Question

A metal rod, 20 mm diameter, is tested in tension (force applied axially). The total extension over a length of 80 mm is 3.04 x 102 mm for a pull of 25 kN. Calculate the normal stress, normal strain and modulus of elasticity (Young's modulus), assuming the rod is linear elastic over the load range.

Answer 1

Answer:stress=79.56MPa

strain=
`3.8* 10^(-4)`

Young Modulus=209.36 GPa

Step-by-step explanation:

Given data

d=20 mm

Length
`\left ( L\right )`=80mm

`\Delta {L}`=
`3.04* 10^(-2)`mm

Load=
`25* 10^(3)`N

`\left ( i\right )`

Stress=
`(Load\ applied)/(cross-section)`

Stress=
`(25* 10^(3))/(314.2)`

Stress=79.56MPa

`\left ( ii\right )`

Strain=
`(Change\ in\ length)/(Length)`

Strain=
`(3.04* 10^(-3))/(80)`

Strain=
`3.8* 10^(-4)`

`\left ( iii\right )`

young modulus
`\left ( E\right )`=
`(Stress)/(Strain)`

E=
`(79.56* 10^(6))/(3.8* 10^(-4))`

E=209.36GPa