Hello All, 1st time posting here! I have a question concerning the staying of flat plates within a Section IV Heating Boiler.

According to the ASME Boiler and Pressure Vessel Code: Section IV Article III for Heating Boilers:

HG-342,2 Load Carried by by Stays.

"The area supported by a stay shall be computed on the basis of the full pitch dimensions with a deduction for the area occupied by the stay."

I'm not sure what the "computed on the basis of the full pitch dimension" refers to? For example, staying a flat plate with an area (supported on all four sides) that is 60" wide by 30" deep. If the full pitch has been determined to be 24", the plate needs to be stayed. Is the idea to calculate the area of 24" x 24" = 576" and "that portion" of the plates "area to stay" would be supported by the one stay located at the pitch dimension(x=24", y=24")? Then, apply that to the rest of the plate, (using the determined stay diameter) at pitch dimensions across the plate? That would require (2) stays to stay the plate?

Thanks!

Bithack,

1- Determine the pitch as per Fig. HG-345.1(a) or Fig. HG-345.1(b) as applicable.

2- Once you determine the pitch, area supported by the staying element shall be pitch square minus area occupied by the stay.(you have to assume the diameter of the stay rod, here. From assumed diameter you shall calculate the area occupied by the stay rod)

3- Force to carry by the saty shall be area determined in step 2 multiplied by design pressure/MAWP.

4- Devide the force calculated in step 3 by the allowable stress of the stay materail at the design temperature.

5- Multiply the result of step 4 by 1.1.

6- The result of the step 5 shall be required least cross sectional area of stay rod.

This is what I believe.

_________________
Thanks & Regards,
Nasir
Welding Engineer
DESCON Engineering Limited.
Lahore, Pakistan.
Mobile # 0092-336-4145402

Thanks for the reply! I found a technical brief describing a similar method that paralleled your steps. I have a follow up question:

Using the following data:

Assumed Stay Diameter: 1.50"
Stay Area: 1.767
Pressure (P): 40
Pitch (p): 24
Steel Pressure Rating: 16,000
Stay Load = P x (p2 - a): 22,969
Min. Required Area of Stay = 1.1 * Stay Load/Steel Pressure Rating: 1.579
Required Stay Diameter: 1.418 (1.5" Diameter would be appropriate)

Using a 60 x 30 plate (see attached Fig. 1), and assuming the plate is supported appropriately on all four sides, the plate would be stayed with (2) 1.50" diameter stays. However if I am not able to use a 1.50" diameter stay due to obstructions, can a lesser diameter stay (1" ?) be doubled up in a row (at some distance apart) to support the plate (see attached Fig.2)? How would you determine the accepted diameters and the dimension "N" between the smaller stays?


Thanks, again.