1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
daser333 [38]
3 years ago
10

A part has been tested to have Sut = 530 MPa, f = 0.9, and a fully corrected Se = 210 MPa. The design requirements call for the

part to be loaded at three different fully-reversed loads and cycled at each one for a set number of cycles. First, it will be loaded at ±350 MPa for 5,000 cycles. Then, it will be loaded at ±260 MPa for 50,000 cycles. Finally, it will be loaded at ±225 MPa until it fails. How many cycles do we expect the part to last at the final loading? Use Miner's method. :g

Engineering
1 answer:
kobusy [5.1K]3 years ago
8 0

Answer:

126984 cycles

Explanation:

Given data :

Sut = 530 MPa

f = 0.9

fully corrected Se = 210 MPa

using Miner's method attached below is the detailed solution of the given problem

when loaded with ± 225 MPa the number of cycles before it fails will be

≈  126984

You might be interested in
2. (Problem 4.60 on main book, diameters different) Water flows steadily through a fire hose and nozzle. The hose is 35 mm diame
Viefleur [7K]

Answer:

coupling is in tension

Force = -244.81 N

Explanation:

Diameter of Hose ( D1 ) = 35 mm

Diameter of nozzle ( D2 ) = 25 mm

water gage pressure in hose = 510 kPa

stream leaving the nozzle is uniform

exit speed and pressure = 32 m/s and atmospheric

<u>Determine the force transmitted by the coupling between the nozzle and hose </u>

attached below is the remaining part of the  detailed solution

Inlet velocity ( V1 ) = V2 ( D2/D1 )^2  

= 32 ( 25 / 35 )^2

= 16.33 m/s

4 0
2 years ago
A 20.0 µF capacitor is charged to a potential difference of 800 V. The terminals of the charged capacitor are then connected to
Sergeu [11.5K]

Answer:

a) Q_initial = 16 * 10^-3 C

b) V_1 = V_2 =  (16/3) * 10^2 V

c)  E = 64/15 J

d)  dE = 32/15 J of decrease

Explanation:

Given:

- Capacitor 1, C_1 = 20.0 uF

- Capacitor 2, C_2 = 10.0 uF

- Charged with P.d V = 800 V

Find:

a) the original charge of the system,

(b) the final potential difference across each capacitor

(c) the final energy of the system

(d) the decrease in energy when the capacitors are connected.

Solution:

a)

- The initial charge in the circuit is the one carried by the first charged capacitor.

                           Q_initial = C_1*V

                           Q_initial = 20*10^-6 * 800

                           Q_initial = 16 * 10^-3 C

b)

- After charging the other capacitor, we know that the total charge is conserved among two capacitor:

                          Q_initial = Q_1 + Q_2

- We also know that potential difference across two capacitor is also same.

                          V_1 = V_2 = Q_1 / C_1 = Q_2 / C_2

- Using the two equations and solve for charge Q_2:

                          Q_2 = Q_1*C_2/C_1

                          Q_2 = Q_1*10/20 = 0.5*Q_1

- using conservation of charge:

                          Q_initial = 1.5*Q_1

                          Q_1 = 16*10^-3 / 1.5 = 10.67*10^-3 C

- Hence the Voltage across each capacitor is:

                          V_2 = V_1 = Q_1 / C_1  

                                            = 10.67*10^-3 / 20*10^-6

                                            = (16/3) * 10^2 V

c)

- The energy in the system is:

                          E = 0.5*C_eq*V^2

Where, C_eq is the equivalent capacitance of paralle circuit.

                           E = 0.5*(20+10)*10^-6 *((16/3) * 10^2)^2

                          E = 64/15 J

d)

- The decrease in energy of the capacitors is:

                           dE = E_initial - E_final

Where, E_initial is due to charging of the C_1 only:

                          dE = 0.5*10^-6*20*800^2 - (64/15)

                          dE = 32/5 - 64/15 = 32/15 J

5 0
3 years ago
Define what a glass transition is and what happens to a polymer because of it
lesya692 [45]

Answer:

 The glass transition occur when the given temperature gets drop below the temperature of the glass transition for the polymers. During the creases of the long range of the given motion, the polymer start changing from the state of hard into the form of brittle.

The range of the glass transition temperature are between the 0 to -150°C. The basic use of the polymer is basically depend upon the main properties of the polymer.

7 0
3 years ago
Problem definition
LekaFEV [45]

Answer:

ummm thats alot

Explanation:

8 0
2 years ago
an inclined manometer is connected to a pitor tube to measure the velocity at the center of a circular duct. If the inclined man
jekas [21]
57.5 m/s
I did 2.3/0.04
I’m not sure if it’s correct though
8 0
3 years ago
Other questions:
  • What's mutual inductance​
    8·1 answer
  • A certain process requires 3.0 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large-diameter supp
    9·1 answer
  • A market research survey has 15 questions and will be sent to 500 people. What is the total cost to conduct survey if it has a $
    12·1 answer
  • 2. The initially velocity of the box and truck is 60 mph. When the truck brakes such that the deceleration is constant it takes
    12·1 answer
  • What are the four basic parts of process plan
    11·1 answer
  • The iron-blade plow allowed humans to increase food production during<br> what age?
    8·1 answer
  • Which of the following are made up of electrical probes and connectors?
    8·1 answer
  • Which of the following justifies the need for an already-certified engineer to continue to take classes?
    15·1 answer
  • 3. What is special about beryllium-copper alloy tools?
    6·2 answers
  • 1. Band saw lower wheel does not require a guard *
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!