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melamori03 [73]
2 years ago
8

Basic Question please help

Engineering
1 answer:
Dvinal [7]2 years ago
7 0
1(A)
2(B)
3(E)
4(C)
5(D)
6(B)
7(C)
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The pads are 200mm long, 150 mm wide and thickness equal to 12mm. 1- Determine the average shear strain in the rubber if the for
lord [1]

Answer:

a) 0.3

b) 3.6 mm

Explanation:

Given

Length of the pads, l = 200 mm = 0.2 m

Width of the pads, b = 150 mm = 0.15 m

Thickness of the pads, t = 12 mm = 0.012 m

Force on the rubber, P = 15 kN

Shear modulus on the rubber, G = 830 GPa

The average shear strain can be gotten by

τ(average) = (P/2) / bl

τ(average) = (15/2) / (0.15 * 0.2)

τ(average) = 7.5 / 0.03

τ(average) = 250 kPa

γ(average) = τ(average) / G

γ(average) = 250 kPa / 830 kPa

γ(average) = 0.3

horizontal displacement,

δ = γ(average) * t

δ = 0.3 * 12

δ = 3.6 mm

5 0
3 years ago
A fatigue test was conducted in which the mean stress was 90 MPa (13050 psi), and the stress amplitude was 190 MPa (27560 psi).
Gwar [14]

Answer:

a) 280MPa

b) -100MPa

c) -0.35

d) 380 MPa

Explanation:

GIVEN DATA:

mean stress \sigma_m = 90MPa

stress amplitude \sigma_a = 190MPa

a) \sigma_m =\frac{\sigma_max+\sigma_min}{2}

    90 =\frac{\sigma_{max}+\sigma_{min}}{2} --------------1

\sigma_a =\frac{\sigma_{max}-\sigma_{min}}{2}

   190 = \frac{\sigma_{max}-\sigma_{min}}{2} -----------2

solving 1 and 2 equation we get

\sigma_{max} = 280MPa

b) \sigma_{min} = - 100MPa

c)

stress ratio=\frac{\sigma_{min}}{\sigma_{max}}

=\frac{-100}{280} = -0.35

d)magnitude of stress range

                      =(\sigma_{max} -\sigma_{min})

                       = 280 -(-100) = 380 MPa

3 0
3 years ago
You will create an array manipulation program that allows the user to do pretty much whatever they want to an array. When launch
enyata [817]

Answer:

Check the explanation

Explanation:

#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 ;

}

*size = *size + 1 ;

for(int i=*size;i>position;i--){

arr[i] = arr[i-1];

}

arr[position] = value ;

}

void print(int arr[], int size){

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

cout<< arr[i] <<" ";

}

cout<<" "<<endl;

}

void remove(int* arr, int* size, int position){

* size = * size - 1 ;

for(int i=position;i<*size;i++){

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

}

}

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

int total = 0 ;

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

if(arr[i] == target)

total += 1 ;

}

return total ;

}

int main()

{

int size;

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

cin>>size;

int arr[size],val;

cout<<"Enter the values to fill the array:"<<endl;

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

cin>>val;

arr[i] = val ;

}

int choice = 5,value,position,target ;

do{

cout<<"Make a selection:"<<endl;

cout<<"1) Insert"<<endl;

cout<<"2) Remove"<<endl;

cout<<"3) Count"<<endl;

cout<<"4) Print"<<endl;

cout<<"5) Exit"<<endl;

cout<<"Choice:";

cin>>choice;

switch(choice){

case 1:

cout << "Enter the value:";

cin>>value;

cout << "Enter the position:";

cin>>position;

insert(arr,&size,value,position);

break;

case 2:

cout << "Enter the position:";

cin>>position;

remove(arr,&size,position);

break;

case 3:

cout<<"Enter the target value:";

cin>>target;

cout <<"The number of times "<<target<<" occured in your array is:" <<count(arr,size,target)<<endl;

break;

case 4:

print(arr,size);

break;

case 5:

cout <<"Thank you..."<<endl;

break;

default:

cout << "Invalid choice..."<<endl;

}

}while(choice!=5);

return 0;

}

Kindly check the attached images below for the code output.

3 0
3 years ago
What is the difference between science and engineering?
jek_recluse [69]

Answer:

The following table is among some of the discrepancies between some of the different topics.

Explanation:

  • Science seems to be the application of structured organized facts which could be interpreted scientifically because when engineering would be a branch of science which deals with either the profession of acquiring including using technological, computational, economical and severity of adverse effects to design and manufacture equipment and devices that seem to be beneficial to mankind.
  • Science would be a set of information and established frameworks whereas the use of these models and information to construct structures as well as mechanisms becomes engineering.
4 0
3 years ago
A 15-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when o
just olya [345]

Answer:

The length of bar will be 2.82 m

Explanation:

Given that

d= 15 mm

r= 7.5 mm

Shear stress = 110 MPa

θ =  30°                                  (30°   = 30°  x π/180°  =0.523 rad)

θ = 0.523 rad

G for steel

G= 79.3 GPa

We know that

\dfrac{\tau}{r}=\dfrac{G\theta }{L}

\dfrac{110}{7.5\times 10^{-3}}=\dfrac{79.3\times 10^3\times 0.523 }{L}

L= 2. 82 m

The length of bar will be 2.82 m

8 0
3 years ago
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