Respuesta :
Answer:
[tex]P=2.66\ N[/tex] is the maximum safe load.
Explanation:
Given:
- diameter of the cylinder, [tex]d= 90\ mm[/tex]
- thickness of the cylinder wall, [tex]t=2\ mm[/tex]
- maximum bearable stress, [tex]\sigma=3\ MPa[/tex]
Firstly we find that:
[tex]\frac{t}{d} =\frac{1}{45} <\frac{1}{20}[/tex]
⇒ This is a thin cylinder.
We know axial stress and hoop stress is given by
[tex]\sigma_a=\frac{P.d}{4t}[/tex]
[tex]\sigma_h=\frac{P.d}{2t}[/tex]
⇒ 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:
[tex]30=\frac{P\times 90}{4\times 2}[/tex]
[tex]P=2.66\ N[/tex] is the maximum safe load.
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
[tex]\sigma=\dfrac{p\times r}{t}[/tex]
[tex]p=\dfrac{\sigma t}{r}[/tex]
Put the value into the formula
[tex]p=\dfrac{3\times10^{6}\times2\times10^{-3}}{45\times10^{-3}}[/tex]
[tex]p=0.133\times10^{6}\ Pa[/tex]
We need to calculate the maximum force
Using formula of maximum force
[tex]p=\dfrac{P}{A}[/tex]
[tex]P=p\times A[/tex]
Here, P = force
p = internal pressure
Put the value into the formula
[tex]F=0.133\times10^{6}\times\pi\times(45\times10^{-3})^2[/tex]
[tex]F=846.11\ N[/tex]
Hence, The maximum force is 846.11 N.
