Question

A steel rod of 25 mm in diameter and 2 meter long is subjected to an axial pull of 45 kN. Find -(i)int
ensity of stress, (ii) the strain, and (iii) the elongation. Take E =2x10^5 N/mm^2.​

Answer 1

Given :

Diameter (D) = 25 mm

Length of the rod (L) = 2 m

Force (F) = 45 KN = 45000 N

To Find :

(i) Intensity of stress,

(ii) The strain, and

(iii) The elongation.

Solution :

(i) Stress (S)   =   $\frac{Force}{Area}$

                      =  $\frac{45000}{\pi (\frac{25}{2})^2 }$

                      =  $\frac{45000 * 7}{22 * (12.5)^2}$

                      = $\frac{315000}{3437.5}$

                      = 91.636 N/mm²

(ii) We know,

        Modulus of elasticity (E) = $\frac{Stress}{Strain}$

⇒          2 × 10⁵                          =  $\frac{91.636}{Strain}$

⇒          Strain                            = 4.5818 × 10⁻⁴

Therefore, strain is 4.5818 × 10⁻⁴.

(iii)      Strain                 = $\frac{Change in Length (l)}{Original Length(L)}$

⇒ 4.5818 × 10⁻⁴            =   $\frac{l}{2}$

∴      l                             =   9.1636 × 10⁻⁴ m

Therefore, the elongation of the rod is  9.1636 × 10⁻⁴ m.

           

Answer 2

Answer:

9.1636

Explanation:

correct answer please