# Lecture 5 Plane Stress Transformation Equations

P4 Stress and Strain
Dr. A.B. Zavatsky
HT08
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.
1
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.
Inclined section
Normal section
P
P
Because the stresses are the same throughout the entire bar, the
stresses on the sections are uniformly distributed.
P
Inclined
section
P
Normal
section
2
y
P
N
θ
x
P
V
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.
y
area
A
y
σθ
x
area (A / cos θ)
τθ
area
A
x
area (A / cos θ)
3
y
τθ
σθ
θ
x
σθ =
Force
N
P cos θ P
=
=
= cos 2θ
Area Area A / cos θ A
σ θ = σ x cos 2θ =
σx
(1+ cos 2θ )
2
σmax = σx occurs when θ = 0&deg;
τθ =
−V
− P sin θ
Force
P
=
=
= − sin θ cos θ
Area Area A / cos θ
A
τ θ = −σ x sin θ cos θ = −
σx
(sin 2θ )
2
τmax = &plusmn; σx/2 occurs when θ = -/+ 45&deg;
4
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.
y
P
σx = P / A
x
Area A
y
σx
y
σx = P/A
x
σx
σx = P/A
x
z
5
Maximum stresses on a bar in tension
P
P
a
σx
σx = σmax = P / A
a
Maximum normal stress,
Zero shear stress
6
Maximum stresses on a bar in tension
P
P
a
b
σx/2
θ = 45&deg;
τmax = σx/2
σx/2
b
Maximum shear stress,
Non-zero normal stress
7
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.
y1
y
x1
σx
θ
σx
x
8
y
σy
τyz
τxy
τzy
σx
Normal stresses σx, σy, σz
(tension is positive)
τyx
τzx τxz
σz
σx x
Shear stresses τxy = τyx,
τxz = τzx, τyz = τzy
z
σy
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
9
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).
Plane stress element in 2D
σx, σy
τxy = τyx
σz = 0
σy
τyx
τxy
y
x
σx
σx
τxy
τyx
σy
10
Stresses on Inclined Sections
y1
σy
τxy
x
σx
σy1 τy1x1
τyx
y
τyx
τx1y1
x1
σx1
θ
σx
σx1
τxy
y
τx1y1
x
τy1x1
σy1
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 uniaxial tension the maximum shear stress occurred when θ = 45 degrees. )
11
Transformation Equations
Forces
Stresses
y
y
y1
y1
θ
σx
x1
τx1y1
σx1
θ
θ
σxA
σx1 A sec θ
x
τxy
τx1y1 A sec θ
x1
θ
x
τxy A
τyx
τyx A tan θ
σy
Forces can be found from stresses if
the area on which the stresses act is
known. Force components can then
be summed.
σy A tan θ
Left face has area A.
Bottom face has area A tan θ.
Inclined face has area A sec θ.
12
y
y1
θ
σxA
τx1y1 A sec θ
σx1 A sec θ
x1
θ
x
τxy A
τyx A tan θ
σy A tan θ
Sum forces in the x1 direction :
σ x1 A sec θ − (σ x A) cos θ − (τ xy A)sin θ − (σ y A tan θ )sin θ − (τ yx A tan θ )cos θ = 0
Sum forces in the y1 direction :
τ x1 y1 A sec θ + (σ x A)sin θ − (τ xy A)cos θ − (σ y A tan θ )cos θ − (τ yx A tan θ )sin θ = 0
Using τ xy = τ yx and simplifying gives :
σ x1 = σ x cos 2 θ + σ y sin 2 θ + 2 τ xy sin θ cos θ
τ x1 y1 = − (σ x − σ y ) sin θ cos θ + τ xy (cos 2 θ − sin 2 θ )
13
Using the following trigonometric identities
cos 2 θ =
1 + cos 2θ
2
sin 2 θ =
1 − cos 2θ
2
sin θ cos θ =
sin 2θ
2
gives the transformation equations for plane stress :
σ x +σ y
σ x −σ y
σ x1 =
+
cos 2θ + τ xy sin 2θ
2
2
HLT, page 108
(
σ x −σ y )
τ x1 y1 = −
sin 2θ + τ xy cos 2θ
2
For stresses on the y1 face, substitute θ + 90&deg; for θ :
σ y1 =
σx +σ y
2
−
σ x −σ y
2
cos 2θ − τ xy sin 2θ
Summing the expressions for x1 and y1 gives :
σ x1 + σ y1 = σ x + σ y
Can be used to find σy1, instead of eqn above.
14
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&deg; clockwise with respect to the original element.
50 MPa
y
80 MPa
80 MPa
x
25 MPa
50 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&deg;.
Substitute numerical values into the transformation equations:
σ x1 =
σx +σ y
+
σ x −σ y
cos 2θ + τ xy sin 2θ
2
2
− 80 + 50
− 80 − 50
σ x1 =
+
cos 2(− 30&deg;) + (− 25)sin 2(− 30&deg;) = −25.9 MPa
2
2
15
σ y1 =
σ x +σ y
−
σ x −σ y
cos 2θ − τ xy sin 2θ
2
2
− 80 + 50
− 80 − 50
σ y1 =
−
cos 2(− 30&deg;) − (− 25)sin 2(− 30&deg;) = −4.15 MPa
2
2
σ x −σ y
sin 2θ + τ xy cos 2θ
τ x1 y1 = −
2
(− 80 − 50) sin 2(− 30&deg;) + (− 25) cos 2(− 30&deg;) = −68.8 MPa
τ x1 y1 = −
2
(
)
y
25.8 MPa
y1
4.15 MPa
+60&deg;
o
-30
x
Note that σy1 could also be obtained
(a) by substituting +60&deg; into the
equation for σx1 or (b) by using the
equation σx + σy = σx1 + σy1
25.8 MPa
x1
4.15 MPa
68.8 MPa
(from Hibbeler, Ex. 15.2)
16
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.
σ x1 =
σx +σ y
2
σ x −σ y
+
σ x −σ y
2
cos 2θ + τ xy sin 2θ
dσ x1
(− 2 sin 2θ ) + τ xy (2 cos 2θ ) = 0
=
dθ
2
dσ x1
= − (σ x − σ y ) sin 2θ + 2τ xy cos 2θ = 0
dθ
2τ xy
θp are principal angles associated with
tan 2θ p =
σ x −σ y
the principal stresses (HLT, page 108)
There are two values of 2θp in the range 0-360&deg;, with values differing by 180&deg;.
There are two values of θp in the range 0-180&deg;, with values differing by 90&deg;.
So, the planes on which the principal stresses act are mutually perpendicular.
17
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.
2τ xy
tan 2θ p =
σ x −σ y
σ x1 =
σ x +σ y
2
+
σ x −σ y
2
cos 2θ + τ xy sin 2θ
⎛σ x −σ y ⎞
R 2 = ⎜⎜
⎝
R
τxy
2θp
sin 2θ p =
(σx – σy) / 2
σ1 =
σx +σ y
2
+
cos 2θ p =
2
2
⎟⎟ + τ xy 2
⎠
σ x −σ y
τ xy
2R
R
σ x −σ y ⎛σ x −σ y ⎞
2
⎜⎜
⎝
2R
⎛ τ xy ⎞
⎟⎟ + τ xy ⎜⎜ ⎟⎟
⎠
⎝ R ⎠
18
Substituting for R and re-arranging gives the larger of the two
principal stresses:
σ1 =
σ x +σ y
2
⎛σ x −σ y
⎜⎜
⎝ 2
+
2
⎞
⎟⎟ + τ xy 2
⎠
To find the smaller principal stress, use σ1 + σ2 = σx + σy.
σ 2 = σ x + σ y − σ1 =
σ x +σ y
2
⎛σ x −σ y
⎜⎜
⎝ 2
−
2
⎞
⎟⎟ + τ xy 2
⎠
These equations can be combined to give:
σ 1, 2 =
σ x +σ y
2
&plusmn;
⎛ σ x −σ y ⎞
⎜⎜
⎝
2
2
⎟⎟ + τ xy 2
⎠
Principal stresses
(HLT page 108)
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.
19
The planes on which the principal stresses act are called the
principal planes. What shear stresses act on the principal planes?
Compare the equations for τ x1 y1 = 0 and dσ x1 dθ = 0
τ x1 y1 =
(
σ
−
x
−σ y )
sin 2θ + τ xy cos 2θ = 0
2
− (σ x − σ y ) sin 2θ + 2τ xy cos 2θ = 0
dσ x1
= − (σ x − σ y ) sin 2θ + 2τ xy cos 2θ = 0
dθ
Solving either equation gives the same expression for tan 2θp
Hence, the shear stresses are zero on the principal planes.
20
y
σy
τyx
σ2
τxy
y
x
σx
θp2
σx
σ1
θp1
x
σ1
τxy
τyx
σ2
σy
Principal Stresses
σ 1, 2 =
σ x +σ y
2
&plusmn;
⎛ σ x −σ y ⎞
⎜⎜
⎝
2
2
⎟⎟ + τ xy 2
⎠
Principal Angles defining the Principal Planes
tan 2θ p =
2τ xy
σ x −σ y
21
Example: The state of plane stress at a point is represented by the
stress element below. Determine the principal stresses and draw the
corresponding stress element.
50 MPa
y
80 MPa
80 MPa
x
Define the stresses in terms of the
established sign convention:
σx = -80 MPa σy = 50 MPa
τxy = -25 MPa
25 MPa
50 MPa
σ 1, 2 =
σ 1, 2
σx +σ y
2
&plusmn;
− 80 + 50
=
&plusmn;
2
σ 1 = 54.6 MPa
⎛σ x −σ y
⎜⎜
⎝ 2
2
⎞
⎟⎟ + τ xy 2
⎠
⎛ − 80 − 50 ⎞
2
⎜
⎟ + (− 25) = −15 &plusmn; 69.6
2
⎝
⎠
σ 2 = −84.6 MPa
2
22
tan 2θ p =
tan 2θ p =
2τ xy
y
σx −σ y
54.6 MPa
2(− 25)
= 0.3846
− 80 − 50
84.6 MPa
o
100.5
84.6 MPa
o
10.5
x
2θ p = 21.0&deg; and 21.0 + 180&deg;
θ p = 10.5&deg;, 100.5&deg;
54.6 MPa
But we must check which angle goes with which principal stress.
σ x1 =
σx +σ y
+
σ x −σ y
cos 2θ + τ xy sin 2θ
2
2
− 80 + 50
− 80 − 50
σ x1 =
+
cos 2(10.5&deg;) + (− 25)sin 2(10.5&deg;) = −84.6 MPa
2
2
σ1 = 54.6 MPa with θp1 = 100.5&deg;
σ2 = -84.6 MPa with θp2 = 10.5&deg;
23
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.
yp
y
σ2
σy
τ
σ1
τ
σx
σx
τ
x
σ1
σ3 = 0
τ
σy
σx, σy, τxy = τyx = τ
xp
zp
σ2
σ1, σ2, σ3 = 0
Usually we take σ1 &gt; σ2 &gt; σ3. Since principal stresses can be compressive as well as tensile, σ3 could be a negative (compressive)
stress, rather than the zero stress.
24
Maximum Shear Stress
To find the maximum shear stress, we must differentiate the transformation equation for shear.
(
σ x −σ y )
τ x1 y1 = −
sin 2θ + τ xy cos 2θ
2
dτ x1 y1
= − (σ x − σ y )cos 2θ − 2τ xy sin 2θ = 0
dθ
⎛σ x −σ y ⎞
⎟
tan 2θ s = − ⎜
⎜ 2τ
⎟
xy
⎝
⎠
There are two values of 2θs in the range 0-360&deg;, with values differing by 180&deg;.
There are two values of θs in the range 0-180&deg;, with values differing by 90&deg;.
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.
25
We can now solve for the maximum shear stress by substituting for
θs in the stress transformation equation for τx1y1.
(σx – σy) / 2
⎛σ x −σ y ⎞
⎟
tan 2θ s = − ⎜
⎟
⎜ 2τ
xy
⎝
⎠
(
σ x −σ y )
τ x1 y1 = −
sin 2θ + τ xy cos 2θ
2
R
2θs
τxy
τ max =
⎛σ x −σ y
⎜⎜
2
⎝
⎛σ x −σ y ⎞
R 2 = ⎜⎜
⎝
cos 2θ s =
2
⎟⎟ + τ xy 2
⎠
2
τ xy
R
sin 2θ s = −
σ x −σ y
2R
2
⎞
⎟⎟ + τ xy 2
⎠
τ min = − τ max
26
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
σ x1 = σ y1 =
2
=σs
y
σy
σs
τmax
τyx
τxy
y
x
σx
σx +σ y
τyx
σy
σs
θs
σx
σs
τxy
τmax
τmax
x
τmax
σs
27
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.
50 MPa
y
80 MPa
80 MPa
x
Define the stresses in terms of the
established sign convention:
σx = -80 MPa σy = 50 MPa
τxy = -25 MPa
25 MPa
50 MPa
τ max =
⎛σ x −σ y ⎞
⎜⎜
⎝
2
2
⎟⎟ + τ xy
⎠
⎛ − 80 − 50 ⎞
2
⎜
⎟ + (− 25) = 69.6 MPa
2
⎝
⎠
2
τ max =
2
σs =
σx +σ y
2
− 80 + 50
σs =
= −15 MPa
2
28
⎛ σ x −σ y ⎞
⎟ = − ⎛⎜ − 80 − 50 ⎞⎟ = −2.6
tan 2θ s = − ⎜
⎜ 2 (− 25) ⎟
⎜ 2τ
⎟
⎠
⎝
xy
⎝
⎠
2θ s = −69.0&deg; and − 69.0 + 180&deg;
y
15 MPa
15 MPa
θ s = −34.5&deg;, 55.5&deg;
55.5
But we must check which angle goes
with which shear stress.
τ x1 y1 =
(
σ
−
x
−σ y )
o
x
-34.5o
15 MPa
sin 2θ + τ xy cos 2θ
15 MPa
69.6 MPa
2
(− 80 − 50) sin 2(− 34.5) + (− 25) cos 2(− 34.5) = −69.6 MPa
τ x1 y1 = −
2
τmax = 69.6 MPa with θsmax = 55.5&deg;
τmin = -69.6 MPa with θsmin = -34.5&deg;
29
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.
σ 1, 2 =
σx +σ y
2
&plusmn;
⎛σ x −σ y
⎜⎜
⎝ 2
2
⎞
⎟⎟ + τ xy 2
⎠
⎛σ x −σ y ⎞
⎜⎜
⎟⎟ + τ xy 2
⎝ 2 ⎠
2
σ1 − σ 2 = 2
σ 1 − σ 2 = 2τ max
σ −σ2
τ max = 1
2
⎛σ x −σ y ⎞
2τ xy
⎟
tan 2θ s = − ⎜
tan 2θ p =
⎜ 2τ
⎟
σ x −σ y
xy
⎝
⎠
−1
tan 2θ s =
= − cot 2θ p
tan 2θ p
30
tan 2θ s + cot 2θ p = 0
cos 2θ p
sin 2θ s
+
=0
cos 2θ s
sin 2θ p
sin 2θ s sin 2θ p + cos 2θ s cos 2θ p = 0
cos (2θ s − 2θ p ) = 0
2θ s − 2θ p = &plusmn; 90&deg;
θ s − θ p = &plusmn; 45&deg;
θ s = θ p &plusmn; 45&deg;
So, the planes of maximum shear stress (θs) occur at 45&deg; to
the principal planes (θp).
31
Original Problem
Principal
Stresses
y
50 MPa
54.6 MPa
y
80 MPa
84.6 MPa
o
100.5
80 MPa
x
84.6 MPa
o
10.5
x
25 MPa
50 MPa
54.6 MPa
σx = -80, σy = 50, τxy = 25
σ1 = 54.6, σ2 = 0, σ3 = -84.6
Maximum Shear
y
15 MPa
15 MPa
τ max =
55.5
o
x
-34.5o
15 MPa
15 MPa
τ max
τ max
σ1 − σ 2
θ s = θ p &plusmn; 45&deg;
2
θ = 10.5&deg; &plusmn; 45&deg;
54.6 − (− 84.6 ) s
=
θ s = −34.5&deg;, 55.5&deg;
2
= 69.6 MPa
69.6 MPa
τmax = 69.6, σs = -15
32