1) A steel bar of 20mm in diameter, when subjected to axial pull of of 50KN, elongates by 0.2mm over a gauge length of 20cm. The diameter of the bar is reduced by 0.004mm. Bulk modulus of the material of the bar is?

A) 88384 MPa   B) 159090 MPa

C) 66288 MPa    D) 210524 MPa


Answer : NA

Given,

P= 50kN

d= 20mm

L= 20cm = 200mm

∆𝐿 = 0.2𝑚𝑚

∆𝑑 = -0.004 mm( negative sign indicates the reduction in diameter when tensile load

applied)

Bulk modulus = (applied stress) / (volumetric strain)


where the applied stress is given by:


applied stress = (applied force) / (cross-sectional area)


and the volumetric strain is given by:


volumetric strain = (change in volume) / (original volume)


First, let's calculate the applied stress:


Applied force = 50 kN = 50,000 N

Cross-sectional area = π/4 * (diameter)^2 = π/4 * (20 mm)^2 = 314.159 mm^2


Applied stress = 50,000 N / 314.16 mm^2 = 159.16 N/mm^2


Next, let's calculate the volumetric strain:


The elongation is 0.2 mm over a gauge length of 20 cm, which is equivalent to 200 mm. Therefore, the strain is:

𝜀𝑥= elongation / original length = 0.2 mm / 200 mm = 0.001

𝐴 = 314.159𝑚𝑚2

Uniaxial stress

𝜎𝑥 = 159.155 N/𝑚𝑚2

Uniaxial strain

𝜀𝑥 = 1x10-3

modulus of elasticity, 

E = 𝜎𝑥 / 𝜀𝑥 = 159.155 N/𝑚𝑚2  / 1x10-3

159154.9431 N/𝑚𝑚2

Poisson’s ratio, 𝜇 = 0.2

We know the relation between

E, K, and 𝜇

E = 3K (1- 2𝜇 )

Bulk modulus

K = 88419.412 N/𝑚𝑚2 = 88419.412 MPa

The answer is 88419MPa. PSC answer is 88384 MPa