Expression of bulk modulus of a cube in terms of strain

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In my text it's given when a cube underwents a uniform unit tensile force be applied in all six faces,
bulk module=1/3(α-2β)

Where α is longitudinal strain and β is lateral strain.is there a derviation for it..?
And it states in x direction there will be increase in length and in y, z direction there will be decrease length.

How can there be decrease in length when all the forces are tensile ....?

Any help appreciated.thanks in advance.
 
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For a given pair of tensile forces (let say along x direction), there is an increase in length along x and decrease along the perpendicular directions (y and z).

But if you have tensile forces applied on all faces, the strain along a give direction will have three contributions. For example, for x direction, there is an increase (described by α ) due to forces along x direction and shrinkage due to forces along y and z directions.
So what will be the total strain along x?
 
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So the total strain along x direction, should decreases the length along x direction because there are two factors responsible for decreases in length and only one responsible for increase in length which is the tensile force in x direction......is it correct..?
 
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http:// http://books.google.co.in/books?id=2IHEqp8dNWwC&pg=PT90&lpg=PT89&ots=TkrGlllFOV&focus=viewport&dq=bulk+modulus+of+a+cube+in+terms+of+longitudinal+and+lateral+strain&output=html_text

In this book i found the derivation but i got a doubt

It given strain of AB due to compressive stress is μσ/E
Where μ is stress and E is young's modulus

μ=β/α where β is lateral strain and α is longitudinal strain.

μσ/E=βσ/Eα but E=1/α therefore μσ/E=μ/(1/β)

1/β , what it is...? I know 1/α=E but what is 1/β is it also young's modules......i didn't get it
 
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So the total strain along x direction, should decreases the length along x direction because there are two factors responsible for decreases in length and only one responsible for increase in length which is the tensile force in x direction......is it correct..?
No, it is not necessary. α and β are not the same thing so 2β can be less than α.
And most of the times it is. So pressure applied on all faces results in volume decrease. And pulling all faces results in volume increase.

The link does not work for me.
 
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In this book i found the derivation but i got a doubt

It given strain of AB due to compressive stress is µs/E
Where µ is stress and E is young's modulus

µ=ß/a where ß is lateral strain and a is longitudinal strain.

µs/E=ßs/Ea but E=1/a therefore µs/E=µ/(1/ß)

1/ß , what it is...? I know 1/a=E but what is 1/ß is it also young's modules......i didn't get it
You have a few errors in this:
1. µ is not stress but Poisson's ratio.
2. This relationship is not right. Look at the units, for one thing

I think there is some confusion here.
If α and β are strains, as you call them, then even the original formula for bulk modulus is not dimensionally correct. A strain is a unitless quantity. The bulk modulus has units of pressure (Pascal). So maybe you check your original formula and the definitions of the quantities in it.
Where did you get it from?
 
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Sorry μ is not stress it's poison ratio only and σ is stress.

And my question is
It is given strain of AB due to compressive stress is μσ/E

Where μ is poison ratio and E is young's modulus
*
μ=β/α where β is lateral strain and α is longitudinal strain.

μσ/E=βσ/Eα but E=1/α therefore μσ/E=σ/(1/β)

1/β , what it is...? I know 1/α=E but what is 1/β is it also young's modules......i didn't get it..

And the source is http://books.google.co.in/books?id=2IHEqp8dNWwC&pg=PT90&lpg=PT89&ots=TkrGlllFOV&focus=viewport&dq=bulk+modulus+of+a+cube+in+terms+of+longitudinal+and+lateral+strain&output=html_text
 
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You keep using the same wrong relationship.
E is not 1/α.

α is longitudinal strain which is a ratio between two lengths, it has no units and cannot be equal to 1/E which has units. (E is in Pa or N/m^2)
If you have a stress σ along the x direction, then the relationship between strain and stress is Hooke's law:
σ = E α
So 1/α = E/σ

Here α=Δx/x

The same stress σ results in compressions along y and z. If we take y direction, for example,
Δy/y=β , again a ratio between two lengths so a pure number.
And Poisson ration is
μ=β/α

If you need 1/β, this will be 1/(μα) = E/(σ μ).

PS. I still don't understand what are you trying to do. These relationships are not in the page linked.
 
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Is this a different problem? There was no share strain in the OP problem.
 
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It's given as separate here in the link, but in my book these two are merged and it is solved in my book with σ=1 so only i wrote α=1/E.....
 
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OK, so what is the problem you are trying now?
Which one in that link?
 
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OK, I think that now I understand your questions.
I'll have to go through all these equations.
I'll get back to you later.
 
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So, you have a relationship between tensile and shear strains. This results from geometry only.
And then you have the other relashionship that relates strainto stress. They are both ok. So what is bothering you here?
 
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I can't get how tensile strain=shear strain/2 and how tensile strain along .BD is =(1+μ)t/E.
 
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The first one is shown step by step on page 78. It follows from the geometry of the problem and the fact that the shear angle is small.

The second one is on page 77. It follows from Hooke's law.
On the previous page they show that the shear stress τ there are tensile stresses along one diagonal and compression stresses along the other diagonal. They even show that the magnitude of these tensile/compression stresses is the same as the shear stress τ.
So if you look along one diagonal, you have stress along the diagonal, which will result in tensile strain s1=t/E. Then you have compression stress perpendicular to the diagonal's direction which will result in compression perpendicular to the diagonal but extension along the diagonal. The compression strain (perpendicular to the diagonal) will be again τ/E but the stress along the diagonal will be s2= μτ/E. So the total tensile stress will be s1+s2.
You can imagine a bar with rectangular cross section with two sets of stresses: one along the bar, puling the ends apart. And one crushing the bar across its length. They bot result in extension along the length of the bar.
 
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Thanks a lot... :-)
 

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