ASEN 3112 - Structures
6
Plane Stress
Transformations
ASEN 3112 Lecture 6 – Slide 1
ASEN 3112 - Structures
Plane Stress State
Recall that in a body in plane stress, the general 3D stress state with
9 components (6 independent) reduces to 4 components (3 independent):
σxx τxy
τyx σyy
τzx τzy
τxz
τyz
σzz
plane stress
σxx τxy
τ yx σ yy
0
0
0
0
0
with τ yx = τxy
Plane stress occurs in thin plates and shells (e.g. aircraft & rocket skins,
parachutes, balloon walls, boat sails, ...) as well as thin wall structural
members in torsion.
In this Lecture we will focus on thin flat plates and associated
two-dimensional stress transformations
ASEN 3112 Lecture 6 – Slide 2
ASEN 3112 - Structures
Thickness dimension
or transverse dimension
Flat Plate in Plane Stress
z
Top surface
y
x
Inplane dimensions: in x,y plane
ASEN 3112 Lecture 6 – Slide 3
ASEN 3112 - Structures
Mathematical Idealization as
a Two Dimensional Problem
y
Midplane
x
Plate
ASEN 3112 Lecture 6 – Slide 4
ASEN 3112 - Structures
Internal Forces, Stresses, Strains
In-plane internal forces
dx dy
pyy
h
y
Thin plate in plane stress
z
dx
x
dy
pxx
h
x
+ sign conventions
for internal forces,
stresses and strains
x
dy
pxy
dx
In-plane stresses
dx dy
y
x
y
σyy
σxx τxy = τ yx
In-plane body forces
dx dy
y
h
x
In-plane strains
dx dy
h
εyy
ε xx γ xy = γ yx
x
bx
by
y
In-plane displacements
dx dy
h
y
y
uy
u
x
x
ASEN 3112 Lecture 6 – Slide 5
ASEN 3112 - Structures
Stress Transformation in 2D
σtt
σyy
(a)
τyx
P
(b)
τxy
σxx
τnt
σnn
P
t
y
τtn
y
n
x
z
z
θ x
Local axes n,t
rotate by θ with
respect to x,y
Global axes
x,y stay fixed
ASEN 3112 Lecture 6 – Slide 6
ASEN 3112 - Structures
Problem Statement
σtt
τtn
P
Plane stress transformation problem:
given σxx , σyy , τ xy and angle θ
express σnn , σ tt and τ nt in terms of the data
t
y
n
z
θ x
This transformation has two major uses:
Find stresses along a given skew direction
Here angle θ is given as data
Find max/min normal stresses, max in-plane shear
and overall max shear
Here finding angle θ is part of the problem
ASEN 3112 Lecture 6 – Slide 7
τnt
σnn
ASEN 3112 - Structures
Analytical Solution
This is also called method of equations in Mechanics of Materials
books. A derivation using the wedge method gives
σnn = σxx cos2 θ + σyy sin2 θ + 2 τxy sin θ cos θ
σ tt = σxx sin 2 θ + σyy cos2 θ − 2 τxy sin θ cos θ
τ nt = −(σxx − σ yy ) sin θ cos θ + τ xy (cos2 θ − sin 2 θ)
ο
ο
For quick checks when θ is 0 or 90 , see Notes. The sum
of the two transformed normal stresses
σ nn + σ tt = σxx + σ yy
is independent of the angle θ: it is called a stress invariant
(mathematically, this is the trace of the stress tensor). A geometric
interpretation using the Mohr's circle is immediate.
ASEN 3112 Lecture 6 – Slide 8
ASEN 3112 - Structures
Double Angle Version
Using double-angle trig relations such as cos 2θ = cos2 θ - sin2 θ and
sin 2θ = 2 sin θ cos θ, the transformation equations may be rewritten as
σnn =
σxx + σ yy
2
+
σxx − σyy
2
cos 2θ + τxy sin 2θ
τ nt = − σxx − σyy sin 2θ + τ xy cos 2θ
2
Here σtt is omitted since it may be easily recovered as σxx + σyy − σnn
ASEN 3112 Lecture 6 – Slide 9
ASEN 3112 - Structures
Principal Stresses: Terminology
The max and min values taken by the in-plane normal stress σnn
when viewed as a function of the angle θ are called principal stresses
(more precisely, principal in-plane normal stresses, but qualifiers
"in-plane" and "normal" are often omitted).
The planes on which those stresses act are the principal planes.
The normals to the principal planes are contained in the x,y plane.
They are called the principal directions.
The θ angles formed by the principal directions and the x axis are
called the principal angles.
ASEN 3112 Lecture 6 – Slide 10
ASEN 3112 - Structures
Principal Angles
To find the principal angles, set the derivative of σnn with respect
to θ to zero. Using the double-angle version,
d σnn
= (σyy − σ xx ) sin 2θ + 2τ cos 2θ = 0
dθ
This is satisfied for θ = θp if
tan 2θ p =
2 τ xy
σxx − σ yy
(*)
It can be shown that (*) provides two principal double angles,
2θp1 and 2θp2 , within the range of interest, which is [0, 360o ] or
[−180 o,180 o ] (range conventions vary between textbooks).
The two values differ by 180o . On dividing by 2 we get the principal
angles θp1 and θ p2 that differ by 90 o . Consequently the
two principal directions are orthogonal.
ASEN 3112 Lecture 6 – Slide 11
ASEN 3112 - Structures
Principal Stress Values
Replacing the principal angles given by (*) of the previous slide
into the expression for σ nn and using trig identities, we get
σ1,2 =
σxx + σ yy
σxx − σ yy
2
2
2
+ τxy2
in which σ1,2 denote the principal normal stresses. Subscripts 1 and 2
correspond to taking the + and − signs, respectively, of the square root.
A staged procedure to compute these values is described in the next
slide.
ASEN 3112 Lecture 6 – Slide 12
ASEN 3112 - Structures
Staged Procedure To Get Principal Stresses
1. Compute
σav =
σxx + σ yy
2
,
R=+
σxx − σ yy
2
2
+ τ xy2
Meaning: σav is the average normal stress (recall that σ xx+ σ yy is an
invariant and so is σ av ), whereas R is the radius of Mohr's circle
described later. This R also represents the maximum in-plane shear
value, as discussed in the Lecture notes.
2. The principal stresses are
σ 1 = σav + R ,
σ 2 = σav − R
3. The above procedure bypasses the computation of principal angles.
Should these be required to find principal directions, use equation (*) of
the Principal Angles slide.
ASEN 3112 Lecture 6 – Slide 13
ASEN 3112 - Structures
Additional Properties
1. The in-plane shear stresses on the principal planes vanish
2. The maximum and minimum in-plane shears are +R and −R,
respectively
3. The max/min in-plane shears act on planes located at +45 and -45
from the principal planes. These are the principal shear planes
4. A principal stress element (used in some textbooks) is obtained
by drawing a triangle with two sides parallel to the principal planes
and one side parallel to a principal shear plane
For further details, see Lecture notes. Some of these properties can be
visualized more easily using the Mohr's circle, which provides a
graphical solution to the plane stress transformation problem
ASEN 3112 Lecture 6 – Slide 14
ASEN 3112 - Structures
Numeric Example
σ2 =10 psi
σyy = 20 psi
(a)
(b)
τ xy = τyx =30 psi
P
y
σxx =100 psi
t
P
principal
directions
θ2= 108.44
σ1 =110 psi
θ1=18.44
x
principal
planes
|τmax |= R=50 psi
(c)
18.44 +45
= 63.44
P
y
principal
planes
n
x
θ
x
principal
planes
x
(d)
plane of max
inplane shear
For computation details see Lecture notes
ASEN 3112 Lecture 6 – Slide 15
45 principal
stress
element
45
P
ASEN 3112 - Structures
Graphical Solution of Example Using Mohr's Circle
(a) Point in plane stress
P
τmax = 50
50
σyy = 20 psi
(a)
2θ2 = 36.88 +180 = 216.88
τ = shear
stress
40
τ xy = τyx =30 psi
30
σxx =100 psi
10
H
Radius R = 50
20
σ2 = 10
0
0
20
40
−10
60
80
100
σ = normal
stress
C
σ1 = 110
−20
y
−30
V
−40
−50
x
2θ1 = 36.88
τmin = −50
(b) Mohr's circle
coordinates of blue points
are
H: (20,30), V:(100,-30), C:(60,0)
ASEN 3112 Lecture 6 – Slide 16
ASEN 3112 - Structures
What Happens in 3D?
This topic be briefly covered in class if time allows,
using the following slides.
If not enough time, ask students to read Lecture notes
(Sec 7.3), with particular emphasis on the computation
of the overall maximum shear
ASEN 3112 Lecture 6 – Slide 17
ASEN 3112 - Structures
General 3D Stress State
σxx τxy
τyx σyy
τzx τzy
τxz
τyz
σzz
There are three (3) principal stresses, identified as
σ1 , σ 2 , σ 3
ASEN 3112 Lecture 6 – Slide 18
ASEN 3112 - Structures
Principal Stresses in 3D (2)
The σ i turn out to be the eigenvalues of the stress matrix.
They are the roots of a cubic polynomial (the so-called
characteristic polynomial)
C(σ) = det
τxy
σxx −σ
τxz
τyx σyy −σ
τyz
τzy
τzx
σzz−σ
3
= −σ + I 1 σ 2 − I 2 σ + I 3 = 0
The principal directions are given by the eigenvectors
of the stress matrix.
Both eigenvalues and eigenvectors can be numerically
computed by the Matlab function eig(.)
ASEN 3112 Lecture 6 – Slide 19
ASEN 3112 - Structures
3D Mohr Circles
(Yes, There Is More Than One)
τ = shear
stress
Overall
+ max shear
Inner Mohr's
circles
Principal stress σ3
σ2
All possible stress states
at the material point lie
on the grey shaded
area between the outer
and inner circles
σ = normal
stress
Principal stress σ 1
Outer Mohr's
circle
The overall maximum shear, which is the radius of the
outer Mohr's circle, is important for assessing strength
safety of ductile materials
ASEN 3112 Lecture 6 – Slide 20
ASEN 3112 - Structures
The Overall Maximum Shear is
the Radius of the Outer Mohr's Circle
If the principal stresses are algebraically ordered as
σ1
then
σ2
σ3
τ overall
= R outer =
max
σ1 − σ3
2
Note that the intermediate principal stress σ2 does not appear.
If they are not ordered it is necessary to use the max function
in a more complicated formula that picks up the largest
of the three radii:
τ overall
= max
max
σ1 − σ2
2
,
σ2 − σ3
2
,
ASEN 3112 Lecture 6 – Slide 21
σ 3 − σ1
2
ASEN 3112 - Structures
Plane Stress in 3D: The 3rd Principal Stress
Consider plane stress but now account for the third dimension.
One of the principal stresses, call it for the moment σ3 , is zero:
σ1 , σ 2 , σ 3 =0
where σ1 and σ2 are the inplane principal stresses obtained
as described earlier in Lecture 6.
The zero principal stress σ3 is aligned with the z axis (the
thickness direction) while σ1 and σ 2 act in the x,y plane:
z
σ3
Thin plate
x
y
σ1
ASEN 3112 Lecture 6 – Slide 22
σ2
ASEN 3112 - Structures
Plane Stress in 3D: Overall Max Shear
Let us now (re)order the principal stresses by algebraic value as
σ1
σ2
σ3
To compute the overall maximum shear 2 cases are considered:
(A) Inplane principal stresses have opposite signs. Then the
zero stress is the intermediate one: σ 2 , and
σ1 − σ3
overall
inplane
τ max = τ max =
2
(B) Inplane principal stresses have the same sign. Then
σ1
If σ1 σ 2 0 and σ 3 = 0, τoverall
=
max
2
σ
overall
If σ3 σ 2 0 and σ 1 = 0, τ max
=− 3
2
ASEN 3112 Lecture 6 – Slide 23
ASEN 3112 - Structures
Plane Stress in 3D: Example
Plane stress example
treated earlier:
τ = shear
stress
σyy = 20 psi
50
τ xy = τyx =30 psi
P
σxx =100 psi
40
30
σ3 = 0
20
10
0
y
overall
τmax
= 55
inplane
τmax
= 50
σ2 = 10
0
20
40
60
80
100
σ = normal
stress
−10
−20
−30
σ1 = 110
−40
x
−50
Yellow-filled circle is the in-plane Mohr's circle
ASEN 3112 Lecture 6 – Slide 24