Sections4-4-5

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Example comapring failure criteria
Stress analysis of a spacecraft structural member
gives the state of stress as shown below. If the
part is made from an alloy with Y = 500 MPa,
check yielding according to Rankine, Tresca
and von Mises criteria. What is its safety factor
for each criterion?
Maximum principal stress Criterion
(Rankine)
0 ⎤
⎡ 200 0
σ = ⎢⎢ 0 100 −30 ⎥⎥ MPa
⎢⎣ 0 −30 −50 ⎥⎦
• The principal stresses are:
σ1 = 200; σ 2 = 105.77; σ3 = −55.77;
• Maximum principal stress is 200MPa
• Factor of safety FS=500/200=2.5
Maximum shear stress Criterion
(Tresca)
• Yield function
f = σe −
Y
2
• Maximum shear stress
σe =
σ1 − σ 3
2
=
200 + 55.78
= 127.89 MPa
2
• Shear stress for uniaxial tension
Y
= 250 MPa
2
f = 127.89 − 250 < 0
• Factor of safety FS=250/127.89=1.95
Distortional Energy Density
Criterion (von Mises)
• Yield function
f =
1
2
( σ1 − σ2 )
2
+ ( σ2 − σ3 ) + ( σ3 − σ1 ) − Y = 224 − 500
2
2
• Factor of safety FS=500/224=2.2
• Why are all the results so close?
σ1 = 200; σ 2 = 105.77; σ3 = −55.77;
Other yield criteria
• For isotropic materials there is usually
substantial difference between yield in
tension and compression. Why?
• For orthotropic materials there is also
differences between the behavior along
the three principal directions and between
shear and normal stresses
• Fortunately, for most applications we have
at least transverse isotropy
4.5.1 Mohr-Coulomb yield criterion
(Charles Augustin de Coulomb, 1736-1806,
Christian Otto Mohr, 1835-1918)
• Yield of materials like rock and concrete does
change with hydrostatic pressure
• Yield function written in terms of cohesion c
and internal friction angle φ as
f = σ 1 − σ 3 + (σ 1 + σ 3 ) sin φ − 2c cos φ
σ1 ≥ σ 2 ≥ σ 3
• What would it look like for plane stress?
From test results
• For tension test
σ 1 = YT , σ 2 = σ 3 = 0,
f =0 ⇒
YT =
2 cos φ
1 + sin φ
• For compression test
σ 3 = −YC , σ 2 = σ 1 = 0, f = 0 ⇒
• Get
c=
YT
2
YT
YC
φ=
⎛ Y ⎞
− tan −1 ⎜⎜ T ⎟⎟
2
⎝ YC ⎠
π
2 cos φ
YC =
1 − sin φ
Simpler treatment for plane stress
• If both stresses are of the same sign, then we
compare the corresponding Y. If they are of
different signs
• Check if the two formulations are the same!
f =
σ1 σ 3
YT
−
YC
−1
Drucker-Prager yield criterion
(Dan Drucker, 1918-2001, William
Prager 1903-1980)
• Generalization of Von Mises to account for
hydrostatic pressure
f = α I1 +
J2 − K
• Gets outer bound of Coulomb Moher with
2sin φ
6c cos φ
α=
,
K=
3 ( 3 − sin φ )
3 ( 3 − sin φ )
• Inner bound with
2sin φ
6c cos φ
α=
,
K=
3 ( 3 + sin φ )
3 ( 3 + sin φ )
Pictorially
Reading assignment
Sections 4.6: Question: For what kind of beam cross-section
there will not be much difference between the moment
causing initial yield and the fully plastic moment?
Source: www.library.veryhelpful.co.uk/ Page11.htm
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