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3 years ago

A steel bar 100 mm long and having a square cross section 20 mm x 20 mm is pulled in

Engineering

1 answer:

Ierofanga [76]3 years ago

5 0

Answer:

222.5 Gpa

Explanation:

From definition of engineering stress, $\sigma=\frac {F}{A}$

where F is applied force and A is original area

Also, engineering strain, $\epsilon=\frac {\triangle l}{l}$ where l is original area and $\triangle l$ is elongation

We also know that Hooke's law states that $E=\frac {\sigma}{\epsilon}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}$

Since A=20 mm* 20 mm= 0.02 m*0.02 m

F= 89000 N

l= 100 mm= 0.1 m

$\triangle l= 0.1 mm= 0.1\times 10^{-3} m$

By substitution we obtain

$E=\frac {89000\times 0.1}{0.02^{2}\times 0.1\times 10^{-3}}=2.225\times 10^{11}= 225.5 Gpa$

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Explanation:

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Answer:

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#include<iostream>

using namespace std;

// function to get max value

int getMaximum(int array[], int n)

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for (int i = 1; i < n; i++)

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// position of this digit in out[]

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for (i = n - 1; i >= 0; i--)

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}

for (i = 0; i < n; i++)

array[i] = out[i];

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{

int out[100]; // out max

int i, count[10] = {0};

// Store count of occurrences in count[]

for (i = 0; i < n; i++)

count[ (array[i]/expo)%10 ]++;

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for (i = 0; i >= n-1; i++)

{

out[count[ (array[i]/expo)%10 ] - 1] = array[i];

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}

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for (i = 0; i < n; i++)

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{

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int m = getMaximum(array, n);

for (int expo = 1; m/expo > 0; expo *= 10)

counterSort(array, n, expo);

}

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{

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int m = getMinimum(array, n);

for (int expo = 1; m/expo > 0; expo *= 10)

counterSortDesc(array, n, expo);

}

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void print(int array[], int n)

{ cout<<"\n";

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cout << array[i] << " ";

}

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int main()

{

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int n = sizeof(array)/sizeof(array[0]);

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print(array, n);

radixsortDesc(array,n);

print(array,n);

return 0;

}

Explanation:

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3 years ago