You have been assigned the task of reviewing the relief scenarios for a specific chemical reactor in your plant. You are current
1 answer:
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4.75cm * 5.22cm is 24.795cm²
but let's do this with m² instead:
0.0475m * 0.0522m = 0.0024795
now we can compare it with the 49780 much easier.
devide. that's the <u>roughly</u> 20,000,000fold of the plot, but it's still squared, so let's take the root
so the representative fraction is 1:4,480.
the exact value is in the screens
have a nice day:)
Answer:
a) Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
b) principal stress :
б1 = ( бx + бy ) / 2 - ( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
Explanation:
Combined normal stress and shear stress sketches attached below
The terms in the sketch are :
бx = tensile stress in x direction
бy = tensile stress in y direction
Txy = y component of shear stress acting on the perpendicular plane to x axis
бn = Normal stress acting on the inclined plane EF
Tn = shear stress acting on the inclined plane EF
A) Normal and shear stresses on inclined plane
Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
B) principal and maximum shear stresses
principal stress :
б1 = ( бx + бy ) / 2 - ( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
Answer:
Code is given as below:
Explanation:
#include <iostream>
using namespace std;
//function prototype declaration
void MilitaryTime(int, int, char, int &, int &);
int main()
{
//declare required variables
int SHour, SMin, MHour, MMin;
char AorP;
//promt and read the hours from the user
cout<<"Enter hours in standard time : ";
cin>>SHour;
//check the hours are valid are not
while(SHour<0 || SHour>12)
{
cout<<"Invalid hours for standard time. "
<<"Try again..."<<endl;
cout<<"Enter hours in standard time : ";
cin>>SHour;
}
//promt and read the minutes from the user
cout<<"Enter minutes in standard time : ";
cin>>SMin;
//check the minutes are valid are not
while(SMin<0 || SMin>59)
{
cout<<"Invalid minutes for standard time. "
<<"Try again..."<<endl;
cout<<"Enter minutes in standard time : ";
cin>>SMin;
}
//promt and read the am or pm from the user
cout<<"Enter standard time meridiem (a for AM p for PM): ";
cin>>AorP;
//check the meridiem is valid are not
while(!(AorP=='a' || AorP=='p' || AorP=='A' || AorP=='P'))
{
cout<<"Invalid meridiem for standard time. "
<<"Try again..."<<endl;
cout<<"Enter standard time meridiem (a for AM p for PM): ";
cin>>AorP;
}
//call function to calculate the military time
MilitaryTime(SHour, SMin, AorP, MHour, MMin);
//fill zeros and display standard time
cout.width(2);
cout.fill('0');
cout<<SHour<<":";
cout.width(2);
cout.fill('0');
cout<<SMin;
if(AorP=='a' || AorP=='A')
cout<<" am = ";
else
cout<<" pm = ";
//fill zeros and display military time
cout.width(2);
cout.fill('0');
cout<<MHour;
cout.width(2);
cout.fill('0');
cout<<MMin<<endl;
system("PAUSE");
return 0;
}
//function to calculate the military time with reference parameters
void MilitaryTime(int SHour, int SMin, char AorP, int &MHour, int &MMin)
{
//check the meredium is am or pm
//and calculate hours
if(AorP=='a' || AorP=='A')
{
if(SHour==12)
MHour = 0;
else
MHour = SHour;
}
else
MHour = SHour+12;
MMin = SMin;
