Question

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 3.7 mm will experience only elastic deformation when a tensile load of 1890 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.

Answer 1

L= 276.4 mm

Step-by-step explanation:

Given that

E= 180 GPa

d= 3.7 mm

F= 1890 N

ΔL= 0.45 mm

We know that ,elongation due to load F in a cylindrical bar is given as follows

`\Delta L =(FL)/(AE)`

`L=(\Delta L* AE)/(F)`

Now by putting the values in the above equation we get

`L=(0.45* 10^(-3)* (\pi)/(4)* (3.7* 10^(-3))^2* 108* 10^9)/(1890)\ m`

L=0.2764 m

L= 276.4 mm

Therefore the length of the specimen will be 276.4 mm