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  • Shear and direct stresses on a shaft

    Discussion in 'Calculations' started by george88b, Nov 21, 2010.

    1. george88b

      george88b New Member

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      hello..

      i have a shaft acting one it 520KN force. i know the torque is 150KN and the diameter of the shaft=200mm. To find the direct stress and shear stresses i will use the formula ?=F/A and ?=Tr/J respectively?

      thnx for ur attention:)
       
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    3. hmck57

      hmck57 Member

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      George88b -

      How is the 520kN force acting on the shaft? Do you have a free body diagram of the shaft? If you are referring to an axial load applied to the end of the shaft then yes the normal or direct stress would be s = F/A. Also, the shear stress due to the applied torque would be t = Tr/J (Note: 150kN is a force - not a torque). This will give you the static stresses acting on your shaft for these two loads - stress concentrations may produce local stresses which are higher.

      It is always a good idea to start with a balanced free body diagram, with all the applied loads and reaction loads included. This will help you to insure you understand how your shaft caries load, and that you haven't overlooked any forces acting on it.
       
    4. lilduka26

      lilduka26 Member

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      It is not so easy as you have mentioned. First of all you have to draw free body diagram of the shaft. Then draw shear and moment diagrams. on moment diagrams you can easily define the point, bigger force applied. find the moment and torque values of the point. this point is the critical point of the shaft in other words it is the weakness point of the shaft. this method is for continous shaft diameter, if your shaft has various diameters you should concern about stress concentrations. you can find useful papers if you write "shaft design" on google and some programs on http://www.mitcalc.com/.

      sincerely
       
    5. rushinpatel

      rushinpatel Active Member

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      use following formula for your shaft calculation,because u do not provide some required detail about shaft,



      Shear Stress in the Shaft
      When a shaft is subjected to a torque or twisting, a shearing stress is produced in the shaft. The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.

      The shear stress in a solid circular shaft in a given position can be expressed as:

      ? = T r / Ip (1)

      where

      ? = shear stress (MPa, psi)

      T = twisting moment (Nmm, in lb)

      r = distance from center to stressed surface in the given position (mm, in)

      Ip = "polar moment of inertia" of cross section (mm4, in4)

      The "polar moment of inertia" is a measure of an object's ability to resist torsion.

      Circular Shaft and Maximum Moment
      Maximum moment in a circular shaft can be expressed as:

      Tmax = ?max Ip / R (2)

      where

      Tmax = maximum twisting moment (Nmm, in lb)

      ?max = maximum shear stress (MPa, psi)

      R = radius of shaft (mm, in)

      Combining (2) and (3) for a solid shaft

      Tmax = (?/16) ?max D3 (2b)

      Combining (2) and (3b) for a hollow shaft

      Tmax = (?/16) ?max (D4 - d4) / D (2c)

      Circular Shaft and Polar Moment of Inertia
      Polar moment of inertia of a circular solid shaft can be expressed as

      Ip = ? R4/2 = ? D4/32 (3)

      where

      D = shaft outside diameter (mm, in)

      Polar moment of inertia of a circular hollow shaft can be expressed as

      Ip = ? (D4 - d4) /32 (3b)

      where

      d = shaft inside diameter (mm, in)

      Diameter of a Solid Shaft
      Diameter of a solid shaft can calculated by the formula

      D = 1.72 (Tmax/?max)1/3 (4)

      Torsional Deflection of Shaft
      The angular deflection of a torsion shaft can be expressed as

      ? = L T / Ip G (5)

      where

      ? = angular shaft deflection (radians)

      L = length of shaft (mm, in)

      G = modulus of rigidity (Mpa, psi)

      The angular deflection of a torsion solid shaft can be expressed as

      ? = 32 L T / (G ? D4) (5a)

      The angular deflection of a torsion hollow shaft can be expressed as

      ? = 32 L T / (G ? (D4- d4)) (5b)

      The angle in degrees can be achieved by multiplying the angle ? in radians with 180/?

      Solid shaft (? replaced)

      ?degrees ? 584 L T / (G D4) (6a)

      Hollow shaft (? replaced)

      ?degrees ? 584 L T / (G (D4- d4) (6b)
       
    6. rushinpatel

      rushinpatel Active Member

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      use following formula for your shaft calculation,because u do not provide some required detail about shaft,



      Shear Stress in the Shaft
      When a shaft is subjected to a torque or twisting, a shearing stress is produced in the shaft. The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.

      The shear stress in a solid circular shaft in a given position can be expressed as:

      ? = T r / Ip (1)

      where

      ? = shear stress (MPa, psi)

      T = twisting moment (Nmm, in lb)

      r = distance from center to stressed surface in the given position (mm, in)

      Ip = "polar moment of inertia" of cross section (mm4, in4)

      The "polar moment of inertia" is a measure of an object's ability to resist torsion.

      Circular Shaft and Maximum Moment
      Maximum moment in a circular shaft can be expressed as:

      Tmax = ?max Ip / R (2)

      where

      Tmax = maximum twisting moment (Nmm, in lb)

      ?max = maximum shear stress (MPa, psi)

      R = radius of shaft (mm, in)

      Combining (2) and (3) for a solid shaft

      Tmax = (?/16) ?max D3 (2b)

      Combining (2) and (3b) for a hollow shaft

      Tmax = (?/16) ?max (D4 - d4) / D (2c)

      Circular Shaft and Polar Moment of Inertia
      Polar moment of inertia of a circular solid shaft can be expressed as

      Ip = ? R4/2 = ? D4/32 (3)

      where

      D = shaft outside diameter (mm, in)

      Polar moment of inertia of a circular hollow shaft can be expressed as

      Ip = ? (D4 - d4) /32 (3b)

      where

      d = shaft inside diameter (mm, in)

      Diameter of a Solid Shaft
      Diameter of a solid shaft can calculated by the formula

      D = 1.72 (Tmax/?max)1/3 (4)

      Torsional Deflection of Shaft
      The angular deflection of a torsion shaft can be expressed as

      ? = L T / Ip G (5)

      where

      ? = angular shaft deflection (radians)

      L = length of shaft (mm, in)

      G = modulus of rigidity (Mpa, psi)

      The angular deflection of a torsion solid shaft can be expressed as

      ? = 32 L T / (G ? D4) (5a)

      The angular deflection of a torsion hollow shaft can be expressed as

      ? = 32 L T / (G ? (D4- d4)) (5b)

      The angle in degrees can be achieved by multiplying the angle ? in radians with 180/?

      Solid shaft (? replaced)

      ?degrees ? 584 L T / (G D4) (6a)

      Hollow shaft (? replaced)

      ?degrees ? 584 L T / (G (D4- d4) (6b)
       
    7. rushinpatel

      rushinpatel Active Member

      Joined:
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      use following formula for your shaft calculation,because u do not provide some required detail about shaft,



      Shear Stress in the Shaft
      When a shaft is subjected to a torque or twisting, a shearing stress is produced in the shaft. The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.

      The shear stress in a solid circular shaft in a given position can be expressed as:

      ? = T r / Ip (1)

      where

      ? = shear stress (MPa, psi)

      T = twisting moment (Nmm, in lb)

      r = distance from center to stressed surface in the given position (mm, in)

      Ip = "polar moment of inertia" of cross section (mm4, in4)

      The "polar moment of inertia" is a measure of an object's ability to resist torsion.

      Circular Shaft and Maximum Moment
      Maximum moment in a circular shaft can be expressed as:

      Tmax = ?max Ip / R (2)

      where

      Tmax = maximum twisting moment (Nmm, in lb)

      ?max = maximum shear stress (MPa, psi)

      R = radius of shaft (mm, in)

      Combining (2) and (3) for a solid shaft

      Tmax = (?/16) ?max D3 (2b)

      Combining (2) and (3b) for a hollow shaft

      Tmax = (?/16) ?max (D4 - d4) / D (2c)

      Circular Shaft and Polar Moment of Inertia
      Polar moment of inertia of a circular solid shaft can be expressed as

      Ip = ? R4/2 = ? D4/32 (3)

      where

      D = shaft outside diameter (mm, in)

      Polar moment of inertia of a circular hollow shaft can be expressed as

      Ip = ? (D4 - d4) /32 (3b)

      where

      d = shaft inside diameter (mm, in)

      Diameter of a Solid Shaft
      Diameter of a solid shaft can calculated by the formula

      D = 1.72 (Tmax/?max)1/3 (4)

      Torsional Deflection of Shaft
      The angular deflection of a torsion shaft can be expressed as

      ? = L T / Ip G (5)

      where

      ? = angular shaft deflection (radians)

      L = length of shaft (mm, in)

      G = modulus of rigidity (Mpa, psi)

      The angular deflection of a torsion solid shaft can be expressed as

      ? = 32 L T / (G ? D4) (5a)

      The angular deflection of a torsion hollow shaft can be expressed as

      ? = 32 L T / (G ? (D4- d4)) (5b)

      The angle in degrees can be achieved by multiplying the angle ? in radians with 180/?

      Solid shaft (? replaced)

      ?degrees ? 584 L T / (G D4) (6a)

      Hollow shaft (? replaced)

      ?degrees ? 584 L T / (G (D4- d4) (6b)
       
    8. SCIYER

      SCIYER Well-Known Member

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      An important aspect is missed out. i.e. Bending stress. Assuming that the shaft is supported at the ends on bearings and considering it to behave like a simply supported beam, one may calculate the direct shear, Torsional shear and the Bending stress.
      The combination all the above three, results in Principal Bending Stresses and Principal Shear Stress.
      It is to be noted that even when a member is subjected to PURE TENSILE LOAD, it results in a shear stress induced at 45 deg to the direction of the load and is called the "Principal Shear Stress".
      One may visit "Mohr's Circle Diagram" for more details.
       
    9. SCIYER

      SCIYER Well-Known Member

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      With respect to my post above, the intention was not to create such large fonts. Don't know what went wrong. I am unable to edit the post too. Hence, no hard feelings.:confused:
       
    10. GarethW

      GarethW Chief Clicker Staff Member

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      Don't worry - now corrected!
       
    11. SCIYER

      SCIYER Well-Known Member

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      Thanks !!!
       

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