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tresset_1 [31]
3 years ago
15

A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima

te tensile stress is 400 MPa and fatigue limit (endurance limit) for R = -1.0 is 320 MPa, what is the amplitude of the largest fully-reversed bending moment that can be applied to this shaft so that fatigue dose not occur?
Engineering
1 answer:
zvonat [6]3 years ago
5 0

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

\sigma_c = \frac{P}{A}

\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}

\sigma_c = 81.5MPa

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}

Where,

\sigma_a=Fatigue limit for comined alternating and mean stress

S_e =Fatigue Limit

\sigma_c=Mean stress (due to static load)

\sigma_u = Ultimate tensile stress

Fs =Security Factor

We can replace the values and assume a security factor of 1, then

\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}

Re-arrenge for \sigma_a

\sigma_a = 254.8Mpa

We know that the stress is representing as,

\sigma_a = \frac{M_c}{I}

Then,

Where M_c=Max Moment

I= Intertia

The inertia for this object is

I=\frac{\pi d^4}{64}

Then replacing and re-arrenge for M_c

M_c = \frac{\sigma_a*\pi*d^3}{32}

M_c = \frac{260.9*10^6*\pi*0.05^3}{32}

M_c = 3201.7N.m

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

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Problem 4.041 SI Refrigerant 134a enters an insulated compressor operating at steady state as saturated vapor at -26oC with a vo
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Answer:

0.0297M^3/s

W=68.48kW

Explanation:

Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  

state 1

X=quality=1

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density 1=α1=5.27kg/m^3  

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Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.

m1=m2

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\frac{(Q1)(\alpha 1) }{\alpha 2} =Q2\\Q2=\frac{(0.18)(5.27) }{31.91} =0.0297M^3/s

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To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation

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W=(0.18)(5.27)(306.9-234.7)

W=68.48kW

the compressor power is 68.48kW

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Answer:

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False

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