Max torque through a 3/8" socket (slightly O/T)

I've seen top quality 1/2" drive torque wrenches listed as going up to 250 ft lbs. The cross sectional area of a 3/8" drive is just over half that of a

1/2" one so logically 125 ft lbs would be the upper limit for a similar material strength. I doubt if most would withstand that much but I've put 75 ft lbs through them easily enough. You might split the difference at 100 ft lbs for a decent quality set.

-- Dave Baker - Puma Race Engines

The message from "James Amor" contains these words:

How long is a piece of string?

I've had some which would snap soon as look at you - and others which would take huge torques.

Having said that - I've snapped off several 3/8" ratchets.

: The cross sectional area of a 3/8" drive is just over half that of a : 1/2" one so logically 125 ft lbs would be the upper limit for a similar : material strength.

Getting to yield at the surface depends on the polar moment of area J = A r^2/2 = A^2 / 2 pi, not the area. So a round bar of half the area will, all other things being equal, take a quarter of the torque load.

Ian

you can buy Norbar SL1s on ebay almost all the time.

Only if you break it by stretching it, if twisting it then it's strengths proportional to the cube of the radius, i.e. 2.37:1

It's the polar section modulus not the polar moment of inertia which is the limit, i.e. 0.208d^3 as opposed to (d^3)/6 for a square section.

Well that's something new I've learned then. I suppose I should have twigged that given it's the 4th power of the wire diameter that governs the spring constant of a coil spring that it wasn't quite so simple as area in the torque capacity of a bar. Having a browse through Google it seems the torsional stress in a bar is proportional to T x radius / diameter^4 which as you say works back to 1/8th of the torque to reach the same shear stress in a bar 1/2 the diameter or 1/4 the area.

-- Dave Baker - Puma Race Engines

: It's the polar section modulus not the polar moment of inertia which is : the limit, i.e. 0.208d^3 as opposed to (d^3)/6 for a square section.

Bugger yes, you are of course quite right.

Ian

Well I was about to say the same but then this nasty memory crept up on me & I looked it up Machinery's handbook.

For the differential drive shaft housing bolts on a P6, I had a long socket which was 3/8 drive, and IIRC, the torque setting for the bolts was

80 ft.lb. But I found myself wondering...

: In article , : James Amor wrote: : > What's the maximum Nm you can safely put through a 3/8" drive socket : > (I'm using a 1/2" drive torque wrench with a 3/8" reducer)? : : For the differential drive shaft housing bolts on a P6, I had a long : socket which was 3/8 drive, and IIRC, the torque setting for the bolts was : 80 ft.lb. But I found myself wondering...

Hmm. 80 ft-lb is 80 * 0.304 * 4.43 = 107Nm.

Nice high tensile steel will let go about 1GPa in direct stress and about 600MPa in shear.

The (right, this time) equation is that shear stress / radius = torque / polar moment, so

Maximum torque = maximum stress * (polar moment / radius) = maximum stress * section modulus.

For a round bar the polar moment is pi r^4 / 2 so the section modulus is pi r^3 / 2.

So 107Nm = 600MPa * pi * r^3 / 2

r^3 = (2 * 107) / (3 * pi * 600 * 10^6) = 3.78*10^-8

so r = 3.36mm ... a quarter inch diameter shaft would have been just about OK!

Ian

PS When someone finds the error in my calculation, please be kind!

Why 3 *pi ?

: For the differential drive shaft housing bolts on a P6, I had a long : socket which was 3/8 drive, and IIRC, the torque setting for the bolts was : 80 ft.lb. But I found myself wondering...

Hmm. 80 ft-lb is 80 * 0.304 * 4.43 = 107Nm. Nice high tensile steel will let go about 1GPa in direct stress and about 600MPa in shear. The (right, this time) equation is that shear stress / radius = torque / polar moment, so Maximum torque = maximum stress * (polar moment / radius) = maximum stress * section modulus. For a round bar the polar moment is pi r^4 / 2 so the section modulus is pi r^3 / 2. So 107Nm = 600MPa * pi * r^3 / 2 r^3 = (2 * 107) / (pi * 600 * 10^6) = 1.14*10^-7 so r = 4.84mm ... a 3/8" diameter shaft would have been right at its limit Ian PS When someone finds the (next) error in my calculation, please be kind!

: On 2 Feb 2005 19:03:28 GMT, Ian Johnston : wrote: : : > 107Nm = 600MPa * pi * r^3 / 2 r^3 = (2 * 107) / (3 * pi * 600 * 10^6) : > = 3.78*10^-8 so r = 3.36mm

: Why 3 *pi ?

That's what we engineers call "misreading your own prvious line" or alternatively "being stupid".

Thanks.

Ian

You're saved by decent alloy steel having a shear stress of ~1200 MPA :-)

: You're saved by decent alloy steel having a shear stress of ~1200 MPA :-)

As high as that? Coo. I don't have my trusty tables handy (Tr: I have lost them somewhere) so I took a wild guess based on 1GPa being a respectable direct yield stress!

Can we agree that 3/8" is probably getting marginal at 80ft-lb, what with stress concentrations round the square drive and all?

Ian

Din Spec is 225Nm/165lbft so unless you've bought an expensive socket set it's probably on the wrong side of marginal :-) It's suprising how far you can twist a 10" 3/8 extension bar & it'll still spring back though.

Of course the shaft - if you're using an extension - is larger than 3/8 AF.

But I was using a simple 1/2 to 3/8 adaptor. And they would have required rather more than 80 ft.lb to undo.

I regularly do bolts to 80lb/ft on a 3/8 drive (stepped down from 1/2" wrench) and it's not let go yet but it is a decent socket system. The mfct supplies a 3/8 torque wrench as well that goes to 960 lb/inch which I make a nice round 80 lb/ft.....