE! m²
Water volume per meter pipeV = A * 1A = #VALUE! m²/mV = #VALUE! m³/m
Pipes. Maximum span between pipe supports for a given maximum bending stress.xlsm
Page 2 of 5
(p/64) *( de^4-di^4)de =di =
m4
(p/4) * di^2di =
Page 3 of 5
Page 4 of 5
Microsoft Equation 3.0
m
m
mm
mm
9.80665 m/s²
Page 5 of 5
[1]
t =
Pipe seamless SSMater: D: Outside diameter
dn =
P=
E =W Y =
Stainless steeldn =sch =
s =
Shedule to be selected: 80S
Note. This example is for a stainless steel pipe.The reference uses for the selection the dimensionsof carbon steel pipesStainless steel
dn =sch =
s =
Carbon steeldn =sch =
s =
de =
t =
de =sb =
t =
Sabin Crocker. Piping Handbook, 4th editionMcGraw Hill N.Y. Pg. 744-745
[4]The American Society of Mechanical Engineers (1995)ASME B31.3 - 2000 Edition, Process Piping, ASME, New York pp 182
Pipe seamless SSASTM A 312 TP 316L
D: Outside diameter 300 NPS
#VALUE! mm
P * dext / ( 2* (sallow * E*W + P* Y) )2 MPa
#VALUE! mm34.53 MPa
11
0.4#VALUE! mm
Stainless steel Stainless steel300 mm dn = 300 mm40S sch = 80S
#VALUE! mm s = #VALUE! mm
Shedule to be selected: 80S
Note. This example is for a stainless steel pipe.The reference uses for the selection the dimensionsof carbon steel pipesStainless steel
300 mm dn = 300 mm20S sch = 30S
#VALUE! mm s = #VALUE! mm
Carbon steel Carbon steel300 mm dn = 300 mm20 sch = 30
#VALUE! mm s = #VALUE! mm
P * dext / ( 2* (sallow * E*W + P* Y) )
Sabin Crocker. Piping Handbook, 4th editionMcGraw Hill N.Y. Pg. 744-745
[4]The American Society of Mechanical Engineers (1995)ASME B31.3 - 2000 Edition, Process Piping, ASME, New York pp 182
Length of a beam to achieve thatthe resulting beam stress has thedefined maximum value
Assumed a simply supported case
sallowed = (2/3) *sy
Microsoft Equation 3.0
Moments at the center of a simple supported beam .[ 5 ], Figure 7
Concentrated load w c at center
M=wc⋅L4
Uniformly distributed load w
M=w⋅L2
8Total moment
M=w⋅L2
8+ w c⋅
L4
( a)
Microsoft Equation 3.0
Bending stress σ
σ=M⋅cI
(b)
M:moment N⋅m c: distance to the neutral axis mI: equatorial inercia moment m 4
Maximum bending stress
σmax=M⋅cmax
I (c )
In case of a pipe
cmax=d2
(d )
and replacing equation ( d ) into (c )
σmax=M⋅(d /2 )I
(e )
Replacing equation
M=w⋅L2
8+ w c⋅
L4
( a)
into equation (e )
σmax=(w⋅L2
8+ w c⋅
L4 )⋅d /2I ( f )
where the maximum stress will have the value corresponding to the givenapplication, and will be denotedsimply as the beam stress σb
σ b=(w⋅L2
8+ w c⋅
L4 )⋅d /2I
σ b=(w⋅L2
16+ wc⋅
L8 )⋅dI ( g)
Beam stressFor the stainless steel A 312 TP316L,ASME B31. 3-2006, Table A-1 givesthe relation between the allowablestress σallow( t <= 300 ºF ) to the yield value σ y as
σ allow=23σ y (n )
Beam stress σ b
Literature [1 ] recommends for theacual application, following securitymargin
σ b=310
⋅σallow (o )
replacing equation
σ allow=23⋅σ y (n )
into equation (o )
σ b=310
⋅23⋅σ y
σ b=15⋅σ y ( p )
Microsoft Equation 3.0
Beam stress σ b
Literature [1 ] recommends for theacual application, following securitymargin
σ b=310
⋅σallow (o )
replacing equation
σ allow=23⋅σ y (n )
into equation (o )
σ b=310
⋅23⋅σ y
σ b=15⋅σ y ( p )
Microsoft Equation 3.0
Deflection at the center of a simple supported beam .[ 5 ], Figure 7
Concentrated load w c at center
y=wc⋅L
3
48⋅E⋅IUniformly distributed load w
y =5⋅w⋅L4
384⋅E⋅ITotal deflection
y=5⋅w⋅L4
384⋅E⋅I+wc⋅L
3
48⋅E⋅I
y=5⋅w⋅L4
384⋅E⋅I+
8⋅wc⋅L3
384⋅E⋅I
y=5⋅w⋅L4+8⋅wc⋅L
3
384⋅E⋅I (q )
y=5⋅w⋅L4+8⋅wc⋅L
3
384⋅E⋅I (q )
In the following application thevalue of the beam stress givenby equation (n) will be used.This value is 20% of the yield stressThe in the standard indicated allowedvalue is 67% of the yield stress.
Microsoft Equation 3.0
Replacing equation
M=w⋅L2
8+ w c⋅
L4
( a)
into equation (e )
σmax=(w⋅L2
8+ w c⋅
L4 )⋅d /2I ( f )
where the maximum stress will have the value corresponding to the givenapplication, and will be denotedsimply as the beam stress σb
σ b=(w⋅L2
8+ w c⋅
L4 )⋅d /2I
σ b=(w⋅L2
16+ wc⋅
L8 )⋅dI ( g)
Microsoft Equation 3.0
from equation
σ b=(w16⋅L2+
wc8
⋅L)⋅dI (g )
σ b⋅Id
=w16
⋅L2+ w c8
⋅L
w16
⋅L2+ w c8
⋅L-σ b⋅Id
=0 (h )
.a . .. . . .. . . .. . . .b . . . .. . . .. . . .cwith
a=w16
( i)
b=wc8
( j)
c=−σ b⋅Id
(k )
equation ( h) becomes a⋅L2+b⋅L+c=0 (m )
σ b=15⋅σ y ( p )
Beam stress σ b
Literature [1 ] recommends for theacual application, following securitymargin
σ b=310
⋅σallow (o )
replacing equation
σ allow=23⋅σ y (n )
into equation (o )
σ b=310
⋅23⋅σ y
σ b=15⋅σ y ( p )
Note that the length obtained fromequation (m) corresponds to the case when the system is assumedto be similar to a simply supportedbeam. A continuos pipe would bebetter represented by sectionscorresponding to a beam fixed atboth ends.
w16
⋅L2+ wc8⋅L-
σb⋅Id
=0 (h )
y = P*L^3/(48*E*I)
y = (5/384)*(w*L^4)/(E*I)
Microsoft Equation 3.0
Beam stress σ b
Literature [1 ] recommends for theacual application, following securitymargin
σ b=310
⋅σallow (o )
replacing equation
σ allow=23⋅σ y (n )
into equation (o )
σ b=310
⋅23⋅σ y
σ b=15⋅σ y ( p )
y=5⋅w⋅L4+8⋅w c⋅L
3
384⋅E⋅I (q )
Page 1 of 3
Page 2 of 3
Microsoft Equation 3.0
Note that the length obtained fromequation (m) corresponds to the case when the system is assumedto be similar to a simply supportedbeam. A continuos pipe would bebetter represented by sectionscorresponding to a beam fixed atboth ends.
Page 3 of 3
Ferritic Steel A term usually applied to a group of stainless steels with a chromium content in the range of 12- 18o and whose structure consists largely of ferrite.Such steels possess good ductility and are easily worked but do not respond to any hardening or tempering processes. Types of applications include automotive trim and architectural cladding.
Austenitic Steels Steels containing high percentages of certain alloying elements such as manganese and nickel which are austenitic at room temperatureand cannot be hardened by normal heat-treatment but do work harden. They are also non-magnetic. Typical examples of austenitic steels include the 18/8 stainless steels and 14% manganese steel.
Martensitic Stainless Steels(400 SERIES WHICH HAVE HIGH CARBON). These grades of stainless have chromium in the range of 11% to 17% as the sole major alloying addition. This is the same as the ferritic grades. However, carbon is added in amounts from 0.10 % to 0.65% to radically change the behavior of the martensitic alloys.
The high carbon enables the material to be hardened by heat treatment.
ASME B31.3-2008Process PipingASME Code for Oressure Piping, B31
t = P*D / ( 2*(S*E*W + P*Y) )
t: Pressure design thicknessP: Internal design gauge pressureD: Outside pipe diameterd: Inside pipe diameterc: sum of mechanical, corrosion and erosion allowancesS: Material stress value. Table A-1 E: Quality facto from table A-1A or A-1BW: Weld joint strength reduction factor perpara. 302.3.5€Y: Coefficient from Table 604.1.1, valid for t <D/6and for materials shown.For t>= D/6, Y = (d + 2*c) / ( D + d + 2*c )
Coefficient YTable 304.1.1Ref. 2 ASME B31.3
ASME B31. 3-2008
t=P⋅dext2⋅[σ⋅E⋅W+P⋅Y ]
(3a )
Microsoft Equation 3.0
Y = 0.4
Equation 3a, ASME B312.3-2008
(b) Equations (3a) and (3b) are not valid fort >= d/6 orP / S*E > 0.385
A term usually applied to a group of stainless steels with a chromium content in the range of 12- 18o and "Y" values for ferritic steels (t < D/6)Table 304.1.1, ASME B31.1-2008, page 20.
Such steels possess good ductility and are easily worked but do not respond to any hardening or tempering processes. 0.40.50.7
Steels containing high percentages of certain alloying elements such as manganese and nickel which are austenitic at room temperatureand cannot be hardened by normal heat-treatment but do work harden. They are also non-magnetic. Typical examples of austenitic steels include the 18/8 stainless steels and 14% manganese steel.
(400 SERIES WHICH HAVE HIGH CARBON). These grades of stainless have chromium in the range of 11% to 17% as the sole major alloying addition. This is the same as the ferritic grades. However, carbon is added in amounts from 0.10 % to 0.65% to radically change the behavior of the martensitic alloys.
Y<482 °C =Y482 to 510 °C =
Y > 510 °C =
st
Spec Grade ksiA 312 TP316L 25
tSpec Grade ºFA 312 TP316L < 100
< 200
16.716,700115.1
syield
at tmin to 100 ºF
sallowed = (2/3) *syield
sallowed =sallowed =sallowed =
st sy sallow
sy sallow st
ksi16.716.7
ksipsiMPa
sallowed
st
Quality factor
E = 1
Weld joint strenght reduction factor W
Weld joint strenght reduction factor W
W = 1
http://www.azom.com/properties.aspx?ArticleID=863
https://www.google.com/fusiontables/DataSource?docid=1wvrlddD8tQHoDc7feP9pqJ2AwImBdcPJpEqV4icDensity of stainless steel
kg/m³
SS 3168000 kg/m³r =
Elasticity modulus190 Gpa205 Gpa
Selected valuehttps://www.google.com/fusiontables/DataSource?docid=1wvrlddD8tQHoDc7feP9pqJ2AwImBdcPJpEqV4ic E = 200 Gpa
E = 2E+11 Pa
Emin =Emax =
http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf
http://www-classes.usc.edu/engr/ce/457/moment_table.pdf
http://www.awc.org/pdf/codes-standards/publications/design-aids/AWC-DA6-BeamFormulas-0710.pdf
Beam simply supported at endsMaximum deflection at beam center a ) Uniformly distributed load w
y=5⋅w⋅L4
384⋅E⋅I
b ) Concentrated load P at the center
y=8⋅P⋅L3
384⋅E⋅Ic ) Both cases, a and b
y=5⋅w⋅L4+8⋅P⋅L3
384⋅E⋅I
Beam fixed at both endsMaximum deflection at beam center a ) Uniformly distributed load w
y=w⋅L4
384⋅E⋅I
b ) Concentrated load P at the center
y=16⋅P⋅L3
384⋅E⋅Ic ) Both cases, a and b
y=w⋅L4+16⋅P⋅L3
384⋅E⋅I
Simple beam. - Uniformly distributed load
Maximum moment M = w * l^2 / 8y = (5/384)*(w*L^4)/(E*I)
Simple beam. - Concentrated load at the center
Maximum moment M = P * l / 4
http://www.awc.org/pdf/codes-standards/publications/design-aids/AWC-DA6-BeamFormulas-0710.pdf
Microsoft Equation 3.0
Beam fixed at both endsMaximum deflection at beam center a ) Uniformly distributed load w
y=w⋅L4
384⋅E⋅I
b ) Concentrated load P at the center
y=16⋅P⋅L3
384⋅E⋅Ic ) Both cases, a and b
y=w⋅L4+16⋅P⋅L3
384⋅E⋅I
M = w * l^2 / 8M = P * l / 4
y = P*L^3/(48*E*I)
Microsoft Equation 3.0
Microsoft Equation 3.0
Beam simply supported at endsMaximum deflection at beam center a ) Uniformly distributed load w
y=5⋅w⋅L4
384⋅E⋅I
b ) Concentrated load P at the center
y=8⋅P⋅L3
384⋅E⋅Ic ) Both cases, a and b
y=5⋅w⋅L4+8⋅P⋅L3
384⋅E⋅I
Simple beam. - Concentrated load at the center
[1] Determination of maximum span between pipe supports using maximum bending stress theoryDr. D.P. Vakharia, Mohd Farroq A.
http://ijrte.academypublisher.com/vol01/no06/ijrte0106046049.pdf
[2] ASME B31.3-2008Process PipingASME Code for Pressure Piping, B31
[3] Sabin Crocker. Piping Handbook, 4th editionMcGraw Hill N.Y. Pg. 744-745
[4] The American Society of Mechanical Engineers (1995)ASME B31.3 - 2000 Edition, Process Piping, ASME, New York pp 182
[5]
http://www.awc.org/pdf/codes-standards/publications/design-aids/AWC-DA6-BeamFormulas-0710.pdf
http://www.awc.org/pdf/codes-standards/publications/design-aids/AWC-DA6-BeamFormulas-0710.pdf
Microsoft Equation 3.0
Microsoft Equation 3.0
Microsoft Equation 3.0
Microsoft Equation 3.0
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