LECTURE DATE:07/12/2009
PREPARED BY:EBRU SEMA KOŞAROĞLU & FARAJ KHALIKOV
STATIC FAILURE THEORIES
DUCTILE BRITTLE
TRESCA VON-MISES MOHR- MAXIMUM
CRITERIAN THEORY COLOUMB NORMAL STRESS
THEORY THEORY(MNST)
MSST:MAXIMUM MDET:MAXIMUM
SHEAR STRESS DISTORTION ENERGY
THEORY THEORY
UNIAXIAL TEST
BRITTLE
𝑆𝑌
DUCTILE
𝑆𝑌
MAXIMUM NORMAL STRESS THEORY
For maximum normal stress theory, the failure occurs when one of the
principal stresses (𝜎1 , 𝜎2 𝑎𝑛𝑑 𝜎3 ) equals to the yield strength.
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𝜎1 > 𝜎2 > 𝜎3
Failure occurs when either 𝜎1 =𝑆𝑡 or 𝜎3 = −𝑆𝑐 ,where 𝑆𝑡 is strength in
tension and 𝑆𝑐 is strenght in compression.
MOHR-COULOMB THEORY
The Coulomb-Mohr theory or internal friction theory assumes that the
critical shearing stress is related to internal friction.
MAXIMUM DISTORTION ENERGY THEORY(VON-MISES THEORY)
The maximum distortion energy theory ,also known as the Von Mises
theory, was proposed by M.T.Huber in 1904 and further developed by R.von
Mises(1913).In this theory failure by yielding occurs when at any point in the
body ,the distortion energy per unit volume in a state of combined stress
becomes equal to that associated with yielding in a simple tension test.
STRAIN ENERGY
Generally strain energy U is obtained by this equation.
1
U= σij εij
2
1
𝜀1 = (𝜎 − 𝜈𝜎2 − 𝜈𝜎3 )
𝐸 1
1
𝜀2 = (−𝜈𝜎1 + 𝜎2 − 𝜈𝜎3 )
𝐸
1
𝜀3 = (−𝜈𝜎1 − 𝜈𝜎2 + 𝜎3 )
𝐸
Then, substituting these three equations in to general strain energy equation:
1 1 1 1 1 1
U= 𝜎1 (𝜎1 − 𝜈𝜎2 − 𝜈𝜎3 )+ 𝜎2 −𝜈𝜎1 + 𝜎2 − 𝜈𝜎3 + 𝜎3 (−𝜈𝜎1 − 𝜈𝜎2 +
2 𝐸 2 𝐸 2 𝐸
𝜎3 )
HYDROSTATIC STRESS
The hydrostatic stress(𝜎 )causes a change in the volume.
2
𝜎1 + 𝜎2 + 𝜎3
𝜎 =
3
Strain energy associated with the hydrostatic stress:
1 3 (1 − 2𝜈) 2
𝑈 = [𝜎 2 + 𝜎 2 + 𝜎 2 − 2𝜈 𝜎 𝜎 + 𝜎 𝜎 + 𝜎 𝜎 = 𝜎
2𝐸 2 𝐸
Then distortional energy 𝑈𝑑 = 𝑈 − 𝑈
1+𝜈
From previous equations: 𝑈𝑑 = 𝜎1 2 + 𝜎2 2 + 𝜎3 2 − 𝜎1 𝜎2 − 𝜎2 𝜎3 − 𝜎3 𝜎1
3𝐸
Then yielding will occur at this condition:
1+𝜈
𝑈𝑑 = 𝑆
3𝐸 𝑌
𝜎𝑒𝑓𝑓 =𝑆𝑌 = 𝜎1 2 + 𝜎2 2 + 𝜎3 2 − 𝜎1 𝜎2 − 𝜎2 𝜎3 − 𝜎3 𝜎1
(𝜎1 − 𝜎2 )2 + (𝜎2 −𝜎3 )2 + (𝜎3 − 𝜎1 )2
𝜎𝑒𝑓𝑓 =
2
For plane stress condition 𝜎3 = 0
𝜎𝑒𝑓𝑓 = 𝜎1 2 + 𝜎2 2 + 𝜎1 𝜎2
If the state stress is in this area then the material
will not yield.
For pure shear condition
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𝜎3 = −𝜎1
𝜎𝑒𝑓𝑓 = 𝜎1 2 + 𝜎3 2 − 𝜎1 𝜎3 = 3𝜎1 2 = 3𝜏𝑚𝑎𝑥 2 =𝑆𝑌
𝜏𝑚𝑎𝑥 = 0.577𝑆𝑌
MAXIMUM SHEAR STRESS THEORY
The maximum shearing stress theory is an outgrowth of the experimental
observation that a ductile material yields as a result of slip or shear along
crystalline planes.
Yielding begins whenever the maximum shear stress in a part equals to
the maximum shear stress in a tension test specimen that beings to yield.
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SY
τmax = SYS = yield strength in shear =
2
σ 1 −σ 3
τmax =
2
Then , 𝑆𝑌 = 𝜎1 − 𝜎3
Elasticity Materials
𝑆𝑌
-brittle 𝜎1 (𝜎𝑥 ) = where N is safety factor
𝑁
(MNST)
-ductile
(MSST) 𝜏𝑚𝑎𝑥 = 𝑆𝑌𝑆 (yield strength in shear )
𝜎1 −𝜎3 𝑆𝑌
=
2 2
𝑆𝑌
𝜎1− 𝜎3 =
𝑁
𝑆𝑌
(MDET) 𝜎𝑒𝑓𝑓 =
𝑁
Strain Energy:
𝜎𝑖
𝜀𝑖
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1
𝜎 𝜀 =𝑈
2 𝑖𝑗 𝑖𝑗
𝑆𝑌 −𝜎 𝑒𝑓𝑓 𝑁
Design Margin=M=
𝑆𝑌
EXAMPLE:A circular shaft of tensile strength 𝑆𝑌 =350 MPa is subjected to a
combined state of loading defined by bending moment M=8 kN.m and torque
T=24kN.m.Calculate the required shaft diameter d in order to achieve a factor
of safety N=2.Use a) the maximum shearing stress theory(MSST-Tresca) b)the
maximum distortion energy theory(MDET –Von Mises)
𝜏𝑥𝑦
𝜎𝑥
𝜏𝑥𝑦
𝜎𝑥
𝑑 𝑀𝑐 𝑀𝑑 𝑇𝑐 𝑇𝑑
c= 𝜎𝑥 = = and 𝜏𝑥𝑦 = =
2 𝐼 2𝐼 𝐽 2𝐽
32𝑀 16𝑇
Then,𝜎𝑥 = and 𝜏𝑥𝑦 =
𝜋𝑑 3 𝜋𝑑 3
We need to find principle stresses:
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a) For maximum shearing stress theory
16
𝜎1,2 = (𝑀 ∓ 𝑀2 + 𝑇 2 )
𝜋𝑑 3
𝑆 32
𝜎1 − 𝜎2 = 𝑌 = 𝜎𝑥 2 + 4𝜏𝑥𝑦 2 = 3 𝑀2 + 𝑇 2
𝑁 𝜋𝑑
Then, substituting the numerical values of M,T, 𝑆𝑌 and N:
d=113.8mm
b) For maximum distortion energy theory
𝑆 16
𝜎𝑒𝑓𝑓 = (𝜎1 − 𝜎2) 2 + 𝜎1 2 + 𝜎2 2 = 𝜎𝑥 2 + 3𝜏𝑥𝑦 2 = 𝑌 = 4𝑀2 + 3𝑇 2
𝑁 𝜋𝑑 3
Then, substituting the numerical values of M,T, 𝑆𝑌 and N:
d=109mm
FRACTURE MODES
OPENING
SLIDING
TEARING
Fracture is defined as the separation of a part into two or more
pieces.The mechanisms of brittle fracture are the concern of fracture
mechanics ,which is based on a stress analysis in the vicinity of a crack or
defect of unknown small radius in a part.
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MULTIPLE FRACTURES
Stress Intensity Factor: In the fracture mechanics approach a stress
intensity factor,𝐾𝐼 , is evaluated. This can be thought of as a measure of the
effective local stress at the crack root.
𝐾𝐼 = 𝛽𝜎 𝜋𝑎
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where
𝜎=normalstress,
𝑎
𝛽 = 𝑔𝑒𝑜𝑚𝑒𝑡𝑟𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝑤𝑖𝑐 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 , 𝑎 = 𝑐𝑟𝑎𝑐𝑘 𝑙𝑒𝑛𝑔𝑡 (or half
𝑤
crack length) ,w=member width(or half width of member)
Fracture Toughness: In a toughness test of a given material, the stress –
intensity factor at which a crack will propagate is measured. This is the critical
stress intensity factor , referred to as the fracture toughness and denoted by
the symbol 𝐾𝐼𝐶 .
𝐾𝐼𝐶
N= (N=factor of safety)
𝐾𝐼
EXAMPLE: For a shape with width 12m , crack length 65mm , thickness 30mm
and applied loading 50MPa.Find the factor of
safety.( 𝐾𝐼𝐶 =28.3MPa 𝑚, 𝑆𝑌 =240MPa)
50MPa
65mm
30mm
12m
240
𝑁1 = = 4.8
50
From the ratio of a/w:
𝑎 65 × 10−3
= = 0.0054
𝑤 12
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𝐾𝐼
From table β≅ 1. 𝛽 =
𝜎 𝜋𝑎
65
𝐾𝐼 = 𝜎 𝜋𝑎=50 𝜋 10−3 =15.97MPa 𝑚
2
𝐾𝐼𝐶 28.3
For safety factor: 𝑁2 = = = 1.77
𝐾𝐼 15.97
𝑁2 is better than 𝑁1 in order to obtain a safe structure; therefore, to have a safe
structure, loading should not be used to calculate safety factor.
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Reference List:
1. Lecture Notes AEE 361, Demirkan ÇÖKER), 2009, “Static Failure Theories”
2. Static Failure theories, Ansel C. UGURAL & Saul K. Fenster, 2007,
“Advanced Strength and Applied Elasticity”, fourth edition.
3. Lecture 5 & 6, The University of Jenessee at Martin School of
Engineering, 2009, “Machine Design Notes”.
4. Static Failure theories, J. Keith Nisbett, 2009, “Shigley’s Mechanical
Engineering Design”, eighth edition.
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