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Andrej [43]
3 years ago
10

A 20-cm-long rod with a diameter of 0.250 cm is loaded with a 5500 N weight. If the diameter decreases to 0.210 cm, determine th

e engineering stress at this load. Assume that the necking point has not been reached. The engineering stress on the rod when it is loaded with a 5500 N weight is ______________________ MPa.
Engineering
1 answer:
ss7ja [257]3 years ago
7 0

Answer:

1561.84 MPa

Explanation:

L=20 cm

d1=0.21 cm

d2=0.25 cm

F=5500 N

a) σ= F/A1= 5000/(π/4×(0.0025)^2)= 1018.5916 MPa

lateral strain= Δd/d1= (0.0021-0.0025)/0.0025= -0.16

longitudinal strain (ε_l)= -lateral strain/ν = -(-0.16)/0.3

(assuming a poisson's ration of  0.3)

ε_l =0.16/0.3 = 0.5333

b) σ_true= σ(1+ ε_l)= 1018.5916( 1+0.5333)

σ_true = 1561.84 MPa

ε_true = ln( 1+ε_l)= ln(1+0.5333)

ε_true= 0.4274222

The engineering stress on the rod when it is loaded with a 5500 N weight is 1561.84 MPa.

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Check the explanation

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#include <iostream>

using namespace std;

void insert(int* arr, int* size, int value, int position){

if(position<0 || position>=*size){

cout<<"position is greater than size of the array"<<endl;

return ;

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*size = *size + 1 ;

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cout<<" "<<endl;

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arr[i] = arr[i+1];

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int count(int arr[], int size, int target){

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for(int i=0;i<size;i++){

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return total ;

}

int main()

{

int size;

cout<<"Enter the initial size of the array:";

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int arr[size],val;

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cout<<"Choice:";

cin>>choice;

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break;

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cin>>position;

remove(arr,&size,position);

break;

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cout<<"Enter the target value:";

cin>>target;

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Kindly check the attached images below for the code output.

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