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DEN102Stress Analysis
Principal Stress
Dr P.H. Wen
Aims1. Recognise the stress and strain tensors ;2. Understand stress state of point and how to
calculate principal stresses and their directions;3. Recognise why principal stresses and their
directions are useful; 4. Understand what a yield criterion is and how it
can be used.
Structure Stress Analysis
• Analytical solutionHand calculation: beam, truss, shaft
• Numerical solutionANASYS, ABAQUS, DYNA, IDEAS
Page 4
8.1. Normal and Shear Stresses in Solid• In general objects, i.e. multi-
dimensional objects, they are generated by exterior forces F, distributed loads or pressures p.
• Exterior loads result in interior forces, which are resultants of interior stresses.
• At an arbitrary cut at S-S (see figure), these stresses act on the surface.
• There are normal stresses σ(orthogonal to the surface) and tangential or shear stresses τ(parallel to the surface).
• On the opposite surface of the cut, stresses occur in the opposite direction.
• The stress vector t at a point P is defined by
• It is a vectorial sum of shear and normal stress:
AAA ddlim
0
FFtrr
r=
∆∆
=→∆
στ
σ
τt
t
Force diagram:
P P=τσFFFFFt +=+=
+==
AAAAnn
dd
dd
ddd
dd ττ
rrrrrr
8.1
Page 5
8.2. General Stress State (Three Dimensional Element)
• ELEMENT
A three dimensional rectangularelement is of differential size:
dx×dy×dz=dV
Positive surfaces +veNegative surfaces -ve
Element is in a state of uniformstress.
xσ
xzτ
yxτxyτyzτ
yσ
zxτzσ
zyτ
x
y
zdx
dz
dy
Page 6
• The stresses depend on the direction of the cut, i.e. the orientation of the surface area A.
• Generally, a Cartesian coordinate system is chosen (see Figure 8.1)
• The total stress state is determined by the normal and shear stresses at six surfaces of an infinitesiminal volume element (see Figure 8.1).
• This element is normally defined with surfaces orthogonal to the coordinate directions.
• The values of all stresses change if the volume element is cut in a different orientation (i.e. coordinate transformation 8.2).
• Positive direct stress are defined as tensile stresses ; negative as compressive stresses .
xσ
xzτ
yxτxyτ
yzτyσ
zxτzσ
zyτ
x
y
z
Figure 8.1
xσ
xzτ
yxτxyτyzτ
yσ
zxτ
zσ
zyτx
y
zFigure 8.2
Page 7
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
zzyzx
yzyyx
xzxyx
στττστττσ
σ
8.3. General Stress State (in Matrix Form)• The normal and shear
stresses are represented by the stress matrix also named stress tensor
• Stresses with non-identical indices are shear stresses, the other are normal stresses; the latter are on the diagonal of the matrix.
Two-dimension
(8.2); ⎥⎦
⎤⎢⎣
⎡=
yyx
xyx
σττσ
σ
Normal stresses
Shear stresses
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
zzyzx
yzyyx
xzxyx
στττστττσ
σxσ
xzτ
yxτxyτyzτ
yσ
zxτzσ
zyτ
x
y
z
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
zzyzx
yzyyx
xzxyx
στττστττσ
σ
Figure 8.3
Three-dimension
Page 8
• The consequence is that the stress matrix is symmetric
• There are hence for each point if the structure 6 unknown stresses to be determined in stress analysis
8.4. Corresponding Shear Stresses
• This can be seen by taking the sum of the moments with respect to point O, i.e.
( )xyyx
xyyx
xyyxO
dV
M
ττ
ττ
ττ
=⇒
=−
=−=∑
0 therefore
0)dydz(dx)dxdz(dy :0
zxxzzyyzyxxy ττττττ === ;;
(8.3a)
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
zyzxz
yzyxy
xzxyx
στττστττσ
σ
xσ
yxxy ττ =
zxxz ττ =
zyyz ττ =
yσ
zσ x
y
z
xzτ
yxτxyτyzτ
zxτ
zyτ
dxdy
yxτxyτ
x
y
O
yσ
xσ
yxτ
xyτ
yσ
xσ
(8.3b)
Figure 8.4
• Shear stresses at two surfaces, which are perpendicular to one another are equal
Page 9
Plane stress state:
• The stress tensor in matrix form is then
or
⎥⎦
⎤⎢⎣
⎡=
yxy
xyx
σττσ
σ
(8.4)
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
00000
yxy
xyx
σττσ
σ
0=== zyzxz σττ
(8.5)
Figure 8.5
8.5. Two Dimensional Element (Plane Stress)
yσ
xσxyτ
dxdy
yxτxyτ
y yσ
xσxyτ
xσ
yxτyσ
x
Page 10
Example 8.0:Illustrate stress state using stress tensor in matrix form (unit= N/mm2)
50
150
70
80
100
30
40
60
30
(a) (b) (c)
Page 11
8.6. Stresses in Straight Bar• Normal exterior forces F leads to
normal stresses σ in the interior (constant over the cross section A)
• Interior normal force N is the resultant forces of normal stresses σ
• If the cut is not perpendicular to the axis of the beam, a normal force Fleads to normal stresses σ and shear stresses τ
]N/m[ 2
AF
=σ (8.6)
F Fcut S-S S
S
F F
σ σ
F F
N N]N[d AAN
A
⋅== ∫ σσ (8.7)
A
F FS
S
θ
F Fσθ
τθ
θθ cosAA =
Non-perpendicular cut S-S (glue)
θθσθθθ
θτ
θσθθθσ
θ
τθ
θθ
cossincossincos/sin
coscoscos/cos 22
====
====
AF
AF
AF
AF
AF
AFn
(8.8)
Fn Fτ
Fτ
F
Fn
θ
θ
Figure 8.6
Figure 8.7
Page 12
]N/m[ 2yI
M
x
bz =σ (8.8)
Undeformed beam
Deformed beam
y
yy
zxy
x z
zx
8.7. Normal Stresses due to Bending Moments
• Bending moment leads to linear distributions of normal stresses
Mb Mbcut S-S S
S
equivalentto
Mbσ
σ
Figure 8.8
σmaxσmax
Page 13
8.8. Shear Stress in Cylinder by Torque
• Normal exterior torque T0 leads to shear stresses τ in the interior
• Interior torque T is the resultant forces of shear stresses τ
T0 cut S-S S
S
JTR
JTr
== max2 ],N/m[ ττ (8.9)
TArJTArrT
AA
=== ∫∫ dd)( 2τ (8.10)
A
T0
T0T
Figure 8.9
τmax
τmax
Page 14
Summary of Today
1. State of Stress at Point (Element)
2. State of Plane Stress
3. Stress Tensor
Example 1 (individual)
A hollow shaft, of external diameter D2 and internal diameter D1, where D1/D2=1/3, is required to transmit a torque of 100kNm and compressive axial load 1500kN as shown in Figure. If D2 is selected as 200mm, illustrate stress state on outer surface of the shaft using stress tensor in matrix form (unit= N/mm2)
D2D1
propeller shaft1500kN
100kNm
Page 16
8.9. Coordinate Transformation• If stresses are known for one coordinate system and
stresses for another coordinate system are to be derived, one uses the following equations (balance of force in local coordinate ):
Therefore
0cos)sin(sin)sin(
sin)cos(cos)cos(:element) of thicknessis ,(direction in 0Fi
x
=−−
−−
×==∑
θθτθθσ
θθτθθσσ
dAdA
dAdAdAdAx
xyy
xyxx
ttc
0sin)sin(cos)sin(
cos)cos(sin)cos(
direction in 0Fiy
=+−
−+
=∑
θθτθθσ
θθτθθστ
dAdA
dAdAdA
y
xyy
xyxyx
θτθσστ
θτθσσσσσ
2cos2sin)(21
2sin2cos)(21)(
21
xyyxyx
xyyxyxx
+−−=
+−++=
dA
xyτxyτ
x
y
O
yσ
xσxyτ
xyτyσ
t
yox
(8.12)
c
xσ
yxτ
yσ
xyτxσ
xyτ
y
(normal) xy
x
θ θ
a
b
Figure 8.10
Page 17
Example 8.1 At a point of steel plate, the state of plane stress (2D) is defined by σ,
where
with respect to the axes xoy. Fine the stresses (normal and shear) for the same point acting on the plane orientated at angle 450 to the x-axis.
50
10050
2N/mm 505050100
⎥⎦
⎤⎢⎣
⎡=σ
50
50
100
045x
Page 18
Example 8.2 At a point of steel plate, the state of plane stress (2D) is defined by σ,
where
in the coordinate . Determine the stress matrix for the same point with respect to the axes , where x lies at 400 anticlockwise from x. Sketch the element in the axes and the stresses acting on its faces.
2N/mm 507575150
⎥⎦
⎤⎢⎣
⎡=σ
yoxxoy
150
5075
040
yox
σ
040x
x
Page 19
Example 8.3Variation of stresses (normal and shear) with rotation angle of normal to the x-axis
xσ
1801651501351201059075604530150θ0
2
2
2
N/mm 20
N/mm 32
N/mm 64
=
−=
=
xy
y
x
τ
σ
σ
64
32
20θ
64
3220
θ
xσyxτ
yxτxσ
Page 20
Example 8.3Variation of stresses (normal and shear) with rotation angle of normal to the x-axis
2
2
2
N/mm 20
N/mm 32
N/mm 64
=
−=
=
xy
y
x
τ
σ
σ
64
32
20θ
64
3220
θ
xσyxτ
θθτθθσ
2cos202sin482sin202cos4816
+−=++=
yx
x
Answer: Stresses acting on the face orientated with θ from (8.12) are
-60
-40
-20
0
20
40
60
80
0 15 30 45 60 75 90 105 120 135 150 165 180
xσ
yxτ
Page 21
8.10. Principal Stresses (Method 1)• The value of the stresses depend on
the choice of the coordinate system: we do not know the highest normal or the highest shear stress.
• For evaluation of yield, failure or fatigue independent (invariant) quantities are required;
• The extreme normal stresses(maximum and minimum values) are obtained via
This leads to
• Two angles are determined by
0=θ
σd
d x
( ) 022cos22sin ** ==+−− yxxyyx τθτθσσyx
xy
σστ
θ−
=∴2
2tan *
*θ
1
2 ; then
2tan
21 **1* πθθθθ
σστ
θ ±==−
= −ppp
yx
xy (8.13)
Figure 8.11
Page 22
• Inserting equation (8.13) into (8.14) leads to the principal stresses
)}(),(min{
)}(),(max{4
)(2
)(
2
1
22
2,1
ppxpx
ppxpx
xyyxyx
θσθσσ
θσθσσ
τσσσσ
σ
=
=
+−
±+
=
(14)
(8.16)
)2/(or ** πθθ ±
ppxyppyxyxppx
pxypyxyxpx
θτθσσσσθσ
θτθσσσσθσ
2sin2cos)(21)(
21)(
2sin2cos)(21)(
21)(
+−++=
+−++= (8.14)
• Substituting equation (8.13) into (8.12) leads to the principal stresses
• Shear stresses
0)()( == ppyxpyx θτθτ (8.15)
Figure 8.12
Page 23
Example 8.4
In a concrete structure a two-dimensionalstress state was computed with
2
2
2
N/mm 20
N/mm 32
N/mm 64
=
−=
=
xy
y
x
τ
σ
σ
Determine:
1. The normal and shear stresses under an angle of 60°
2. The principle stresses and the principles directions
3. Sketch the element corresponding to principal stresses
4. In which direction do you expect fractures (concrete)?
64
32
20*θ
Page 24
8.11. Principal Stresses (Method 2)• An easier way to compute principle
stresses is obtained by the mathematical scheme to compute Eigen value (natural values) of a matrix:
• Therefore:
22
2
2
12
2
1
4)(
2)(
4)(
2)(
στσσσσ
λ
στσσσσ
λ
=+−
−+
=
=+−
++
=
xyyxyx
xyyxyx
0detdet =−
−=−
λσττλσ
λyxy
xyxIσ
( )( )( ) ( ) 0
022
2
=−++−
=−−−
xyyxyx
xyyx
τσσσσλλ
τλσλσ
Example 8.5
Determine principal stresses by method 2.
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡10101010
yxy
xyx
σττσ
(8.17)
(8.18)
Page 25
)(21)(
21)()( 21 σσσσθσθσ +=+== yxssxsx
8.13. Principal Shear Stress• The extreme shear stresses
(maximum and minimum values) are obtained via
This leads to
Two angles are determined by
0=θ
τd
d yx
( ) 02sin22cos **** =−−− θτθσσ xyyx
xy
yx
τσσ
θ2
2tan ** −−=
*θ
1
2 ; then
2tan
21 ****1** πθθθθ
τσσ
θ ±==−
−= −sss
xy
yx
(8.20)eq.(8.14)] [from 22
2122
(max)σστ
σστ −
±=+⎟⎟⎠
⎞⎜⎜⎝
⎛ −±= xy
yxyx
**θ
)( syx θτ )( sx θσ
(8.19)
Figure 8.15
Page 26
• Maximal shear stress is obtained when an angle of 45° with respect to the principle directions is chosen
• Maximal shear stress is equal to
1σ
2σ
1σ
2σ
( ) ( )2121max 21
21 σσσσστ +=−±= (8.21)
450
maxτσ σ
maxτσ σ
Figure 8.16
Page 27
8.14. Linear Elasticity (Hooke’s Law)• In one dimension we have
• E is Young’s modulus and describes the stiffness of the material (i.e. it depends only on the material) as long as it is elastic
• Steel: E = 210.109 N/m²• Aluminium E = 70.109 N/m²• Beyond the elastic limit (yield
stress), the material reacts plastically.
• In elastic state, material can be loaded and unloaded without remaining strains, i.e. the loading procedure is reversible
• Material may show linear or non-linear elastic behaviour
εσ ⋅= E (8.22)
Stre
ss σ
=F/A
200
400
600
500
10 15 20 25
L0
∆L ∆L ∆L
Strain: 0 / LL∆=ε
L0 = originallength
E
Yield stress: yieldσ
Figure 8.18
Page 28
8.15. Multi-dimensional Linear Elasticity• In the multi-dimensional case,
stresses in one direction generate strains also in the other directions; we have (2D):
ν is the Poisson coefficient• In linear elasticity, it is sufficient to
describe the material by two parameters, e.g. Young’s modulus and Poisson coefficient; but you can use as well other parameters
( )
( )
G
E
E
xyxy
yxy
yxx
τγ
σνσε
νσσε
=
+−=
−=
1
1
(8.28) • Shear strains are related to shear stresses via (G is the shear modulus)
)1(2 ν+=
EG (8.29)
Figure 8.19
Page 29
• Natural values:
• The natural values are the principal values
(2D)
(3D)
8.16. Strain Matrix and Strain Principle Values• In analogy to stresses, the strains
are grouped into a strain matrix(or strain tensor)
(2 Dimension) (3 Dimension)
• The strain matrix is as well symmetric
• Principle strains can be computed by
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=
zzyzx
yzyyx
xzxyx
εγγγεγγγε
2/2/2/2/2/2/
ε (8.23) 2/
2/⎥⎦
⎤⎢⎣
⎡=
yyx
xyx
εγγε
ε
zyyzzxxzyxxy γγγγγγ === ;;
4)(
2)(
4)(
2)(
22
2
22
1
xyyxyx
xyyxyx
γεεεεε
γεεεεε
+−−
+=
+−+
+=
(8.25)
(8.24)
02/
2/det =
−−
λεγγλε
yxy
xyx
( )( )( ) ( ) 04/
04/22
2
=−++−
=−−−
xyyxyx
xyyx
γεεεελλ
γλελε
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
3
2
1
000000
εε
εε
(8.26);0
0
2
1⎥⎦
⎤⎢⎣
⎡=
εε
ε
2211 and λελε ==
(8.27)
Page 30
Example 8.7
Principle directions of a beam subjected to a central unit force
Compression
Tension
2N/mm 5040
40100⎥⎦
⎤⎢⎣
⎡−
=σ
A stress (matrix) tensor for the plane stress elasticity is given by
The Young’s modulus and Poisson’s ratio of this material are E=200×103N/mm2 and ν=1/3 respectively.
(1) Draw a square element with the above stress components acting on its sides;
(2) Calculate the strain tensor in matrix form; (3) Determine the principal stresses and their directions;(4) Determine the principal strains.
Page 31
8.17. Equivalent Stress and structure failure• To determine failure, the multi-
dimensional stress state should be taken into account;
• Although the one-dimensional stress limit is not reached, the structure may fail when subjected to multi-dimensional stresses;
• Multi-dimensionality of the stress state might improve or weaken.
• To obtain simple measures to evaluate multi-dimensional stress states, the concept of equivalent stress is used, i.e. a single value is obtained from the stress matrix, which is then compared to the one-dimensional limit stress (yield stress, ultimate stress, etc.)
n is the safety factor
• Simple approaches to determine the equivalent stress (the choice between them depends on the material):
– Normal stress hypothesis:Failure occurs when maximal normal stress is reached
– Shear stress hypothesis (2D):Failure occurs when the maximal shear stress is reached (Tresca)
– Energy hypothesis (2D):Failure occurs if the maximal elastic energy is reached (von Mises)
n/Limitequiv σσ ≤ (8.30)
1equiv σσ =
( ) ( ) 22max21equiv 42 xyyx τσστσσσ +−==−=
222
2122
21equiv
3 xyyxyx τσσσσ
σσσσσ
+−+=
−+=
(8.32)
(8.33)
(8.31)
Example (individual)
A hollow shaft, of external diameter D2 and internal diameter D1, where D1/D2=1/3, is required to transmit a torque of 100kNm and axial load 1500kN (compressive). If D2 is selected as 200mm, ultimate stress σLimit=300N/mm2, safety factor n=2.5 and von Mises criterion is considered, check whether the shaft is safe.
D2D1
propeller shaft1500kN
100kNm