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Trava [24]
2 years ago
3

A multi-plate clutch is to transmit 12 kW at 1500 rev/min. The inner and outer radii for the plates are to be 50 mm and 100 mm r

espectively. The maximum axial spring force is restricted to lkN. Calculate the necessary number of pairs of surfaces if ll = 0-35 assuming constant ‘vyear. What will be the necessary axial force?
Engineering
1 answer:
xenn [34]2 years ago
8 0

Answer:

The uniform pressure for the necessary axial force  is  W = 945 N

The uniform wear for the necessary axial force is  W = 970.15 N

Explanation:

Solution

Given that:

r₁ = 0.1 m

r₂ = 0.05m

μ = 0.35

p = 12 N or kW

N = 1500 rpm

W = 1000 N

The angular velocity is denoted as  ω= 2πN/60

Here,

ω = 2π *1500/60 = 157.07 rad/s

Now, the power transferred becomes

P = Tω this is the equation (1)

Thus

12kW = T * 157.07 rad/s

T = 76.4 N.m

Now, when we look at the uniform condition, we have what is called the torque that is frictional which acts at the frictional surface of the clutch dented as :

T = nμW R this is the equation (2)

The frictional surface of the mean radius is denoted by

R =2/3 [(r₁)³ - (r₂)³/(r₁)² - (r₂)²]

=[(0.1)³ - (0.05)³/[(0.1)² - (0.05)²]

R is =0.077 m

Now, we replace this values and put them into the equation (2)

It gives us this, 76. 4 N.m = n * 0.35* 1000 N * 0.077 m

n = 2.809 = 3

The number of pair surfaces is = 3

Secondly, we determine the uniform wear.

So, the mean radius is denoted as follows:

R = r₁ + r₂/ 2

=0.1 + 0.05/2

=0.075 m

Now, we replace the values and put it into the equation (2) formula

76. 4 N.m = n *0.35* 1000 N * 0.075 m

n= 2.91 = 3

Again, the number of pair surfaces = 3

However, for the uniform pressure with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.077 m

W = 945 N

Also, for the uniform wear with regards to the number of clutch plates is 3 we can derive the necessary axial force from the equation (2)

76. 4 N.m = 3 * 0.35 * W *0.075 m

W = 970. 15 N

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Answer:

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Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

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The design factor for metals is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}

The design factor for plastic and ceramics is:

DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039

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DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5

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The transition time ∆t corresponds to the energy that is ∆E

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