Find out bulk modulus, Mechanical Engineering

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 = 1o = 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

Posted Date: 1/21/2013 2:41:05 AM | Location : United States







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