Question for you engineers or tool experts?

[XJ1100]

Holy ancient thread resurrection, Batman!

Sorry to say, my post is correct. What you are thinking is that the yielding material does indeed limit the torque that can be applied. Say, if you want 100 ft-lbs and the extension yields at 80 ft-lbs - Then the fastener can only get 80 ft-lbs....but, note that as the extension is yielding you have 80 ft-lbs AT BOTH ENDS, on the torque wrench AND on the bolt. Please re-read my post and consider it carefully.

No need to be sorry...I miss understood your post when I read it earlier.

I retract my earlier statements...Ironcrow is correct.

 

[kartracer55]

I'm wondering what size this little hex bit is to be twisting with only 90 inch pounds applied to it.

2-3 inch long 5/32

 

[NotStock]

I will have to respectfully disagree with you on the items in bold. If the material stays in the elastic region these statements are true. But if the material goes past its yield point into the plastic region then the statements are not true. Once the material starts to yield it will no longer transmit 100% of the torque that is applied.

Just because the material enters the plastic region does not mean it will be unable to reach a static state at that torque input. Ultimate is higher than yield on a stress strain diagram, and past SU its busted, so no confusing torque input in that scenario.

One thing that we just found surprising though.... using a universal joint can change your torque a HUGE amount. We were using an Apex universal socket, and were seeing about a 30% or more variation on the torque actually applied to the fastener, depending on which direction the cross-bar in the universal joint was facing compared to the bend in the U-joint. Obviously when this is occurring, more angle on the U-joint the higher the effect. Using it straight made no difference.
Phil

cool to see that effect measured in this application. The output changes dramatically vs input depending on the angle of the u joint. The wikipedia graph shows the relationship.

http://en.wikipedia.org/wiki/File:UJoint1.png

 

[Wakefield]

So the wrench turns more degrees to wind up the extension and long socket to reach the final torque? Also the wrench ends up turned more degrees than the degrees at the fastener? Takes a tiny bit of time for the extension to wind up?
I would think slow is better-remember the torque stick absorbing some of the fast jerking of the impact gun.

 

[Steinmetz]

You are right that you do lose some torque when you use a long socket or extension on the torque wrench.
I had a way to figure it out written down back when I used to wrench professionally. Send me 20$ and it might make it worth my while to dig through my garage and find it for you.
In the mean time I did a quick search for "torque loss equasion". You may find the answer to your question in this rambling post.

By what mechanism? Explain.

 

[Steinmetz]

Is there an easy way to calculate how torsion affects torque being transfered on a long hex socket?

Heres the situation... I need to torque 5 socket cap screws to 95 inch lbs. There are two methods to do this, one involving a standard hex socket, the other with a long hex socket. To use the standard hex socket means quite a bit more work in this process, and the time adds up because these screws are replaced and retorqued quite often.

As I go to torque these screws, I can see the hex stock actually twisting on me, and im guessing this is somehow affecting the torque being transfered through. So is there an easy way for me to calculate how much torque is actually being lost because of it? When I am home I can do it the correct way with the standard hex socket, but at the track I simply dont have the time to go through all this so I would like to know if, instead of setting the torque wrench to 95 in/lbs, I can set it to 95+X in/lbs to compensate for anything being lost through the long socket.

Also, If anybody could help me understand something a bit better it would be great... Is this length of hex stock in the socket absorbing a certain percentage of the torque proportionally as applied torque is increased? Along those same lines, is there a point where this begins to happen or does it only become more apparant at higher loads? Or, does it reach a certain point where it simply cant transfer anymore torque and begins to twist like this?

Thanks

Jim

The wrench achieves static torsional equilibrium when the prescribed torque is reached. No "correction" is necessary.

 

[andywander]

Imagine if, instead of using the torque wrench, extension and driver(socket?) to tighten a fastener, you simply coupled it to another torque wrench.

You will always read the same value on both torque wrenches, even if you twist the extension and/or socket in knots. (that is, assuming both wrenches are properly calibrated)

 

[larry_g]

The wrench achieves static torsional equilibrium when the prescribed torque is reached. No "correction" is necessary.

You realize your answering a question posted 6+ years ago? I do concure that you are correct. Goes back to the old physics thing--" For every force there is an equal and opposite force."

Enjoy the read here

lg
no neat sig line

 

[Steinmetz]

You realize your answering a question posted 6+ years ago? I do concure that you are correct. Goes back to the old physics thing--" For every force there is an equal and opposite force."

Enjoy the read here

lg
no neat sig line

Oops.