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1 year ago

What are the general rules for press fit allowances

Engineering

1 answer:

Keith_Richards [23]1 year ago

4 0

Explanation:

As a general rule of thumb, the large the diameter of a bearing, bushing or pin, the larger the tolerance range,” Brieschke points out. “The inverse is true for smaller-diameter pieces.”

Mike Brieschke, vice president of sales at Aries Engineering, says a 0.25-inch-diameter metal dowel that is press-fit into a mild steel hole usually has an interference of ±0.0015 inch. Parts in noncritical assemblies tend to have looser tolerances

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A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis

kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

$V_1=0.3m^3,T_1=298 K$

$V_2=0.1m^3,T_1=1273 K$

Mass of air=1 kg

For polytropic process $pv^n=C$ ,n is the polytropic constant.

$Tv^{n-1}=C$

$T_1v^{n-1}_1=T_2v^{n-1}_2$

$298\times .3^{n-1}_1=1273\times .1^{n-1}_2$

n=2.32

Work in polytropic process given as

w= $\dfrac{P_1V_1-P_2V_2}{n-1}$

w= $mR\dfrac{T_1-T_2}{n-1}$

Now by putting the values

w= $1\times 0.287\dfrac{289-1273}{2.32-1}$

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

5 0

3 years ago

Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h

yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ; B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

8 0

3 years ago

Suggest appropriate materials for a cylinder head and give four reason for your choice

ycow [4]

Answer:

head and give four reason for your choice

7 0

3 years ago

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

Allisa [31]

Answer:

theoretical fracture strength = 16919.98 MPa

Explanation:

given data

Length (L) = 0.28 mm = 0.28 × 10⁻³ m

radius of curvature (r) = 0.002 mm = 0.002 × 10⁻³ m

Stress (s₀) = 1430 MPa = 1430 × 10⁶ Pa

solution

we get here theoretical fracture strength s that is express as

theoretical fracture strength = $s_{0} \times \sqrt{\frac{L}{r} }$ .............................1

put here value and we get

theoretical fracture strength = $1430 \times 10^6\times \sqrt{\frac{0.28\times 10^{-3}}{0.002\times 10^{-3}} }$

theoretical fracture strength = $16919.98 \times 10^6$

theoretical fracture strength = 16919.98 MPa

3 0

3 years ago

Which of the following explains why material properties present challenges for engineers?

Maurinko [17]

Answer:

Explanation:

They are altered by variables such as temperature hence making materials challenging when dealing with them.

4 0

3 years ago