Failure Theories (5.3-5.8, 5.14)

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Failure Theories(5.3-5.8, 5.14)

• MAE 316 Strength of Mechanical Components
• NC State University Department of Mechanical and
  Aerospace Engineering

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Example
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Static Loading

• Failure theories
• For a given stress state (s1, s2, s3) use
  properties from a simple tension test (Sy, Sut)
  to assess the strength.

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Static Loading

• For what values of T and F will the material fail
  if the yield strength is Sy?
• If T 0
• For T and F non-zero, how do we calculate an
  equivalent or effective stress to assess the
  strength?

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Ductile vs Brittle
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Maximum Normal Stress Theory (5.8)

• Failure occurs when maximum principal stress
  exceeds the ultimate strength.
• Primarily applies to brittle materials
• Principal stresses s1, s2, s3
• Define sa, sb, sc where
• sa max (s1, s2, s3)
• sc min (s1, s2, s3)
• sb is the value in between

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Maximum Normal Stress Theory (5.8)

• For failure
• sa Sut if sa gt 0 (where Sut is ultimate
  strength in tension)
• sc -Suc if sc lt 0 (where Suc is ultimate
  strength in compression)
• Case I Uni-axial tension (bar, sx so P/A, sy
  txy 0)
• s1 so, s2 0 s3 0 (plane stress)
• sa so, sb 0, sc 0
• Failure when so Sut ?

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Maximum Normal Stress Theory (5.8)

• Case 2 Pure torsion (shaft, txy Tr/J, sx sy
  0)
• s1 txy, s2 -txy s3 0 (plane stress)
• sa txy, sb 0, sc -txy
• Failure when t xy Sut or t xy Suc ?
• Does not agree with experimental data.
• Experimental data would show that failure occurs
  when t 0.6 Sut.

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Maximum Shear Stress Theory (5.4)

• Tresca yield criterion
• Failure occurs when the maximum shear stress
  exceeds the yield strength (max shear stress in a
  tension test is Sy/2).
• Applies to ductile materials.

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Maximum Shear Stress Theory (5.4)

• Case I Uni-axial tension (bar, sx so P/A, sy
  txy 0)
• s1 so, s2 0 s3 0 (plane stress)
• sa so, sb 0, sc 0
• tmax so/2 Sy/2
• Failure when so Sy ?
• Case 2 Pure torsion (shaft, txy Tr/J, sx sy
  0)
• s1 txy, s2 -txy s3 0 (plane stress)
• sa txy, sb 0, sc -txy
• tmax txy
• Failure when t xy Sy/2 ?

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Distortion Energy Theory (5.5)

• von Mises theory
• Failure occurs when
• Applies to ductile materials.

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Distortion Energy Theory (5.5)

• Case I Uni-axial tension (bar, sx so P/A, sy
  txy 0)
• s1 so, s2 0 s3 0 (plane stress)
• so2 so2 2Sy2
• Failure when so Sy ?
• Case 2 Pure torsion (shaft, txy Tr/J, sx sy
  0)
• s1 txy, s2 -txy s3 0 (plane stress)
• (txy txy)2 txy2 txy2 2Sy2
• 6 txy2 2Sy2
• Failure when t xy 0.577Sy ?
• Agrees very closely with experiments!
• Section 5.14 in the textbook summarizes failure
  theories.

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Example

• Find the minimum allowable diameter, with a
  factor of safety of 2, using both Tresca and von
  Mises formulas. Assume Sy 50,000 psi, P 500
  lbs, T 1000 in-lb, and L 5 in.

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Stress Concentration Factor(3.13)

• MAE 316 Strength of Mechanical Components
• NC State Department of Mechanical and Aerospace
  Engineering

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Examples
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Stress Concentration Factor (3.13)

• Consider the following two stress analysis
  problems

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Stress Concentration Factor (3.13)

• For a plate with a hole, the maximum stress
  occurs around the hole.

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Stress Concentration Factor (3.13)

• Maximum stress is defined using a stress
  concentration factor, Kt.

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Example

• For a plate with w 2.0 in. and t 1.0 in.
  subject to a 50,000 lb axial load, find the
  maximum stress for d 0, 0.1 in., 0.5 in., and
  1.0 in.

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Stress Concentration Factor (3.13)

• What if the hole is elliptical?

(hole)
(crack)
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Stress Concentration Factor (3.13)

• This suggests that structures with sharp cracks
  could not sustain any level of applied stress
  without failure.
• This cannot be correct fracture mechanics
  analysis will resolve this.

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Stress Concentration Factor (3.13)

• Other types of stress concentrations (Appendix
  A-15)
• Plate with fillet
• Plate with notch
• Shaft with fillet
• Grooved shaft

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Fracture Mechanics(5.12, 5.14)

• MAE 316 Strength of Mechanical Components
• NC State University Department of Mechanical and
  Aerospace Engineering

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Fracture Mechanics (5.12)
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Fracture Mechanics (5.12)

• For a sharp crack, Kt ? 8, smax ? 8.
• The conclusion is that P gt 0 will lead to
  failure, but this is not reasonable.

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Stress Intensity Factor (5.12)

• Figure 5-23 Crack deformation types (a) mode I,
  opening (b) mode 2, sliding (c) mode III,
  tearing

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Stress Intensity Factor (5.12)

• In fracture mechanics, design analysis is based
  not on stress, but stress intensity factor.
• Stress intensity modification factors vary
  depending on load and geometry.
• Refer to Figures 5-25 through 5-29 in the
  textbook.

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Stress Intensity Factor (7.3)

• a/w ?
• 0 (w?8) 1.12
• 0.2 1.37
• 0.4 2.11
• 0.5 2.83

Table 7.1 Case B

• So, for the cracked plate shown previously

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Stress Intensity Factor (5.12)

• So, for the cracked plate shown previously

Figure 5-26
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Fracture Toughness (5.12)

• Failure will occur when K KIc (KIc is fracture
  toughness, a material property).
• Failure means the crack extends unstably and
  the structure fractures (i.e. breaks).

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Fracture Toughness (5.12)

• In fracture mechanics, factor of safety can also
  be expressed as

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Fracture Toughness (5.12)

• Two different analysis methods
• To design conservatively for safety, we must do
  both analyses.

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Example

• The cracked plate shown below is made of 4340
  steel has Sy 240 ksi and KIc 50 ksi(in)1/2.
  Find the maximum allowable load, P, that can be
  applied to the beam without failure.
• Given b 1 in, h 2 in, L 24 in, a 0.25 in

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Example

• Find the stress intensity factor for a plate with
  a center crack if the average normal stress in
  the plate is 10 ksi.
• Given 2a 3 in and 2b 10 in, d4 in

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Example

• Find the stress intensity factor for a plate with
  an edge crack if the average normal stress in the
  plate is 10 ksi.
• Given a 3 in and b 10 in

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A Real-Life Example of Fracture
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A Real-Life Example of Fracture