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```					                                                               31-Oct-11

FAILURE THOERIES FOR STATIC

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

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

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

31-Oct-11                     Machine Design, Eng. Mahmoud Refaat   15

Tensile Test
•

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Mohr’s Circle for Tensile Test
•

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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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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
•

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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

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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

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Element A
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

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Element A
• Look at a stress element
• Sum up stresses due to all the
•      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
           
                      

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

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

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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

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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
• 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

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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

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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
           
                      

•
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

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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

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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
• Find:
– For a factor of safety of
F.S = 2.8, what should
the diameter of the
shaft (d) be?

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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

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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
– 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

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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
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

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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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