Question

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5 mm will experience only elastic deformation when a tensile load of 2170 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.

Answer 1

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

$\epsilon = \deltaL /L$ (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

$\sigma = E/ \epsilon$ (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

$L= \delta L/ \epsilon$

$\epsilon =\sigma /E$  

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

$\sigma=F/A$

$\sigma=\frac{F} {\pi*d^2/4}$

$\sigma=\frac{2170 N}{\pi*4.5 mm^2/4}$

$\sigma= 136000000 Pa= 136 Mpa$  

Then de specific defotmation:

$\epsilon =136 MPa / 108 GPa = 1.26*10^{-3}$

Finally the maximum specimen lenght for a elongation 0f 0.45 mm is:

$L= 0.45 mm/ 1.26*10^{-3} = 358 mm = 0.358 m$