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svetoff [14.1K]
3 years ago
14

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

6 MN/mm per bolt. The bolts are preloaded to 75 percent of proof strength. Assume the external load is equally distributed to all the bolts. The bolts are M6 × 1 class 5.8 with rolled threads. A fluctuating external load is applied to the entire joint with Pmax = 60 kN and Pmin = 20 kN.(a) Determine the yielding factor of safety.(b) Determine the overload factor of safety.(c) Determine the factor of safety based on joint separation.(d) Determine the fatigue factor of safety using the Goodman criterion.
Engineering
1 answer:
rjkz [21]3 years ago
4 0

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN,  Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

n_{p} = \frac{S_{p}A_{t}  }{CP_{max}+ F_{i}  }  ------ ( 1 )

Sp = 380 MPa,   At = 20.1 mm^2,   C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = \frac{380*20.1}{0.277*7500*5728.5} = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

n_{L} =  \frac{S_{p}A_{t}-F_{i}   }{C(P_{max} )}  ------- ( 2 )

Sp =  380 MPa,   At =  20.1 mm^2, C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

input values into equation 2 above

hence : n_{L} = 0.9191n_{L}  = 0.9191

<u>C)  Determine the factor of safety based on joint separation</u>

n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }

Fi =  5728.5 N,  Pmax = 7500 N,  C = 0.277,

input values into equation above

Hence n_{0} = 1.056

<u>D)  Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

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The Java program that accepts a matrix of M × N order and then interchanges diagonals of the matrix is given below:

<h3>Steps:  </h3>
  • 1. We can only interchange diagonals for a square matrix.
  • 2. Therefore, we would have to create a square matrix of size [M × M].
  • 3. We would check whether the matrix is a square matrix or not. If the matrix is square then follow step 3 else terminate the program.
  • 4. Apply logic for interchange diagonal of the matrix some logic is given below.

<h3>Java Code</h3>

//  Java Program to Accept a Matrix of Order M x N &

//  Interchange the Diagonals

import java.util.Scanner;

public class InterchangeDiagonals {

   public static void main(String[] args)

   {

       // declare variable

       int m, n, i, j, temp;

       // create a object of scanner class

       Scanner sc = new Scanner(System.in);

       System.out.print("Enter number of rows ");

       // take number of rows

       m = sc.nextInt();

       System.out.print("Enter number of columns ");

       // take number of columns

       n = sc.nextInt();

       // declare a mxn order array

       int a[][] = new int[m][n];

       // if block it's execute when m is equals to n

       if (m == n) {

           System.out.println(

               "Enter all the values of matrix ");

           // take the matrix inputs

           for (i = 0; i < m; i++) {

               for (j = 0; j < n; j++) {

                   a[i][j] = sc.nextInt();

               }

           }

           System.out.println("original Matrix:");

           // print the original matrix

           for (i = 0; i < m; i++) {

               for (j = 0; j < n; j++) {

                   System.out.print(a[i][j] + " ");

               }

               System.out.println("");

           }

          // perform interchange

           for (j = 0; j < m; j++) {

               temp = a[j][j];

               a[j][j] = a[j][n - 1 - j];

               a[j][n - 1 - j] = temp;

           }

           System.out.println(

               " after interchanging diagonals of matrix ");

           // print interchanged matrix

           for (i = 0; i < m; i++) {

               for (j = 0; j < n; j++) {

                   System.out.print(a[i][j] + " ");

               }

               System.out.println("");

           }

       }

       // else block it's only execute when m is not equals

       // to n

       else {

           System.out.println("Rows not equal to columns");

       }

   }

}

Read more about java programming here:

brainly.com/question/18554491

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