Whats wrong with my derivation?

[caspernorth]

force = stress x area [for a volume stress or normal stress and we are going to find out the work done ]
Force = bulk modulus x strain x area
F= K.s.a
F = K.(dv/V).a {change in volume by original volume}
work done dW = F.dx
dW = k.(dv/V).a.dx
integrating;
W = ∫ ∫k.(dv/V).a.dx
= ∫ ∫k.(dv/X).dx {area / volume = length}
= ∫k.(dv/X) ∫ dx
= ∫k.(dv/X)x
= k(v/X)x
Is this right, or what's wrong in my calculation [in t.book they've taken a.dx as volume change and then integrated. why not in this method}
my textbooks says its 1/2(Kv^/V)
v = change in volume
V = original volume
x =Change in length
X= Original lenth
K = bulk modulusthis is not a homework but a doubt of mine

 

[tiny-tim]

hi caspernorth!

i'm a bit confused as to what this is about , but the following line looks wrong to me …

dW = k.(dv/V).a.dx

i don't like seeing two d's on the RHS

shouldn't there be a relation between v and x, something like x = v/A so dx = dv/A, and you end up with a vdv ?

 

[Studiot]

Things changed whilst I was thinking about this one, Hello Tiny Tim.

The book is correct because the force varies linearly from zero to F so averages 1/2F.

However, like Tiny Tim I am struggling to follow quite what you are doing.

 

[AlephZero]

I think part of the confusion is that you start off with "dV" being a finite change in volume (related to the amount of strain) and your "dx" is an infinitesimal change in length.

But when you integrate, you seem to be treating dv as an infinitesimal change of volume.

You can't use the "dv" to mean two different things, but as the other answers said its hard to follow exactly what you were trying to do.