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insens350 [35]
3 years ago
9

Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe

red from a torque (T) of 100 Nmat one end (fixed at the other end). By neglecting the outside pressureand radial stresses, choose a proper 2D element and showcase the non-zero stresses on the element. Then, determine all three principal stresses and the maximum shear stress at that point.
Engineering
1 answer:
NemiM [27]3 years ago
6 0

Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

Maximum shear stress = 14.116mpa

Explanation:

di = 75 = 0.075

wall thickness = 0.1 = 0.0001

internal pressure pi = 150 kpa = 150 x 10³

torque t = 100 Nm

finding all values

∂1 = 150x10³x0.075/2x0,0001

= 0.5625 = 56.25mpa

∂2 = 150x10³x75/4x0.1

= 28.12mpa

T = 16x100/(πx75x10³)²

∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]

= 1/2[84.37±√791.2969+5.827396]

= 1/2[84.37±28.33]

∂1 = 1/2[84.37+28.33]

= 56.301mpa

∂2 = 1/2[84.37-28.33]

= 28.07mpa

This is a 2 d diagram donut is analyzed in 2 direction.

So ∂3 = 0mpa

∂max = 56.301-28.07/2

= 14.116mpa

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Explanation:

Solution:

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Continuous insulation helps building to withstand the test of time without the need to upgrade and repair

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4 years ago
In c the square root of a number N can be approximated by repeated calculation using the formula NG = 0.5(LG + N/LG) where NG st
DanielleElmas [232]

Answer:

Following are the program to the given question:

#include <stdio.h>//header file

double square_root(double N, double initialGuess)//defining a method square_root that takes two variable in parameters

{

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while(1)//use loop to calculate square root value

{

NG = 0.5 * (LG + N / LG);//using given formula

diff = NG - LG;//calculating difference

if(diff < 0)//use if to check difference is less than 0

diff = -diff;//decreaing difference

if(diff < 0.005)//use if that check difference is less than 0.005

break;//using break keyword  

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{

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}

}

return NG;//return value

}

int main()//defining main method

{

double ans, n,initialguess = 1.0;//defining double variable

n = 4;//use n to hold value

ans = square_root(n, initialguess);//calculating the square root value and print its value

printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number

n = 120.5;//use n to hold value

ans = square_root(n, initialguess);//calculating the square root value and print its value

printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number

n = 36.01;//use n to hold value

ans = square_root(n, initialguess);//calculating the square root value and print its value

printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number

n = 0.25;//use n to hold value

ans = square_root(n, initialguess);//calculating the square root value and print its value

printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number

printf("\nEnter a number: ");//print message

scanf("%lf", &n);//input value

ans = square_root(n, initialguess);//calculating the square root value and print its value

printf("square_root(%lf) = %lf \n", n, ans);//print calculated value with number

}

Output:

Please find the attachment file.

Explanation:

  • In this code, a method "square_root" is declared that takes two variable "N, initialGuess" in its parameters, inside the method a three double variable is declared.
  • It uses the given formula and uses the diff variable to hold its value and uses two if to check its value is less than 0 and 0.005 and return its calculated value.
  • In the main method, three double variables are declared that use the "n" to hold value and "ans" to call the method that holds its value and print its value.

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Answer:

(i) 3.5385 ohm, 3.768 ohm (ii) 39.89 A (III) 4773.857 W (vi) 348 var (vii) 0.9973 (viii) 4.1796°

Explanation:

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(VII) cos\Phi =0.9973

\Phi =4.1796^{\circ}

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