Find out bulk modulus:

A steel bar of 25 mm diameter was tested on a gauge length of 250 cm in tension and in torsion. A tensile load of 50 kN generated an extension of 0.13 mm and a torque of 200 N-m produced in twist, Find out :

(a) the modulus of elasticity (b) the modulus of rigidity (c) The Poisson's ratio (d) the bulk modulus

Solution

d = 25 mm = 0.025 m

A = (π/4 ) d ²  = 4.9 × 10- 4 m²

J =  (π /32) d ⁴  = 3.8 × 10- 8 m4

l = 25 cm = 0.25 m

P = 50 kN = 50 × 103 N

Δ l = 0.13 mm = 0.13 × 10^{- 3} m

T = 200 N-m

θ = 1^o = 0.017 radians

Δ l =  Pl  /A E

 0.13 × 10^{- 3} = (50 × 10 × 0.15)/ (4.9 × 10^{- 4} × E)

∴          E = 2 × 1011 N/m² = 2 × 105 N/mm²

                  T/ J = G θ/ l

200 / (3.8 × 10^{- 8} = (G × 0.017 )/ 0.25

G = 0.77 × 1011 N/m²  = 0.8 × 105 N/mm²

E = 2G (1 + μ)

⇒         2 × 10⁵  =2 × 0.8 × 10⁵ (1 + μ)

∴          μ= 0.25

E = 3K (1 - 2μ)

2 × 10⁵  = 3 × K (1 - 2 × 0.25)

K = 1.33 × 10⁵ N/mm²