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Hw ch7
Hw ch7
Hw ch7
Hw ch7
Hw ch7
Hw ch7
Hw ch7
Hw ch7
Hw ch7
Hw ch7
Hw ch7
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Hw ch7

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  • 1. Solved ProblemsProblem 7.124:The compressed-air tank AB has a 250mm outside diameter and an 8mmwall thickness. It is fitted with a collar by which a 40-KN force F is appliedat B in the horizontal direction. Knowing that the gage pressure inside thetank is 5-MPa. Determine the maximum normal stress and the maximumshearing stress at point K.Solution:Assume: neglect of the collar dimensions. ( ) ( )
  • 2. Using Mohr Circle: √ ( ) ,
  • 3. Problem 7.120: A torque of the magnitude T= 12 KN. m is applied to the end of a tankcontaining compressed air under a pressure of 8 MPa. Knowing that thetank has a 180-mm inner diameter and 12-mm wall thickness, determinethe maximum normal stress and the maximum shearing stress in the tank.Solution: = 90mm, t=12mm, ,Torsion:C1=90mm, C2=102mm, ( )Using Mohr’s Circle: √ ( )
  • 4. Problem 7.121:The tank shown has a 180-mm inner diameter and 12-mm wall thickness.Knowing that the tank contains compressed air under a pressure of 8-MPa, determine the magnitude T of the applied torque for which themaximum normal stress is 75-MPa.Solution: ,Torsion:C1=90mm, C2=102mm, ( )Using Mohr’s Circle:
  • 5. Problem 7.104: The unpressurized cylindrical storage tank shown has a 5-mm wallthickness and is made of steel having 450 MPa, ultimate strength intension. Determine the max h to which it can be filled with water if thefactor of safety of 4 is desired. (Specific weight of water=9810 N/m2.).Solution:
  • 6. ,Solution: ( ) ,Using Mohr’s Circle:
  • 7. Since,| | , stress point lies in 4th quadratic Equation of 4thquadratic boundary is:R= 34.612 MPa √ ( )Torsion:C=16-mm,
  • 8. Solution: √ ( ) | |Since,| | stress point lies in 4thquadratic Equation of 4th quadratic boundary is: | |
  • 9. Solution: ( ) , √ ( ) √ √ √ ( ) ( )
  • 10. Problem 7.85:The 44-mm diameter shaft AB is made of a grade of steel forwhich the yield strength is 250 MPa. Using the maximum shearingstress principle criterion, determine the magnitude of the force Pfor which yield occurs when T= 1.5 KN. m.Solution: ( ) √ ( ) ………………………………………………………………………….. (1) √ ( ) ……………………………………………………………………………. (2)
  • 11. Subtract (2) from (1): √ = 174 MPa, ( )

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