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MAVERICK [17]
3 years ago
14

g A part made from annealed AISI 1018 steel undergoes a 20 percent cold-work operation. (a) Obtain the yield strength and ultima

te strength before and after the cold-work operation. Determine the percent increase in each strength. (b) Determine the ratios of ultimate strength to yield strength before and after the cold-work operation. What does the result indicate about the change of ductility of the part
Engineering
1 answer:
nikdorinn [45]3 years ago
6 0

Answer:

A) - Yield strength before operation = 32 kpsi

- Ultimate Strength before operation = 49.5 kpsi

- Yield strength after operation = 61.854 kpsi

- Ultimate Strength after operation = 61.875 kpsi

- Percentage increase of yield strength = 93.29%

- Percentage increase of ultimate strength = 25%

B) ratio before operation = 1.55

Ratio after operation = 1

Explanation:

From online values of the properties of this material, we have;

Yield strength; S_y = 32 kpsi

Ultimate Strength; S_u = 49.5 kpsi

Modulus; m = 0.25

Percentage of cold work; W_c = 0.2

S_o = 90 kpsi

A) Let's calculate the strain(ε) from the formula;

A_o/A = 1/(1 - W_c)

A_o/A = 1/(1 - 0.2)

A_o/A = 1.25

Thus, strain is;

ε = In(A_o/A)

ε = In(1.25)

ε = 0.2231

Yield strength after the cold work operation is;

S'_y = S_o(ε)^(m)

Plugging in the relevant values;

S'_y = 90(0.2231)^(0.25)

S'_y = 61.854 kpsi

Percentage increase of yield strength = S'_y/(S'_y - S_u) × 100% = (61.854 - 32)/32 × 100% = 93.29%

Ultimate strength after the cold work operation is;

S'_u = S_u/(1 - W_c)

S'_u = 49.5/(1 - 0.2)

S'u = 61.875 kpsi

Percentage increase of ultimate strength = S'_u/(S'_u - S_u) × 100% = (61.875 - 49.5)/49.5 × 100% = 25%

B) Ratio of ultimate strength and yield strength before cold work operations is;

S_u/S_y = 49.5/32

S_u/S_y = 1.547

Ratio of ultimate strength and yield strength after cold work operations is;

S'_u/S'_y = 61.875/61.854 = 1

The ratio after the operation is less than before the operation, thus the ductility reduced.

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Answer:

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3 years ago
while performing a running compression test how should running compression compare to static compression
algol [13]

Answer:

The idle speed of a running compression should be between 50-75 PSI and that is about half of the static compression.

Explanation:

The Running or Dynamic compression is used to determine how well the cylinder in an engine  is absorbing air, reserving it for the proper length of time, and releasing it to the exhaust. The static or cranking compression test is used to check the sealing of the cylinder. Before performing the running compression test, the static compression test is first performed to rule out other issues like bent valves.

The standard value for the static compression is given by;

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2 years ago
Carbon resistors often come as a brown cylinder with colored bands. These colored bands can be read to determine the manufacture
alexandr1967 [171]

Answer:

a) 4.7 kΩ, +/- 5%

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Explanation:

a) If the resistor has the following combination of color bands:

1) Yellow = 1st digit = 4

2) Violet = 2nd digit = 7

3) Red = multiplier = 10e2

4) Gold = tolerance = +/- 5%

this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.

b) Repeating the process for the following combination of color bands:

1)  Red = 1st digit = 2

2) Black = 2nd digit = 0

3) Green = multiplier = 10e5

4) Nothing = tolerance = +/- 20%

This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.

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Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at  

Replacing by the values already known for vo, vf and a, and solving for t, we get:

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Answer:

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Explanation:

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