Totally off topic....... puzzle

[Steve Bailey]

I think it's 5.610 Lbs

That's not a torque.

 

[elevmike]

3. Does it matter where you grab the torque wrench when you're tightening the nut?

A: NO. If you grab-on at the end (assuming a parallel alignment), you only need to apply the actual required pounds of force; 2.952-Lbs of Force. The indicator will show 2.952 Ft-Lbs of Torque.

However, if you grab-on closer to the head, then you will need to apply more actual pounds of force. The closer to the head, the more actual force is required. However, the indicator will still indicate the "effective torque". You need to apply whatever force is necessary to produce an indication of 2.952 Ft-Lbs of torque. The indication is based on the assumption that the user will grab-on at the handle.

As you grab-on closer to the head you have to apply more force. If you grab-on at the common point (point "B") then you have to apply 3.936-Lbs of Force (3.936-Lbs x .75-Ft = 2.952 Ft-Lbs). However, the indicator will show Zero Ft-Lbs because no force is being applied through the torque wrench.

Terry,

Torque = Weight (force applied) x Distance (the distance from the bolt). Torque wrenches are calabrated in such a manner so that the shaft arcs accross the entire distance from the head to the indicator. If you pull on the torque wrench below the indicator, the arc of the shaft will be concentrated between the pull point and the bolt. This will cause the indicated torque to be less then the actual torque applied.

 

[Bob O]

I will take a shot at this..

2.25 ft-lb if inline or 1.57 ft-lb at 30deg.

3) I don’t know how torque wrenches are calibrated but I don’t think on a “click” type wrench it would matter where you put your hand but it may on the “beam type”.


Bob O.

 

[Steve Bailey]

A torque wrench measures the torque at its pivot point. When the pivot point of the torque wrench is aligned with the point where you really want to measure torque, it doesn't matter where you apply the force.

The puzzle is about what happens when the pivot point of the torque wrench is offset from the point where you want to measure torque.

 

[Jiri Toman]

460 / 600 V

 

[elevmike]

A torque wrench measures the torque at its pivot point. When the pivot point of the torque wrench is aligned with the point where you really want to measure torque, it doesn't matter where you apply the force.

What type of torque wrench are you using? With the ol' style torque wrenches, (the type with the indicator scale up near the handle, and the long needle seated at the pivot/head), it matters very much if you want to get an accurate reading. Im not sure about the new-fangled electronic ones, or the clickers.

 

[Steve Bailey]

Actually Mike, if you can get your brain around why there is some variation in the measured torque on the beam-type torque wrench, depending on where you grab it, you can understand how the effect is magnified when the point of measurment is offset from the point you're really concerned about.

 

[Terry Woods]

Steve...

Don't spill the beans just yet...

 

[Terry Woods]

Ok... one more time...

If the nut is required to be torqued to 2.952 Ft-Lbs...

And the combo arm is 1.75-Ft (9" + 12" = 21" = 1.75') then, without a doubt, the force that must be applied is...

2.952 Ft-Lbs / 1.75-Ft = 1.686 Lbs of force.


Now comes the tricky part...

If 1.686-Lbs of force is applied to the "system" then the nut will be torqued to 2.952 Ft-Lbs.
At that point, the nut is not moving... the metric wrench is not moving... and still there is 1.686-Lbs of force applied to the torque wrench.

If the metric wrench is not moving then the torque head is stationary... and still there is 1.686-Lbs of force applied to the torque wrench.

Since 1.686-Lbs of force is being applied to a 12" torque wrench whose head is not moving... the torque developed at the head of the torque wrench is... duh... 1.686 Ft-Lbs... do ya think? This is what the indicator should show.


Now, regarding the 30-degree bit...

At first I though Steve was talking about the handle of the torque wrench being elevated relative to the torque plane. I think I see now that he was talking about the handle being non-parallel within the torque plane.

In the first case, the distance between the nut and the handle was 1.75-Ft.

In this new case, the distance between the nut and handle is reduced.

So now, the new torque value showing up, on the indicator, will have to be increased because the effective length of the arm is reduced.

You can do the Trigonometry... I'm tired.

 

[Terry Woods]

Steve...?

 

[Steve Bailey]

Looks like Terry gets it. Do you see the point about where you apply the force to the torque wrench?

Mike's point about the classic torque wrench is this: That type of torque wrench actually measures the deflection of a cantilevered beam of a known cross section. The scale is calibrated in the amount of torque required to produce the deflection. The scale is most accurate when the force is applied at the end of the beam. The error introduced by not applying the force at the expected point magnified when the head of the torque wrench is offset from the nut being tightened.

 

[Thomas Sullens]

fun

1 - 4 nt-m = 2.95 ft/lbs
2 - yes
2a - 1.8255 change to 1.74
3 - no

Is 2a right
i see I missed #3

 

[Jerobi]

hello everybody! I've just been born here in the site. This sure caught my curiosity. Is it proven that these puzzle can gauge one's intelligence? I am also new to the PLC world. I wasn't born yesterday but I guess I've been sleeping since then.
I accidentally hit the correct answer on the sec. roll. but then I figured it out on the 4th or 5th maybe. Did that make me dumb?

 

[Rod]

Jerobi,

No dumber than the rest of us. Welcome to the forum.

Pull up a chair, your favorite PLC and enjoy.

Rod (The CNC Dude)

 

[Bob O]

Something isn’t sounding right here. If this was a piece of pipe with a 30 deg. bend then I would agree you would use the effective length of 20.3” but since we are looking for the reading on the torque wrench I am not sure this is correct. I still think that you would find the effective length (1cos30) and treat the system still in parallel.

Thanks,
Bob O.