Question

A brass rod with a length of 1.50 m and a cross-sectional area of 2.50 cm2 is fastened end to end to a nickel rod with length L and cross-sectional area 0.880 cm2 . The compound rod is subjected to equal and opposite pulls of magnitude 3.46×10⁴ N at its ends.

a.Find the length L of the nickel rod if the elongations of the two rods are equal.

b.What is the stress in the brass rod?

c.What is the stress in the nickel rod?

d.What is the strain in the brass rod?

e.What is the strain in the nickel rod?

Answer 1

NOTE :

Stress is defined as the force per unit area of a material.

i.e. Stress = force / cross sectional area:

............................(1)

where,

σ = stress,

F = force applied, and

A= cross sectional area of the object.

Strain is defined as extension per unit length.

Strain has no units.

Strain = extension / original length :

   .................................(2)

where,

= strain,

l₀ = the original length

e = extension = (l-lo), and

l = stretched length

And the Young Modulus (E) is defined as

Putting the value of stress and strain , we get

...............................(3)

E of brass = 9.0 x 10¹⁰Pa and

E of nickel = 21 x 10¹⁰ Pa

Answer :

Given, length of brass rod = 1.5 m

cross - sectional area of brass rod = 2.50 cm² = 2.50 x 10^-4 m²

length of nickel rod = L

cross - sectional area of nickel rod = 0.880 cm² = 0.880 x 10^-4 m²

Force (F) = 3.46 x 10⁴ N

(a)

From equation (3), we can write

So,

and

Since, elongation in brass rod = elongation in nickel rod

i.e.

so, 23.07 x 10^-4 = 18.72 x 10^-4 x L

or, L = 1.23 m

(b) Stress in brass rod :

(c) Stress in nickel rod :

(d) We know that ,

  

or,   

So, strain in brass rod is :

  

(e) Strain in nickel rod is

  

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