Boiler Design – issue 4765

[DiogenesParticipant @diogenes]

re. WS – shouldn’t that be working pressure rather than test pressure? ..D x 110 rather than D x 220?

[MEinThailandParticipant @meinthailand]

Yield Point Method – AMBSC Code Uses it!

The Australian Miniature Boiler Safety Committee(AMBSC) Code Part 1 Copper boilers, on page 13 of Issue 8 – 2012 provides a formula for calculating stay pitches and diameters.

The formula uses the maximum allowable stress as that for annealed copper, 26,000 kPa (3,771 psi).

The code does not give a formula for the thickness of boiler shells but if it uses the value for annealed copper for stay design, wouldn’t it be incongruous if it used the UTS value of hard copper, plus a 6 to 10 times ‘safety factor’?

Perhaps someone with an inside knowledge of the AMBSC Code could clarify?

[Diogenes]

On Diogenes Said:

re. WS – shouldn’t that be working pressure rather than test pressure? ..D x 110 rather than D x 220?

No. The maximum stress a boiler will (normally) be under is during the hydrostatic test = 2 x Working Pressure.

[DiogenesParticipant @diogenes]

Apologies, I misunderstood the ‘W’..

[MEinThailand]

On MEinThailand Said:

The UTS Method Explained – Part 3 – Safety Factors

The factor of safety (FoS) in engineering, particularly in the design of steam boilers, is a measure of how much stronger a system is than it needs to be for its intended load. It is defined as the ratio of the ultimate strength (or failure stress) of a material to the allowable stress (or Working Stress – WS) applied in operation.

Mathematically, it is expressed as:

Factor of Safety = Ultimate Strength / Allowable Stress

Example 1
From the article “The Yield Pont Method ” by Les Smith and Alan Brown, published in Model Engineers & Workshop magazine in Volume 134, Issue 4765, June 2025 edition, page 26.

Taking a known published design for a G.W.R. boiler for a 5” gauge loco from Page 150 of The Model Steam Locomotive as an example, where:
Diameter, D = 4.75″
Working Pressure, WP = 110psi
Test Pressure = 2 x WP = 220psi
Shell Thickness, T = 0.092″

Using Barlow’s Formula. S = (P x D) / (T x 2)
WS (Stress) = (220 x 4.75) / (0.092 x 2)
WS = 5,679 psi

Ultimate Strength for annealed copper = Yield Point for annealed copper (to avoid permanent distortion) = 4,830 psi

Factor of Safety = Ultimate Strength / Allowable Stress
Factor of Safety = 4,830 / 5,679
Factor of Safety = 0.850

Yet the UTS Method proclaims a FOS of 8 (Average)

Clearly the UTS Method is both incorrect and misleading, because it does not allow for the Yield Point of annealed copper.

What you explain is how YOU back calculate a FOS in a given case against a given material property.  What you don’t explain is why passing yield or proof stress is deemed a failure.  The fact remains as proven by others on here and best part of a century of practice that either method results in a similar and safe design when used with the appropriate Factor of Uncertainty.  I use my term deliberately having explained how this mis-named FOS is actually used.

Martin

[Martin Johnson 1]

On Martin Johnson 1 Said:

What you explain is how YOU back calculate a FOS in a given case against a given material property.  What you don’t explain is why passing yield or proof stress is deemed a failure.

Which is exactly where we started 150+ posts ago.

[JasonBModerator @jasonb]

You don’t really need any qualifications to see that the FOS is just about cancelled out by the larger stress figure being used in the UTS equasion as the eight is above the line and the stress below which is why the two methods give a very similar result and why the UTS has served us well for years.

All you need to do is divide top and bottom by eight to get rid of the FOS and what does that leave at the bottom……………….5000 rather than 4830 of the YP method.

[duncan webster 1Participant @duncanwebster1]

If we believe MEinT’s figure for proof stress 4830 psi (please stop calling it yield), then boiler shells built to all the other criteria will all take on a permanent swelling at 2* pressure test, where they could be at 8000 psi (JasonB’s working stress of 4000 psi * 2). I’ve never seen one, and I’ve been involved in model engineering for 40+ years. It would be obvious, as the end near the tube plate is reinforced and so would not permanently deform. I’ve found another figure here which is for C12200, which claims to be equivalent to C106, which is what we in UK will use. It quotes 0.5% proof stress at 10,000psi. My figure was 0.2%, so I’d expect it to be a bit lower. As others say it wouldn’t matter if you did get some permanent set, but UK national standards are built around staying below proof stress.

If we adopt MEinT’s figure, all we achieve is more expensive boilers, not more safe. I very much hope that subsequent instalments of this saga will remain in the unused pile.

 

[JasonBModerator @jasonb]

I think you will be Ok there Duncan, chatting with Neil even he and Diane were unsure of publishing it.

[JasonBModerator @jasonb]

It should also be bourne in mind that this is a calculation for teh MINIMUM thickness and there are a lot of other factors that will go into arriving at the final tube thickness, to name a few.

– use Factor – will the boiler be a structiural part in which case you may add 20-25%

– Joint Factor – will there be a butt joint or tig welded copper joint, piercings such as bushes and domes add 20%+

– Construction considerations – will you need thicker tube where the barrel meets the firebox, will the barrel be unwrapped to form the firebox sides 10-20%

– nearest Available – Always go up in size to the next available which is becoming less as 13g has all but gone

All of these are going to increase the minimum calculated using the UTS method by more than the 3.5% difference between UTS and YS method so what’s the point in changing?

[duncan webster 1Participant @duncanwebster1]

You dont need to increase the thickness for domes, bushes etc, you just add local compensation, all detailed in BS and Aus code. For structures, which I’m more familiar with, a properly performed full penetration weld is reckoned to be as strong as parent material,  but Tubal Cain states a reduction to 80% for but straps in silver soldered boilers

[JasonBModerator @jasonb]

Ah, I was thinking of Haining but that was under steel boilers where 1 is used for an unpierced seamless shell and adjusted for holes and joints. Copper he only mentions joints, 0.8 (80%) for soldered or welded and down to 0.5 (50%) for rivited but unlikely to see that used now.

Would “localised compensation” be a thickening plate or similar?

 

[duncan webster 1Participant @duncanwebster1]

If its a small bush, the bush itself is usually enough (but needs checking) for a dome hole it can be a doubling plate around the hole. Best get a copy of the code.

Traction engines  are another thing altogether,  working to a lower acceptable stress for thd bits that ard loaded by the engine parts would be one way.

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