Acylindrical 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.

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:

(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:

(2)

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

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:

Then de specific defotmation:

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

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )