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saw5 [17]
3 years ago
6

A bar of steel has the minimum properties Se = 40 kpsi, S = 60 kpsi, and S-80 kpsi. The bar is subjected to a steady torsional s

tress of 15 kpsi and an alternating bending stress of 25 kpsi Find the factor of safety guarding against a static failure, and either the factor of safety guard- ing against a fatigue failure or the expected life of the part. For the fatigue analysis use:
(a) Modified Goodman criterion.

(b) Gerber criterion.

(C) ASME-elliptic criterion.

Engineering
1 answer:
nlexa [21]3 years ago
5 0

Answer:

(a) Modified Goodman criterion:

Factor of safety against fatigue failure =  1.0529

(b) Gerber criterion:

Factor of safety against fatigue failure = 1.31

(c) ASME-elliptic criterion:

Factor of safety against fatigue failure = 1.315

Explanation:

See the attached file for the calculation.

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Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the
vekshin1

Answer:

Re=100,000⇒Q=275.25 \frac{W}{m^2}

Re=500,000⇒Q=1,757.77\frac{W}{m^2}

Re=1,000,000⇒Q=3060.36 \frac{W}{m^2}

Explanation:

Given:

For air      T_∞=25°C  ,V=8 m/s

  For surface T_s=179°C

     L=2.75 m    ,b=3 m

We know that for flat plate

Re⇒Laminar flow

Re>30\times10^5⇒Turbulent flow

<u> Take Re=100,000:</u>

 So this is case of laminar flow

  Nu=0.664Re^{\frac{1}{2}}Pr^{\frac{1}{3}}

From standard air property table at 25°C

  Pr= is 0.71  ,K=26.24\times 10^{-3}

So    Nu=0.664\times 100,000^{\frac{1}{2}}\times 0.71^{\frac{1}{3}}

Nu=187.32   (\dfrac{hL}{K_{air}})

187.32=\dfrac{h\times2.75}{26.24\times 10^{-3}}

     ⇒h=1.78\frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                           =275.25 \frac{W}{m^2}

<u> Take Re=500,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 500,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=1196.18  ⇒h=11.14 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =11.14(179-25)

                           = 1,757.77\frac{W}{m^2}

<u> Take Re=1,000,000:</u>

So this is case of turbulent flow

  Nu=0.037Re^{\frac{4}{5}}Pr^{\frac{1}{3}}

Nu=0.037\times 1,000,000^{\frac{4}{5}}\times 0.71^{\frac{1}{3}}

Nu=2082.6  ⇒h=19.87 \frac{W}{m^2-K}

heat transfer rate =h(T_∞-T_s)

                             =19.87(179-25)

                           = 3060.36 \frac{W}{m^2}

7 0
3 years ago
Explain how a CO2 cartridge powers the dragster you will be building. A good website to use is How Stuff Works. (howstuffworks.c
andrezito [222]

Answer

C02 cartridges are compressed.

Explanation:

The C02 Cartridge will be used to power our dragsters by the compression inside of the bottle. Since C02 is a compressed gas inside of a small container, The pressure when released my provide enough force to move the dragster at high speeds. An example of this would be like an air soft gun. some air soft guns are used by small C02 Cartridges, which let out a strong burst of compressed C02, which fires the projectiles at very high speeds.

5 0
3 years ago
Read 2 more answers
Consider a 8-m-long, 8-m-wide, and 2-m-high aboveground swimming pool that is filled with water to the rim. (a) Determine the hy
Stolb23 [73]

Answer:

The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.

<u>Explanation:</u>

To calculate force on the walls of swimming pool whose dimensions are given as <em>8-m-long, 8-m-wide, and 2-m-high</em>. We know that formula for hydrostatic force is \text {hydrostatic force}=\text {pressure} \times \text {area,}=\rho g h \times(l \times h)  

\equiv \rho g h^{2} l, we know ρ=density of fluid=1000 g / c m^{3},

g=acceleration due to gravity=9.81 m / s^{2}, h=height of the pool=2 m and l=length of the pool=8 m.  

hydrostatic force on each wall=1000 \times 9.81 \times 2^{2} \times 8 = 313920 N.

<em>The distance at which hydrostatic force is acted is half of the height of the swimming pool. </em>

At 1 m from the ground this hydrostatic force is acted on each wall.  

The force is <em>quadruple if the height of the walls of the pool is doubled</em> this is because, the<em> height is doubled and taken as h=4 m</em> and substitute in the equation =\rho g h^{2} l = 1000 \times 9.81 \times 4^{2} \times 8 = 1255680 N. This is 4 times 313920 N.

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3 years ago
Is A fine by the EPA may be imposed on the employer or
goldfiish [28.3K]

Answer:

No false

Explanation:

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3 years ago
Is santa real or nah is santa real or nah
Elena L [17]

Answer:

nah

Explanation:

3 0
3 years ago
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