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3 years ago

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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.

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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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

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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 $(T_∞-T_s)$

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<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 $\frac{W}{m^2}$

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