Continue to Site

Eng-Tips is the largest engineering community on the Internet

Intelligent Work Forums for Engineering Professionals

  • Congratulations MintJulep on being selected by the Eng-Tips community for having the most helpful posts in the forums last week. Way to Go!

Flat Plate Stay Diameter

Status
Not open for further replies.

bithack

Mechanical
Jan 17, 2009
8
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."

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?
 
Replies continue below

Recommended for you

i see that your 24x24 will need a bolt in the center, so more than 2 needed. there are also rules to stay to edge distance, so you will be short when you apply the edge rules.
figure a circle then put the stay in the center;
deduct the stay dia.; that's the area you need to stay.
 
Using the following calculations based on full pitch dimensions:

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?

Any one have ideas on this?
 
 http://files.engineering.com/getfile.aspx?folder=f5ff6431-71f7-4ce6-aacc-7d4f104d0edd&file=Stayed_Plate.pdf
NO,. you do not doubleup w/o calcs.
I see you want to use smaller dis. stays so you calc. the stay power instead of plate stay dia. needs.
reverse the calc and calc stay power then you will get the area so you apply it to your plate, 3/4 stays: you may end up wit 6 stays, 1 in may be 4 pcs.
 
My calcs from stay area are coming up with 8 stays if using 3/4" diameter stays and 6 with 1" stays, and they are not agreeing with my required diameter calcs?? Something is off; is it possible the calcs are referring to including the staying of the (already) supported edges of the plate?

The plate is 60" (Length) by 30" (width). The thickness of the plate requires a 24" pitch. I cannot go less than this dimension in the Length direction due to obstructions. Could you supply the stay power calc equation you are referring to so that I can check my calcs?
 
Hi, I think there is a rule of max. not more than 15 times
the dia. of the stay, Im I right?
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor