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

-- 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

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

: 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.....