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Andrews [41]
3 years ago
10

A) A cross-section of a solid circular rod is subject to a torque of T = 3.5 kNâ‹…m. If the diameter of the rod is D = 5 cm, wha

t is the maximum shear stress? include units.B) The maximum stress in a section of a circular tube subject to a torque is Tmax = 37 MPa . If the inner diameter is Di = 4.5 cm and the outer diameter is Do = 6.5 cm , what is the torque on the section? include units.
Engineering
1 answer:
Alisiya [41]3 years ago
7 0

Answer:

\tau_{max}  = 142.6 MPa

T = 1536.8 N m

Explanation:

Given data:

Torque = 3.5 k N m = 3.5*10^3 N.m

Diameter D = 5 cm = 0.05 m

a) from torsional equation we have

\frac[T}{J_{solid}} = \frac{\tau_{max}}{D/2}

\frac{T}{\pi/32 D^4} = \frac{\tau_{max}}{D/2}

solving for \tau_{max}

\tau_{max} = \frac{16 T}{\pi D^3} =\frac{16 \times 3.5*10^3}{\pi 0.05^3}

\tau_{max}  = 142.6 MPa

B)

\tau = 37 MPa = 37 \times  10^6 Pa

D_i = 4.5 cm = 0.045 m

D_o = 6.5 cm = 0.065 m

\frac{T}{J_{hollow}} = \frac{\tau_{max}}{D_o /2}

\frac{T}{(\pi/32) (0.065^4 - 0.045^4)} =\frac{37*10^6}{0.065/2}

T = 1536.8 N m

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#include<stdio.h>

#include<stdlib.h>

#include<unistd.h>

#include<sys/types.h>

#include<string.h>

#include<pthread.h>

//#include<sys/wait.h>

int main(int argc, char** argv)

{

int fd1[2];

int fd2[2];

int fd3[2];

int fd4[2];

char message[] = "abcd";

char input_str[100];

pid_t p,q;

if (pipe(fd1)==-1)

{

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 return 1;

}

if (pipe(fd2)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd3)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

if (pipe(fd4)==-1)

{

 fprintf(stderr, "Pipe Failed" );

 return 1;

}

p = fork();

if (p < 0)

{

 fprintf(stderr, "fork Failed" );

return 1;

}

// child process-1

else if (p == 0)

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 // Concatenate a fixed string with it

 int k = strlen(concat_str);

 int i;

 for (i=0; i<strlen(fixed_str); i++)

 {

  concat_str[k++] = fixed_str[i];

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 concat_str[k] = '\0';//string ends with '\0'

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 close(fd1[1]);

 wait(NULL);

//.......................................................................

 close(fd2[1]);

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 {

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 {

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 }

 close(fd2[0]);//close reading end of pipe 2

 exit(0);

}

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{

 close(fd1[1]);//Close writting end of first pipe

 char concat_str[100];

 read(fd1[0], concal_str, strlen(concat_str)+1);

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 {

  //if not then

  write(fd2[1], "invalid",sizeof(concat_str));

  return 0;

 }

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 {

  //if yes then

  write(fd2[1], "valid",sizeof(concat_str));

  close(fd2[1]);

  q=fork();//create chile process 2

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   close(fd3[1]);

   wait(NULL);//wait till child_process_2 send ACK

   //...........................................................

   close(fd4[1]);

   read(fd4[0],concat_str,100);

   close(fd4[0]);

   if(sctcmp(concat_str,"ack")==0)

   {

    printf("Messageof child process_1 is received by child process_2");

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    printf("Messageof child process_1 is not received by child process_2");

   }

  }

  else

  {

   if(p<0)

   {

    printf("Chiile_Procrss_2 not cheated");

   }

   else

   {

     

    close(fd3[1]);//Close writing end of first pipe

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    write(fd4[1], "ack",sizeof(concat_str));

     

   }

  }

 }

 close(fd2[1]);

}

}

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