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vredina [299]
2 years ago
10

Compute the strain-hardening exponent (n) for an alloy in which a true stress of 415 MPa produces a true strain of 0.10; assume

a value of 1035 MPa for K.

Engineering
1 answer:
Naddika [18.5K]2 years ago
4 0

Answer:

See the attached picture.

Explanation:

See the attached picture.

You might be interested in
A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. If the
Bad White [126]

Answer:

=>> 167.3 kpa.

=>> 60° from horizontal face.

Explanation:

So, we are given the following data or parameters or information which is going to assist us in solving this kind of question;

=>> "A soil element is subjected to a minor principle stress of 50 kPa on a plane rotated 20 ° counterclockwise from vertical. "

=>>"If the deviator stress is 120 kPa and the shear strength parameters are a friction angle of 30° and a cohesion of 5 kPa."

The orientation of this plane with respect to the major principle stress plane = 50 tan^2 (45 + 30/2) + 10 tan ( 45 + 30/2).

magnitude of the stresses on the failure plane = 167.3 kpa.

The orientation of this plane with respect to the major principle stress plane => x = 60 cos 60° = 30kpa.

y = 60 sin 60° = 30√3 = sheer stress.

the orientation of this plane with respect to the major principle stress plane.

Theta = 45 + 15 = 60°.

5 0
2 years ago
Lydia is the CEO for a large pharmaceutical manufacturer. Her company is in the final stages of FDA
weqwewe [10]
OSHA inspections are generally unannounced. In fact, except in four exceptional circumstances when advance notice may be given.


It is a criminal offense for any person to give unauthorized advance notice of an OSHA inspection.
5 0
2 years ago
A sample of normally consolidated clay was subjected to a CU triaxial compression test that was carried out until the specimen f
QveST [7]

Answer:

Check the explanation

Explanation:

Given

1) CU traixial compression test

2) Devatoric stress at failure  = бd = 50 kN/m^2

3) Confining pressure at failure  = бd = 48 kN/m^2

4) Pore pressure at failure  = u = 18 kN/m^2

5) Unconfined compression stress  = q = 20 kN/m^2

6) Undrained cohesion  = q/2 = 20 kN/m^2

To find:

1) Effective and total stress strength failure envelope

Kindly check the attached image below .

7 0
2 years ago
A shipment of rebar that weighs 745 kg would weigh roughly how much in pounds​
Andre45 [30]

Answer:

Dont no but will check

Explanation:

6 0
2 years ago
Twenty distinct cars park in the same parking lot every day. Ten of these cars are US-made, while the other ten are foreign-made
Zina [86]

Answer:

Total no. of ways to line up cars is 20! = 2.43 c 10^18

Probability that the cars alternate is 0.00001 or 0.001%

Explanation:

Since, the position of a car is random.Therefore, number ways in which cars can line up is given as:

<u>No. of ways = 20! = 2.43 x 10^18</u>

For the probability that cars alternate, two groups will be formed, one consisting of US-made 10 cars and other containing 10 foreign made. The number of favorable outcomes for this can be found out as the arrangements of 2! between these groups multiplied by the arrangements of 10! for each group, due to the arrangements among the groups themselves.

Favorable Outcomes = 2! x 10! x 10!

Thus the probability of event will be:

Probability = Favorable Outcomes/Total No. of Ways

Probability = (2! x 10! x 10!)/20!

<u>Probability = 0.00001 = 0.001%</u>

4 0
2 years ago
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