Hello,


Cannot understand Carr Mc Master 'spring constants' in their spring
section.

They state 'the spring constant is the number of pounds force required
to compress
the spring one inch'

The spring constants being given seem to be far too large.
A long 36 inch spring being made out of 0.013 inch wire has a spring
constant
of 4.8. ie it takes 4.8 pounds to compress the spring one inch.

Where am I going wrong.


TIA.

Jack

Where am I going wrong.

Skipping, or sleeping through, high school physics?

Try googling "Hooke's Law" or "F = -kx".

wrote: (clip)it takes 4.8 pounds to compress the
spring one inch. Where am I going wrong.
^^^^^^^^^^^^^^^^
I don't see anything wrong with a spring constant of about 5 lb/in on a
spring like that. We're talking about a coil spring in compression or
tension, right? You don't state the diameter of the coil, or the turn
spacing, but it seems reasonable to me that if you compress a spring about
3% of its length, you might expect to exert about 5 lb.


TIA.

Jack

You neglected to say that the spring is question has an
OD of .094"

And this section of the catalog is for continous length which states:

Springs have open ends and can be cut to the length you need. Great for
manufacturing, utility, and maintenance jobs. To determine the length
(in inches) to which to cut your spring, take the spring constant and
divide it by the number of coils per inch. Then take this value and
divide it by your desired spring rate in lbs./inch.

From what I can tell, this means that the spring constant is NOT the K
value from your textbook but the spring value for ONE turn. DIVIDE by
the number of turns to get the spring constant from the textbook.

Don't complain to me, I just read the fine print!!!

wrote:
Hello,


Cannot understand Carr Mc Master 'spring constants' in their spring
section.

They state 'the spring constant is the number of pounds force required
to compress
the spring one inch'

The spring constants being given seem to be far too large.
A long 36 inch spring being made out of 0.013 inch wire has a spring
constant
of 4.8. ie it takes 4.8 pounds to compress the spring one inch.

Where am I going wrong.


TIA.

Jack

RoyJ wrote:
You neglected to say that the spring is question has an
OD of .094"

And this section of the catalog is for continous length which states:

Springs have open ends and can be cut to the length you need. Great for
manufacturing, utility, and maintenance jobs. To determine the length
(in inches) to which to cut your spring, take the spring constant and
divide it by the number of coils per inch. Then take this value and
divide it by your desired spring rate in lbs./inch.

From what I can tell, this means that the spring constant is NOT the K
value from your textbook but the spring value for ONE turn. DIVIDE by
the number of turns to get the spring constant from the textbook.

Don't complain to me, I just read the fine print!!!


Thanks for that, I am inclined to say that they are using the term
"Spring Constant"
in a not so very scientifically correct manner :-|

The spring is no good for me as it has too many turns per inch.

Cheers


wrote:
Hello,


Cannot understand Carr Mc Master 'spring constants' in their spring
section.

They state 'the spring constant is the number of pounds force required
to compress
the spring one inch'

The spring constants being given seem to be far too large.
A long 36 inch spring being made out of 0.013 inch wire has a spring
constant
of 4.8. ie it takes 4.8 pounds to compress the spring one inch.

Where am I going wrong.


TIA.

Jack

wrote:

Hello,


Cannot understand Carr Mc Master 'spring constants' in their spring
section.

They state 'the spring constant is the number of pounds force required
to compress
the spring one inch'

The spring constants being given seem to be far too large.
A long 36 inch spring being made out of 0.013 inch wire has a spring
constant
of 4.8. ie it takes 4.8 pounds to compress the spring one inch.

Where am I going wrong.


TIA.

Jack



Spring constants are dependent on the wire size, coil diameter, number
of coils and material.

A coil spring is really a torsion spring formed in a spiral. When
compressed or extended beyond it's rest length the wire is being
twisted, not bending.

Smaller or fewer coils means that the wire is shorter and must be
twisted more to compress the spring the same distance. Therefore it
would have a higher spring constant.

Fred

Hooke's Law is also only valid within the elastic limit of the spring.
If you take a small spring and stretch it an inch, it will probably
never recover.
Bugs

ff wrote:


A coil spring is really a torsion spring formed in a spiral. When
compressed or extended beyond it's rest length the wire is being
twisted, not bending.

Hey, thanks Fred! I never stopped to think about coil springs working
that way, I just stopped at "Pull it and it stretches, pull it twice as
hard and it stretches twice as far."

That hit me about the same way as the first time someone pointed out to
me that the "working fluid" in my car's engine isn't gasoline, it's air.
The gasoline just heats it and makes it expand.

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"Truth exists; only falsehood has to be invented."

In article ,
says...
ff wrote:


A coil spring is really a torsion spring formed in a spiral. When
compressed or extended beyond it's rest length the wire is being
twisted, not bending.

Hey, thanks Fred! I never stopped to think about coil springs working
that way, I just stopped at "Pull it and it stretches, pull it twice as
hard and it stretches twice as far."


And to carry this a bit further, the wire in a typical torsion spring is
primarily subject to bending, i.e., it's a coiled up beam.

Ned Simmons

"Ned Simmons" wrote: And to carry this a bit further, the wire in a typical
torsion spring is primarily subject to bending, i.e., it's a coiled up
beam.
^^^^^^^^^^^^^
Let's unscramble this a little. A "torsion spring" is an elastic bar
subject to twisting--not bending. A "coil spring" is a torsion spring that
has been wound into a helix for compactness. A "leaf spring" is a beam in
bending.

In article ,
says...

"Ned Simmons" wrote: And to carry this a bit further, the wire in a typical
torsion spring is primarily subject to bending, i.e., it's a coiled up
beam.
^^^^^^^^^^^^^
Let's unscramble this a little. A "torsion spring" is an elastic bar
subject to twisting--not bending. A "coil spring" is a torsion spring that
has been wound into a helix for compactness. A "leaf spring" is a beam in
bending.


A torsion *bar* is an elastic bar subject to twisting, a torsion
*spring* is a helical beam - like this...
http://www.sterlingspring.com/torsion_springs.html

Or a beam coupling...
http://www.bervina.com/kpl/special.html

Ned Simmons

Leo Lichtman wrote:
"Ned Simmons" wrote: And to carry this a bit further, the wire in a typical
torsion spring is primarily subject to bending, i.e., it's a coiled up
beam.
^^^^^^^^^^^^^
Let's unscramble this a little. A "torsion spring" is an elastic bar
subject to twisting--not bending. A "coil spring" is a torsion spring that
has been wound into a helix for compactness. A "leaf spring" is a beam in
bending.


Which means I suppose that a "clock spring" is just an extra long "leaf
spring"?

Jeff

--
Jeffry Wisnia

(W1BSV + Brass Rat '57 EE)

"Truth exists; only falsehood has to be invented."