Distortion-Energy Theory
Postulate: Yielding will occur when the distortionenergy per unit volume equals the distortionenergy per unit volume in a uniaxial tension
specimen stressed to its yield strength.
Strain Energy
Strain Energy
σ
The strain energy per unit
volume is given by the equation
σi
1
1
1
U = σ1ε1 + σ 2 ε 2 + σ 3ε 3
2
2
2
U
εi
Units
[ ]
[
[U ] = lb in 2 ⋅ [in in ] = lb − in in 3
]
ε
The strain energy in a tensile
test specimen is the area under
the stress-strain curve.
Strain Energy
(Elastic Stress-Strain Relationship)
An expression for the strain energy per unit
volume in terms of stress only can be obtained
by making use of the stress-strain relationship
Algebraic Format
1
ε1 = (σ1 − νσ 2 − νσ3 )
E
1
ε 2 = (σ 2 − νσ1 − νσ 3 )
E
1
ε 3 = (σ 3 − νσ1 − νσ 2 )
E
Matrix Format
é 1 − ν − ν ù ì σ1 ü
ì ε1 ü
ï ï 1ê
ï ï
ú
íε 2 ý = ê− ν 1 − ν ú íσ 2 ý
ïε ï E ê − ν − ν 1 ú ïσ ï
ë
î 3þ
ûî 3 þ
Strain Energy
(Stress Form of Equation)
1
1
1
U = σ1ε1 + σ 2ε 2 + σ 3ε 3
2
2
2
1 ì1
ü
= σ1 í (σ1 − νσ 2 − νσ3 )ý
2 îE
þ
1 ì1
ü
+ σ 2 í (σ 2 − νσ1 − νσ3 )ý
2 îE
þ
1 ì1
ü
+ σ3 í (σ 3 − νσ1 − νσ 2 )ý
2 îE
þ
[
]
1 2
U=
σ1 + σ 22 + σ32 − 2ν(σ1σ 2 + σ 2σ3 + σ3σ1 )
2E
Distortion and Hydrostatic
Contributions to Stress State
σ2
σ3
Principal Stresses
Acting on Principal
Planes
σh
σ2 − σh
σ1
σh
=
+
σh
σ1 + σ 2 + σ 3
σh =
3
σ1 − σ h
σ3 − σ h
Distortional Stresses
Hydrostatic Stress
The distortional stress components are often called the
deviatoric stress components.
Physical Significance
(Hydrostatic Component)
σh
The hydrostatic stress causes a
change in the volume.
σh
σh
σ1 + σ 2 + σ 3
σh =
3
The cube gets bigger in tension,
smaller in compression.
σ h = Ke
K ≡ Bulk Modulus
e ≡ volumetric strain
Physical Significance
(Distortional Stresses)
σ2 − σh
σ1 − σ h
σ3 − σ h
These unequal stresses act to
deform or distort the material
element.
There is no change in volume,
but there is a change in shape.
These stresses try to elongate
or compress the material more
in one direction than in
another.
Strain Energy Associated with the
Hydrostatic Stress
[
]
1 2
2
2
U=
σ1 + σ2 + σ3 − 2ν(σ1σ 2 + σ 2σ3 + σ3σ1 )
2E
1 2
Uh =
σ h + σ 2h + σ2h − 2ν(σ h σ h + σ h σ h + σ h σ h )
2E
1
3σ 2h − 6ν ⋅ σ 2h
=
2E
This term is equal to the strain
3 (1 - 2ν ) 2
energy per unit volume from the
Uh =
σh
hydrostatic stress components.
2 E
[
[
]
]
Distortional Strain Energy
The distortional strain energy is equal to the difference between
the total strain energy and the hydrostatic strain energy.
Ud = U − Uh
[
]
[
]
1 2
σ1 + σ 22 + σ32 − 2ν(σ1σ 2 + σ 2σ3 + σ3σ1 )
=
2E
2
3 (1 − 2ν ) (σ1 + σ 2 + σ3 )
−
2 E
9
1 2
σ1 + σ 22 + σ32 − 2ν(σ1σ 2 + σ 2σ3 + σ3σ1 )
=
2E
æ σ12 + σ1σ 2 + σ1σ3 ö
÷
1 (1 − 2ν ) ç 2
ç + σ 2 + σ1σ 2 + σ 2σ3 ÷
−
2 3E ç 2
÷
ç + σ3 + σ1σ 3 + σ 2σ3 ÷
ø
è
Distortional Strain Energy
(Continued)
Ud = U − Uh
[
]
1 2
=
σ1 + σ 22 + σ 32 − 2ν(σ1σ 2 + σ 2σ3 + σ3σ1 )
2E
1 (1 − 2ν ) 2
−
σ1 + σ 22 + σ32 + 2(σ1σ 2 + σ 2σ3 + σ3σ1 )
2 3E
(
)
[
1+ ν 2
Ud =
σ1 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ 3σ1
3E
]
Distortional Strain Energy in Tension
Test Specimen
Postulate: Yielding will occur when the distortionenergy per unit volume equals the distortionenergy per unit volume in a uniaxial tension
specimen stressed to its yield strength.
[
1+ ν 2
σ1 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ 3σ1
Ud =
3E
1+ ν 2
Ud =
Sy
3E
]
Hamrock, Fig. 3.1
Distortion Energy Failure Theory
Equating the distortional strain energy at the point under
consideration to the distortional strain energy in the tensile
test specimen at the yield point yields
[
1+ ν 2
Ud =
σ1 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ 3σ1
3E
1+ ν 2
=
Sy
3E
S2y = σ12 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ 3σ1
σ eff = Sy
σ eff = σ12 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ 3σ1
]
Alternate Forms of Effective Stress
σ eff = σ12 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ 3σ1
Form 1
(σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2
Form 2
σ eff =
2
The effective stress is commonly referred
to as the von Mises stress, after Dr. R. von Mises
who contributed to the theory.
Plane Stress Condition
σ3 = 0
Sy
σ2
σ eff = σ12 + σ 22 − σ1σ 2
- Sy
σ eff =
(σ1 − σ 2 )2 + σ 22 + σ12
2
Sy
- Sy
• As long as the stress state falls within the shaded area,
the material will not yield.
• The surface, blue line, at which the material just begins
to yield is called the yield surface.
σ1
σ 3 = −σ1
Pure Shear Condition
Sy
τ1,3
σ2
σ3
σ1
σ3
- Sy
45° S
y
σ1
- Sy
Mohr’s Circle
for Pure Shear
σ eff = σ12 + σ 32 − σ1σ 3
= 3σ12 = 3τ 2max = Sy
This is an important result.
τ max = 0.577 ⋅ Sy = Sys
Yield Surface in 3-D Stress State
Hamrock, Fig. 6.9
Other Names for Distortion
Energy Theory
σ eff =
(σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2
2
τ1/3
People came up with the same
equation using different starting
points.
•Shear Energy Theory
•Von Mises-Hencky Theory
•Octahedral-Shear-Stress Theory
τ
τ1/2
τ 2/3
σ3
σ2
σ1 > σ 2 > σ 3
σ1 σ
Assignment
• Show that the two forms of the equation for the effective stress
are equal.
• Show that the effective stress for a hydrostatic stress state is
zero.
• Compute the effective stress at the critical location in the
stepped shaft loaded in tension (previous assignment). The
yield strength of the material is 30 ksi. Will the material yield
at the critical location?
σ eff = σ12 + σ 22 + σ 32 − σ1σ 2 − σ 2 σ 3 − σ 3σ1
σ eff =
(σ1 − σ 2 )2 + (σ 2 − σ 3 )2 + (σ 3 − σ1 )2
2
Assignment
(Continued)
In the rear wheel suspension of the Volkswagen “Beetle” the spring motion was
provided by a torsion bar fastened to an arm on which the wheel was mounted.
See the figure for more details. The torque in the torsion bar was created by a
2500-N force acting on the wheel from the ground through a 300-mm lever
arm. Because of space limitations, the bearing holding the torsion bar was
situated 100-mm from the wheel shaft. The diameter of the torsion bar was 28mm. Find the von Mises stress in the torsion bar at the bearing.
Hamrock, Fig. 6.12
0
