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saw5 [17]
3 years ago
6

A bar of steel has the minimum properties Se = 40 kpsi, S = 60 kpsi, and S-80 kpsi. The bar is subjected to a steady torsional s

tress of 15 kpsi and an alternating bending stress of 25 kpsi Find the factor of safety guarding against a static failure, and either the factor of safety guard- ing against a fatigue failure or the expected life of the part. For the fatigue analysis use:
(a) Modified Goodman criterion.

(b) Gerber criterion.

(C) ASME-elliptic criterion.

Engineering
1 answer:
nlexa [21]3 years ago
5 0

Answer:

(a) Modified Goodman criterion:

Factor of safety against fatigue failure =  1.0529

(b) Gerber criterion:

Factor of safety against fatigue failure = 1.31

(c) ASME-elliptic criterion:

Factor of safety against fatigue failure = 1.315

Explanation:

See the attached file for the calculation.

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Which of the following is the class of ingredient which is the base for most baked items?
Ierofanga [76]

Answer:flour

Explanation:

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3 years ago
Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin
Andre45 [30]

Answer:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Please refer to attachment for step by step solution of the question.

5 0
3 years ago
Given a square matrix [A], write a single line MATLAB command that will create a new matrix [Aug] that consists of the original
Liono4ka [1.6K]

Answer:

Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:

Aug=[A eye(size(A))]

Above command is tested in MATLAB command window and is attached in figure below

8 0
3 years ago
The dot plot shows the number of cupcakes bought by each person who came to a bake sale.
Maslowich

The true statement about the dot plot is 1 has 4 and 0 dots.

Explanation:

  • after creating a Bar chart 4 is the right answer.
  • The highest among st all the other plots is 1 but 4 shows 3.
  • Taking an average from all the data points 3 comes to the right answer.
  • Median the central Mid-point is 3.
  • Mode also comes to 3.
  • It is skewed on the right due to the 1st one.
  • Skewed shows the data point either increasing or in decreasing.
  • There a to Bi- histograms which has two ups and downs.
  • The true statement has to be as Mean=Median=Mode is 3.

6 0
3 years ago
Think of the differences between circuit-switching and packet-switching paradigms in the Internet core design. Assume an Interne
dem82 [27]

Answer:

0.264 ; 0.079

Explanation:

Given that:

Sample size, n = 100

Probability of being active, p = 1% = 1/100 = 0.01

Using the binomial probability relation :

P(x =x) = nCx * p^x * (1 - p)^(n - x)

Probability that more than 1 user will be active

P(x > 1) = 1 - [p(x=0) + p(x = 1)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x > 1) = 1 - [0.366 + 0.370]

P(x > 1) = 0.264

2.)

Probability that more than 2 user will be active

P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]

P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366

P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370

P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185

P(x > 1) = 1 - [0.366 + 0.370 + 0.185]

P(x > 1) = 0.079

7 0
3 years ago
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