MadModder

Gallery, Projects and General => Neat Stuff => Topic started by: Chuck in E. TN on April 12, 2015, 07:05:38 PM

Title: Nut insert for repair
Post by: Chuck in E. TN on April 12, 2015, 07:05:38 PM

 John(Doubleboost),
Very neat repair you demonstrated in Sunday Nite Nightcap # 66 (

0iPSQ&index=1). Is there a formula or rule of thumb that lead you to choose 7mm to bore for a M4 nut? Any correlation that can be used for such repairs using SAE nuts?
All the best to you and Debs.
Chuck

Title: Re: Nut insert for repair
Post by: studders on April 13, 2015, 12:16:39 AM

I've used that method a few times in wood, never occurred to me to use it in aluminumium. One to remember.

Title: Re: Nut insert for repair
Post by: studders on April 13, 2015, 11:26:21 AM

Quote from: mklotz on April 13, 2015, 11:15:54 AM

To do this we need to calculate the diameter of a nut ACROSS THE POINTS .........

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Doubtless that was the very method John used to arrive at 7mm.

I favour the 'guess it and whack it' approach.

Title: Re: Nut insert for repair
Post by: awemawson on April 13, 2015, 11:42:57 AM

I claim prior invention on this as per my posting in my Fanuc Wire EDM rebuild thread in October 2014  :lol:


(Actually the idea was suggested to me by a poster on the MIG welding forum)

Title: Re: Nut insert for repair
Post by: mklotz on April 13, 2015, 12:06:02 PM

Quote from: studders on April 13, 2015, 11:26:21 AM

I favour the 'guess it and whack it' approach.

I'm sure you do.

Title: Re: Nut insert for repair
Post by: bertie_bassett on April 13, 2015, 04:41:41 PM

Quote from: mklotz on April 13, 2015, 11:15:54 AM

To do this we need to calculate the diameter of a nut ACROSS THE POINTS given the conventional width of the nut measured ACROSS THE FLATS.  That diameter, suitably adjusted for an interference fit, will be the size of the recess into which the nut is driven.

Take a look at the canonical circular arc diagram shown below.  In this diagram the desired diameter,  d, will be twice the radius, r.  Similarly, the width of the nut, w, will be twice the apothem, a, shown in the diagram.

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Remember to drill slightly smaller than d so the points of the nut have something to bite into.

Could you not just measure the nut??

Title: Re: Nut insert for repair
Post by: sparky961 on April 13, 2015, 04:57:36 PM

:poke: I'm sure I heard John go through that formula on the video.  I just couldn't understand him. :poke:

Title: Re: Nut insert for repair
Post by: krv3000 on April 13, 2015, 05:04:33 PM

I think he added ten to it to get the factor of 7

Title: Re: Nut insert for repair
Post by: doubleboost on April 13, 2015, 05:26:25 PM

Sorry Lads
No magic formula I measured across the flats and just went with that
The flat bottomed hole it the important bit

I will have a play with different size nuts & post a bit video

Any further DRO installs will be made from steel
 John

Title: Re: Nut insert for repair
Post by: vtsteam on April 13, 2015, 10:50:12 PM

Quote from: bertie_bassett on April 13, 2015, 04:41:41 PM

Quote from: mklotz on April 13, 2015, 11:15:54 AM

To do this we need to calculate the diameter of a nut ACROSS THE POINTS given the conventional width of the nut measured ACROSS THE FLATS.  That diameter, suitably adjusted for an interference fit, will be the size of the recess into which the nut is driven.

Take a look at the canonical circular arc diagram shown below.  In this diagram the desired diameter,  d, will be twice the radius, r.  Similarly, the width of the nut, w, will be twice the apothem, a, shown in the diagram.

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Remember to drill slightly smaller than d so the points of the nut have something to bite into.

Could you not just measure the nut??

 :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol:


And then of course apply the technical term "slightly smaller than" to arrive at an exact value.

Title: Re: Nut insert for repair
Post by: RobWilson on April 14, 2015, 12:19:26 PM

 :smart:

Quote from: vtsteam on April 13, 2015, 10:50:12 PM

 :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol: :lol:


And then of course apply the technical term "slightly smaller than" to arrive at an exact value.

 :lol:  PRICELESS !   :lol:


Rob  :coffee:

Title: Re: Nut insert for repair
Post by: RobWilson on April 15, 2015, 01:09:28 PM

Quote from: mklotz on April 13, 2015, 11:15:54 AM

To do this we need to calculate the diameter of a nut ACROSS THE POINTS given the conventional width of the nut measured ACROSS THE FLATS.  That diameter, suitably adjusted for an interference fit, will be the size of the recess into which the nut is driven.

Take a look at the canonical circular arc diagram shown below.  In this diagram the desired diameter,  d, will be twice the radius, r.  Similarly, the width of the nut, w, will be twice the apothem, a, shown in the diagram.

d = 2*r
w = 2*a

From the diagram, we have,

a = r * cos(x)

so:

2*a = w = 2*r*cos(x) = d * cos(x)

Solving for d,

d = w/cos(x)

For a hexagonal nut, x will be 30 degrees so,

cos(x) = sqrt(3)/2 = 0.866

Finally,

d = w/0.866 = 1.1547 * w

Remember to drill slightly smaller than d so the points of the nut have something to bite into.

Just checked above with  my book " Math to build on a book for those who build "  close enough  ,but  the book did not cover the "slight smaller factor"  :coffee:


Rob