MEP 209: Design of Machine elements LEC 4

MPE 209: Dr Mohamed Elfarran

Static failure theories
• Why do parts fail?
• You may say “Parts fail because their stresses exceed
their strength”
• Then what kind of stresses cause the failure: Tensile?
Compressive? Shear?
• Answer may be: It depends.
• It depends on the material and its relative strength in
compressive, tension, and shear.
• It also depends on the type loading (Static, Fatigue,
Impact) and
• presence of the cracks in the material
Page 2

• The failure may be elastic or fracture
• Elastic failure results in excessive
deformation, which makes the machine
component unfit to perform its function
satisfactorily
• Fracture results in breaking the component
into two parts
Page 3

• Question: How do one compare stresses
induced to the material properties?
– Generally machine parts are subjected to combined
loading and to find material properties under real
loading condition is practically not economical
– Thus, material properties are obtained from simple
tension/torsion test
– These data like Syt, Sut etc are available in form of
table (Design Data Book)
Page 4

• Theories of failure provide a relationship between the
strength of machine component subjected to complex
state of stress with the material properties obtained
from simple test (Tensile)
Strength of machine
component subjected
to complex state of
stress
Strength of
standard component
subjected to
uniaxial state of
stress
Page 5

• Loads are assumed to not vary over time
• Failure theories that apply to:
– Ductile materials
– Brittle materials
• Why do we need different theories ??
Stress-strain curve of a ductile material Stress-strain curve of a brittle material
Page 6

Static failure theories: Tension test
Why nearly 0o ??
Failure along
principal shear stress
plane
Failure along
principal normal
stress plane
Cast iron has C
between 2.1% to 4%
and Si between 1%
and 3% C contents
less than 2.1% are
steels.
Page 7

Static failure theories:
Compression test
Why doesn’t it fail ?? Why does it fail ??
Why nearly 45o ??
Does not
“fail”
Shear failure
Page 8

Failure along principal shear stress plane Failure along principal normal stress plane
Page 9

• In general, ductile, isotropic materials are limited by
their shear strengths.
• Brittle materials are limited by their tensile strengths.
• If cracks are present in a ductile material, it can
suddenly fracture at nominal stress levels well below
its yield strength, even under static loads.
• Static loads are slowly applied and remain constant
with time.
• Dynamic loads are suddenly applied (impact), or
repeatedly varied with time (fatigue), or both.
Page 10

• In dynamic loading, the distinction between
failure mechanisms of ductile and brittle
materials blurs.
• Ductile materials often fail like brittle
materials in dynamic loading.
Page 11

Accepted failure theories that apply to ductile materials:
Accepted failure theories that apply to brittle materials:
Page 12
• Maximum normal stress theory (even material)
• Maximum normal stress theory (uneven material)
• Coulomb-Mohr theory
• Modified Mohr theory
• To
istortion energy the
tal strain energy theory
• D ory
• Pure shear-stress theory
• Maximum shear-stress theory
• Maximum normal stress theory (limited application)

Maximum Normal Stress Theory
• Credited to the English scientist and educator
W. J. M. Rankine (1802–1872)
Page 13

• This predicts that failure of machine component, subjected to complex loading,
occurs if the maximum normal principal stress tends to exceeds the uniaxial
tensile yield (ductile) or the ultimate tensile strength (brittle) of the material
• Providing a square failure boundary with Sy as the principal stress for ductile
materials
• Note: not a safe theory for ductile materials
Exercise: Draw
failure envelop
for two
dimension
case?
Page 14

• Let 1, 2, 3 are the principal stresses at the
critical point in component due to applied
loading, and let 1> 2> 3
• Then failure occurs when
𝜎1 ≥ 𝑆 𝑦𝑡 𝑜𝑟 𝜎1 ≥ 𝑆 𝑢𝑡
If we want to include uncertainty of data available and loads
acting on the component, i.e., factor of safety, then to avoid
failure:
𝜎1 =
𝑆 𝑦𝑡
𝐹𝑆
𝑜𝑟 𝜎1 =
𝑆 𝑢𝑡
𝐹𝑆
Page 15

Maximum Shear-Stress Theory
• The oldest failure theory, originally proposed
by the great French scientist C. A. Coulomb
(1736–1806)
• Tresca modified it in 1864, and J. J. Guest
• Validated by experiments around 1900
• For these reasons the maximum-shear-stress
theory is sometimes called the Tresca-Guest
theory
Page 16

Maximum Shear-Stress Theory
This theory states that a material subjected to any combination of loads will fail (by
yielding or fracturing) whenever the maximum shear stress exceeds the shear strength
(yield or ultimate) of the material. The shear strength, in turn, is usually assumed to be
determined from the standard uniaxial tension test.
Exercise: Draw
hexagonal
failure envelop
for two
dimesion case?
Page 17

This theory states that failure occurs when:
Ductile materials: maximum shear-stress theory
𝑆 𝑦𝑠 = 0.5𝑆 𝑦
(Failure occurs when maximum
shear stress
exceeds the shear stress at
yield in pure tension)
Mohr’s circle:
pure tension
𝑆𝐹 =
𝑆 𝑦𝑠
𝜏 𝑚𝑎𝑥
𝜏 𝑚𝑎𝑥 ≤ 𝑆 𝑦𝑠
Page 18

Ductile materials: maximum shear-stress theory
Providing a hexagonal failure envelope that is more conservative than
the distortion energy theory
Page 19

Ductile materials
Total strain energy U:
Elastic range
assuming stress strain
curve is linear upto
yield point
𝑈 =
1
2
𝜎𝜀
𝑈 =
1
2
𝜎𝜀 =
1
2
(𝜎1 𝜀1 + 𝜎2 𝜀2 + 𝜎3 𝜀3)
Principal stresses and
strains
𝜀1 =
1
𝐸
(𝜎1 − 𝜐𝜎2 − 𝜐𝜎3
𝜀1 =
1
𝐸
(𝜎2 − 𝜐𝜎3 − 𝜐𝜎1
𝜀1 =
1
𝐸
(𝜎3 − 𝜐𝜎1 − 𝜐𝜎2
where
Page 20

Ductile materials: total strain energy
Using previous expressions, total energy is:
𝑈 =
1
2
𝜎𝜀 =
1
2𝐸
[𝜎1
2
+ 𝜎2
2
+ 𝜎3
2
− 2𝜐(𝜎1 𝜎2 + 𝜎2 𝜎3 + 𝜎3 𝜎1)
which can be expressed as 𝑈 = 𝑈ℎ + 𝑈 𝑑
Hydrostatic energy Deformation energy
𝑈ℎ =
3
2
(1 − 2𝜐)
𝐸
𝜎ℎ
2
𝜎ℎ =
𝜎1 + 𝜎2 + 𝜎3
3
Obtained by setting:
𝑈ℎ = 𝑈(𝜎1 = 𝜎2 = 𝜎3 = 𝜎ℎ)
𝑈 𝑑 =
1 + 𝜐
3𝐸
[𝜎1
2 + 𝜎2
2 + 𝜎3
2
−(𝜎1 𝜎2 + 𝜎2 𝜎3 + 𝜎3 𝜎1)]
Obtained by setting:
𝑈 𝑑 = 𝑈 − 𝑈ℎ
Page 21

Ductile materials: distortion energy theory
𝑈 𝑑 =
1 + 𝜐
3𝐸
[𝜎1
2 + 𝜎2
2 + 𝜎3
2 − (𝜎1 𝜎2 + 𝜎2 𝜎3 + 𝜎3 𝜎1
If uniaxial yield stress state (failure state):
𝜎1 = 𝑆 𝑦
𝜎2 = 0
𝜎3 = 0
Therefore:
𝑈 𝑑 =
1 + 𝜐
3𝐸
𝑆 𝑦
2 Using uniaxial yield
stress state (failure
state)
Page 22

𝑈 𝑑 =
1 + 𝜐
3𝐸
[𝜎1
2 + 𝜎2
2 + 𝜎3
2 − 𝜎1 𝜎2 + 𝜎2 𝜎3 + 𝜎3 𝜎1 ]
For any other state of stresses:
Failure criterion is obtained by setting:
1 + 𝜐
3𝐸
[𝜎1
2
+ 𝜎2
2
+ 𝜎3
2
− (𝜎1 𝜎2 + 𝜎2 𝜎3 + 𝜎3 𝜎1] ≤
1 + 𝜐
3𝐸
𝑆 𝑦
2
Distortion energy:
uniaxial stress at
yield
Distortion energy:
any other state of
stresses
𝜎1
2 + 𝜎2
2 + 𝜎3
2 − 𝜎1 𝜎2 − 𝜎2 𝜎3 − 𝜎3 𝜎1 ≤ 𝑆 𝑦
2
Page 23

𝑆 𝑦
2 = 𝜎1
2 + 𝜎2
2 + 𝜎3
2 − 𝜎1 𝜎2 − 𝜎2 𝜎3 − 𝜎3 𝜎1
For a 2D stress where 𝜎2= 0, 𝑆 𝑦
2 = 𝜎1
2 + 𝜎3
2 − 𝜎1 𝜎3
• The 2D distortion
energy equation is
described in an ellipse
• The interior of the
ellipse show the biaxial
safe stress sage against
yielding under static
loads
Page 24

Von Mises effective stress
𝑆 𝑦
2 = [𝜎1
2 + 𝜎2
2 + 𝜎3
2 − 𝜎1 𝜎2 − 𝜎2 𝜎3 − 𝜎3 𝜎1] ≡ 𝜎′ 2
Definition:
𝜎′ = 𝜎1
2
+ 𝜎2
2
+ 𝜎3
2
− 𝜎1 𝜎2 − 𝜎2 𝜎3 − 𝜎3 𝜎1
(Yield surface)
von Mises effective stress
von Mises effective stress: uniaxial stress that would create the
same distortion energy as is created by actual combination of
applied stresses
Page 25

Distortion Energy Theory (DET)
• The failure theory based on distortion
energy is also known as von Mises-Hencky
𝜎′
≤ 𝑆 𝑦
𝑆𝐹 =
𝑆 𝑦
𝜎′
Yield strength of
the material
von Mises
effective
stress
Page 26

Example: pure shear load
as in case of torsional failure Mohr’s circle:
pure shear𝜎1= max : 𝜎3 =- max and 𝜎2 = 0
𝑆 𝑦
2
= 𝜎1
2
+ 𝜎2
2
+ 𝜎3
2
− 𝜎1 𝜎2
− 𝜎2 𝜎3 − 𝜎3 𝜎1
Using:
𝑆 𝑦
2 = 3𝜏2
𝑚𝑎𝑥
𝑆 𝑦𝑠 ≡ 𝜏 𝑚𝑎𝑥 =
1
3
𝑆 𝑦 = 0.577𝑆 𝑦
Maximum stress before failure, in this case, is: 𝑆 𝑦𝑠 =
1
3
𝑆 𝑦 = 0.577𝑆 𝑦
Page 27

Static failure theories: experimental verifications
Ductile & brittle materials
Page 28

MEP 209: Design of Machine elements LEC 4

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