Standard vs Custom Battery Packs - Decoding the Power Play
UDA 5 - P.pdf
1. 1
Lecture 5
Plane Stress
Transformation Equations
Stress elements and plane stress.
Stresses on inclined sections.
Transformation equations.
Principal stresses, angles, and planes.
Maximum shear stress.
2. 2
Normal and shear stresses on inclined sections
To obtain a complete picture of the stresses in a bar, we must
consider the stresses acting on an “inclined” (as opposed to a
“normal”) section through the bar.
P
P
Normal section
Inclined section
Because the stresses are the same throughout the entire bar, the
stresses on the sections are uniformly distributed.
P Inclined
section
P Normal
section
3. 3
P
V
N
P
θ
x
y
The force P can be resolved into components:
Normal force N perpendicular to the inclined plane, N = P cos θ
Shear force V tangential to the inclined plane V = P sin θ
If we know the areas on which the forces act, we can calculate
the associated stresses.
x
y
area
A
area A
( / cos )
θ
σθ
x
y
τθ
area
A
area A
( / cos )
θ
5. 5
Introduction to stress elements
Stress elements are a useful way to represent stresses acting at some
point on a body. Isolate a small element and show stresses acting on all
faces. Dimensions are “infinitesimal”, but are drawn to a large scale.
x
y
z
σ =
x P A
/
σx
x
y
σx
σ =
x P A
/
P σx = A
P /
x
y
Area A
6. 6
Maximum stresses on a bar in tension
P
P
a
σx = σmax = P / A
σx
a
Maximum normal stress,
Zero shear stress
7. 7
P
P
a
b
Maximum stresses on a bar in tension
b
σx/2
σx/2
τmax = σx/2
θ = 45°
Maximum shear stress,
Non-zero normal stress
8. 8
Stress Elements and Plane Stress
When working with stress elements, keep in mind that only
one intrinsic state of stress exists at a point in a stressed
body, regardless of the orientation of the element used to
portray the state of stress.
We are really just rotating axes to represent stresses in a
new coordinate system.
x
y
σx σx
θ
y1
x1
9. 9
y
z
x
Normal stresses σx, σy, σz
(tension is positive)
σx
σx
σy
σy
σz
τxy
τyx
τxz
τzx
τzy
τyz
Shear stresses τxy = τyx,
τxz = τzx, τyz = τzy
Sign convention for τab
Subscript a indicates the “face” on which the stress acts
(positive x “face” is perpendicular to the positive x direction)
Subscript b indicates the direction in which the stress acts
Strictly σx = σxx, σy = σyy, σz = σzz
10. 10
When an element is in plane stress in the xy plane, only the x and
y faces are subjected to stresses (σz = 0 and τzx = τxz = τzy = τyz = 0).
Such an element could be located on the free surface of a body (no
stresses acting on the free surface).
σx σx
σy
σy
τxy
τyx
τxy
τyx
x
y
Plane stress element in 2D
σx, σy
τxy = τyx
σz = 0
11. 11
Stresses on Inclined Sections
σx σx
σy
τxy
τyx
τxy
τyx
x
y
x
y
σx1
θ
y1
x1
σx1
σy1
σy1
τx y
1 1
τy x
1 1
τx y
1 1
τy x
1 1
The stress system is known in terms of coordinate system xy.
We want to find the stresses in terms of the rotated coordinate
system x1y1.
Why? A material may yield or fail at the maximum value of σ or τ. This
value may occur at some angle other than θ = 0. (Remember that for uni-
axial tension the maximum shear stress occurred when θ = 45 degrees. )
12. 12
Transformation Equations
y
σx1
x
θ
y1
x1
σx
σy
τx y
1 1
τxy
τyx
θ
Stresses
y
σ θ
x1 A sec
x
θ
y1
x1
σxA
σ θ
y A tan
τ θ
x y
1 1 A sec
τxy A
τ θ
yx A tan
θ
Forces
Left face has area A.
Bottom face has area A tan θ.
Inclined face has area A sec θ.
Forces can be found from stresses if
the area on which the stresses act is
known. Force components can then
be summed.
13. 13
y
σ θ
x1 A sec
x
θ
y1
x1
σxA
σ θ
y A tan
τ θ
x y
1 1 A sec
τxy A
τ θ
yx A tan
θ
( ) ( ) ( ) ( ) 0
cos
tan
sin
tan
sin
cos
sec
:
direction
in the
forces
Sum
1
1
=
−
−
−
− θ
θ
τ
θ
θ
σ
θ
τ
θ
σ
θ A
A
A
A
A
σ
x
yx
y
xy
x
x
( ) ( ) ( ) ( ) 0
sin
tan
cos
tan
cos
sin
sec
:
direction
in the
forces
Sum
1
1
1
=
−
−
−
+ θ
θ
τ
θ
θ
σ
θ
τ
θ
σ
θ
τ A
A
A
A
A
y
yx
y
xy
x
y
x
( ) ( )
θ
θ
τ
θ
θ
σ
σ
τ
θ
θ
τ
θ
σ
θ
σ
σ
τ
τ
2
2
1
1
2
2
1
sin
cos
cos
sin
cos
sin
2
sin
cos
:
gives
g
simplifyin
and
Using
−
+
−
−
=
+
+
=
=
xy
y
x
y
x
xy
y
x
x
yx
xy
15. 15
Example: The state of plane stress at a point is represented by the
stress element below. Determine the stresses acting on an element
oriented 30° clockwise with respect to the original element.
80 MPa 80 MPa
50 MPa
x
y
50 MPa
25 MPa
Define the stresses in terms of the
established sign convention:
σx = -80 MPa σy = 50 MPa
τxy = -25 MPa
We need to find σx1, σy1, and
τx1y1 when θ = -30°.
Substitute numerical values into the transformation equations:
( ) ( ) ( ) MPa
9
.
25
30
2
sin
25
30
2
cos
2
50
80
2
50
80
2
sin
2
cos
2
2
1
1
−
=
°
−
−
+
°
−
−
−
+
+
−
=
+
−
+
+
=
x
xy
y
x
y
x
x
σ
θ
τ
θ
σ
σ
σ
σ
σ
16. 16
( )
( ) ( ) ( ) ( ) MPa
8
.
68
30
2
cos
25
30
2
sin
2
50
80
2
cos
2
sin
2
1
1
1
1
−
=
°
−
−
+
°
−
−
−
−
=
+
−
−
=
y
x
xy
y
x
y
x
τ
θ
τ
θ
σ
σ
τ
( ) ( ) ( ) MPa
15
.
4
30
2
sin
25
30
2
cos
2
50
80
2
50
80
2
sin
2
cos
2
2
1
1
−
=
°
−
−
−
°
−
−
−
−
+
−
=
−
−
−
+
=
y
xy
y
x
y
x
y
σ
θ
τ
θ
σ
σ
σ
σ
σ
Note that σy1 could also be obtained
(a) by substituting +60° into the
equation for σx1 or (b) by using the
equation σx + σy = σx1 + σy1
+60°
25.8 MPa
25.8 MPa
4.15 MPa
x
y
4.15 MPa
68.8 MPa
x1
y1
-30
o
(from Hibbeler, Ex. 15.2)
17. 17
Principal Stresses
The maximum and minimum normal stresses (σ1 and σ2) are
known as the principal stresses. To find the principal stresses,
we must differentiate the transformation equations.
( ) ( )
( )
y
x
xy
p
xy
y
x
x
xy
y
x
x
xy
y
x
y
x
x
d
d
d
d
σ
σ
τ
θ
θ
τ
θ
σ
σ
θ
σ
θ
τ
θ
σ
σ
θ
σ
θ
τ
θ
σ
σ
σ
σ
σ
−
=
=
+
−
−
=
=
+
−
−
=
+
−
+
+
=
2
2
tan
0
2
cos
2
2
sin
0
2
cos
2
2
sin
2
2
2
sin
2
cos
2
2
1
1
1
θp are principal angles associated with
the principal stresses
There are two values of 2θp in the range 0-360°, with values differing by 180°.
There are two values of θp in the range 0-180°, with values differing by 90°.
So, the planes on which the principal stresses act are mutually perpendicular.
18. 18
θ
τ
θ
σ
σ
σ
σ
σ
σ
σ
τ
θ
2
sin
2
cos
2
2
2
2
tan
1 xy
y
x
y
x
x
y
x
xy
p
+
−
+
+
=
−
=
We can now solve for the principal stresses by substituting for θp
in the stress transformation equation for σx1. This tells us which
principal stress is associated with which principal angle.
2θp
τxy
(σx – σy) / 2
R
R
R
R
xy
p
y
x
p
xy
y
x
τ
θ
σ
σ
θ
τ
σ
σ
=
−
=
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
=
2
sin
2
2
cos
2
2
2
2
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
−
+
+
=
R
R
xy
xy
y
x
y
x
y
x τ
τ
σ
σ
σ
σ
σ
σ
σ
2
2
2
1
19. 19
Substituting for R and re-arranging gives the larger of the two
principal stresses:
2
2
1
2
2
xy
y
x
y
x
τ
σ
σ
σ
σ
σ +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
+
+
=
To find the smaller principal stress, use σ1 + σ2 = σx + σy.
2
2
1
2
2
2
xy
y
x
y
x
y
x τ
σ
σ
σ
σ
σ
σ
σ
σ +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
−
+
=
−
+
=
These equations can be combined to give:
2
2
2
,
1
2
2
xy
y
x
y
x
τ
σ
σ
σ
σ
σ +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
±
+
=
Principal stresses
To find out which principal stress goes with which principal angle,
we could use the equations for sin θp and cos θp or for σx1.
20. 20
The planes on which the principal stresses act are called the
principal planes. What shear stresses act on the principal planes?
Solving either equation gives the same expression for tan 2θp
Hence, the shear stresses are zero on the principal planes.
( )
( )
( ) 0
2
cos
2
2
sin
0
2
cos
2
2
sin
0
2
cos
2
sin
2
0
and
0
for
equations
the
Compare
1
1
1
1
1
1
=
+
−
−
=
=
+
−
−
=
+
−
−
=
=
=
θ
τ
θ
σ
σ
θ
σ
θ
τ
θ
σ
σ
θ
τ
θ
σ
σ
τ
θ
σ
τ
xy
y
x
x
xy
y
x
xy
y
x
y
x
x
y
x
d
d
d
d
24. 24
The two principal stresses determined so far are the principal
stresses in the xy plane. But … remember that the stress element
is 3D, so there are always three principal stresses.
y
x
σx
σx
σy
σy
τ
τ
τ
τ
σx, σy, τxy = τyx = τ
yp
zp
xp
σ1
σ1
σ2
σ2
σ3 = 0
σ1, σ2, σ3 = 0
Usually we take σ1 > σ2 > σ3. Since principal stresses can be com-
pressive as well as tensile, σ3 could be a negative (compressive)
stress, rather than the zero stress.
25. 25
Maximum Shear Stress
( )
( )
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
−
=
=
−
−
−
=
+
−
−
=
xy
y
x
s
xy
y
x
y
x
xy
y
x
y
x
d
d
τ
σ
σ
θ
θ
τ
θ
σ
σ
θ
τ
θ
τ
θ
σ
σ
τ
2
2
tan
0
2
sin
2
2
cos
2
cos
2
sin
2
1
1
1
1
To find the maximum shear stress, we must differentiate the trans-
formation equation for shear.
There are two values of 2θs in the range 0-360°, with values differing by 180°.
There are two values of θs in the range 0-180°, with values differing by 90°.
So, the planes on which the maximum shear stresses act are mutually
perpendicular.
Because shear stresses on perpendicular planes have equal magnitudes,
the maximum positive and negative shear stresses differ only in sign.
26. 26
2θs
τxy
(σ
x
–
σ
y
)
/
2
R
( ) θ
τ
θ
σ
σ
τ
τ
σ
σ
θ
2
cos
2
sin
2
2
2
tan
1
1 xy
y
x
y
x
xy
y
x
s
+
−
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
−
=
We can now solve for the maximum shear stress by substituting for
θs in the stress transformation equation for τx1y1.
R
R
R
y
x
s
xy
s
xy
y
x
2
2
sin
2
cos
2
2
2
2
σ
σ
θ
τ
θ
τ
σ
σ
−
−
=
=
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
=
max
min
2
2
max
2
τ
τ
τ
σ
σ
τ −
=
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
= xy
y
x
27. 27
Use equations for sin θs and cos θs or τx1y1 to find out which face
has the positive shear stress and which the negative.
What normal stresses act on the planes with maximum shear stress?
Substitute for θs in the equations for σx1 and σy1 to get
s
y
x
y
x σ
σ
σ
σ
σ =
+
=
=
2
1
1
x
y
σs
θs
σs
σs
σs
τmax
τmax
τmax
τmax
σy
σx σx
σy
τxy
τyx
τxy
τyx
x
y
28. 28
Example: The state of plane stress at a point is represented by the
stress element below. Determine the maximum shear stresses and
draw the corresponding stress element.
80 MPa 80 MPa
50 MPa
x
y
50 MPa
25 MPa
Define the stresses in terms of the
established sign convention:
σx = -80 MPa σy = 50 MPa
τxy = -25 MPa
( ) MPa
6
.
69
25
2
50
80
2
2
2
max
2
2
max
=
−
+
⎟
⎠
⎞
⎜
⎝
⎛ −
−
=
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
=
τ
τ
σ
σ
τ xy
y
x
MPa
15
2
50
80
2
−
=
+
−
=
+
=
s
y
x
s
σ
σ
σ
σ
30. 30
Finally, we can ask how the principal stresses and maximum shear
stresses are related and how the principal angles and maximum
shear angles are related.
2
2
2
2
2
2
2
1
max
max
2
1
2
2
2
1
2
2
2
,
1
σ
σ
τ
τ
σ
σ
τ
σ
σ
σ
σ
τ
σ
σ
σ
σ
σ
−
=
=
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
=
−
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
±
+
=
xy
y
x
xy
y
x
y
x
p
p
s
y
x
xy
p
xy
y
x
s
θ
θ
θ
σ
σ
τ
θ
τ
σ
σ
θ
2
cot
2
tan
1
2
tan
2
2
tan
2
2
tan
−
=
−
=
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
−
=
32. 32
80 MPa 80 MPa
50 MPa
x
y
50 MPa
25 MPa
Original Problem
σx = -80, σy = 50, τxy = 25
84.6 MPa
84.6 MPa
54.6 MPa
x
y
54.6 MPa
10.5
o
100.5
o
Principal Stresses
σ1 = 54.6, σ2 = 0, σ3 = -84.6
15 MPa
15 MPa
15 MPa
x
y
15 MPa
69.6 MPa
-34.5o
55.5o
Maximum Shear
τmax = 69.6, σs = -15
°
°
−
=
°
±
°
=
°
±
=
5
.
55
,
5
.
34
45
5
.
10
45
s
s
p
s
θ
θ
θ
θ
( )
MPa
6
.
69
2
6
.
84
6
.
54
2
max
max
2
1
max
=
−
−
=
−
=
τ
τ
σ
σ
τ
33. 1
Mohr’s Circle for Plane Stress
Transformation equations for plane stress.
Procedure for constructing Mohr’s circle.
Stresses on an inclined element.
Principal stresses and maximum shear stresses.
Introduction to the stress tensor.
34. 2
Stress Transformation Equations
σx σx
σy
τxy
τyx
τxy
τyx
x
y
x
y
σx1
θ
y1
x1
σx1
σy1
σy1
τx y
1 1
τy x
1 1
τx y
1 1
τy x
1 1
( ) θ
τ
θ
σ
σ
τ
θ
τ
θ
σ
σ
σ
σ
σ
2
cos
2
sin
2
2
sin
2
cos
2
2
1
1
1
xy
y
x
y
x
xy
y
x
y
x
x
+
−
−
=
+
−
+
+
=
If we vary θ from 0° to 360°, we will get all possible values of σx1 and τx1y1
for a given stress state. It would be useful to represent σx1 and τx1y1 as
functions of θ in graphical form.
35. 3
To do this, we must re-write the transformation equations.
( ) θ
τ
θ
σ
σ
τ
θ
τ
θ
σ
σ
σ
σ
σ
2
cos
2
sin
2
2
sin
2
cos
2
2
1
1
1
xy
y
x
y
x
xy
y
x
y
x
x
+
−
−
=
+
−
=
+
−
Eliminate θ by squaring both sides of each equation and adding
the two equations together.
2
2
2
1
1
2
1
2
2
xy
y
x
y
x
y
x
x τ
σ
σ
τ
σ
σ
σ +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
=
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ +
−
Define σavg and R
2
2
2
2
xy
y
x
y
x
avg R τ
σ
σ
σ
σ
σ +
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ −
=
+
=
36. 4
Substitue for σavg and R to get
( ) 2
2
1
1
2
1 R
y
x
avg
x =
+
− τ
σ
σ
which is the equation for a circle with centre (σavg,0) and radius R.
This circle is usually referred to as
Mohr’s circle, after the German civil
engineer Otto Mohr (1835-1918). He
developed the graphical technique for
drawing the circle in 1882.
The construction of Mohr’s circle is
one of the few graphical techniques
still used in engineering. It provides
a simple and clear picture of an
otherwise complicated analysis.
37. 5
Sign Convention for Mohr’s Circle
x
y
σx1
θ
y1
x1
σx1
σy1
σy1
τx y
1 1
τy x
1 1
τx y
1 1
τy x
1 1
( ) 2
2
1
1
2
1 R
y
x
avg
x =
+
− τ
σ
σ
σx1
τx1y1
R
2θ
σavg
Notice that shear stress is plotted as positive downward.
The reason for doing this is that 2θ is then positive counterclockwise,
which agrees with the direction of 2θ used in the derivation of the
tranformation equations and the direction of θ on the stress element.
Notice that although 2θ appears in Mohr’s circle, θ appears on the
stress element.
38. 6
Procedure for Constructing Mohr’s Circle
1. Draw a set of coordinate axes with σx1 as abscissa (positive to the
right) and τx1y1 as ordinate (positive downward).
2. Locate the centre of the circle c at the point having coordinates σx1
= σavg and τx1y1 = 0.
3. Locate point A, representing the stress conditions on the x face of
the element by plotting its coordinates σx1 = σx and τx1y1 = τxy. Note
that point A on the circle corresponds to θ = 0°.
4. Locate point B, representing the stress conditions on the y face of
the element by plotting its coordinates σx1 = σy and τx1y1 = −τxy.
Note that point B on the circle corresponds to θ = 90°.
5. Draw a line from point A to point B, a diameter of the circle passing
through point c. Points A and B (representing stresses on planes
at 90° to each other) are at opposite ends of the diameter (and
therefore 180° apart on the circle).
6. Using point c as the centre, draw Mohr’s circle through points A
and B. This circle has radius R.
(based on Gere)
40. 8
Stresses on an Inclined Element
1. On Mohr’s circle, measure an angle 2θ counterclockwise from
radius cA, because point A corresponds to θ = 0 and hence is
the reference point from which angles are measured.
2. The angle 2θ locates the point D on the circle, which has
coordinates σx1 and τx1y1. Point D represents the stresses on the
x1 face of the inclined element.
3. Point E, which is diametrically opposite point D on the circle, is
located at an angle 2θ + 180° from cA (and 180° from cD). Thus
point E gives the stress on the y1 face of the inclined element.
4. So, as we rotate the x1y1 axes counterclockwise by an angle θ,
the point on Mohr’s circle corresponding to the x1 face moves
counterclockwise through an angle 2θ.
(based on Gere)
43. 11
Maximum Shear Stress
σx σx
σy
τxy
τyx
τxy
τyx
x
y
A (θ=0)
σx1
τx1y1
c
R
B (θ=90)
A
B
2θs
σs
τmax
τmin
x
y
σs
θs
σs
σs
σs
τmax
τmax
τmax
τmax
Note carefully the
directions of the
shear forces.
44. 12
Example: The state of plane stress at a point is represented by the stress
element below. Draw the Mohr’s circle, determine the principal stresses and
the maximum shear stresses, and draw the corresponding stress elements.
80 MPa 80 MPa
50 MPa
x
y
50 MPa
25 MPa
σ
τ
15
2
50
80
2
−
=
+
−
=
+
=
=
y
x
avg
c
σ
σ
σ
c
A (θ=0)
A
B (θ=90)
B
( )
( ) ( )
6
.
69
25
65
25
15
50
2
2
2
2
=
+
=
+
−
−
=
R
R
R
σ1
σ2
MPa
6
.
84
MPa
6
.
54
6
.
69
15
2
1
2
,
1
2
,
1
−
=
=
±
−
=
±
=
σ
σ
σ
σ R
c
τmax
MPa
15
MPa
6
.
69
s
max
−
=
=
=
=
c
R
σ
τ
45. 13
84.6 MPa
84.6 MPa
54.6 MPa
x
y
54.6 MPa
10.5o
100.5
o
σ
τ
c
A (θ=0)
B (θ=90)
R
80 MPa 80 MPa
50 MPa
x
y
50 MPa
25 MPa
σ1
σ2
2θ2
2θ1
°
=
°
=
°
=
°
+
=
°
=
=
−
=
5
.
10
5
.
100
201
180
0
.
21
2
0
.
21
2
3846
.
0
15
80
25
2
tan
2
1
1
2
2
θ
θ
θ
θ
θ
2θ
46. 14
σ
τ
c
A (θ=0)
B (θ=90)
R
80 MPa 80 MPa
50 MPa
x
y
50 MPa
25 MPa
2θ2
2θ
15 MPa
15 MPa
15 MPa
x
y
15 MPa
69.6 MPa
-34.5o
55.5
o
2θsmax
°
=
°
=
°
+
=
°
=
5
.
55
0
.
111
90
0
.
21
2
0
.
21
2
max
max
2
s
s
θ
θ
θ
2θsmin
taking sign convention into
account
°
−
=
°
−
=
−
−
=
°
=
5
.
34
0
.
69
)
0
.
21
90
(
2
0
.
21
2
min
min
2
s
s
θ
θ
θ
τmax
τmin
47. 15
80 MPa 80 MPa
50 MPa
x
y
50 MPa
25 MPa
A (θ=0)
σ
τ
B (θ=90)
25.8 MPa
25.8 MPa
4.15 MPa
x
y
4.15 MPa
68.8 MPa
x1
y1
-30
o
Example: The state of plane stress at a point is represented by the stress
element below. Find the stresses on an element inclined at 30° clockwise
and draw the corresponding stress elements.
-60°
-60+180°
C (θ = -30°)
C
D (θ = -30+90°)
D
2θ2
σx1 = c – R cos(2θ2+60)
σy1 = c + R cos(2θ2+60)
τx1y1= -R sin (2θ2+60)
σx1 = -26
σy1 = -4
τx1y1= -69
2θ
θ = -30°
2θ = -60°
48. 16
σ
τ
A (θ=0)
B (θ=90)
Principal Stresses σ1 = 54.6 MPa, σ2 = -84.6 MPa
But we have forgotten about the third principal stress!
Since the element is in plane stress (σz = 0),
the third principal stress is zero.
σ1 = 54.6 MPa
σ2 = 0 MPa
σ3 = -84.6 MPa
σ1
σ2
σ3
This means three
Mohr’s circles can
be drawn, each
based on two
principal stresses:
σ1 and σ3
σ1 and σ2
σ2 and σ3
52. 20
From our analyses so far, we know that for a given stress system,
it is possible to find a set of three principal stresses. We also know
that if the principal stresses are acting, the shear stresses must be
zero. In terms of the stress tensor,
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛
zz
zy
zx
yz
yy
yx
xz
xy
xx
σ
τ
τ
τ
σ
τ
τ
τ
σ
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎝
⎛
3
2
1
0
0
0
0
0
0
σ
σ
σ
In mathematical terms, this is the process of matrix diagonaliza-
tion in which the eigenvalues of the original matrix are just the
principal stresses.
53. 21
80 MPa 80 MPa
50 MPa
x
y
50 MPa
25 MPa
Example: The state of plane stress at a point is represented by the
stress element below. Find the principal stresses.
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
50
25
25
80
y
yx
xy
x
M
σ
τ
τ
σ
We must find the eigenvalues of
this matrix.
Remember the general idea of eigenvalues. We are looking
for values of λ such that:
Ar = λr where r is a vector, and A is a matrix.
Ar – λr = 0 or (A – λI) r = 0 where I is the identity matrix.
For this equation to be true, either r = 0 or det (A – λI) = 0.
Solving the latter equation (the “characteristic equation”)
gives us the eigenvalues λ1 and λ2.
54. 22
6
.
54
,
6
.
84
0
4625
30
0
)
25
)(
25
(
)
50
)(
80
(
0
50
25
25
80
det
2
−
=
=
−
+
=
−
−
−
−
−
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
−
−
λ
λ
λ
λ
λ
λ
λ
So, the principal stresses are –84.6 MPa and
54.6 MPa, as before.
Knowing the eigenvalues, we can find the eigenvectors. These can be
used to find the angles at which the principal stresses act. To find the
eigenvectors, we substitute the eigenvalues into the equation (A – λI ) r
= 0 one at a time and solve for r.
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
−
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
−
−
0
0
6
.
54
50
25
25
6
.
54
80
0
0
50
25
25
80
y
x
y
x
λ
λ
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛−
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
−
−
−
1
186
.
0
186
.
0
0
0
64
.
4
25
25
6
.
134
y
x
y
x
is one eigenvector.
