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vodka [1.7K]
3 years ago
9

A sleeve made of SAE 4150 annealed steel has a nominal inside diameter of 3.0 inches and an outside diameter of 4.0 inches. It i

s pressed onto a solid shaft with a medium drive force fit. Specify the class of fit, dimensions for the shaft and sleeve, and calculate the stress in the shaft and sleeve after installation. Use the basic hole system. Is the design acceptable
Engineering
2 answers:
LuckyWell [14K]3 years ago
8 0

Answer:

Design is acceptable.

Explanation:

Class of fit: Interference (Medium Drive Force Fits constitute a special kind of Interference Fits and these are the tightest fits where accuracy is important).

So, minimum shaft diameter will be larger than the maximum hole diameter.

Medium Drive Force Fits are FN 2 Fits.

As per standard ANSI B4.1 :

Desired Tolerance: FN 2

Tolerance TZone: H7S6

Max Shaft Diameter: 3.0029

Min Shaft Diameter: 3.0022

Max Hole Diameter:3.0012

Min Hole Diameter: 3.0000

Max Interference: 0.0029

Min Interference: 0.0010

Stress in the shaft and sleeve can be taking note of as the compressive stress which can be determined using load/interference area.

Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.

irga5000 [103]3 years ago
4 0

Answer:

Class of fit:

Interference (Medium Drive Force Fits constitute a special type of Interference Fits and these are the tightest fits where accuracy is important).

Here minimum shaft diameter will be greater than the maximum hole diameter.

Medium Drive Force Fits are FN 2 Fits.

As per standard ANSI B4.1 :

Desired Tolerance: FN 2

Tolerance TZone: H7S6

Max Shaft Diameter: 3.0029

Min Shaft Diameter: 3.0022

Max Hole Diameter:3.0012

Min Hole Diameter: 3.0000

Max Interference: 0.0029

Min Interference: 0.0010

Stress in the shaft and sleeve can be considered as the compressive stress which can be determined using load/interference area.

Design is acceptable If compressive stress induced due to selected dimensions and load is less than compressive strength of the material.

Explanation:

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Answer:

Bimetallic strip is a strip consisting of two metals of different coefficients of expansion welded together so that it buckles on heating : used in thermostats.

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2 years ago
Which of the following is not true about manufacturing employment in the U.S?
MakcuM [25]

Explanation:

If I guessed, I'd say the answer is choice (a).

The change in skills required to perform new tasks in manufacturing, along with import competition and a decline in mobility, have contributed to the decline of employment rate for manufacturing since 2000

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Storing parts outside doesn't cause any environmental risks as long as the items are covered.
telo118 [61]

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False

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3 0
3 years ago
Read 2 more answers
can someone help me with this engineering mechanics homework, please? I tried to solve it, but I got so confused.​
marishachu [46]

Explanation:

Sum of forces in the x direction:

∑Fx = ma

Rx − 250 N = 0

Rx = 250 N

Sum of forces in the y direction:

∑Fy = ma

Ry − 120 N − 300 N = 0

Ry = 420 N

Sum of forces in the z direction:

∑Fz = ma

Rz − 50 N = 0

Rz = 50 N

Sum of moments about the x axis:

∑τx = Iα

Mx + (-50 N)(0.2 m) + (-120 N)(0.1 m) = 0

Mx = 22 Nm

Sum of moments about the y axis:

∑τy = Iα

My = 0 Nm

Sum of moments about the z axis:

∑τz = Iα

Mz + (250 N)(0.2 m) + (-120 N)(0.16 m) = 0

Mz = -30.8 Nm

6 0
4 years ago
Two common methods of improving fuel efficiency of a vehicle are to reduce the drag coefficient and the frontal area of the vehi
qaws [65]

Answer:

\Delta V = 209.151\,L, \Delta C = 217.517\,USD

Explanation:

The drag force is equal to:

F_{D} = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot A

Where C_{D} is the drag coefficient and A is the frontal area, respectively. The work loss due to drag forces is:

W = F_{D}\cdot \Delta s

The reduction on amount of fuel is associated with the reduction in work loss:

\Delta W = (F_{D,1} - F_{D,2})\cdot \Delta s

Where F_{D,1} and F_{D,2} are the original and the reduced frontal areas, respectively.

\Delta W = C_{D}\cdot \frac{1}{2}\cdot \rho_{air}\cdot v^{2}\cdot (A_{1}-A_{2})\cdot \Delta s

The change is work loss in a year is:

\Delta W = (0.3)\cdot \left(\frac{1}{2}\right)\cdot (1.20\,\frac{kg}{m^{3}})\cdot (27.778\,\frac{m}{s})^{2}\cdot [(1.85\,m)\cdot (1.75\,m) - (1.50\,m)\cdot (1.75\,m)]\cdot (25\times 10^{6}\,m)

\Delta W = 2.043\times 10^{9}\,J

\Delta W = 2.043\times 10^{6}\,kJ

The change in chemical energy from gasoline is:

\Delta E = \frac{\Delta W}{\eta}

\Delta E = \frac{2.043\times 10^{6}\,kJ}{0.3}

\Delta E = 6.81\times 10^{6}\,kJ

The changes in gasoline consumption is:

\Delta m = \frac{\Delta E}{L_{c}}

\Delta m = \frac{6.81\times 10^{6}\,kJ}{44000\,\frac{kJ}{kg} }

\Delta m = 154.772\,kg

\Delta V = \frac{154.772\,kg}{0.74\,\frac{kg}{L} }

\Delta V = 209.151\,L

Lastly, the money saved is:

\Delta C = \left(\frac{154.772\,kg}{0.74\,\frac{kg}{L} }\right)\cdot (1.04\,\frac{USD}{L} )

\Delta C = 217.517\,USD

4 0
4 years ago
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