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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

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siniylev [52]

Answer:

Between 35°– 45°

Explanation:

In the vertical position, Point the flame in the direction of travel. Keep the flame tip at the correct height above the base metal. An angle of 35°–45° should be maintained between the torch tip and the base metal. This angle may be varied up or down to heat or cool the weld pool if it is too narrow or too wide

4 0
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A battery is an electromechanical device. a)- True b)- False
lilavasa [31]

Answer:

b)False

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3 years ago
Which of the eight diagnostic steps for locating an engine performance problem is performed first?
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Answer:

D. Perform a thorough visual inspection.

4 0
3 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0
2 years ago
) A shaft encoder is to be used with a 50 mm radius tracking wheel to monitor linear displacement. If the encoder produces 256 p
andrey2020 [161]

Answer:

number of pulses produced =  162 pulses

Explanation:

give data

radius = 50 mm

encoder produces = 256 pulses per revolution

linear displacement = 200 mm

solution

first we consider here roll shaft encoder on the flat surface without any slipping

we get here now circumference that is

circumference = 2 π r .........1

circumference = 2 × π × 50

circumference = 314.16 mm

so now we get number of pulses produced

number of pulses produced = \frac{linear\ displacement}{circumference} × No of pulses per revolution .................2

number of pulses produced = \frac{200}{314.16} × 256

number of pulses produced =  162 pulses

5 0
3 years ago
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