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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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pashok25 [27]

Answer:

Modulus of  elasticity is dependent on the elemental constitution of  1080  steel and various heat treatment don't affect this elemental composition  but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel

Explanation:

Generally the underlying physical reason why when we conduct various heat treatments on 1018 steel we expect the modulus of elasticity to stay about the same for every heat treatment is  

 

Modulus of  elasticity is dependent on the elemental constitution of  1080  steel and various heat treatment don't affect this elemental composition  but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel

6 0
3 years ago
Filler metals range in diameter from 1/16" to 3/8"*<br> O<br> true<br> O False
vesna_86 [32]

Answer:

O False

Explanation:

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6 0
2 years ago
Which of the following characteristics would not give animals an advantage in the ocean?
Taya2010 [7]
I believes it’s long body hair
3 0
2 years ago
Consider a circular grill whose diameter is 0.3 m. The bottom of the grill is covered with hot coal bricks at 961 K, while the w
soldi70 [24.7K]

Answer:

Step 1

Given

Diameter of circular grill,   D = 0.3m

Distance between the coal bricks and the steaks,  L = 0.2m

Temperatures of the hot coal bricks,  T₁ = 950k

Temperatures of the steaks, T₂ = 5°c

Explanation:

See attached images for steps 2, 3, 4 and 5

4 0
2 years ago
Calculate the equivalent capacitance of the three series capacitors in Figure 12-1 A) 0.060 uF B) 0.8 uF C) 0.58 uF D) 0.01 uF
Naddik [55]

Answer:

The correct answer is option (A) 0.060 uF

Note: Kindly find an attached image of the complete question below

Sources: The complete question was well researched from Quizlet.

Explanation:

Solution

Given that:

C₁ = 0.1 μF

C₂ =0.22 μF

C₃ = 0.47 μF

In this case, C₁, C₂ and C₃ are in series

Thus,

Their equivalent becomes:

1/Ceq = (1/C₁ + 1/C₂ +1/C₃

1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]

1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]

1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)

1/Ceq =[(0.1724)/(0.01034)]

1/Ceq = [(16.67)]

1/Ceq =(1/16.67) = 0.059μf

Ceq = 0.059μf ≈ 0.060μf

Therefore the equivalent capacitance of the three series capacitors is 0.060μf

4 0
3 years ago
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