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

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

Step-by-step 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=(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`