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mylen [45]
3 years ago
14

A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of

24 kpsi. After some calculation, the designer has selected a standard 1/2 in diameter (d) bar. The factor of safety (n) is ____. Round the answers to three significant digits.
Engineering
1 answer:
schepotkina [342]3 years ago
5 0

Answer:

n = 2.36

Explanation:

The stress experimented by the circular bar is:

\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)

\sigma = 10.186\,kpsi

The safety factor is:

n = \frac{24\,kpsi}{10.186\,kpsi}

n = 2.36

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Svetllana [295]

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

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Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

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V \propto[1+e]

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Shrinkage limit is lowest water content

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Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

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Zarrin [17]

Answer:

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