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					                                                               31-Oct-11




            FAILURE THOERIES FOR STATIC
                     LOADING




                 What is Failure?
• Failure – any change in a machine part which
  makes it unable to perform its intended
  function.

• We will normally use a yield failure criteria for
  ductile materials. The ductile failure theories
  presented are based on yield.


31-Oct-11           Machine Design, Eng. Mahmoud Refaat   14




                                                                      1
                                                                         31-Oct-11




                        Failure Theories
• Static failure
      – Ductile
      – Brittle
      – Stress concentration

• Recall
      – Ductile
            • Significant plastic
              deformation between
              yield and fracture
      – Brittle
            • Yield ~= fracture


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                             Tensile Test
•




31-Oct-11                     Machine Design, Eng. Mahmoud Refaat   16




                                                                                2
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            Mohr’s Circle for Tensile Test
•




31-Oct-11                    Machine Design, Eng. Mahmoud Refaat   17




                  Static Ductile Failure
• Two primary theories for static ductile failure
      – Von Mises criterion
            • Maximum Distortion-energy Theory
            • MDE
      – Maximum Shear Stress criterion
            • MSS (TRESCA)




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                                                                               3
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                 Static Ductile Failure
• Max Shear Stress criterion
      – Material yields (fails) when:

                           Sy
            1)    max 
                           2                 or
            2)    1   3   S y
                                                  Sy            Sy
      – Factor of Safety:          F .S                   
                                             1   3        2 max




31-Oct-11                       Machine Design, Eng. Mahmoud Refaat     19




       Maximum Shear Stress Criteria
•




31-Oct-11                       Machine Design, Eng. Mahmoud Refaat     20




                                                                                    4
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                     Static Ductile Failure
• Von Mises criterion
      – Let the Mises stress (e, equivalent stress) be:


                  e 
                            1
                            2
                              
                               1   2 2   2   3 2   1   3 2   
      – Then failure (yield) occurs when:                          e  Sy
                                              Sy
      – Factor of Safety:            F .S 
                                              e

            • Typically, 1.25  F .S  4
            • Want a margin of error but not completely overdesigned

31-Oct-11                          Machine Design, Eng. Mahmoud Refaat            21




                    Which theory to use?
•   Look at a plot of the principal
    stresses
      – B vs. A
      – The non-zero principal stresses
•   Failure occurs when the principal
    stresses lie outside the enclosed area
•   Shape of area depends on the failure
    theory
•   Data points are experimental results
•   MSS
      – Slightly more conservative
      – Easier to calculate
•   MDE
      – More accurate
      – If not specified, use this one!




31-Oct-11                          Machine Design, Eng. Mahmoud Refaat            22




                                                                                              5
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       Comparison of MDE, MSS,MNS
•




31-Oct-11                          Machine Design, Eng. Mahmoud Refaat   23




        Ductile failure theory example
• Given:
      – Bar is AISI 1020 hot-rolled
        steel
            • A DUCTILE material
      – F = 0.55 kN
      – P = 8.0 kN
      – T = 30 Nm
• Find:
      – Factor of safety ()
• Two areas of interest:
      – A
            • Top – where max normal
              stress is seen (bending!)
      – B
            • Side – where max shear
              stress is seen



31-Oct-11                          Machine Design, Eng. Mahmoud Refaat   24




                                                                                     6
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                                Element A
• Consider the types of loading we
  have
• Axial?
      – Yes – due to P
• Bending?
      – Recall that bending produces 
        and , depending on the element
        of interest
      – Yes – due to M ( at A,  at B)
      –
• Torsion?
      – Yes – due to T

31-Oct-11                      Machine Design, Eng. Mahmoud Refaat   25




                                Element A
• Calculate stresses due to each load
• Axial:
                      P     P       4P
               x               
                      A  D  D 2
                              2
                          
                           4  
                               
• Bending:
                         FL  D 
                               
               x 
                    My
                              2   32 FL
                     I     D 4      D 3
                          
                           64  
                                
• Shear:        xy  0

• Torsion:                  T  D 
                                 
                 xz 
                       Tc
                                2   16T
                       J      D 4  D 3
                              32 
                                   
                                   

31-Oct-11                      Machine Design, Eng. Mahmoud Refaat   26




                                                                                 7
                                                                                        31-Oct-11




                                        Element A
• Look at a stress element
• Sum up stresses due to all the
  loads
•      4 P 32 FL 4 PD  32 FL
     x                        
            D 2       D3                   D3
•
              16T
      xz 
              D 3
• x = 95.5 MPa
• xz = 19.1 MPa


31-Oct-11                               Machine Design, Eng. Mahmoud Refaat        27




                                        Element A
• Draw Mohr’s Circle with the
  stresses that we calculated
      – x = 95.5 MPa
      – xz = 19.1 MPa
      – x at (x, xz)
            • (95.5, 19.1)
      – y at (y, zx)
            • (y, -xz)
            • (0, -19.1)
      – Find C
            •   x   y   95.5  0 
               2 ,0    2 ,0   47.8,0
                         
                                    

      – Find radius
            • R      x  Cx 2   xz 2      95.5  47.82  19.12    51.4


31-Oct-11                               Machine Design, Eng. Mahmoud Refaat        28




                                                                                               8
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 Out of Plane Maximum Shear for Biaxial State of Stress
   • Case 1                        Case 2                            Case 3
      – 1,2 > 0                     2,3 < 0                          1 > 0, 3 < 0

                                     1 = 0                            2 = 0
      – 3 = 0
                                                                                3
      – max   1
                                                                      
                                       max  3                           max  1
                  2                               2                               2




31-Oct-11                    Machine Design, Eng. Mahmoud Refaat                          29




                              Element A
• Find principal stresses
      – 1 = C + R
             • 99.2 MPa
      – 2 = C - R
             • -3.63 MPa
      –     Think about 3-D Mohr’s Circle!
      –     This is Case #3…
      –     We want 1 > 2 > 3
      –     Assign 2 = 0 and 3 = -3.63 MPa
• No failure theory was given, so
  use MDE


31-Oct-11                    Machine Design, Eng. Mahmoud Refaat                          30




                                                                                                      9
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                                        Element A
• Find the von Mises stress (e)

            e 
                    1
                    2
                        
                       1   2 2   2   3 2   1   3 2   
            e 
                    1
                    2
                        
                      99.2  02  0  3.632  99.2  3.632          
             e  101MPa


• Sy for our material = 331 MPa
• Calculate the factor of safety
      –            Sy       331
                              3.28 For yield
                 e         101


31-Oct-11                              Machine Design, Eng. Mahmoud Refaat    31




                                        Element B
• Consider the types of loading we have
• Axial?
      – Yes – due to P
• Bending?
      – Recall that bending produces  and ,
        depending on the element of interest
      – Yes – due to M ( at A,  at B)
• Torsion?
      – Yes – due to T




31-Oct-11                              Machine Design, Eng. Mahmoud Refaat    32




                                                                                         10
                                                                               31-Oct-11




                                     Element B
• Calculate stresses due to each load
• Axial:
                          P     P      4P
                 x                
                          A  D 2  D 2
                              
                               4 
                                  
• Bending:
      – Use equation for round solid cross-section

                 xy 
                         VQ 4V
                                
                                     4 F   16F
                          Ib 3 A     D 2  3D 2
                                     4 
                                   3      
                                          
• Shear:          xy    0

                                T  D 
                                     
• Torsion:         xy   
                           Tc
                                    2   16T
                           J      D 4  D 3
                                 
                                  32  
                                       

31-Oct-11                           Machine Design, Eng. Mahmoud Refaat   33




                                     Element B
• Look at a stress element
• Sum up stresses due to all the loads
•
                4P
      x 
•              D 2
               16F 16T
       xy               19.1  .002
               3D 2 D 3
• x = 25.5 MPa
• xy = 19.1 MPa
      – Note small contribution of shear stress
        due to bending



31-Oct-11                           Machine Design, Eng. Mahmoud Refaat   34




                                                                                     11
                                                                                                            31-Oct-11




                                         Element B
• Draw Mohr’s Circle with the
  stresses that we calculated
      – x = 25.5 MPa
      – xy = 19.1 MPa
      – x at (x, xy)
            • (25.5, 19.1)
      – y at (y, yx)
            • (y, -xy)
            • (0, -19.1)
      – Find C
            •   x   y   25.5  0 
               2 ,0    2 ,0   12.8,0
                         
                                    

      – Find radius
            •
                R    x  Cx 2   xz 2      25.5  12.82  19.12    22.96


31-Oct-11                               Machine Design, Eng. Mahmoud Refaat                            35




 Out of Plane Maximum Shear for Biaxial State of Stress
   • Case 1                                     Case 2                            Case 3
      – 1,2 > 0                                  2,3 < 0                          1 > 0, 3 < 0

                                                  1 = 0                            2 = 0
      – 3 = 0
                                                                                             3
      – max   1
                                                                                   
                                                    max  3                           max  1
                     2                                       2                                 2




31-Oct-11                               Machine Design, Eng. Mahmoud Refaat                            36




                                                                                                                  12
                                                                                         31-Oct-11




                                      Element B
• Find principal stresses
      – 1 = C + R
             • 35.8 MPa
      – 2 = C - R
             • -10.2 MPa
      –     Think about 3-D Mohr’s Circle!
      –     This is Case #3…
      –     We want 1 > 2 > 3
      –     Assign 2 = 0 and 3 = -10.2 MPa
• No failure theory was given, so
  again use MDE


31-Oct-11                             Machine Design, Eng. Mahmoud Refaat           37




                                      Element B
• Find the von Mises stress (e)

            e 
                      1
                      2
                          
                         1   2 2   2   3 2   1   3 2       
            e 
                    1
                    2
                          
                      35.8  02  0  10.22  35.8  10.22                
             e  41.8MPa
• Sy for our material = 331 MPa
• Calculate the factor of safety
      –          Sy       331
                             7.91 For yield
                 e       41.8


31-Oct-11                             Machine Design, Eng. Mahmoud Refaat           38




                                                                                               13
                                                                      31-Oct-11




                    Example, concluded
• We found the factors of
  safety relative to each
  element, A and B
      – A – 3.28
      – B – 7.91
• A is the limiting factor of
  safety
      – F.S = 3.3




31-Oct-11                  Machine Design, Eng. Mahmoud Refaat   39




                                     Quiz
• Given:
      – Shaft of st50 subject to
        loading shown
• Find:
      – For a factor of safety of
        F.S = 2.8, what should
        the diameter of the
        shaft (d) be?




31-Oct-11                  Machine Design, Eng. Mahmoud Refaat   40




                                                                            14
                                                                              31-Oct-11




Solution
• First, we need to find the forces
  acting on the shaft
      – Torque on shaft from pulley at B
              • TB = (300-50)(4) = 1000 in·lb
      – Torque on shaft from pulley at C
              • TC = (360-27)(3) = 1000 in·lb
      – Shaft is in static equilibrium
      – Note that shaft is free to move
        along the x-axis (bearings)
• Draw a FBD
      – Reaction forces at points of
        attachment to show constrained
        motion


31-Oct-11                          Machine Design, Eng. Mahmoud Refaat   41




                                   Equilibrium
•   Use statics to solve for reactions forces
      –     RAy = 222 lb
      –     RAz = 106 lb
      –     RDy = 127 lb
      –     RDz = 281 lb
•   OK, now we know all the forces. The
    problem gives us a factor of safety, but unlike
    our last example, we aren’t told specific
    places (elements) at which to look for
    failure!
•   We are going to have to calculate stresses
•   What do we need?
      – Axial forces, bending moments, and torques
      – We need to find our moments… HOW?
      – Shear-Moment diagrams will give us the
        forces and moments along the shaft.
      – Failure will likely occur where the max values
        are seen




31-Oct-11                          Machine Design, Eng. Mahmoud Refaat   42




                                                                                    15
                                                                     31-Oct-11




        Torsion and moment diagrams
• Let’s look at torsion and
  how it varies across the
  shaft
      – We calculated the torques
        at B and C to be 1000 in·lb
        each
      – Plot that along the shaft
        and we see that max
        torque occurs at B and C
        (and all points between)



31-Oct-11                 Machine Design, Eng. Mahmoud Refaat   43




        Torsion and moment diagrams
• Now let’s look at the
  moments
• We have a 3-D loading
      – How are we going to do
        the V-M diagrams?
      – Look at one plane at a time
• Moment in the x-y plane
      – From geometry you can
        calculate the values of the
        moment at B and C




31-Oct-11                 Machine Design, Eng. Mahmoud Refaat   44




                                                                           16
                                                                                                  31-Oct-11




          Torsion and moment diagrams
• Moment in the x-z plane
      – Failure is going to occur at either
        B or C, since these are locations
        where maximum moments are
        seen
      – But we have moments in both
        planes
      – To find the max bending stresses,
        we must find the total maximum
        moment
      – Just as we would vectorally add
        the two force components to
        find the force magnitude, we can
        vectorally add the two moment
        components to find the moment
        magnitude
              M  M xy  M xz
                                  2    2




31-Oct-11                             Machine Design, Eng. Mahmoud Refaat                    45




               Calculate the max moment


•   We found the following:
      –     MB x-y = 1780 in·lb
      –     MB x-z = 848 in·lb
      –     MC x-y = 762 in·lb
      –     MC x-z = 1690 in·lb
•                                                M  M xy  M xz
                                                                2           2
    Calculating the magnitudes with
      – MB = 1971.7 in·lb
      – MC = 1853.8 in·lb
•   Since the overall max moment is at B, we will expect failure there, and use M B in our
    stress calculations
•   If we had been told the location of interest, we would essentially start here.



31-Oct-11                             Machine Design, Eng. Mahmoud Refaat                    46




                                                                                                        17
                                                                                         31-Oct-11




                Calculate the stresses at B
• Bending stress ( and )
      – We know from experience that  is the predominant stress,
        so essentially we will look for failure at an element at the top
        of the shaft                                    My
                                                                          
              • M = 1971                                                       I
              • Plug in known values                                      y
                                                                             d
              • max = (20x103)/d3                                           2
                                                                             d 4
• Torsional stress                             Tc
                                                                          I
                                                                            64
      – T = 1000                               J
                                               d
                                            c
      –  = (5.1x103)/d3                       2
                                               d 4
                                            J
                                                32

31-Oct-11                           Machine Design, Eng. Mahmoud Refaat             47




                                   Mohr’s Circle
•   Let’s look at our stress element
•   Now construct Mohr’s circle
      –     C at (10 x 103)/d3
      –     R = (11.2 x 103)/d3
      –     1 = (21.2 x 103)/d3
      –     3 = (-1.2 x 103)/d3
•   Use TRESCA to find the diameter.




31-Oct-11                           Machine Design, Eng. Mahmoud Refaat             48




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