High Tensile Steel Question

[HopperParticipant @hopper]

What RPM will it be doing? Doesn't seem like it would be high enough to generate centrifugal force sufficient to tear 7/16 diameter 8.8 bolts asunder?

If I were making them from scratch here in Oz, I would use SAE 4140 steel, a chro-mo alloy used for general high tensile work in most jobbing shops. UK equivalent is EN19. It is tough stuff.

[Tim StevensParticipant @timstevens64731]

I am not encouraged by a detailed peer at the drawing. In particular – the bolts in question are drawn with a sharp corner where the head meets the shaft. Really?

Tim

[SillyOldDufferModerator @sillyoldduffer]

Duncan's suggestion of establishing the mass of the counter-balance was a good one. If the mass and rotational speed are known, it's easy to calculate the centripetal force on the bolts:

F = mv²/r

m is mass,

v is velocity

r is radius, distance from the axle's axis

An ordinary nasty M20 bolt will take about 9000kg force. 8.8 high-tensile bolts about twice that, and 12.9 bolts three times.

Not done the sums because the weight has to be CAD extracted but I suspect a 5" scale-model traction engine doesn't put an enormous strain on the bolts compared with a full-size engine worked hard every day for years on end. Traction engines are low RPM.

High-tensile is probably specified on the full-size engine because the extra strength resists fatigue stress, especially if torqued properly. The strength of bolt needed to hold a static load is much lower than one subjected to alternating stresses.

Doing the sums would reveal if a pair of ordinary bolts capable of holding 18 tons are good enough, or 8.8 at 36 tons are needed, or even the 54 tons provided by 12.9 bolts.

The breaking strain of an ordinary bolt gives an idea of how much damage the counter-balance could do if the bolts failed. An 18 ton punch in the face would hurt!

Dave

 

Edited By SillyOldDuffer on 27/12/2022 12:30:36

[Howard LewisParticipant @howardlewis46836]

As an aside, some the tractor pulling boys used to use Perkins V8.540 engines that had been tuned up.

The balance weight bolts were not as large as M20, but were always , even on "untuned" production engines always VERY carefully bedded and torqued.

Apparently, at about 5,000 rpm (Nearly twice normal 2,600 rpm rated speed ) the forged balance weights would start to straighten out, before the bolts failed.

In a traction engine context, even full size, I cannot see such high speeds being encountered.

Having said that, rest assured that a balance weight becoming detached does an awful lot of damage!

Howard

[Michael GilliganParticipant @michaelgilligan61133]

Based on your very reasonable logic, Dave … The Wensleydale should be fine

… provided that Tim’s point about stress-raisers is properly addressed.

MichaelG.

[HopperParticipant @hopper]

Posted by SillyOldDuffer on 27/12/2022 12:29:25:

Duncan's suggestion of establishing the mass of the counter-balance was a good one. If the mass and rotational speed are known, it's easy to calculate the centripetal force on the bolts:

F = mv²/r

m is mass,

v is velocity

r is radius, distance from the axle's axis

An ordinary nasty M20 bolt will take about 9000kg force. 8.8 high-tensile bolts about twice that, and 12.9 bolts three times.

Not done the sums because the weight has to be CAD extracted…

CAD extracted? CAD extracted? Does nobody know how to do back-of-fag-packet geometry anymore?

We know from the drawing the main radius on the counterweight is 3 3/4" so we can calculate the area of a circle with that radius then halve it. Then deduct the square cutout in the middle that is 2 1/2 x 1 1/4". Then knock off maybe 10 per cent for those rounded corners with the smaller radius as a guesstimate. (or simply leave it on as a safety factor). Then multiply by the 1" thickness to give us the total volume in cubic inches. Then multiply by weight of material per cubic inch which is readily available. Simples. But it's 11pm here so I'm off to bed and leave you to it.

CAD extracted indeed.

[Howard LewisParticipant @howardlewis46836]

To give an example, which may help::

A 1/2 inch bolt in W range takes about 9 tons to put it just past the yield point.

So a 3/8 bolt in the same material would yield at about 1.77 tons

Consequently, a M10 fixing, in good quality steel should be safe with a 1 ton tensile load.

To my simple mind, you would need to spin a 3.75 inch "diameter" balance weight pretty fast to generate a 2 ton centrifugal load, (Assumes two bolts per balance weight )

Howard

[JasonBModerator @jasonb]

Posted by SillyOldDuffer on 27/12/2022 12:29:25:

An ordinary nasty M20 bolt will take about 9000kg force. 8.8 high-tensile bolts about twice that, and 12.9 bolts three times.

We are not suggesting using the whole bolt to keep the weights in place, use the bolts shank dia for the flange and turn down to M12 or 7/16" W

[old martParticipant @oldmart]

When the drawings were made, high tensile meant stronger than mild steel or wrought iron. 8.8 would have been classified as high tensile and now 10.9 and 12.9 would be even better choices if you can machine them satisfactorily. The advice about inside rads is excellent, it will greatly reduce the danger of stress cracking. Any washers should be thick and have an internal chamfer on the bolt head side to ensure thea the rad s not loaded. If you can, have at least 1 1/2 thread diameters in the threaded hole.

[Howard LewisParticipant @howardlewis46836]

A sharp corner will be a stress raiser, leading to fatigue failure.

The bore into which the bolt enters should have a chamfered / counterbored entrance to prevent a foul between the radius and the edge of the hole.

The load in the bolt will not be constant, but an alternating one…

There will be the slightly varying load from centrifugal force (The crankshaft will not rotate with a constant angular velocity ) to which will be added the loads imposed by accelerating and decelerating the rotating masses at each end of the stroke as the crankshaft rotates.

The higher the rotational speed, the greater each of these forces will become.

Howard.

[duncan webster 1Participant @duncanwebster1]

So in the absence of any volunteers, I've applied my lack of skill to produce a 3D model

The dimensions on the drawing don't quite work. If the 3.75 rad is centred on the top line and you use the 1.5" rad, the overall width is ever s slightly less than 6.375. This has properties thus

and then done some sums. I arbitrarily used 1000 rpm, is that fast enough?

 

13.5 MPa is 0.87 tons/sq.in, so we are into congealed monkey snot territory ( as a previous section leader of mine would have put it). What puts a different complexion on it is the 3/8 sq drive dog. Torsion of non round shapes is a bit involved. I've got to take the dog out now, I'll see if I fancy some more sums later. However, this result is not entirely surprising for a scale model, Linear dimensions are halved, so volume and hence mass of weight goes down by factor 8. Radius to cg goes down by 2, so for the same speed, force goes down by factor 16. Bolt diameter is halved so area goes down by 4. Hence stress goes down by factor 4, we are certainly not into grade 12.9 bolts.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Edited By duncan webster on 27/12/2022 15:57:55

Edited By duncan webster on 27/12/2022 15:59:03

Edited By duncan webster on 27/12/2022 15:59:45

[SillyOldDufferModerator @sillyoldduffer]

Posted by Hopper on 27/12/2022 12:44:59:

Posted by SillyOldDuffer on 27/12/2022 12:29:25:

…

Not done the sums because the weight has to be CAD extracted…

 

CAD extracted? CAD extracted? Does nobody know how to do back-of-fag-packet geometry anymore?

We know from the drawing the main radius on the counterweight is 3 3/4" so we can calculate the area of a circle with that radius then halve it. Then deduct the square cutout in the middle that is 2 1/2 x 1 1/4". Then knock off maybe 10 per cent for those rounded corners with the smaller radius as a guesstimate. (or simply leave it on as a safety factor). Then multiply by the 1" thickness to give us the total volume in cubic inches. Then multiply by weight of material per cubic inch which is readily available. Simples. But it's 11pm here so I'm off to bed and leave you to it.

CAD extracted indeed.

I have no shame!

All in favour of 'weight of a round cow' estimates, but the counter-balance shape is more complicated than Hopper's simplification, and once the model is created the CAD package does all the necessary maths.

Assuming I drew it correctly the weight is:

Told the material is StCst10 GOST 1412-85 cast-iron, Solid Edge says the mass is 2.086kg and the volume 306741 cubic mm.

The centre of mass is shown by the purple dot. It's at coordinates X=0, Y=-17.04mm, Z=-12.70mm

Dave

PS.  Duncan got did first while I was typing.  Relieved to see our weights are within 200g.  The difference is because we've drawn the weight differently.  One or both of us has misinterpreted the 2D drawing – I expect it's me…

Edited By SillyOldDuffer on 27/12/2022 16:05:03

Edited By SillyOldDuffer on 27/12/2022 16:09:47

[JasonBModerator @jasonb]

I would think 500rpm would be more that the max a 1/2 scale TE would see as they run a bit faster than original the smaller you get.

[Les RileyParticipant @lesriley75593]

I'm not sure about RPM but I agree that 500 sounds like a maximum. I must put my rev counter on my 4" and see what that does, though I never rev it hard in neutral as I am sure it would go faster than I feel comfortable stood nearby!

I have drilled the weights for M12 but not yet counterbored them. If I was using cap screws the head size would be 18mm diameter so I guess 11/16 is about right.

The weights are a meaty lump, I put one on the kitchen scales and it is 1723 grams.

[duncan webster 1Participant @duncanwebster1]

Posted by SillyOldDuffer on 27/12/2022 16:03:41:

Posted by Hopper on 27/12/2022 12:44:59:

Posted by SillyOldDuffer on 27/12/2022 12:29:25:

…

Not done the sums because the weight has to be CAD extracted…

CAD extracted? CAD extracted? Does nobody know how to do back-of-fag-packet geometry anymore?

We know from the drawing the main radius on the counterweight is 3 3/4" so we can calculate the area of a circle with that radius then halve it. Then deduct the square cutout in the middle that is 2 1/2 x 1 1/4". Then knock off maybe 10 per cent for those rounded corners with the smaller radius as a guesstimate. (or simply leave it on as a safety factor). Then multiply by the 1" thickness to give us the total volume in cubic inches. Then multiply by weight of material per cubic inch which is readily available. Simples. But it's 11pm here so I'm off to bed and leave you to it.

CAD extracted indeed.

I have no shame!

All in favour of 'weight of a round cow' estimates, but the counter-balance shape is more complicated than Hopper's simplification, and once the model is created the CAD package does all the necessary maths.

Assuming I drew it correctly the weight is:

Told the material is StCst10 GOST 1412-85 cast-iron, Solid Edge says the mass is 2.086kg and the volume 306741 cubic mm.

The centre of mass is shown by the purple dot. It's at coordinates X=0, Y=-17.04mm, Z=-12.70mm

Dave

PS. Duncan got did first while I was typing. Relieved to see our weights are within 200g. The difference is because we've drawn the weight differently. One or both of us has misinterpreted the 2D drawing – I expect it's me…

Edited By SillyOldDuffer on 27/12/2022 16:05:03

Edited By SillyOldDuffer on 27/12/2022 16:09:47

you put the holes in, I didn't, on the basis that the holes would be full of bolt which is also going round. Perhaps I'm not as useless as I thought

[Clive FosterParticipant @clivefoster55965]

One point that must never be forgotten when re-purposing high tensile bolts is that the higher the tensile strength the greater the crack sensitivity. The critical length at which a crack expands unstoppably is very short for 12.9 bolts. The minimum radii at section changes is surprisingly large and the surface finish needs to be good if premature cracking is to be avoided.

It has been suggested, with some validity, that the casual and home shop machinist is unlikely to be able to produce threaded fasteners that can be reliably assumed to be significantly stronger than 8.8.

Over simplifying as a thought experiment.

Starting with a 12.9 bolt and reducing the materials safe tensile loading to that of an 8.8 bolt by insufficient attention to surface finish and section change radii produces a fastener that must, in practice, be considered much weaker. The crack propagation rate and critical length remain essentially the same as that of the original material so the overload sensitivity is much greater making failure much more likely.

Its remarkably easy to overload modest size fasteners when tightening. Although I use a torque wrench on vehicle and similar work where a workshop manual exists I rarely do on more casual work. Despite having the lists of recommended torque for different classes of fastener.

Clive

[duncan webster 1Participant @duncanwebster1]

Now I've done some more sums, easier than I thought. According to this, grade 8,8 M12 bolts have a proof stress of 600 MPa, recommended torque 65 lb.ft (88 Nm). EN1a has proof 230 MPa, so recommended torque would be 88*230/600=33.7 Nm. Applying this to a 3/8 square drive gives a shear stress of 187 MPa, which is too high for EN1a by a long way. There's no need to tighten them this tight obviously, 15 Nm would be OK

I'd still use M12 capheads, drill and tap say M5 concentric with the hex (not necessarily concentric with the shank, been there), make an adaptor with M5 spike, the 10 diameter to fit the hex socket, then 10 square (or 3/8 if you're really picky). Then you can tighten the capheads with a proper key, and screw in the dummies with a dab of low strength loctite. If you really want solid and a higher torque, use EN16 and adjust tightening torque appropriately. Limiting feature is still the square, 33 Nm is OK

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