Ncert question..plz i’m a bit confused..plz help

Question 9.8:

A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

which modulous should i use here , young’s modulous or sheer modulous of copper....i’m really very confused with the answer written below..i found it in a website...But this answer doesn’t match the book answer...According to ncert book answer ,Young’s modulous is used in this case...plz help...& plz tell me where to use Young’s modulous & sheer modulous...

answer;; (This answer is found in a website)

Length of the piece of copper, l = 19.1 mm = 19.1 × 10^–3 m

Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10^–3 m

Area of the copper piece:

A = l × b

= 19.1 × 10^–3 × 15.2 × 10^–3

= 2.9 × 10^–4 m²

Tension force applied on the piece of copper, F = 44500 N

Sheer Modulus of elasticity of copper, η = 42 × 10⁹ N/m²

Modulus of elasticity, η = stress/strain = F/A/strain

Therefore, strain = F/An

= 44500/2.9* 10^-4 * 42 * 10^9

= 3.65 × 10^–3

I know my post is very long..but plz help me understanding the concept well..

thanks in advance

Priyanka , 10 Years ago

Grade 12th pass