find the young modulus of brass rod of diameter 25mm and of length 250mm which is subjected to a tensile load of 50 KN when the extension of the rod is equal to0.3 mm
Answers
Answer: ≈ 8.5 × 10¹⁰ N/m²
Physical Properties of the Brass rod as given in the question :-
- Length, L = 250 mm
- Radius, R = Diameter / 2 = 25/2 mm
We have some other variables given in the question like :-
- Load of 50 kN hanging on the brass rod which results in elongation of the rod to 0.3 mm.
For finding the Young Modulus of the brass rod we will need Stress on the rod, Ratio of elongation to total length.
Let's find the stress on the rod which equals to:
- Stress(Pressure) = Force / Area
⇒ Stress = 50,000 / (Area of cross-section)
Because the rod is shaped like a cylinder then the area of its cross-section will be πR² ( R = radius)
⇒ Stress = 50000 / (π (25/2 mm)² )
Because, Force is given in newton, then we must convert the unit of area into m² in order to get the answer in pascal.
- 1 mm = 0.001 m
⇒ Stress = 50,000 / 22/7 ( 25/2 × 10⁻³ m)²
take, π = 22/7
⇒ Stress = 50000 / ( 22/7 × 625 / 4 × 10⁻⁶ )
⇒ Stress = 50000 / ( 11/14 × 625 × 10⁻6 )
⇒ Stress = 50000 × 14 × 10⁶ / 625×11
⇒ Stress = 80 × 14 × 10⁶ / 11
⇒ Stress = 1120/11 × 10⁶ Pa
Now, that we have Stress, Let's find Strain (ratio of elongation to original length) ,
⇒ Strain = (Elongation) / (Original length)
⇒ Strain = 0.3 / 250
⇒ Strain = 3 / 2500
Hence,
⇒ Young Modulus, Y = Stress / Strain
⇒ Y = 1120/11 × 10⁶ / 3/2500
⇒ Y = 1120/11 × 2500/3 × 10⁶
⇒ Y = 101.8 × 833.3 × 10⁶
⇒ Y = 84829.94 × 10⁶
⇒ Y ≈ 8.5 × 10¹⁰ Pa
Hence, The Young Modulus of the brass rod is ≈ 8.5 × 10¹⁰ Pa or N/m²