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nikdorinn [45]
3 years ago
14

A 10-mm steel drill rod was heat-treated and ground. The measured hardness was found to be 290 Brinell. Estimate the endurance s

trength, Se, in MPa if the rod is used in rotating bending.
Engineering
1 answer:
grandymaker [24]3 years ago
8 0

Answer:

the endurance strength  S_e = 421.24  MPa

Explanation:

From the given information; The objective is to estimate the endurance strength, Se, in MPa .

To do that; let's for see the expression that shows the relationship between the ultimate tensile strength and Brinell hardness number .

It is expressed as:

200 \leq H_B \leq 450

S_{ut} = 3.41 H_B

where;

H_B = Brinell hardness number

S_{ut} =  Ultimate tensile strength

From ;

S_{ut} = 3.41 H_B; replace 290 for H_B ; we have

S_{ut} = 3.41 (290)

S_{ut} = 988.9 MPa

We can see that the derived value for the ultimate tensile strength when the Brinell harness number = 290 is less than 1400 MPa ( i.e it is 988.9 MPa)

So; we can say

S_{ut} < 1400

The Endurance limit can be represented by the formula:

S_e ' = 0.5 S_{ut}

S_e ' = 0.5 (988.9)

S_e ' = 494.45 MPa

Using Table 6.2 for parameter for Marin Surface modification factor. The value for a and b are derived; which are :

a = 1.58

b =  -0.085

The value of the surface factor can be calculate by using the equation

k_a = aS^b_{ut}

K_a = 1.58 (988.9)^{-0.085

K_a = 0.8792

The formula that is used to determine the value of  k_b for the rotating shaft of size factor d = 10 mm is as follows:

k_b = 1.24d^{-0.107}

k_b = 1.24(10)^{-0.107}

k_b = 0.969

Finally; the the endurance strength, Se, in MPa if the rod is used in rotating bending is determined by using the expression;

S_e =k_ak_b S' _e

S_e= 0.8792×0.969×494.45

S_e = 421.24  MPa

Thus; the endurance strength  S_e = 421.24  MPa

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Explanation:

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Answer:

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Explanation:

#include<iostream>

using namespace std;

//Function Declarations

void add();

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//Main Code Displays Menu And Take User Input

int main()

{

  int choice;

  cout << "\nMenu";

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  cout << "\n\nEnter your choice: ";

 

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  cout << "\n";

 

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              break;

             

      case 2: sub();

              break;

             

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              break;

     

      case 0: cout << "Exited";

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      default: cout << "Invalid";      

  }

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}

//Addition Of Matrix

void add()

{

  int rows1,cols1,i,j,rows2,cols2;

 

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  //Taking First Matrix

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  //Printing 1st Matrix

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

  //Taking Second Matrix

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  //Displaying second Matrix

  cout << "\n";

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m2[i][j] << " ";

      cout << "\n";

  }

  //Displaying Sum of m1 & m2

  if(rows1 == rows2 && cols1 == cols2)

  {

      cout << "\n";

      for(i=0;i<rows1;i++)

      {

          for(j=0;j<cols1;j++)

              cout << m1[i][j]+m2[i][j] << " ";

          cout << "\n";  

      }

  }

  else

      cout << "operation is not supported";

     

  main();

 

}

void sub()

{

  int rows1,cols1,i,j,k,rows2,cols2;

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

 

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

 

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

 

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

  cout << "\n";

  //Displaying Subtraction of m1 & m2

  if(rows1 == rows2 && cols1 == cols2)

  {

      for(i=0;i<rows1;i++)

      {

          for(j=0;j<cols1;j++)

              cout << m1[i][j]-m2[i][j] << " ";

          cout << "\n";  

      }

  }

  else

      cout << "operation is not supported";

     

  main();

 

}

void mul()

{

  int rows1,cols1,i,j,k,rows2,cols2,mul[10][10];

  cout << "\nmatrix1 # of rows: ";

  cin >> rows1;

 

  cout << "\nmatrix1 # of columns: ";

  cin >> cols1;

 

   int m1[rows1][cols1];

 

  for(i=0;i<rows1;i++)

      for(j=0;j<cols1;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m1[i][j];

          cout << "\n";

      }

  cout << "\n";

  for(i=0;i<rows1;i++)

  {

      for(j=0;j<cols1;j++)

          cout << m1[i][j] << " ";

      cout << "\n";

  }

     

  cout << "\nmatrix2 # of rows: ";

  cin >> rows2;

 

  cout << "\nmatrix2 # of columns: ";

  cin >> cols2;

 

  int m2[rows2][cols2];

 

  for(i=0;i<rows2;i++)

      for(j=0;j<cols2;j++)

      {

          cout << "\nEnter element (" << i << "," << j << "): ";

          cin >> m2[i][j];

          cout << "\n";

      }

  cout << "\n";

  //Displaying Matrix 2

  for(i=0;i<rows2;i++)

  {

      for(j=0;j<cols2;j++)

          cout << m2[i][j] << " ";

      cout << "\n";

  }

     

  if(cols1!=rows2)

      cout << "operation is not supported";

  else

  {

      //Initializing results as 0

      for(i = 0; i < rows1; ++i)

  for(j = 0; j < cols2; ++j)

  mul[i][j]=0;

// Multiplying matrix m1 and m2 and storing in array mul.

  for(i = 0; i < rows1; i++)

  for(j = 0; j < cols2; j++)

  for(k = 0; k < cols1; k++)

  mul[i][j] += m1[i][k] * m2[k][j];

// Displaying the result.

  cout << "\n";

  for(i = 0; i < rows1; ++i)

      for(j = 0; j < cols2; ++j)

      {

      cout << " " << mul[i][j];

      if(j == cols2-1)

      cout << endl;

      }

      }  

  main();

 }

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3 years ago
Which of the following Identifies the challenges faced by scientists experimenting with using artificial photosynthesis as
bezimeni [28]

Answer:

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Explanation:

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8 0
2 years ago
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