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High tensile bolts


Gromit
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The number before the decimal point is the Ultimate Tenisle Strength (UTS) of the bolt in hundreds of N/mm2.

The number after the decimal point is the percentage of the UTS that the bolt will yield at in tens of percentage points. ("Yield" means stretch beyond it's elastic limit.)

So, for a grade 8.8 bolt, the UTS is 800N/mm2 and the yield stress is 80% of that which is 640 N/mm2.

Your turn - what is the yield stress of a 10.9 bolt?

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:ph34r::ph34r::ph34r: The only bit Lewis has right there is you should have nuts of the same grade or higher to mobilise the full strength of the bolt. His strength calculation is a bit off.... :ph34r::ph34r::ph34r::ph34r:

8.8tonnes / mm2 is equvenlent (on this planet) to 86,328N/mm2. I think that material is called unobtainium!! :lol:;) (well, Osmium and diamond are stronger, but only in compression)

Ah, Lewis my good man, I see you have posted before I could finish. Yet I notice nobody has replied for a 10.9.... tut tut..... chop chop fellows!

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:ph34r::ph34r::ph34r: The only bit Lewis has right there is you should have nuts of the same grade or higher to mobilise the full strength of the bolt. His strength calculation is a bit off.... :ph34r::ph34r::ph34r::ph34r:

Well, thats more than I usually get right :rolleyes::lol:

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Bish,

what does this mean in the real world? does an 8.8 bolt breaking at 800n/mm2 and start to elongate at 640, or does it elongate at 800 and take another 640 before it breaks?

if this is tensile strength how can you have mm2 as a unit? i imagine that to measure breaking strain you would stick one of a bolt to an immovable object and connect the other end to a hydraulic pressure cell- in which case does the bolt break when the cell shows 800nm or when teh cell shows 800nm x the surface ares of the bolt in mm2 (which for an M8 bolt is around 28mm2) whcih would give a UTS of 22,4000N

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Bish,

what does this mean in the real world? does an 8.8 bolt breaking at 800n/mm2 and start to elongate at 640, or does it elongate at 800 and take another 640 before it breaks?

if this is tensile strength how can you have mm2 as a unit? i imagine that to measure breaking strain you would stick one of a bolt to an immovable object and connect the other end to a hydraulic pressure cell- in which case does the bolt break when the cell shows 800nm or when teh cell shows 800nm x the surface ares of the bolt in mm2 (which for an M8 bolt is around 28mm2) whcih would give a UTS of 22,4000N

The 8.8 will start to deform at 640N/mm2 and break at 800N/mm2. The unit is a measure of the stress in the bolt and is saying you can apply 640N per sq. mm of cross-sectional area. A newton (N) is approximatley equal to 9.81kg, so 640N/mm2 is about 65kg/mm2. I would tend to use t/cm2 for work (naval architecture) as for the kind of stuff we are doing it gives nice figures to work with and most of the technical steel stock books have the section properties shown in cm, cm2 etc.

Edited to add yield is basically the figure you need to look at.

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The unit is a measure of the stress in the bolt and is saying you can apply 640N per sq. mm of cross-sectional area.

I take it then that this is only along the length of the bolt?

What about the shear force, for example where a JATE ring is attached by an 8.8 to the chassis rail?

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I take it then that this is only along the length of the bolt?

What about the shear force, for example where a JATE ring is attached by an 8.8 to the chassis rail?

Correct, the allowable shear will be lower than the tensile strength. The typical figure that springs to mind is around 60% of the tensile but it does vary slighty.

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I seem to remember Mohr's circle says that shear loading will be about half the strength of a tensile load, but then you've additional friction in the joint from the clamping force with how a JATE ring is mounted too so slightly more in reality.

Since there are two shear planes with the way a JATE ring is loaded (one at each end), confusingly I'd estimate the maximum (yield) force at about the same as the yield strength of the bolt.

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Good thread, takes me back 20 years since I last did physics

A newton (N) is approximatley equal to 9.81kg,

I had to think about it but this should be the other way around 1kg approx 9.81N.

I remember when asked how the unit of a Newton was invented/determined my physics teacher told us it was the force of the apple which hit Issac's head :rolleyes:

I once had a beam fitted with a jointing plate, the engineer had specified 10.8 bolts - the steel erector used 8.8 and the plate slipped and the beam dropped in the middle when loaded! Shows the grade really does make the difference

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I had to think about it but this should be the other way around 1kg approx 9.81N.

You're quite right, brain must not be functioning properly and typed without thinking, wrote down the wrong explantion but did use the 1kg to 9.81N to do the conversion. Feeling suitably embarassed.

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cm? Steve that's disgraceful!!! Whatever happened to SI units!!!!! :P:P (Although I do agree that section properties for some bizzare reason are quoted in cm... :unsure: )

As far as shear goes, the magic number you need to look at is the square root of 2. The allowable shear stress is the yield stress divided by root 2 - about 70% of yield.

You also need to check the shear plane(s) to see if the threaded portion of the bolt is crossing it or if it is on the shank. Asa a rule of thumb, the threaded cross sectional area of a bolt is about 80% of the gross area - example M20 bolt, gross area = 100 PI, threaded area = 80 PI = 251mm2. Checked in a grown up bolt book (Steel Designer's Manual, the bible), net area is actually 245mm2. -2% error. Not bad for a rule of thumb. For smaller diameters, 75% is a better approximation. - M12 75% = 84.8mm2, actual is 84.3mm2.

With a jate ring you have two shear planes, probably one in the threads and one not. It is standard practice, unless you can guarantee that the shank will be in the shear plane to always err on the conservative side and use the threaded area in a shear calc.

Don't forget that the bolt is only one part of the connection. You also need to make sure all the other components are up to the task, something that is often overlooked when people start bolting things through paper thin chassis rails....

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Cool tech Bish

I suspected my college lecturer was full of sh*t

That explains why the big end and main bearing bolts are torqued to yield

No it does not explain anything about why bolts are torqued to yield.

In bolts subject to cyclic tensile loads, fatigue is the usual mode of failure.

Depending upon the stiffness of the joint, fatigue failure is unlikely if the bolt pre-tension is 2 to 5 (depending on what the loading characteristics are like) times the applied tensile load.

It is all about reducing the range of the fluctuating part of the stress, not the maximum stress, to increase fatigue strength.

If the pre-tension is too low (equal to, or less than the applied load), the stress fluctuation in the bolt is proportional to the cyclic load. With high pre-tension, most of the fluctuation is seen in the change of the compression in the joint (much less in the bolt).

Increasing the flexibility of the bolt also reduces the range of fluctuating stress, which is why some bolts are necked down in the body/shank. Body necked bolts/studs are stronger in fatigue than full diameter body bolts/studs and considerably stronger than bolts with enlarged bodies.

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I suspected my college lecturer was full of sh*t

The quite often are, during my national diploma I had a nice argument with one lecture about force/pressure/area with simple hydraulic systems, he had sucsefully confused it with the drawing pin analergy where the force is constant where as hydraulic systems the pressure is constant. Same lecturer also though the differance between the ecconomy of petrol and diesels was that petrols sucked in more fuel on over run?????

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cm? Steve that's disgraceful!!! Whatever happened to SI units!!!!! :P:P (Although I do agree that section properties for some bizzare reason are quoted in cm... :unsure: )

As far as shear goes, the magic number you need to look at is the square root of 2. The allowable shear stress is the yield stress divided by root 2 - about 70% of yield.

You also need to check the shear plane(s) to see if the threaded portion of the bolt is crossing it or if it is on the shank. Asa a rule of thumb, the threaded cross sectional area of a bolt is about 80% of the gross area - example M20 bolt, gross area = 100 PI, threaded area = 80 PI = 251mm2. Checked in a grown up bolt book (Steel Designer's Manual, the bible), net area is actually 245mm2. -2% error. Not bad for a rule of thumb. For smaller diameters, 75% is a better approximation. - M12 75% = 84.8mm2, actual is 84.3mm2.

With a jate ring you have two shear planes, probably one in the threads and one not. It is standard practice, unless you can guarantee that the shank will be in the shear plane to always err on the conservative side and use the threaded area in a shear calc.

Don't forget that the bolt is only one part of the connection. You also need to make sure all the other components are up to the task, something that is often overlooked when people start bolting things through paper thin chassis rails....

Stop showing off mr Bishop! :ph34r: Thats like me telling people to just knock up a winch mount. :rolleyes:

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Stop showing off mr Bishop! :ph34r: Thats like me telling people to just knock up a winch mount. :rolleyes:

Take a look at this then B) B) B) - give me a welder and I'll conquer the world!!! :P:lol::lol:

Walfy - I could do loads of calcs, but why don't you just bung loads of semtex or whatever at it and save the hassle? Alternatively, you could ask Dirty Diesel to drive over it in his wagon..... :lol:

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