Failure Theories
(5.3-5.8, 5.14)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical and Aerospace Engineering
1
Failure Theory
Example
2
Failure theory
Static Loading
Failure theories


For a given stress state (σ1, σ2, σ3) use properties from a simple tension
test (Sy, Sut) to assess the strength.
y
diameter = d
T
A
F
x

Tr 16T

J d 3
y
A
3
Failure Theory
x

F 4F

A d 2
Static Loading

For what values of T and F will the material fail if the yield
strength is Sy?

If T = 0
 d Sy
4F
 x  2  Sy  F 
(yield load)
d
4
2
For T and F non-zero, how do we calculate an equivalent
or effective stress to assess the strength?

4
Failure Theory
Ductile vs Brittle
5
Column Design
Maximum Normal Stress Theory (5.8)

Failure occurs when maximum principal stress exceeds the
ultimate strength.

Primarily applies to brittle materials
σy

Principal stresses: σ1, σ2, σ3

Define σa, σb, σc where



6
τxy
y
σa = max (σ1, σ2, σ3)
σc = min (σ1, σ2, σ3)
σb is the value in between
x
Failure Theory
σx
Maximum Normal Stress Theory (5.8)
For failure



σa = Sut if σa > 0 (where Sut is ultimate strength in tension)
σc = -Suc if σc < 0 (where Suc is ultimate strength in
compression)
Case I: Uni-axial tension (bar, σx = σo = P/A, σy = τxy =
0)




7
σ1 = σo, σ2 = 0 & σ3 = 0 (plane stress)
σa = σo, σb = 0, σc = 0
Failure when σo = Sut 
Failure Theory
Maximum Normal Stress Theory (5.8)
Case 2: Pure torsion (shaft, τxy = Tr/J, σx = σy = 0)




σ1 = +τxy, σ2 = -τxy & σ3 = 0 (plane stress)
σa = +τxy, σb = 0, σc = -τxy
Failure when τ xy = Sut or τ xy = Suc 

Does not agree with experimental data.

Experimental data would show that failure occurs when
τ ≈ 0.6 Sut.
8
Failure Theory
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.
 max
9
Sy
1
 ( a   c ) 
2
2
Failure Theory
Maximum Shear Stress Theory (5.4)

Case I: Uni-axial tension (bar, σx = σo = P/A, σy = τxy = 0)





σ1 = σo, σ2 = 0 & σ3 = 0 (plane stress)
σa = σo, σb = 0, σc = 0
τmax = σo/2 = Sy/2
Failure when σo = Sy 
Case 2: Pure torsion (shaft, τxy = Tr/J, σx = σy = 0)




10
σ1 = +τxy, σ2 = -τxy & σ3 = 0 (plane stress)
σa = +τxy, σb = 0, σc = -τxy
τmax = τxy
Failure when τ xy = Sy/2 
Failure Theory
Distortion Energy Theory (5.5)

von Mises theory

Failure occurs when
( 1   2 ) 2  ( 1   3 ) 2  ( 2   3 ) 2  2S y

Applies to ductile materials.
11
Failure Theory
2
Distortion Energy Theory (5.5)

Case I: Uni-axial tension (bar, σx = σo = P/A, σy = τxy = 0)




Case 2: Pure torsion (shaft, τxy = Tr/J, σx = σy = 0)






σ1 = σo, σ2 = 0 & σ3 = 0 (plane stress)
σo2 + σo2 = 2Sy2
Failure when σo = Sy 
σ1 = +τxy, σ2 = -τxy & σ3 = 0 (plane stress)
(τxy + τxy)2 + τxy2 + τxy2 = 2Sy2
6 τxy2 = 2Sy2
Failure when τ xy = 0.577Sy 
Agrees very closely with experiments!
Section 5.14 in the textbook summarizes failure theories.
12
Failure Theory
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.
y
Stress point
d
T
x
P
13
Failure Theory
+
Stress Concentration Factor
(3.13)
MAE 316 – Strength of Mechanical Components
NC State Department of Mechanical and Aerospace Engineering
14
Stress Concentration Factor
Examples
15
Stress Concentration Factor
Stress Concentration Factor (3.13)

Consider the following two stress analysis problems:
h
P
b
P
P P
 
A bh
h
d
P
b
P
 avg 
16
P
P

A (b  d )h
Stress Concentration Factor
Stress Concentration Factor (3.13)

For a plate with a hole, the maximum stress occurs
around the hole.
17
Stress Concentration Factor
Stress Concentration Factor (3.13)

Maximum stress is defined using a stress concentration
factor, Kt.
 max  Kt avg
18
Stress Concentration Factor
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.
19
Stress Concentration Factor
Stress Concentration Factor (3.13)

What if the hole is elliptical?
σo
Kt  1  2
b
a
 max  K t avg
σmax
2a
2b
b
a
1 3
2 5
10 21
 
σo
20
Kt
Stress Concentration Factor
(hole)
(crack)
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.
σo
σo
 o  , max  
21
Stress Concentration Factor
Stress Concentration Factor (3.13)

Other types of stress concentrations (Appendix A-15)




22
Plate with fillet
Plate with notch
Shaft with fillet
Grooved shaft
Stress Concentration Factor
Fracture Mechanics
(5.12, 5.14)
MAE 316 – Strength of Mechanical Components
NC State University Department of Mechanical and Aerospace Engineering
23
Fracture Mechanics
Fracture Mechanics (5.12)
y
P
b
x
h
L
Mc
 bending 

I
( PL)(h / 2) 6 PL
 2
3
bh /12
bh
Failure will occur when:
 bending
24
bh2 S y
6 PL
 Sy  2  Sy  P 
bh
6L
Fracture Mechanics
Fracture Mechanics (5.12)
y
P
crack
 max  K t bending
b
a
x
h
 max
6 PL
 Kt
bh 2
L

For a sharp crack, Kt → ∞, σmax → ∞.

The conclusion is that P > 0 will lead to failure, but this is
not reasonable.
25
Fracture Mechanics
Stress Intensity Factor (5.12)
Figure 5-23 Crack deformation types: (a) mode I,
opening; (b) mode 2, sliding; (c) mode III, tearing
26
Fracture Mechanics
Stress Intensity Factor (5.12)

In fracture mechanics, design analysis is based not on stress,
but stress intensity factor.
K    a
where
  stress intensity modification factor
  normal stress
a  crack length (or half length)


Stress intensity modification factors vary depending on load
and geometry.
Refer to Figures 5-25 through 5-29 in the textbook.
27
Fracture Mechanics
Stress Intensity Factor (5.12)

So, for the cracked
plate shown previously
 6 PL 
K  1.12  2   a
 bh 
Figure 5-26
29
Fracture Mechanics
Fracture Toughness (5.12)

Failure will occur when K ≥ KIc (KIc is fracture toughness,
a material property).
K  K Ic
 6 PL 
1.12  2   a  K Ic
 tb 
tb 2 K Ic
P
1.12  a (6 L)

“Failure” means the crack extends unstably and the
structure fractures (i.e. breaks).
30
Fracture Mechanics
Fracture Toughness (5.12)

In fracture mechanics, factor of safety can also be
expressed as
K Ic
K Ic
n

K
  a
31
Fracture Mechanics
Fracture Toughness (5.12)

Two different analysis methods
Stress Analysis Material

Traditional Design

Sy
Fracture Mechanics Design
K
K Ic
To design conservatively for safety, we must do both
analyses.
32
Fracture Mechanics
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
y
P
crack
b
a
x
L
33
Fracture Mechanics
h
Example

Find the stress intensity factor for a plate with a center
crack if the average normal stress in the plate is 10 ksi.

34
Given: 2a = 3 in and 2b = 10 in, d=4 in
Fracture Mechanics
Example

Find the stress intensity factor for a plate with an edge
crack if the average normal stress in the plate is 10 ksi.

35
Given: a = 3 in and b = 10 in
Fracture Mechanics
36
Fracture Mechanics
37
Fracture Mechanics
A Real-Life Example of Fracture
38
Fracture Mechanics
A Real-Life Example of Fracture
39
Fracture Mechanics