A rotating beam is subject to an alternating stress of 48 kpsi and a mean stress of 24 kpsi. The ultimate strength of the material is 100 kpsi and the fully correctly endurance limit is 40 kspi. Calculate the factor of safety using the Goodman fatigue failure theory to show this is finite life. Then, calculate the life of the beam in number of cycles.
Answers
Answer:
goodman = 0.694
life of beam = 211597
Explanation:
alternating stress = 48 kpsi
mean stress = 24 kpsi
ultimate strength = 100 kpsi
endurance limit = 40 kpsi
goodman:
= [tex]\frac{mean stress}{ultimate stress} +\frac{alternating stress}{endurance limit} =\frac{1}{N}[/tex]
= [tex]\frac{24}{100} +\frac{48}{40} =\frac{1}{N}[/tex]
= 0.24 + 1.2 = [tex]\frac{1}{N}[/tex]
N = 1/1.44
N = 0.694
2. check attachment for diagram
Log(N)-3/3 = log90 - log48/log90 - log40
Log(N)-3/3 = 0.77517
Log N = 5.325509
N = 10^(5.325509)
N = 211597
Related Questions
**Can someone please help me solve these problems step by step, thank you!
1. If a 12-inch diameter driving Gear A has 100 tooth which turns CCW with a torque of 200 ft-lb and 100 rpm is meshed with Gear B, 25 tooth, turn along with Gear C, 50 tooth as they are a compound gear and Gear D, 150 tooth meshed with C.
-->sketch and annotate the gear train.
-->what direction does gear ratio B, C, and D turn?
-->what is the gear ratio for B, C, and D?
-->what is the angular velocity for gears B, C, and D?
-->what is the torque for gears B, C, and D?
-->what is the diameter of gears B, C, and D?
2. In a pulley system, pulley A is moving 1000 rpm and has a diameter of 18inches. There are two pulleys with different sizes, B moving ot 1575 rpm and C moving at 1850 rpm attached on a single axle.
-->sketch and annote the pulley system.
-->what is the diameter of pulley B and pulley C?
Answers
Answer:
The answer is "[tex]\bold{11.428 \ in \ \ and \ \ 9.729 \ in}[/tex]".
Explanation:
The second question solution can be defined as follows:
[tex]N_A= 1000 \ rpm\\\\N_B=1575 \ rpm\\\\N_C=1850 \ rpm\\\\D_A= 18 \ in \\\\[/tex]
In point 1:
Calculating the value of the linear velocity is the same:
[tex]\to V_A=V_B\\\\\to \pi D_A N_A=\pi D_BN_B\\\\\to D_B= \frac{D_A \times N_A}{N_B}[/tex]
[tex]=\frac{18 \times 1000}{1575}\\\\=11.428 \ in \ \ \text{Diameter of pulley B}[/tex]
In point 2:
[tex]\to V_A=V_C\\\\\to \pi D_A N_A=\pi D_C N_C\\\\\to D_C=\frac{D_A \times N_A}{N_C}\\\\[/tex]
[tex]=\frac{1000 \times 18}{1850}\\\\=9.729 \ in \ \ \text{Diameter of pulley B}[/tex]
which systems engineering support discipline has the goal to ensure that support considerations are an integral part of the system design requirements so that the system can be cost-effectively supported throughout its life cycle
Answers
Answer:what's the question
Explanation:
Which statement describes a possible limitation on a experimental design? A. Collecting samples to analyze is expensive B. The experiment has a very simple procedure C. The subject of the experiment has been studied by other scientist's D. There is a large number of constants
Answers
Answer:
A. Collecting samples to analyze is expensive
Explanation:
Experimental research designs may be referred to as an analytical approach which is based on establishing a scientific hypothesis through by subjecting observations or subjects to certain treatment carried out in a controlled environment. In a typical experimental research design, we have two groups which are the constant group or observation and the variable group which are the actual experimetal group which are subjected to treatment and whose behavior are compared to that of the constant group. From the options given, one actual limitation of the experimetal research design is the associated cost particularly those which require longer time to study and analyse. Other limitations include ; difficulty in measuring human response and skepticism over applicability of findings in a natural environment.
Answer:
collecting samples to analyze is very expensive is correct^
Explanation:
just took the quiz, thanks btw
what is the power output of a „U reactor if it takes 30 days to use up 2kg of
fuel Given that energy release per fission is 200MeV and Avogadro's number = 6,023 x 10°
per kilomole
Answers
Answer:
The power output of the reactor is approximately 63.387 MW
Explanation:
The energy released per one fission reaction = 200 MeV
The Avogadro's number, [tex]N_A[/tex] = 6.023 × 10²³
The mass of fuel used = 2 kg
The mass of one mole of uranium 235, U²³⁵ = 235 grams = 0.235 kg
Therefore, the number of moles of U²³⁵ in 2 kg = 2 kg·mol⁻¹/(0.235 kg) = 400/47 moles ≈ 8.51 moles
The number of uranium atoms in 400/47 moles of U²³⁵ = (400/47) × 6.023 × 10²³ ≈ 5.126 × 10²⁴ atoms of U²³⁵
The energy released per fission of an atom = 200 MeV = 200 × 10⁶ eV × 1.602177 × 10⁻¹⁹ J/(eV) ≈ 3.204354 × 10⁻¹¹ J
The energy, 'E', released by the 5.126 × 10²⁴ atoms of U²³⁵ in the 2 kg of U²³⁵ is given as follows;
E = 3.204354 × 10⁻¹¹ J/atom × 5.126 × 10²⁴ atoms = 1.643 × 10¹⁴ Joules
The power, P = Energy/Time
The time = 30 days = 30 × 24 × 60 × 60 = 2,592,000 seconds
∴ P = (1.643 × 10¹⁴ Joules)/(2,592,000 s) = 63.3873457 megawatts ≈ 63.387 MW
The power output of the reactor, P ≈ 63.387 MW
Summarize three reasons for the failure of the Tacoma Narrows Bridge, and identify the steps in the engineering design process they relate to.
Answers
Answer:
1) Perform
2) Prototype
Explanation:
The prototype testing phase of the design process to test the readiness of the design for the intended use
The prototype which is a first or early model of the project made so as to ensure proper functioning of the concept is to pass through general function test as well as design performance evaluation in a testing or lab setting
Other tests include, excessive stress test, testing to failure and other design tests.
Therefore;
The problem faced by Galloping Gertie the failed bridge across the Tacoma Narrows, was caused because the engineers did not perform tests on the prototype in the engineering design process.