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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.

1 Answer

1 vote

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

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