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xeze [42]
3 years ago
11

A machine part will be cycled at ±350 MPa for 5 (103) cycles. Then the loading will be changed to ±260 MPa for 5 (104) cycles. F

inally, the load will be changed to ±225 MPa. How many cycles of operation can be expected at this final stress level? For the part, S = 530 MPa, f = 0.9, and has a fully corrected endurance strength of Se = 210 MPa.

Engineering
1 answer:
dimulka [17.4K]3 years ago
6 0

Answer:

a) The number of cycles of operation at 225 MPa is 184,100 cycles

b) The number of cycles of operation at 225 MPa is 33,600 cycles

Explanation:

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motion and power

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An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change i
Harman [31]

Answer:2058.992KJ

Explanation:

Given data

Mass of object\left ( m\right )=521kg

initial velocity\left ( v_0\right )=90m/s

Final velocity\left ( v\right )=14m/s

kinetic energy of body is given by=\frac{1}{2}mv^{2}

change in kinectic energy is given by substracting  final kinetic energy from initial kinetic energy of body.

Change in kinetic energy=\frac{1}{2}\times m\left ( V_0^{2}-V^2\right )

Change in kinetic energy=\frac{1}{2}\times521\left ( 90^{2}-14^2\right )

Change in kinetic energy=2058.992KJ

7 0
3 years ago
Air at 7 deg Celcius enters a turbojet engine at a rate of 16 kg/s and at a velocity of 300 m/s (relative to engine). Air is hea
Savatey [412]

To solve this problem we will start using the given temperature values and then transform them to the Kelvin scale. From there through the properties of the tables, described for the Air, we will find the entropy. With these three data we can perform energy balance and find the speed of the fluid at the exit, which will finally help us find the total force:

V_i = 300m/s

T_1 = 7\°C = 280K

T_2 = 427\°C = 700K

\dot{m} 16kg/s

\dot{Q}_{in} = 15000kJ/s

Using the tables for gas properties of air we can find the enthalpy in this two states, then

T_1 = 280K \rightarrow h_1 = 280.13kJ/Kg\cdot K

T_2 = 700K \rightarrow h_2 = 713.27kJ/Kg\cdot K

Applying energy equation to the entire engine we have that

\frac{\dot{Q}_{in}}{\dot{m}}+h_1 +\frac{1}{2} V^2_{in} = h_2 + \frac{1}{2} V^2_{out}

\frac{15000}{16}*10^3+280.13*10^3+\frac{1}{2} 300^2 = 713.27*10^3+\frac{1}{2} V^2_{out}

V_{out} = 1048.198m/s

Finally the force in terms of mass flow and velocity is

F = \dot{m} (V_{out}-V_{in})

F = 16(1048.198-300)

F= 11971.168N

Therefore the thrust produced by this turbojet engine is  11971.168N

4 0
3 years ago
Key length is designed to provide desired factor of safety<br> a. True<br> b. False
zhuklara [117]

Answer: true

Explanation:

A key is a machine element that us used to connect the element of a rotating machine to a shaft. It should be noted that the key hinders the relative rotation that may take place between the two parts.

Key length is designed to provide desired factor of safety. It should also be noted that the factor of safety shouldn't be much and the key length is typically limited to the hub length.

8 0
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