LECTURE
Third Edition
FAILURE CRITERIA:
MULTIAXIAL STRESS STATES
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering
23
Chapter
7.5 – 7.6,
7.9
by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
The Stress Transformation
Equations for Plane Stress
„
Slide No. 1
ENES 220 ©Assakkaf
Example 7
At a point in structural member subjected
to plane stress there are normal and
shearing stresses on horizontal and
vertical planes through the point, as shown
in Fig. 24. Use Mohr’s circle to determine
(a) the principal stresses and maximum.
shearing stress at the point, (b) the normal
and shearing stresses on the inclined
plane AB shown in the figure.
1
Slide No. 2
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
The Stress Transformation
Equations for Plane Stress
„
Example 7 (cont’d)
y
120MPa
B
640
A
640
B
0
64
20 Mpa
1160
x
640
A
θ
80 Mpa
Fig.24
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Mohr’s Circle for Plane Stress
„
Example 7 (cont’d)
The given values for use in
drawing Mohr’s circle are:
σ x = 20 MPa
σ y = 120 MPa
σ z = σ p3 = 0
τ
C
R = 94.34
V (20,80)
C
τ xy = -80 MPa
C=
Slide No. 3
ENES 220 ©Assakkaf
σ
2θ = −2(64 ) = −1280
H (120,−80)
20 + 120
= 70 MPa
2
R = radius = (120 − 70) 2 + (− 80 ) = 94.34
2
2
Slide No. 4
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Mohr’s Circle for Plane Stress
„
Example 7 (cont’d)
The principal stresses and
maximum shearing stress
can be computed as
α = sin −1
τ
C
ENES 220 ©Assakkaf
80
= 580
94.34
V (20,80)
σ p2
(a )
580
σ p1 = C + R = 70 + 94.34 = 164.34 MPa
σ p1
1280
σ
σ p 2 = C − R = 70 − 94.34 = −24.34 MPa
(σ n ,τ nt )
σ p3 = 0
τ p = R = 94.34 MPa
H (120,−80)
R = 94.34
Since σ p1 and σ p1 have opposite sign,
τ max = τ p = 94.34 MPa
Slide No. 5
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Mohr’s Circle for Plane Stress
„
Example 7 (cont’d)
The normal and shearing
stress on inclined plane
can be computed as
( b)
α = sin −1
τ
C
580
= 163.8 MPa
τ nt = R sin 6 = 94.34 sin 6 = 9.86 MPa
0
60
128
0
σ p1
σ
(σ n ,τ nt )
σ n = C + R cos 60 = 70 + 94.34 cos 60
0
80
= 580
94.34
V (20,80)
σ p2
ENES 220 ©Assakkaf
R = 94.34
H (120,−80)
3
Slide No. 6
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Generalized Hooke’s law
– Hooke’s law can be extended to include
the biaxial and triaxial states of stress that
often encounter in engineering
applications.
– Let’s consider the differential element of
the material subjected to biaxial state of
normal stress (Figure 26)
E=
σ
ε
Slide No. 7
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Generalized Hooke’s Law
y
σy
Fig.25
τ yz
τ zy
σz
τ yx
τ zx
x
τ xy
τ xz
z
σx
τ xy = τ yx
τ yz = τ zy
(29)
τ zx = τ xz
4
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 8
ENES 220 ©Assakkaf
Generalized Hooke’s Law
Fig.26
σy
dy
σx
dx
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 9
ENES 220 ©Assakkaf
Generalized Hooke’s Law
– Shearing stresses have not been shown in
the differential element of Fig. 26 because
they do not produce changes in the lengths
of sides of the element.
– They only produce distortion of the element
(angle changes), that can contributes to
the angular strain.
5
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 10
ENES 220 ©Assakkaf
Generalized Hooke’s Law
– The deformation of that element in the
direction of the normal stresses, for a
combined loading, can be determined by
computing the deformations resulting from
the individual stresses separately and
adding the values obtained algebraically.
– This procedure is based on the principle of
superposition.
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 11
ENES 220 ©Assakkaf
Generalized Hooke’s Law
– When applying the principle of
superposition, the following conditions
must be satisfied:
• Each effect is linearly related to the load that
produced it.
• The effect of the first load does not scientifically
change the effect of the second load.
6
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 12
ENES 220 ©Assakkaf
Generalized Hooke’s Law
– The first condition is satisfied if the
stresses do not exceed the proportional
limit of the material.
– The second condition is also satisfied if the
deformations small so that the small
changes in the areas of the faces of the
element do not produce significant
changes in the stresses.
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 13
ENES 220 ©Assakkaf
Generalized Hooke’s Law
– Consider an element of material in the
shape of cube as shown in Figure 27.
– We may assume the side of the cube to be
equal to unity, since it is always possible to
select the side of cube as a unit length.
– Under the given multiaxial loading, the
element will deform into a rectangular
parallelepiped of sides equal, respectively
to
7
Slide No. 14
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Generalized Hooke’s Law
σy
1+ ε x
Fig.27
1+ ε y
y
σx
σz
(a)
1+ ε z
1
1
x
z
1
(b)
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 15
ENES 220 ©Assakkaf
Generalized Hooke’s Law
1+ ε x
1+ ε y
(30)
1+ ε z
Where εx, εx, and εz denote the values of the
normal strain in the directions of the three
coordinate axes of Figure 27b.
8
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 16
ENES 220 ©Assakkaf
Generalized Hooke’s Law
– Considering first the effect of the stress
component σx, we know that this stress will
cause a strain equal to σx/E in the xdirection, and strains equal to (-ν σx/E) in
each of the y and z directions.
– In a similar manner, if σy is applied
separately, it will cause a strain equal to
σy/E in the y- direction, and strains equal to
(-ν σy/E) in the other directions.
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 17
ENES 220 ©Assakkaf
Generalized Hooke’s Law
– Finally, the stress component σz will cause
a strain equal to σz/E in the z- direction,
and strains equal to (-ν σz/E) in each of the
x and y directions.
– Combining the results obtained, we
conclude that the components of strain
corresponding to the given multiaxial
loading are
9
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
General State of Strain
[
]
[
]
[
]
1
σ x −ν (σ y + σ z )
E
1
ε y = σ y −ν (σ x + σ z )
E
1
ε z = σ z − ν (σ x + σ y )
E
εx =
(31)
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 18
ENES 220 ©Assakkaf
Slide No. 19
ENES 220 ©Assakkaf
Generalized Hooke’s Law
– It can be shown based on the previous
discussion and on Eq. 31 that if these
equations are solved in terms of the
strains, we could have two formulations or
expressions for the components of normal
stresses.
– These two general formulations are
provided in the next two slides.
10
Slide No. 20
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
State of Stress
σx =
E
(ε x +νε y )
1 −ν 2
(32)
σy =
E
(ε y +νε x )
1 −ν 2
These equations can be used to calculate normal stresses
from measured or computed normal strains.
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 21
ENES 220 ©Assakkaf
General State of Stress
[
]
[
]
[
]
E
(1 −ν )ε x + ν (ε y + ε z )
(1 + ν )(1 − 2ν )
E
(1 −ν )ε y + ν (ε x + ε z )
σy =
(1 + ν )(1 − 2ν )
E
(1 −ν )ε z + ν (ε x + ε y )
σz =
(1 + ν )(1 − 2ν )
σx =
(33)
11
Slide No. 22
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Useful Relationship between G, E and ν
E
G=
2(1 + ν )
(34)
Slide No. 23
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Generalized Hooke’s Law for Shearing
Stress and Strain in Isotropic Materials
E
γ xy
2(1 + ν )
E
τ yz = Gγ yz =
γ yz
2(1 + ν )
E
τ zx = Gγ zx =
γ zx
2(1 + ν )
τ xy = Gγ xy =
(35)
12
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 24
ENES 220 ©Assakkaf
Example 8
At a point on the surface of a structural
steel (E = 200 GPa and G = 76 GPa)
machine part subjected to a biaxial state of
stress, the measured strains were εx =
+750 µm/m, εy = +350 µm/m, and γxy = 560 µrad. Determine the stresses σx, σy,
and τxy at the point.
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 25
ENES 220 ©Assakkaf
Example 8 (cont’d)
The given values are:
εx = +750 µm/m, εy = +350 µm/m, γxy = -560
µrad, E = 200 GPa, and G = 76 GPa
Using Eq. 34,
G=
E
1 E

⇒ ν =  − 2
2(1 + ν )
2 G

1  200

− 2 = 0.3158
ν= 
2  76

13
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 26
ENES 220 ©Assakkaf
Example 8 (cont’d)
Using Eq. 32
σx =
E
200 ×109
(
)
ε
νε
[750 + 0.3158(350)]×10−6
+
=
x
y
2
1 −ν 2
1 − (0.3158)
= 191.17 ×10 6 N/m 2 = 191.2 MPa (T)
σy =
200 × 109
E
(
)
+
=
ε
νε
[350 + 0.3158(750)]×10−6
y
x
2
1 −ν 2
1 − (0.3158)
= 130.37 ×10 6 N/m 2 = 130.4 MPa (T)
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 27
ENES 220 ©Assakkaf
Example 8 (cont’d)
The shearing stress τxy can be computed
from the the following relationship:
τ xy = Gγ xy
= 76 × 109 (−560 ×10 −6 )
= 42.56 ×10 6 N/m 2 = 42.6 MPa
14
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 28
ENES 220 ©Assakkaf
Example 9
Determine the state of strain that
corresponds to the following state of stress
at a point in a steel (E = 30,000 ksi and ν =
0.30) machine part: σx = 15,000 psi, σy =
5000 psi, σz = 7500 psi, τxy = 5500 psi, τyz
= 4750 psi, and τzx = 3200 psi
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 29
ENES 220 ©Assakkaf
Example 9 (cont’d)
The given values are as follows:
σ x = 15.0 ksi, σ y = 5.0 ksi, σ z = 7.5 ksi
τ xy = 5.5 ksi, τ yz = 4.75 ksi, τ zx = 3.2 ksi
E = 30,000 ksi, and ν = 0.30
Using Eqs. 31, the strains are
εx =
[
]
1
1
σ x −ν (σ y + σ z ) =
[15 − 0.3(5 + 7.5)] = 375 µin/in
30,000
E
15
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 30
ENES 220 ©Assakkaf
Example 9 (cont’d)
[
]
[
]
1
1
σ y − ν (σ x + σ z ) =
[5 − 0.3(15 + 7.5)] = −58.3 µin/in
30,000
E
1
1
ε z = σ z −ν (σ x + σ y ) =
[7.5 − 0.3(15 + 5)] = 50 µin/in
30,000
E
εy =
Using Eq. 35
τ yz = Gγ yz =
E
2(1 + ν )
γ yz ⇒ γ yz =
τ yz
E
2(1 + ν )
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 31
ENES 220 ©Assakkaf
Example 9 (cont’d)
2(1 + ν )
2(1 + 0.3) )
(5.5) = 477 µrad
τ xy =
30,000
E
2(1 + ν )
2(1 + 0.3) )
(4.75) = 412 µrad
=
τ yz =
30,000
E
γ xy =
γ yz
γ zx =
2(1 + ν )
2(1 + 0.3) )
(3.2) = 277 µrad
τ zx =
30,000
E
16
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 32
ENES 220 ©Assakkaf
Strain Measurement and Rosette
Analysis
– Electrical resistance strain gages provide
accurate measurements of normal strain.
– The gage may consist of a length of 0.001
in-diameter wire arranged as shown in
Figure 28 and cemented between two
pieces of paper.
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 33
ENES 220 ©Assakkaf
Strain Measurement and Rosette
Analysis
Fig.28
17
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 34
ENES 220 ©Assakkaf
Strain Measurement and Rosette
Analysis
– The wire or foil gage is centered to the
material for which the strain is to be
determine.
– As the material is strained, the wires are
lengthened or shortened.
– This lengthening and shortening will cause
changes in the electrical resistance.
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 35
ENES 220 ©Assakkaf
Strain Measurement and Rosette
Analysis
– The change in resistance can be measured
and calibrated to provide normal strain
– Shearing strains are often obtained by
measuring normal strains in two or three
different directions.
– The shearing strains can be computed
from normal strain data using the following
equations:
18
Slide No. 36
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
Strain Measurement and Rosette
Analysis
„
ε a = ε x cos 2 θ a + ε yx sin 2 θ a + γ xy sin θ a cos ϑa
ε b = ε x cos 2 θ b + ε yx sin 2 θ b + γ xy sin eθ b cosθ b
y
a
(36)
ε c = ε x cos 2 θ c + ε yx sin 2 θ c + γ xy sin θ c cosθ c
b
θc
θb
a
θa
x
Slide No. 37
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Strain Measurement and Rosette
Analysis
– Rosette Types
Gage
Gage
450
Gage
(a) 450 Rosette
Fig.29
Gage
600
600
Gage
Gage
600
(a) Delta Rosette
19
Slide No. 38
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Strain Measurement and Rosette
Analysis
– In-plane strains and their Orientations
tan 2θ p =
ε p1 , ε p 2 =
γ xy
εx −εy
εx +εy
2
2
 ε x − ε y   γ xy 
 + 

± 
 2   2 
2
 ε x − ε y   γ xy 
 + 

 2   2 
2
(37)
2
γ p = 2 
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 39
ENES 220 ©Assakkaf
Strain Measurement and Rosette
Analysis
– The principal strain εz = εp3 can be
determined from the measured data. From
Eqs. 31 and 33,
εz = −
=−
ν
E
(σ
ν
1 −ν
x
+σ y )= −
(ε
x
+εy)
ν  E 
[(ε x +νε y ) + (ε y +νε x )]
E 1 −ν 2 
(38)
20
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 40
ENES 220 ©Assakkaf
Strain Measurement and Rosette
Analysis
– Out-of-plane principal Strain
εz = −
ν
1 −ν
(ε
x
+εy)
(39)
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 41
ENES 220 ©Assakkaf
Example 10
At a point on the free surface of an
aluminum alloy (E = 73 GPa and ν = 0.33)
machine part, the strain rosette shown in
Fig. 30 was used to obtain the following
normal strain data: εa = +780µ, εb = +345µ,
εc = -332µ. Determine (a) the strain
components εx, εy, γxy and (2) the principal
strains and maximum shearing strain.
21
Slide No. 42
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Example 10 (cont’d)
600
Gage c
Gage b
y
600
600
x
Gage a
Figure.30. Delta Rosette
Slide No. 43
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Example 10 (cont’d)
The given values are as follows:
ε a = ε x = +780µ , ε b = +345µ (θ = −600 ), ε c = −332 µ (θ = 600 )
(a ) ε n = ε x cos 2 θ n + ε y sin 2 θ + γ xy sin θ n cosθ n
ε b = 780 cos 2 (− 600 ) + ε y sin 2 (− 600 ) + γ xy sin (− 600 )n cos(− 600 ) = +345µ
ε c = 780 cos 2 (600 ) + ε y sin 2 (600 ) + γ xy sin (600 )n cos(600 ) = −332µ
Solving above eqaution, yields
ε x = ε a = +780 µ
ε y = −251µ
γ xy = −782µrad
22
Slide No. 44
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Example 10 (cont’d)
ε p1 , ε p 2 =
εx +εy
2
2
 ε x − ε y   γ xy 
 + 

± 
 2   2 
2
2
=
780 − 251
 780 + 251   − 782 
± 
 +

2
2

  2 
2
Therefore,
ε p1 = 264.5 + 647 = 911.5µ
ε p 2 = 264.5 − 647 = −383µ
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
Slide No. 45
ENES 220 ©Assakkaf
Example 10 (cont’d)
Using Eq. 39, the third principal strain is
ε p3 = ε z = −
ν
1 −ν
(ε
x
+εy )=
0.33
(780 − 251) = −261µ
1 − 0.33
γ p = γ max = 2(647) = 1294 µrad
1
2
θ p = tan −1
γ xy
− 782
1
= tan −1
= 18.59 0
780 + 251
εx −εy 2
23
Slide No. 46
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Example 11
The strain rosette shown in Figure 31 was
used to obtain normal strain data at a point
on the free surface of a 2024-T4 aluminum
alloy structural component. The gage
readings were εa = +525µ, εb = +450µ, and
εc = +1425µ. Determine
(a) the strain components εx, εy, γxy at the
point.
Slide No. 47
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Example 11 (cont’d)
(b) the stress components σx, σy, and γxy.
(c) the principal stresses and maximum
shearing stress at the point (in SI units).
y
Gage c
450 450
Gage b
450
x
Gage a
Figure.31
24
Slide No. 48
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
Example 11 (cont’d)
„
The given values are as follows:
ε a = ε x = +525µ , ε b = +450 µ (θ = 450 ), ε c = +1425µ (θ = −450 )
(a ) ε n = ε x cos 2 θ n + ε y sin 2 θ + γ xy sin θ n cos θ n
ε b = 525 cos 2 (450 ) + ε y sin 2 (450 ) + γ xy sin (450 )n cos(450 ) = +450µ
ε c = 525 cos 2 (− 450 ) + ε y sin 2 (− 450 ) + γ xy sin (− 450 )n cos(− 450 ) = +1425µ
Solving above eqaution, yields
ε x = ε a = +525µ
ε y = +1350 µ
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
Multiaxial Stress States
„
γ xy = −975µrad
Slide No. 49
ENES 220 ©Assakkaf
Example 11 (cont’d)
(b) For 2024 - T4 aluminum alloy : E = 73 GPa and G = 28 GPa
Eq. 34 gives ν =
1 E
 1  73 
− 2 =  − 2 = 0.3036

2 G
 2  28 
Eqs 32 give
[
]
[
]
σx =
E
73 ×109
[525 + 0.3036(1350)]×10 −6 = 75.2 MPa (T)
ε x + νε y =
2
2
1 −ν
1 − (0.3036 )
σy =
E
73 × 109
ε
+
νε
=
[1350 + 0.3036(525)]×10 −6 = 121.4 MPa (T)
y
x
2
1 −ν 2
1 − (0.3036 )
τ xy = Gγ xy = 28 ×109 (− 975 ×10 −6 ) = −27.3 MPa
25
Slide No. 50
LECTURE 22. FAILURE CRITERIA: MULTIAXIAL STRESS STATES (7.5 – 7.6, 7.9)
ENES 220 ©Assakkaf
Multiaxial Stress States
„
Example 11 (cont’d)
(c) Eq. 22a gives
σ x +σ y
2
σ x −σ y 
75.2 + 121.4
 75.2 − 121.4 
2
 + τ xy2 =
± 
± 
 + (− 27.3)
2
2
2
2




σ p1 = 98.27 + 35.76 = 134 MPa
σ p1, p 2 =
2
σ p 2 = 98.27 − 35.76 = 62.5 MPa
σ p3 = σ z = 0
τ p = 35.8 MPa (note ≠ τ max because σ p1 and σ p 2 have same sign)
1
[σ max − σ min ] = 1 [134 − 0] = 67 MPa
2
2
2
τ
1
1
2(− 27.3)
xy
= tan −1
= 24.9 0
θ p = tan −1
2
75.2 − 121.4
σ x −σ y 2
τ max =
26