Extension changes torque?

[SuzukiGS750EZ]

Hey guys. Doing head gaskets on my car this weekend. I'll need an extension to clear the heads for the bolts. Adding a 6" or 8" extension will change the torque applied to the bolt, right? What's the formula for this? Are torque wrenches calculated with a certain length of socket or is it straight to the square drive? I think I heard if it's a linear extension like a crows foot it's effects it more than a right angle to the wrench.

 

[91bronc300]

At a right angle to the wrench the extension doesn't change the torque at all under static conditions i.e. using your hand. With an impact it will but not with a torque wrench you pull by hand.

 

[Steinmetz]

An extension coincident with the axis of rotation of the wrench will have no effect on the torque reading. If you add any extension that projects outwardly from the axis of rotation so that the effective moment arm is changed, the torque reading will be different from the torque applied.

 

[SuzukiGS750EZ]

Would explain an experiment I did today. 6" extension torquing an upper intake. With an without extension it beeped the same.

 

[A_Pmech]

Remember Newton's Third Law: For every action there must be an equal and opposite reaction.

 

[Haveblue]

For hand torqueing, no, extentions will not effect torque,,,on the other hand, extensions will affect an impact, as it is hammering against the clearances between the connections. One thing you need to watch for is good clean bolt holes, with no oil, or coolant in them. A blind hole with fluid in it will seem to torque down fine, but after the fluid creeps up the threads, it loses its torque. Blow out the bolt holes, and use a small amount of oil on the threads if specs call for it!

 

[joe_padavano]

Just so we're all clear, an extension like this will not change the torque reading for a hand torque wrench.

This will also not change the torque:

This, however, will change the reading:

 

[woodstockva]

Good info Joe....Thanks! I always like to see diagrams

 

[MonoxieChild]

Holding the Torque wrench Correctly can have a impact on Torque readings though. I remember learning that in shop class. haha

 

[Steinmetz]

Just so we're all clear, an extension like this will not change the torque reading for a hand torque wrench.

This will also not change the torque:

This, however, will change the reading:

Correct and correct again.

 

[Steinmetz]

Holding the Torque wrench Correctly can have a impact on Torque readings though. I remember learning that in shop class. haha

That's why torque wrenches are often fitted with a pivot at the handle.

 

[DSLTRK]

This will also not change the torque:

Not true. The image above will provide two different torque ratings.

 

[zkling]

For extensions co-axial to the drive square.

Hand torque wrench, value will not be changed. Why, as APMech said above look at the reaction force. In a steady load the force cannot dissipate along the length of the extension*. Thus input = output torque. Now what can happen is if the extension(s) are long enough you will get a cork screw type effect, but that is a whole nother discussion.

*under normal operating conditions, provided you have not hit the yield point of the extension(s).

Impact wrench, value can and will usually be changed. Why, transmissibility. It has to do with how the extension(s) which act as torsion springs can transmit the repeated vibrations through the extensions and to the drive end. If it wasn't for this, torque sticks would not be possible.

Anyone that tells you differently, ask them why.

Not true. The image above will provide two different torque ratings.

Why?

 

[alwaysFlOoReD]

Why?

I'm going to guess;
a^2+b^2=c^2
The lever length is slightly longer with the right angle extension.

Richard

 

[DSLTRK]

I'm going to guess;
a^2+b^2=c^2
The lever length is slightly longer with the right angle extension.

Richard

Correct, BUT also, since the force vector isn't perpendicular with the lever length, you must also use the law of sines to determine the correct amount of force to apply to the torque wrench.
You can also use the law of sines to find the factor to multiply to the read torque wrench value.

 

[anti-everything1990]

It will change it since an extension can twist. The effect is small enough that it wont matter

 

[GCncsuHD]

One thing you need to watch for is good clean bolt holes, with no oil, or coolant in them. A blind hole with fluid in it will seem to torque down fine, but after the fluid creeps up the threads, it loses its torque. Blow out the bolt holes, and use a small amount of oil on the threads if specs call for it!

IF specs call for it. The largest majority of torque specs are assuming "dry" fasteners. Applying any sort of lubricant to the fastener reduces the needed torque to get the same tension in the bolt/clamping force (which is generally what you are really after). Heck, even loctite acts as a lubricant until it sets up. A good rule of thumb is oiled threads reduce the needed torque spec by roughly 20%, though this is dependent on the material being used and the lubricant being used. Something torqued at 100 ft lbs may be just right, but with oil on the threads you may have applied too much tension to the bolt, potentially causing it to yield or fail, or damage the bolted components.

Not true. The image above will provide two different torque ratings.

No it will not. If the torque wrench and the extension shown are at any angle other than 90 degrees then it would.

 

[larry_g]

Had this discussion a few years back. http://www.garagejournal.com/forum/showthread.php?t=4751

lg
no neat sig line

 

[DSLTRK]

No it will not. If the torque wrench and the extension shown are at any angle other than 90 degrees then it would.

Ok

 

[Steinmetz]

Correct, BUT also, since the force vector isn't perpendicular with the lever length, you must also use the law of sines to determine the correct amount of force to apply to the torque wrench.
You can also use the law of sines to find the factor to multiply to the read torque wrench value.

Not true. If the adaptor is perpendicular, as shown, there is no correction.

 

[DSLTRK]

Not true. If the adaptor is perpendicular, as shown, there is no correction.

I'll just leave this here>

http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

 

[Steinmetz]

I'll just leave this here>

http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

Nothing presented there supports your view that a perpendicular adapter requires a torque correction.

Try out your theory on this little applet. It may also be of benefit to others on this thread.

http://www.cncexpo.com/TorqueAdapter.aspx

 

[A_Pmech]

I'll just leave this here>

http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

If you draw a free body diagram of a wrench with an extension perpendicular to the shaft you'll see that there is no change in torque applied to the fastener.

You're forgetting that the force vector applied is perpendicular to the wrench body, not parallel with the hypotenuse of the triangle: Force application / wrench measurement pivot / fastener. They cancel.

 

[Coach James]

They do not cancel. Quantities may divide out or their vector sum may be zero, but they do not cancel, but I do get what you mean. The "c" word is banned in my classroom.

Coach

 

[A_Pmech]

They do not cancel. Quantities may divide out or their vector sum may be zero, but they do not cancel, but I do get what you mean. The "c" word is banned in my classroom.

Coach

"Cancel" is banned? What about algebraic fractions?

Your point is well taken though.

I was bored while stuck on hold:

 

[The Ratchet Man]

This thread is great. I've learned 18 new words, and that the result of the corrective calculation for the letter "L" is 0.

Seriously though, interesting stuff. I will never look at a torque wrench the same.

 

[DSLTRK]

If you draw a free body diagram of a wrench with an extension perpendicular to the shaft you'll see that there is no change in torque applied to the fastener.

I'm talking about what the torque wrench reads and what the fastener is actually torqued at. The torque read and the actual torque will not be the same, as you just figured on the paper.

You have an error calculating the normal vector force.

 

[Chuck122]

basically, the shape of the lever does not influence the applied torque. you can extend the force vector so that a line perpendicular to the vector axis will intersect with the point at which you are trying to find the applied torque. the length(l) or that line is the actual lever arm. thus, l*(the force applied)=torque
in the case where the extension is perpendicular to the torque wrench, the torque wrench and l are parallel. also, the force vector is parallel to the extension. So, l=the length
of the torque wrench and the torque reading is not influenced by the extention

 

[Chuck122]

And this is what happens if the extension is not perpendicular to the torque wrench

 

[er3456df]

If you draw a free body diagram of a wrench with an extension perpendicular to the shaft you'll see that there is no change in torque applied to the fastener.

You're forgetting that the force vector applied is perpendicular to the wrench body, not parallel with the hypotenuse of the triangle

I disagree about the direction of the force- the wrench will be rotating around the fastener, and regardless of the direction of your applied force, your hand will be moving along the tangent- NOT perpendicular to the handle. I think doing what you describe would not only waste effort and side-load the fastener, it would feel just plain awkward. I'd bet that you'd automatically pull along the tangent without even thinking about it.

BUT ALSO- the other guy made a good point- the torque wrench is measuring the torque at the joint between the two- not at the fastener, and not at your hand.

I can't wrap my head around how to draw a free-body for that, though. Would love to see one.

 

[DSLTRK]

If I didn't feel this topic was this important, I wouldn't go this far to prove my point.

This is important to know, especially when torquing head studs where access is limited and these extensions are frequently used.
As I've said before, the value the torque wrench is set at will not indicate the applied torque to the fastener. The longer the extension, as well as the angle it makes to the normal of the torque wrench will affect the actual torque applied.

 

[A_Pmech]

I see your error.

You're not decomposing the vector "Applied Force" acting on the handle in the extension scenario. You're assuming that the applied force pulling on the handle is normal to imaginary position vector "C", the hypotenuse of the wrench / extension system, which it is not.

The vector "Applied Force" acting on the handle of the wrench acts normal to it and at some angle θ from the decomposed component vector "Normal Force". When you decompose "Applied Force" you will see that, per my earlier example:

(Applied Force)(B) = (Applied Force)(cos θ)(C)

Thus, when negating rounding error in computing θ and C:

(100)(12) = (100)(cos 26.57˚)(13.42)

Where:

Applied Force = Force applied normal to the wrench handle
B = Lever arm of the plain wrench
θ = Angle of vector component of "Applied Force" acting normal to imaginary line C.
C = Imaginary position vector of "Applied Force" with origin at the reaction point.

Thus, the reaction torque around the metering pivot of the wrench is the equal by example.

If I didn't feel this topic was this important, I wouldn't go this far to prove my point.

This is important to know, especially when torquing head studs where access is limited and these extensions are frequently used.
As I've said before, the value the torque wrench is set at will not indicate the applied torque to the fastener. The longer the extension, as well as the angle it makes to the normal of the torque wrench will affect the actual torque applied.

Nicely done!

 

[DSLTRK]

I see your error.

You're not decomposing the vector "Applied Force" acting on the handle in the extension scenario. You're assuming that the applied force pulling on the handle is normal to imaginary position vector "C", the hypotenuse of the wrench / extension system, which it is not.

The vector "Applied Force" acting on the handle of the wrench acts normal to it and at some angle θ from the decomposed component vector "Normal Force". When you decompose "Applied Force" you will see that, per my earlier example:

(Applied Force)(B) = (Applied Force)(cos θ)(C)

Thus, when negating rounding error in computing θ and C:

(100)(12) = (100)(cos 26.57˚)(13.42)

Where:

Applied Force = Force applied normal to the wrench handle
B = Lever arm of the plain wrench
θ = Angle of vector component of "Applied Force" acting normal to imaginary line C.
C = Imaginary position vector of "Applied Force" with origin at the reaction point.

Thus, the reaction torque around the metering pivot of the wrench is the equal by example.

You keep forgetting that the torque wrench is the measurement tool. Not the hand pulling it. I can pull on a torque wrench at 45* from the normal, 6 inches from the end, and I can also tell you it would take a hell of a lot more force than if I just pulled it perpendicular to the handle, at the end of the handle.

In my example, I choose a torque value of 200ft-lbs. With a two foot long torque wrench, it will "click" when a force of 100 lbs is normal to the end of the wrench, regardless if you pull out, in, or where you pull on the handle. If you look, my math accounts for that.

I used simple trigonometry to relate the torque values, the apparent(what the torque wrench reads), and the actual(what the fastener is actually torqued at). Then I simply reversed the process to find the unknown> If I want the fastener torque to be 200ft-lbs, what should the wrench be set at?

This is Physics 201 all over again.

 

[Chuck122]

You seem to forget that the force applied on the torque wrench is perpendicular to said wrench, no matter what extensions is used. Thus, the extensions make no difference

 

[Chuck122]

You keep forgetting that the torque wrench is the measurement tool. Not the hand pulling it. I can pull on a torque wrench at 45* from the normal, 6 inches from the end, and I can also tell you it would take a hell of a lot more force than if I just pulled it perpendicular to the handle, at the end of the handle.

In my example, I choose a torque value of 200ft-lbs. With a two foot long torque wrench, it will "click" when a force of 100 lbs is normal to the end of the wrench, regardless if you pull out, in, or where you pull on the handle. If you look, my math accounts for that.

I used simple trigonometry to relate the torque values, the apparent(what the torque wrench reads), and the actual(what the fastener is actually torqued at). Then I simply reversed the process to find the unknown> If I want the fastener torque to be 200ft-lbs, what should the wrench be set at?

This is Physics 201 all over again.

it is normal to the torque wrench, not the imaginary line between the fastener and the handle of the torque wrench. although you could use that imaginary line, yo would have to decompose the force vector to know what force is applied in each axis (x & y) and the distance between each decomposed vector and the fastener. if you get your trig right, the applied torque will match that of the torque wrench

 

[DSLTRK]

That's okay, you do what you feel is the proper way.

 

[A_Pmech]

In the interest of ending the argument:

 

[DSLTRK]

In the interest of ending the argument:

Do you really think companies would have people calculate the actual torque?
As you saw in my example, the actual torque differed by only a few ft-lbs.

 

[SuzukiGS750EZ]

So to conclude, as long as whatever it is I'm using is at a right angle to my wrench, it's not changing the end value of torque. As soon as I add something in line with the wrench which adds a length off the head horizontally, it needs calculation.

 

[fhemm20]

I love watching engineers proving a point.

Next week we will debate properties of a vacuum

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