A cylinder 250 mm in diameter has a wall thickness of 5mm and is full of a fluid at atmospheric pressure. Its ends are closed by rigid plates and an axial compressive force of 80KN is applied to the cylinder so that the pressure of the fluid rises by 90 kPa. Calculate the bulk modulus of the fluid. Take for the cylinder material E=200GPa and μ=0.25

Data:

D = 250 mm

t = 5 mm

Force P = 80 KN

Pressure P = 90 KPa

$E = 200 Gpa = 200 \times 10^3 \ N/mm^2$

M = 0.25

Internal pressure = $90 \ Kpa = \frac{90 \times 1000}{1000^2} = 0.09 \ N/mm^2$

Longitudinal stress $\sigma l = \frac{P}{\pi \ d t} - \frac{Pd}{4t} = \frac{80 \times 10^3}{\pi \times 25 \times 5} - \frac{0.09 \times 250}{4 \times 5} $

= 19.247 $N/mm^2$ (Compressive)

Circumferential stress or Hoop stress $\sigma n = \frac{pd}{2t} = \frac{0.09 \times 250}{2 \times 5} = 2.25 \ N/mm^2 $ (tensile)

Longitudinal strain $\sigma l = \frac{1}{E} (El - \mu \ \sigma n) = \frac{1}{2 \times 10^5} [-19.247 – 2.25 \times 0.25]$ (Compressive in nature $\therefore$ using negative sign)

$El = -9.90475 \times 10^{-5}$

Circumferential strain = $En = \frac{1}{E} [ \sigma n - \mu \sigma l] = \frac{1}{2 \times 10^5} (2.25 – (- 19.247 \times 0.25)]$

$= 3.53087 \times 10^{-5}$

Volumetric strain/strain of the capacity Ev:

$Ev = 2 \ En + El = 2 \times 3.53087 \times 10^{-5} – 9.9047 \times 10^{-5}$

=$ -2.843 \times 10^{-5}$

Volumetric strain of fluid =$ \frac{P}{K}$

$2.843 \times 10^{-5} = \frac{0.09}{K}$

$\therefore$ K = 3165.67 $N/mm^2$