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The plug has a diameter of 30 mm and fits within a rigid sleeve having an inner diameter of 32 mm. Both the plug and the sleeve are 50 mm long. Determine the axial pressure p that must be applied to the top of the plug to cause it to contact the sides of the sleeve. Also, how far must the plug be compressed downward in order to do this? The plug is made from a material for which E = 5 MPa, n = 0.45.

Respuesta :

Answer:

P=740 KPa

Δ=7.4 mm

Explanation:

Given that

Diameter of plunger,d=30 mm

Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

[tex]\varepsilon _t=\dfrac{D-d}{d}[/tex]

[tex]\varepsilon _t=\dfrac{32-30}{30}[/tex]

[tex]\varepsilon _t=0.0667[/tex]

We know that

[tex]n=-\dfrac{\epsilon _t}{\varepsilon _{long}}[/tex]

[tex]\varepsilon _{long}=-\dfrac{\epsilon _t}{n}[/tex]

[tex]\varepsilon _{long}=-\dfrac{0.0667}{0.45}[/tex]

[tex]\varepsilon _{long}=-0.148[/tex]

So the axial pressure

[tex]P=E\times \varepsilon _{long}[/tex]

[tex]P=5\times 0.148[/tex]

P=740 KPa

The movement in the sleeve

[tex]\Delta =\varepsilon _{long}\times L[/tex]

[tex]\Delta =0.148\times 50[/tex]

Δ=7.4 mm

The axial pressure P is 740 Pa and the movement of the plug is 7.4 mm.

How do you calculate the pressure and movement of the plug?

Given that the diameter of the plug D is 30 mm and the inner diameter of the rigid sleeve d is 32 mm. The length L of the plug and sleeve is 50 mm.

The plug is made from a material for which E = 5 MPa, n = 0.45.

The lateral strain can be calculated as given below.

[tex]\varepsilon_l = \dfrac {d-D}{D}[/tex]

[tex]\varepsilon _l = \dfrac { 32-30}{30}[/tex]

[tex]\varepsilon _l = 0.067[/tex]

We know that the lateral strain can be given as,

[tex]n =- \dfrac {\varepsilon _i}{\varepsilon _{long}}[/tex]

[tex]\varepsilon_ {long} = -\dfrac {\varepsilon _l}{n}[/tex]

[tex]\varepsilon _{long} = -\dfrac {0.067}{0.45}[/tex]

[tex]\varepsilon _{long} = -0.148[/tex]

The axial pressure P can be calculated as given below.

[tex]P = E\times \varepsilon _{long}[/tex]

[tex]P = 5\times 10^3 \times 0.148[/tex]

[tex]P = 740 \;\rm Pa[/tex]

The movement of the plug is calculated as given below.

[tex]\Delta D = \varepsilon _{long} \times L[/tex]

[tex]\Delta D = 0.148 \times 50[/tex]

[tex]\Delta D = 7.4 \;\rm mm[/tex]

Hence we can conclude that the axial pressure P is 740 Pa and the movement of the plug is 7.4 mm.

To know more about axial pressure, follow the link given below.

https://brainly.com/question/356585.

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