The elastic limit of brass is 379,MPa. What s | vedclass

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(B) The stress $\sigma$ is defined as the force $F$ divided by the cross-sectional area $A$. To remain within the elastic limit,the stress must not ex

Vedclass · Accepted Answer

$1.16$

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(B) The stress $\sigma$ is defined as the force $F$ divided by the cross-sectional area $A$. To remain within the elastic limit,the stress must not exceed the elastic limit $\sigma_{max} = 379\,MPa = 379 	imes 10^6\,Pa$.The formula for stress is $\sigma = \frac{F}{A} = \frac{F}{\frac{\pi}{4}d^2}$.Given $F = 400\,N$ and $\sigma = 379 	imes 10^6\,Pa$,we have:$379 	imes 10^6 = \frac{400}{\frac{\pi}{4}d^2}$.Rearranging for $d^2$:$d^2 = \frac{4 	imes 400}{\pi 	imes 379 	imes 10^6} = \frac{1600}{1190.7 	imes 10^6} \approx 1.3437 	imes 10^{-6}\,m^2$.Taking the square root:$d = \sqrt{1.3437} 	imes 10^{-3}\,m \approx 1.159 	imes 10^{-3}\,m$.Converting to $mm$:$d \approx 1.16\,mm$.

The elastic limit of brass is $379\,MPa$. What should be the minimum diameter (in $mm$) of a brass rod if it is to support a $400\,N$ load without exceeding its elastic limit?

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