find the youngs modulus of the roda composite rod

Question

Find the Youngs modulus of the rod:

A composite rod is made by joining a copper rod end to end with a second rod of different material but of same cross-section. At 25oC, the composite rod is 1 m in length of which the length of copper rod is 30 cm. At 125oC, the length of the composite rod increases by 1.91 mm. While the composite rod is not permitted to expand through holding it among two rigid walls, it is found that the length of constituents does not change with rise in temperature. Find the Young's modulus and coefficient of linear expansion of the second rod. For copper, α = 1.7 × 10^{- 5 o}C^-1, and Y = 1.3 × 10¹¹ Nm^{- 2}.

Solution

For copper rod α₁ = 1.7 × 10^{- 5 o}C- 1, and Y₁ = 1.3 × 10¹¹ Nm^-2, l1 = 30 cm = 0.3 m. Let l₂ be initial length of the rod of other material at 25^oC, and α₂ and Y₂ be coefficient of linear expansion and Young's modulus for the material of this rod. Then, at initial temperature θ1 = 25^oC, l₁ + l₂ = 1 m,

l₂ = 1 - l₁ = 1 - 0.3 = 0.7 m

Let the final temperature θ₂ = 125^oC, the increases in the length of cooper rod and that of other material be Δl₁ and Δl₂, respectively.

Then,

 Δl₁ + Δl₂ = 1.91 mm = 1.91 × 10^-3  m                                                        . . . (A)

Now,

Δl₁ = l₁ α₁ (θ₂ - θ₁) = 0.3 × 1.7 × 10^{- 5} × (125 - 25) = 0.51 × 10^-3 m

Also,

Δl₂ = l₂ α₂ (θ₂ - θ₁) = 0.7 × α₂ × (125 - 25) = 70 α₂

Substituting for Δl₁ and Δl₂ in Eq. (A), we get

0.51 × 10^{- 3} + 70 α₂ = 1.91 × 10^-3

or, α₂  = 2.0 × 10- 5 ^oC ^-1

Now,

Y =   (F/A)/ (Δl/C) = FL/ A Δl ⇒ F = Y A Δl/l

Let A be the cross-section of each of the two rods, thus, tension produced in copper rod F₁ = Y₁AΔl₁ /l₁, and tension generates in the rod of other material, F₂ = Y₂ A Δl₂/ l₂. The tension developed in the two rods will be same, i.e. F₁= F₂.

Therefore,

(Y₁ A Δl₁)/l₁ = (Y₂ A Δl₂)/l₂

or        Y₂  = Y₁ l₂Δl₁/l₁Δl₂ ⇒ Y₂  = 1.3 × 10¹¹ × 0.7×0.51×10^-3/0.3 × 1.40 × 10^-3

or       Y₂  = 1.105 × 10¹¹  Nm^-2