Axial Deformation 1” = C


The axial deformation is important and to find out you just need to know that Normal stress = σ = p /A

Normal strain = e = δ  /L

Deformation can be obtained after knowing the function of A, L, P, E.

What is Hook’s Law?

According to the hook’s law

Stress ∝ Strain or Stress and strain are proportional to each other.

σ = E .e where E is modulus of elasticity and it is constant.

Now,  P /A = E x δ / L

  • PL/AE = δ

Or δ = PL/AE

At the condition of the constant value of A, P, L and E, the  weight of the bar is avoided or neglected.

What are the various cases when constants vary?

At the time of first case when P varies

At this condition,  the complete weight of the body acts at the exact point C.G. Moreover, the value of its weight varies in other points except C.G. This is for a bar that suspends freely. Now,  consider a part of this length as dy. Let that A, E and P do not vary, then the formula  δ = PL / AE is applicable here.


Axial Deformation 2” = C


Y=specific weight

Dδ = (Yy A) dy / AE

= YL2 / 2E

= WL / 2 AE

δ= WL / 2 AE

Case 2 at the time Varying A


Axial Deformation 3” = C


BC = b2 – b1 / 2

ED = y – b1 / 2

AC = L

If you take the similar triangles ABC and AED, then you can easily see that

BC / ED = AC / AD

(b2 – b1)2 / 2 (y-b1) = L / x

y = (b2 – b1) / L x + b1

For the elemental area you can apply for the formula δ = PL / AE.

Now, Dδ = p dx / (yt) E is clear for the elemental area and thus


Axial Deformation 4” = C


3rd case when P and A vary

Solid ABCD has the volume =

1/2 (x + b1) ty

Now, the solid ABCD has the weight = 1/2  ( x + b1) yt x y

dδ =


Axial Deformation 5” = C


If you go through the similar triangle AEE’ and ADD’, then

x = b1 + (b2 –b1) /L .Y


Axial Deformation 6” = C


Axial Deformation 7” = C


Now, through the above explanation you can easily get that how deformation takes place and in the different conditions you will get the result in a different way. Thus, be careful about finding out the solution in a proper way.


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