Young's modulus for filled metal? 1

[dculp1]

I have a tin material (E = 13.28e6 psi, density = 0.268 lb/in^3) that is uniformly filled with 0.030" diameter steel shot (E = 30.0e6 psi, density = 0.280 lb/in^3). If the steel is 30% by weight of the total and assuming the surfaces of the steel shot are well-bonded to the tin, what value should I assume for the resulting modulus E?

(As a first guess I have adjusted the modulii according the the relative volumes occupied by the tin and steel, for which the resulting modulus is 13.28e6 psi. Is this correct? If so, is there any theoretical or empirical justification? If not, is there a better approach?)

Thanks,
Don Culp

 

[NickE]

I would try considering this a metal particulate rienforced composite.

then the modulus is:

             Ef * Em
E=E(perp)= ------------
            Ef(1-f)+f*Em

Where:

E(perp) == modulus perpendicular to fibre axis
Ef == modulus of the fibre or rienforcement
Em == modulus of the matrix
f == volume fraction matrix

Then your answer would be:
21.5e6

That again assumes a high strength bond between the two materials.

Nick
I love materials science!

 

[dculp1]

Nick --

Your equation appears to be for an orthotropic material. My material is isotropic on a macro scale (more akin to concrete) -- i.e., no fiber axis. Therefore, would your equation still apply?

If your equation applies then my calculations give a result of 21.96e6 psi. This uses f = .709 (i.e., for each pound of material the tin matrix weighs 0.7 pounds ==> 2.612 in^3; the reinforcing steel weighs 0.3 pounds ==> 1.071 in^3; the total volume is the sum = 3.683 pounds). Is this the f value that you used?

 

[NickE]

Yes, thats how we derived the equation in class many years ago. When in tension perpendicular to the fibre axis the fibre behaves as a particulate. I'll convert to text and get the derivation into a future post.

As far as the maths:

My Bad, I botched the weight to volume calc. (Ouch, I do real math so little that I would have no chance at a qualifier right now.)

then I would get what you did:

21.96e6psi


Nick
I love materials science!

 

[CoryPad]

Nick's equation is wrong. The volume fraction f in the denominator should be switched with 1-f.

Don,

Since your data is in weight fraction, you should use the following equations:

E_c = {E_rE_m}/{E_mv_r + E_rv_m}

v_r = w_r[ρ]_c/[ρ]_r

v_m = w_m[ρ]_c/[ρ]_m

[ρ]_c = {[ρ]_r[ρ]_m}/{w_mv_r + w_rv_m}

where

E = Young's modulus
v = volume fraction
w = weight fraction
[ρ] = mass density

and subscripts

r = reinforcement
m = matrix
c = composite

Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[NickE]

Cory- you are right, I took these notes 8 years ago and havent used them since. looking back I see that we defined f as volume fraction reinforcement, not matrix.

d is the length in the tension direction.
d(f) is the length of all the reinforcement if it was packed together
d(f-1) is the length of the remaining matrix

? == deflection due to tension
? == strain
? == stress
E == Modulus
Same subscripts as Cory


Derivation:

?c = ?r + ?m

      ?c
?c = ----
       d

        ?m
?m = ---------
       d(1-f)

       ?r
?r = ------
      d(f)
So:
?cd = ?rfd + ?m(1-f)d

The length d cancels out.

?c = ?rf + ?m(1-f)

And in an Iso-Stress (stress is distributed equally among all components):
       ?
?c = -----
       Ec

Which also applies to the matrix and reinforcement Giving:
 (I skipped the simple steps)

  1        f      (1-f)
------ = ----- + -------
  Ec      Er       Em

And flipping:
             Er Em
Ec =  ---------------------
          fEm + (1-f) Er

(By the way, how did you get sub-scripts to carry over?)

Nick
I love materials science!

 

[NickE]

And greek letters... forgot to hit preview.....

D == deflection due to tension (del)
e == strain (epsilon)
s == stress (sigma)
E == Modulus


try this:

Dc = Dr + Dm

      Dc
ec = ----
       d

        Dm
em = ---------
       d(1-f)

       Dr
er = ------
      d(f)
So:
ecd = erfd + em(1-f)d

The length d cancels out.

ec = erf + em(1-f)

And in an Iso-Stress (stress is distributed equally among all components):
       s
ec = -----
       Ec

Which also applies to the matrix and reinforcement Giving:
 (I skipped the simple steps)

  1        f      (1-f)
------ = ----- + -------
  Ec      Er       Em

And flipping:
             Er Em
Ec =  ---------------------
          fEm + (1-f) Er

Nick
I love materials science!

 

[CoryPad]

You need to use TGML. For subscripts, here is a sample of what you need to type:

[ignore]D_c = D_r + D_m[/ignore]

which will display as

D_c = D_r + D_m



Regards,

Cory

Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

 

[dculp1]

Nick --

I believe I understand the derivation of your equation. However, I think it may apply to my situation.

If we divide your final equation by Er:

Ec = Em / [ f (Em/Er) + (1 - f)]

This equation shows that as Er decreases Ec decreases. If Er = 0 (e.g., if the reinforcing material were air) then Ec = 0, regardless of f (except if f = 0). In this case the composite material could not sustain any load. This would be the case if the composite material were built up in layers:

Matrix layer 1
Fiber layer 1
Matrix layer 2
Fiber layer 2
etc.

However, if the fibers are uniformly dispersed (as in my situation) then replacing them by air (Er = 0) should, in reality, still allow the composite material to sustain substantial load, so that Ec would not equal 0. Thus it seems like the (1 - f) factor should somehow appear in the numerator, something like:

Ec = Em (1 - f)^k where k = fit constant

so that as f increases Ec will decrease.

Hence, it seems that your equation may not apply to my situation, at least in the limit.

On the other hand, if Er is infinite then your equation gives:

Ec = Em / (1 - f)

At least qualitatively this seems reasonable -- i.e., as f increases Ec also increases.

Any thoughts?

Don Culp