Question

8–6. Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress in the cylinder does not exceed 3 MPa. Each piston has a radius of 45 mm and the cylinder has a wall thickness of 2 mm.

Answer 1

Answer:

$P=2.66\ N$ is the maximum safe load.

Explanation:

Given:

• diameter of the cylinder, $d= 90\ mm$

• thickness of the cylinder wall, $t=2\ mm$

• maximum bearable stress, $\sigma=3\ MPa$

Firstly we find that:

$\frac{t}{d} =\frac{1}{45}$

⇒ This is a thin cylinder.

We know axial stress and hoop stress is given by

$\sigma_a=\frac{P.d}{4t}$

$\sigma_h=\frac{P.d}{2t}$

⇒ Axial load will always be larger than the circumferential load under while other parameters are same.

So we find the load in case of axial stress:

$30=\frac{P\times 90}{4\times 2}$

$P=2.66\ N$ is the maximum safe load.

Answer 2

Answer:

The maximum force is 846.11 N.

Explanation:

Given that,

Stress = 3 MPa

Radius = 45 mm

Thickness = 2 mm

We need to calculate the internal pressure

Using formula of internal pressure

$\sigma=\dfrac{p\times r}{t}$

$p=\dfrac{\sigma t}{r}$

Put the value into the formula

$p=\dfrac{3\times10^{6}\times2\times10^{-3}}{45\times10^{-3}}$

$p=0.133\times10^{6}\ Pa$

We need to calculate the maximum force

Using formula of maximum force

$p=\dfrac{P}{A}$

$P=p\times A$

Here, P = force

p = internal pressure

Put the value into the formula

$F=0.133\times10^{6}\times\pi\times(45\times10^{-3})^2$

$F=846.11\ N$

Hence, The maximum force is 846.11 N.