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 m^{2}
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/m^{2} = 2 × 105 N/mm^{2}
T/ J = G θ/ l
200 / (3.8 × 10^{- 8} = (G × 0.017 )/ 0.25
G = 0.77 × 1011 N/m^{2} = 0.8 × 105 N/mm^{2}
E = 2G (1 + μ)
⇒ 2 × 10^{5} =2 × 0.8 × 10^{5} (1 + μ)
∴ μ= 0.25
E = 3K (1 - 2μ)
2 × 10^{5} = 3 × K (1 - 2 × 0.25)
K = 1.33 × 10^{5} N/mm^{2}