Railway Preservation News
https://www.rypn.org/forums/

Curved Stayed Surfaces
https://www.rypn.org/forums/viewtopic.php?f=1&t=34850		 Page 1 of 1
Author:  Paul Boschan [ Mon Apr 08, 2013 5:38 pm ]
Post subject:  Curved Stayed Surfaces
For Steamer Dave:
The purpose of this post is to describe the rules for determining the maximum allowable working pressure (MAWP) for stayed curved surfaces (a locomotive wrapper sheet for example) found in the ASME Code, Section I, paragraph PFT-23. Essentially, the MAWP of a wrapper sheet is the lower pressure determined by the rules in PFT-23.1 or PFT-23.2.

Step 1: Per PFT-23.1, determine the MAWP of the wrapper using the efficiency of the ligaments between radial stay bolts without taking credit for the holding power of the bolts. Ligaments with an equal pattern are addressed in PG-52.2.1.
E=(p-d)/p
Where E = the efficiency; p = longitudinal pitch (distance from stay bolt center to stay bolt center measuring from front to back); d = diameter of the stay bolt hole. As an example we’ll assume p = 4” and d = 1” which gives E = 0.75.

Step 2: Use the formula in PG-27.2.2 to determine the MAWP of the wrapper using the efficiency calculated in Step 1.
P=(2SE(t-c))/(D-2y(t-c))
Where P = MAWP; S = maximum allowable stress of the wrapper material (in simple terms, the allowable stress is the nominal ultimate tensile strength (UTS) of the plate divided by the factor of safety); E = efficiency determined in Step 1; t = thickness of the wrapper plate; c = corrosion allowance (if any); D = outside diameter of wrapper sheet; y = 0.4 (this is a temperature coefficient from Note 6 in PG-27.4.6, for carbon steel at or below 900° F y = 0.4).
Assume:
S = 20000psi (SA-516-70 nominal UT of 70000/3.5. 3.5 is the ASME FS, for FRA it is 4)
E = 0.75
t = 0.5”
c = .125”
D = 60”
P=(2×20000×0.75(0.5-.125))/(60-2×0.4(0.5-.125))
P = 252psi
Here is an example using data one might encounter when calculating an existing wrapper sheet on a locomotive in FRA regulated service:
S = 12500psi (the FRA allows one to assume 50000psi UTS when the actual nominal UTS of the steel in unknown. 50000psi divided by the FRA required factor of safety of 4 equals 12500)
E = 0.75
t = 0.5”
c = .125”
D = 60”
P=(2×12500×0.75(0.5-.125))/(60-2×0.4(0.5-.125))
P = 118psi

Step 3: Determine the MAWP per PG-46 using 1.3 for the value of C. the value p = the area supported by a staybolt. In a wrapper sheet where the distance between the stay bolts increases from row to row, there will not be an even pitch. Use the total area supported by a single stay bolt in place of p2.
P=(t^2 SC)/p^2
Using the previous example:
P=(〖0.5〗^2×20000×1.3)/28
Here, 28 square inches replaces p2.
P=232psi
Using the FRA example:
P=(〖0.5〗^2×12500×1.3)/28
P=145psi

Step 4: Determine the MAWP using the following formula:
P_1=(A_1 S)/A_2
Where P1 = pressure corresponding to the strength of the stay; A1 = cross-sectional area of the stay (in the example I used the net cross-sectional area of a 1” diameter stay with a .188” diameter telltale hole); A2 = maximum area supported by a stay; S = allowable stress of the stay. For SA-36 steel round bar, S = 16,600 at 400° F (~58,000 UTS / 3.5)
P_1=(.758×16600)/28
P1 = 449psi

Step 5: Add the value obtained in Step 2 (225psi) to the lower value obtained in either Step 3 or Step 4 (232psi from Step 3) to get a final MAWP of 457psi.

Step 6: The MAWP for a locomotive wrapper sheet is the lesser of the amount from Step 5 or the amount from the following equation:
P=StE/R-∑(s×sina)
Where P = MAWP; S = allowable stress of the wrapper; E = efficiency as determined in Step 1; R = radius of the wrapper sheet; s = transverse spacing of the crown stays in the crown sheet; a = angle a radial crown stays makes with the vertical axis of the boiler.
The Greek letter Sigma means, “the sum of.” Before proceeding, a list of all the crown stay angles must be made and the sin of each found. For our example we’ll assume the crown stay angles are 3°, 10°, 20°, 32° and 45°. The sin of each, respectively is .052, .174, .342, .530, .707. These numbers multiplied by the transverse pitch of the crown stays (4”) are .208, .696, 1.37, 2.12, 2.83. Added together these last numbers equal 7.22.
P=(20000×0.5×.75)/(30-7.22)
P = 329psi
Since the value determined in Step 6 (329psi) is lower than the value determined in Step 5 (457psi), the MAWP of the wrapper sheet in the example is 329psi. Note I did not show an FRA example for Steps 4, 5 or 6. For these steps, substitute an appropriate allowable stress to meet the FRA requirements.
Page 1 of 1 All times are UTC - 5 hours [ DST ]
Powered by phpBB © 2000, 2002, 2005, 2007 phpBB Group
http://www.phpbb.com/