I was checking out my torque wrench. I measure it to be
429mm between the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm.
The indicator is a little short of the outer pivot---the
deflection at the outer pivot for 150lbf is around 41mm.

I found the formula for deflection: delta = F L^3 / (3 E I)

E is the flexural modulus --- I'll use 180 GPa (Young modulus for steel)
F is force applied,
L the beam length
I is the areal moment of inertia, I = pi/2 r^4

I am going to use the torque instead of force, T=FL, i.e.

delta = T L^2 / (3 E I)
= 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)

'units' calculates that as 30mm, a little too short of 41mm measured.
Am I making a stupid error somewhere?

On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski
wrote:

I was checking out my torque wrench. I measure it to be
429mm between the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm.
The indicator is a little short of the outer pivot---the
deflection at the outer pivot for 150lbf is around 41mm.

I found the formula for deflection: delta = F L^3 / (3 E I)

E is the flexural modulus --- I'll use 180 GPa (Young modulus for steel)
F is force applied,
L the beam length
I is the areal moment of inertia, I = pi/2 r^4

I am going to use the torque instead of force, T=FL, i.e.

delta = T L^2 / (3 E I)
= 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)

'units' calculates that as 30mm, a little too short of 41mm measured.
Am I making a stupid error somewhere?

First thing I'd look at is the assumption (implied by the 3EI term)
that the wrench acts like an ideal cantilever.

--
Ned Simmons

On 9/2/2014 11:30 PM, Ned Simmons wrote:
On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski
wrote:

I was checking out my torque wrench. I measure it to be
429mm between the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm.
The indicator is a little short of the outer pivot---the
deflection at the outer pivot for 150lbf is around 41mm.

I found the formula for deflection: delta = F L^3 / (3 E I)

E is the flexural modulus --- I'll use 180 GPa (Young modulus for steel)
F is force applied,
L the beam length
I is the areal moment of inertia, I = pi/2 r^4

I am going to use the torque instead of force, T=FL, i.e.

delta = T L^2 / (3 E I)
= 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)

'units' calculates that as 30mm, a little too short of 41mm measured.
Am I making a stupid error somewhere?

First thing I'd look at is the assumption (implied by the 3EI term)
that the wrench acts like an ideal cantilever.

There's also room for error in the assumption for E, since you can't be
sure what sort of steel was used for the beam.

If you doubt the accuracy of your wrench, clamp it in a vise and hang
some weights off it. It's better to know how it actually responds to a
load rather than how it "should" respond.

On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski
wrote:

I was checking out my torque wrench. I measure it to be
429mm between the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm.
The indicator is a little short of the outer pivot---the
deflection at the outer pivot for 150lbf is around 41mm.

I found the formula for deflection: delta = F L^3 / (3 E I)

E is the flexural modulus --- I'll use 180 GPa (Young modulus for steel)
F is force applied,
L the beam length
I is the areal moment of inertia, I = pi/2 r^4

I am going to use the torque instead of force, T=FL, i.e.

delta = T L^2 / (3 E I)
= 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)

'units' calculates that as 30mm, a little too short of 41mm measured.
Am I making a stupid error somewhere?

First, your value for Young's modulus is the one for stainless steel.
Assuming it's any alloy other than a high-nickel grade, try 210. Any
grade of steel will be in between those two figures, 180 and 210, but
210 is probably very close for any steel grade used for the beam of a
torque wrench.

I am not going to do the calculations. g However, your formula for
area moment of inertia doesn't look right. I don't have my statics
book handy but I think it should be I = pi/4*r^4.

Good luck.

--
Ed Huntress

On Tue, 02 Sep 2014 23:52:35 -0400, Ed Huntress wrote:

On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski
wrote:

I was checking out my torque wrench. I measure it to be 429mm between
the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm.
The indicator is a little short of the outer pivot---the deflection at
the outer pivot for 150lbf is around 41mm.

I found the formula for deflection: delta = F L^3 / (3 E I)

E is the flexural modulus --- I'll use 180 GPa (Young modulus for steel)
F is force applied,
L the beam length I is the areal moment of inertia, I = pi/2 r^4

I am going to use the torque instead of force, T=FL, i.e.

delta = T L^2 / (3 E I)
= 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)

'units' calculates that as 30mm, a little too short of 41mm measured. Am
I making a stupid error somewhere?

First, your value for Young's modulus is the one for stainless steel.
Assuming it's any alloy other than a high-nickel grade, try 210. Any
grade of steel will be in between those two figures, 180 and 210, but
210 is probably very close for any steel grade used for the beam of a
torque wrench.

OK, thanks---it's chrome-plated rusty-spotted rod so you're probably
right. The change from that is in the wrong direction, though


I am not going to do the calculations. g

Here's a great tip: just drop this expression in Google---yes, the Google
search field---

150 lbf ft * (429mm)^2/(3 * 180 GPa * (pi/2)* (12.4mm/2)^4) , in mm

and good old G responds with 29.8625457 millimeters---COOL, HEH?

I got lazy and started dropping the multiplications ('*') because 'units'
allows me to, but Google requires at least some to recognize arithmetical
expressions

However, your formula for
area moment of inertia doesn't look right. I don't have my statics book
handy but I think it should be I = pi/4*r^4.


OK, you got me there--you're good. I think I inadvertently used Iz.
Still, Google sez

150 lbf ft (429mm)^2/(3 * 210 GPa (pi/4) (12.4mm/2)^4) , in mm

is 51.1929355 millimeters. Now it's too much

On Thu, 4 Sep 2014 21:28:48 +0000 (UTC), Przemek Klosowski
wrote:

On Tue, 02 Sep 2014 23:52:35 -0400, Ed Huntress wrote:

On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski
wrote:

I was checking out my torque wrench. I measure it to be 429mm between
the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm.
The indicator is a little short of the outer pivot---the deflection at
the outer pivot for 150lbf is around 41mm.

I found the formula for deflection: delta = F L^3 / (3 E I)

E is the flexural modulus --- I'll use 180 GPa (Young modulus for steel)
F is force applied,
L the beam length I is the areal moment of inertia, I = pi/2 r^4

I am going to use the torque instead of force, T=FL, i.e.

delta = T L^2 / (3 E I)
= 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)

'units' calculates that as 30mm, a little too short of 41mm measured. Am
I making a stupid error somewhere?

First, your value for Young's modulus is the one for stainless steel.
Assuming it's any alloy other than a high-nickel grade, try 210. Any
grade of steel will be in between those two figures, 180 and 210, but
210 is probably very close for any steel grade used for the beam of a
torque wrench.

OK, thanks---it's chrome-plated rusty-spotted rod so you're probably
right. The change from that is in the wrong direction, though

I hoped it would all balance out with the pi/4 factor, but I didn't
try to do it in my head, so I really didn't know which way it would
go.



I am not going to do the calculations. g

Here's a great tip: just drop this expression in Google---yes, the Google
search field---

150 lbf ft * (429mm)^2/(3 * 180 GPa * (pi/2)* (12.4mm/2)^4) , in mm

and good old G responds with 29.8625457 millimeters---COOL, HEH?

OOH! I LIKE that! Thanks for the tip!

I'll get some use out of that.


I got lazy and started dropping the multiplications ('*') because 'units'
allows me to, but Google requires at least some to recognize arithmetical
expressions

However, your formula for
area moment of inertia doesn't look right. I don't have my statics book
handy but I think it should be I = pi/4*r^4.


OK, you got me there--you're good. I think I inadvertently used Iz.
Still, Google sez

150 lbf ft (429mm)^2/(3 * 210 GPa (pi/4) (12.4mm/2)^4) , in mm

is 51.1929355 millimeters. Now it's too much

Yeah, backing into it, it takes a value of 260 to make it work, and
that's just not possible with any steel. In general, the Young's
modulus for steel falls into a very narrow range, except for
high-nickel alloys (including stainless), which are lower, and
high-speed steel with tungsten, which is higher. Your torque-wrench
beam is most likely a plain-carbon grade, which should be in the range
of 205 - 210.

Assuming your measurements are right, I don't know where else to go
with it.

--
Ed Huntress

On Thu, 04 Sep 2014 18:09:58 -0400, Ed Huntress wrote:

On Thu, 4 Sep 2014 21:28:48 +0000 (UTC), Przemek Klosowski
wrote:

On Tue, 02 Sep 2014 23:52:35 -0400, Ed Huntress wrote:

On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski
wrote:

I was checking out my torque wrench. I measure it to be 429mm between
the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm.
The indicator is a little short of the outer pivot---the deflection at
the outer pivot for 150lbf is around 41mm.

I found the formula for deflection: delta = F L^3 / (3 E I)

E is the flexural modulus --- I'll use 180 GPa (Young modulus for
steel)
F is force applied,
L the beam length I is the areal moment of inertia, I = pi/2 r^4

I am going to use the torque instead of force, T=FL, i.e.

delta = T L^2 / (3 E I)
= 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)

'units' calculates that as 30mm, a little too short of 41mm measured.
Am I making a stupid error somewhere?

First, your value for Young's modulus is the one for stainless steel.
Assuming it's any alloy other than a high-nickel grade, try 210. Any
grade of steel will be in between those two figures, 180 and 210, but
210 is probably very close for any steel grade used for the beam of a
torque wrench.

OK, thanks---it's chrome-plated rusty-spotted rod so you're probably
right. The change from that is in the wrong direction, though

I hoped it would all balance out with the pi/4 factor, but I didn't try
to do it in my head, so I really didn't know which way it would go.



I am not going to do the calculations. g

Here's a great tip: just drop this expression in Google---yes, the
Google search field---

150 lbf ft * (429mm)^2/(3 * 180 GPa * (pi/2)* (12.4mm/2)^4) , in mm

and good old G responds with 29.8625457 millimeters---COOL, HEH?

OOH! I LIKE that! Thanks for the tip!

I'll get some use out of that.


I got lazy and started dropping the multiplications ('*') because
'units'
allows me to, but Google requires at least some to recognize
arithmetical expressions

However, your formula for
area moment of inertia doesn't look right. I don't have my statics
book handy but I think it should be I = pi/4*r^4.


OK, you got me there--you're good. I think I inadvertently used Iz.
Still, Google sez

150 lbf ft (429mm)^2/(3 * 210 GPa (pi/4) (12.4mm/2)^4) , in mm

is 51.1929355 millimeters. Now it's too much

Yeah, backing into it, it takes a value of 260 to make it work, and
that's just not possible with any steel. In general, the Young's modulus
for steel falls into a very narrow range, except for high-nickel alloys
(including stainless), which are lower, and high-speed steel with
tungsten, which is higher. Your torque-wrench beam is most likely a
plain-carbon grade, which should be in the range of 205 - 210.

Assuming your measurements are right, I don't know where else to go with
it.

Given that the dependence is on diameter^4, it wouldn't take much of a
measurement error to throw it off.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

On Thu, 04 Sep 2014 18:18:32 -0500, Tim Wescott
wrote:

On Thu, 04 Sep 2014 18:09:58 -0400, Ed Huntress wrote:

On Thu, 4 Sep 2014 21:28:48 +0000 (UTC), Przemek Klosowski
wrote:

On Tue, 02 Sep 2014 23:52:35 -0400, Ed Huntress wrote:

On Wed, 3 Sep 2014 03:10:11 +0000 (UTC), Przemek Klosowski
wrote:

I was checking out my torque wrench. I measure it to be 429mm between
the pivots, and made out of a 12.4mm steel rod,
that is marked as 150 lb ft when the indicator deflects by 38mm.
The indicator is a little short of the outer pivot---the deflection at
the outer pivot for 150lbf is around 41mm.

I found the formula for deflection: delta = F L^3 / (3 E I)

E is the flexural modulus --- I'll use 180 GPa (Young modulus for
steel)
F is force applied,
L the beam length I is the areal moment of inertia, I = pi/2 r^4

I am going to use the torque instead of force, T=FL, i.e.

delta = T L^2 / (3 E I)
= 150 lbf ft (429mm)^2/(3 180 GPa (pi/2) (12.4mm/2)^4)

'units' calculates that as 30mm, a little too short of 41mm measured.
Am I making a stupid error somewhere?

First, your value for Young's modulus is the one for stainless steel.
Assuming it's any alloy other than a high-nickel grade, try 210. Any
grade of steel will be in between those two figures, 180 and 210, but
210 is probably very close for any steel grade used for the beam of a
torque wrench.

OK, thanks---it's chrome-plated rusty-spotted rod so you're probably
right. The change from that is in the wrong direction, though

I hoped it would all balance out with the pi/4 factor, but I didn't try
to do it in my head, so I really didn't know which way it would go.



I am not going to do the calculations. g

Here's a great tip: just drop this expression in Google---yes, the
Google search field---

150 lbf ft * (429mm)^2/(3 * 180 GPa * (pi/2)* (12.4mm/2)^4) , in mm

and good old G responds with 29.8625457 millimeters---COOL, HEH?

OOH! I LIKE that! Thanks for the tip!

I'll get some use out of that.


I got lazy and started dropping the multiplications ('*') because
'units'
allows me to, but Google requires at least some to recognize
arithmetical expressions

However, your formula for
area moment of inertia doesn't look right. I don't have my statics
book handy but I think it should be I = pi/4*r^4.


OK, you got me there--you're good. I think I inadvertently used Iz.
Still, Google sez

150 lbf ft (429mm)^2/(3 * 210 GPa (pi/4) (12.4mm/2)^4) , in mm

is 51.1929355 millimeters. Now it's too much

Yeah, backing into it, it takes a value of 260 to make it work, and
that's just not possible with any steel. In general, the Young's modulus
for steel falls into a very narrow range, except for high-nickel alloys
(including stainless), which are lower, and high-speed steel with
tungsten, which is higher. Your torque-wrench beam is most likely a
plain-carbon grade, which should be in the range of 205 - 210.

Assuming your measurements are right, I don't know where else to go with
it.

Given that the dependence is on diameter^4, it wouldn't take much of a
measurement error to throw it off.

Yeah, that's what I was thinking, too.

--
Ed Huntress