Plane Stress Transformations - Home | University of · PDF file · 2015-09-09Plane...
Transcript of Plane Stress Transformations - Home | University of · PDF file · 2015-09-09Plane...
ASEN 3112 - Structures
Plane Stress State
Recall that in a body in plane stress, the general 3D stress state with9 components (6 independent) reduces to 4 components (3 independent):
σσ
σ
ττττ τ
τxx xy xz
yy
zz
yz
zyzx
yx
σxx τxy
τxy
τyx
τyx
σyy
00
0 0 0
plane stress
with =
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 associatedtwo-dimensional stress transformations
ASEN 3112 Lecture 6 – Slide 2
ASEN 3112 - Structures
Flat Plate in Plane Stress
x
y
z
Inplane dimensions: in x,y plane
Thi
ckne
ss d
imen
sion
or t
rans
vers
e di
men
sion
Top surface
ASEN 3112 Lecture 6 – Slide 3
ASEN 3112 - Structures
Mathematical Idealization as a Two Dimensional Problem
x
y
Midplane
Plate
ASEN 3112 Lecture 6 – Slide 4
ASEN 3112 - Structures
x
y
z
x
x x
x
y
In-plane stresses
σσ
xxyyτ = τxy yx
y
In-plane strains
hh
h
εε
xxyy
γ = γxy yx
y
y
In-plane displacements
uu
xy
h
y
y
x
x
In-plane internal forces
Thin plate in plane stresspxx pxy
pyy
+ sign conventionsfor internal forces,stresses and strains
dx
dx
dy
dy
dx dy
dx dy dx dy
h
In-plane body forces
bb
xy
dx dy
dxdy
Internal Forces, Stresses, Strains
ASEN 3112 Lecture 6 – Slide 5
ASEN 3112 - Structures
Stress Transformation in 2D
x
y
zx
ytn
z
P σxx
τxy
τyx
σyy
σnnτnt
τtn
σtt
θ
Global axes x,y stay fixed
Local axes n,t rotate by θ with respect to x,y
P
(a) (b)
ASEN 3112 Lecture 6 – Slide 6
ASEN 3112 - Structures
Problem Statement
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
Plane stress transformation problem: given σ , σ , τ and angle θexpress σ , σ and τ in terms of the data
xx yy xy
nn tt nt
x
ytn
z
σnnτnt
τtn
σtt
θ
P
ASEN 3112 Lecture 6 – Slide 7
ASEN 3112 - Structures
Analytical Solution
2
2
2 2
2
2σ = σ cos θ + σ sin θ + 2 τ sin θ cos θσ = σ sin θ + σ cos θ − 2 τ sin θ cos θτ = −(σ − σ ) sin θ cos θ + τ (cos θ − sin θ)
xx
xx
xx
yy
yy
yy
xy
xy
xy
nn
tt
nt
σ + σ = σ + σ xx yynn
For quick checks when θ is 0 or 90 , see Notes. The sum of the two transformed normal stresses
is independent of the angle θ: it is called a stress invariant(mathematically, this is the trace of the stress tensor). A geometricinterpretation using the Mohr's circle is immediate.
ο ο
tt
This is also called method of equations in Mechanics of Materials books. A derivation using the wedge method gives
ASEN 3112 Lecture 6 – Slide 8
ASEN 3112 - Structures
Double Angle Version
σ = cos 2θ + τ sin 2θ
τ = − sin 2θ + τ cos 2θ
xy
xy
nn
nt
2 2σ + σ σ − σ xx xxyy yy+
2σ − σ xx yy
Using double-angle trig relations such as cos 2θ = cos θ - sin θ andsin 2θ = 2 sin θ cos θ, the transformation equations may be rewritten as
2 2
Here σ is omitted since it may be easily recovered as σ + σ − σ xx yy nntt
ASEN 3112 Lecture 6 – Slide 9
ASEN 3112 - Structures
Principal Stresses: Terminology
The max and min values taken by the in-plane normal stress σ 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 arecalled the principal angles.
nn
ASEN 3112 Lecture 6 – Slide 10
ASEN 3112 - Structures
Principal Angles
To find the principal angles, set the derivative of σ with respectto θ to zero. Using the double-angle version,
nn
nnxx
xy
xx
yy
yyp
p1
p1
p2
p2
p
d σd θ
= (σ − σ ) sin 2θ + 2τ cos 2θ = 0
This is satisfied for θ = θ if
tan 2θ = 2 τσ − σ
It can be shown that (*) provides two principal double angles,2θ and 2θ , within the range of interest, which is [0, 360 ] or[−180 ,180 ] (range conventions vary between textbooks). The two values differ by 180 . On dividing by 2 we get the principal angles θ and θ that differ by 90 . Consequently thetwo principal directions are orthogonal.
o
oo
o
o
(*)
ASEN 3112 Lecture 6 – Slide 11
ASEN 3112 - Structures
Principal Stress Values
nn
Replacing the principal angles given by (*) of the previous slideinto the expression for σ and using trig identities, we get
22
σ = + τ xy1,2
1,2
2σ + σ xx yy
2σ − σ xx yy
in which σ denote the principal normal stresses. Subscripts 1 and 2correspond to taking the + and − signs, respectively, of the square root.
A staged procedure to compute these values is described in the nextslide.
ASEN 3112 Lecture 6 – Slide 12
ASEN 3112 - Structures
Staged Procedure To Get Principal Stresses
1. Compute
Meaning: σ is the average normal stress (recall that σ + σ is an invariant and so is σ ), 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 = σ + R , σ = σ − 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.
22
σ = , R = + + τ xyav
1 av 2 av
av
av
2σ + σ xx
xx
yy
yy
2σ − σ xx yy
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 bevisualized more easily using the Mohr's circle, which provides agraphical solution to the plane stress transformation problem
ASEN 3112 Lecture 6 – Slide 14
ASEN 3112 - Structures
Numeric Example
x
x
y
x
yt
n
P σ =100 psixx
xyτ = τ =30 psiyx
σ = 20 psiyy
θ
1θ =18.44
2θ = 108.44
x
|τ |= R=50 psimax
18.44 +45 = 63.44
principal planes
principaldirections
principal planes
principal planes
(a) (b) (c)
P
(d)
σ =110 psi1
2σ =10 psi
P
plane of maxinplane shear
principalstress
element
P 45
45
For computation details see Lecture notes
ASEN 3112 Lecture 6 – Slide 15
ASEN 3112 - Structures
Graphical Solution of Example Using Mohr's Circle
x
y
P σ =100 psixx
xyτ = τ =30 psiyx
σ = 20 psiyy(a)
σ = normal stress
τ = shear stress
0 20 40 60 80 100
50
40
30
20
10
0
−10
−20
−30
−40
−50
H
V
C
σ = 1101
σ = 102
coordinates of blue points areH: (20,30), V:(100,-30), C:(60,0)
τ = −50min
τ = 50max
Radius R = 50
12θ = 36.88
2θ = 36.88 +180 = 216.882
(b) Mohr's circle
(a) Point in plane stress
ASEN 3112 Lecture 6 – Slide 16
What Happens in 3D?
ASEN 3112 - Structures
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
General 3D Stress State
ASEN 3112 - Structures
σ
σ , σ , σ
σσ
ττττ τ
τxx xy xz
yy
zz
yz
zyzx
yx
There are three (3) principal stresses, identified as
1 2 3
ASEN 3112 Lecture 6 – Slide 18
Principal Stresses in 3D (2)ASEN 3112 - Structures
The σ turn out to be the eigenvalues of the stress matrix.They are the roots of a cubic polynomial (the so-calledcharacteristic polynomial)
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(.)
σ −σσ −σ
σ −σ
τ τ τ τ τ
τxx xy xz
yy
zz
yz
zyzx
yxC(σ) = det
i
= −σ + I σ − I σ + I = 0 1 22
33
ASEN 3112 Lecture 6 – Slide 19
3D Mohr Circles (Yes, There Is More Than One)
ASEN 3112 - Structures
All possible stress states at the material point lieon the grey shadedarea between the outerand inner circles
σ = normal stress
τ = shear stress
Principal stress σ 1
Outer Mohr's circle
Inner Mohr's circles
σ 2Principal stress σ 3
Overall+ max shear
The overall maximum shear, which is the radius of theouter Mohr's circle, is important for assessing strength safety of ductile materials
ASEN 3112 Lecture 6 – Slide 20
The Overall Maximum Shear isthe Radius of the Outer Mohr's Circle
ASEN 3112 - Structures
2
If the principal stresses are algebraically ordered as
then
Note that the intermediate principal stress σ 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:
2 31σ σ σ
31σ − σ2
τ = R = overalloutermax
21σ − σ2
32σ − σ2
13σ − σ2
τ = max , , overallmax
ASEN 3112 Lecture 6 – Slide 21
ASEN 3112 - Structures
σ , σ , σ =01 2
where σ and σ are the inplane principal stresses obtainedas described earlier in Lecture 6.
1 2
3
Consider plane stress but now account for the third dimension. One of the principal stresses, call it for the moment σ , is zero:3
Plane Stress in 3D: The 3rd Principal Stress
The zero principal stress σ is aligned with the z axis (the thickness direction) while σ and σ act in the x,y plane:1 2
3
x
y
z
1σ 2σ
3σThin plate
ASEN 3112 Lecture 6 – Slide 22
Plane Stress in 3D: Overall Max ShearASEN 3112 - Structures
Let us now (re)order the principal stresses by algebraic value as
(A) Inplane principal stresses have opposite signs. Then the zero stress is the intermediate one: σ , and 2
To compute the overall maximum shear 2 cases are considered:
31σ − σ2
1 2 31σ
2
(B) Inplane principal stresses have the same sign. Then
If σ σ 0 and σ = 0, τ = overallmax
23σ
2If σ σ 0 and σ = 0, τ = − overall
max3 1
τ = τ = overall inplanemax max
2 31σ σ σ
ASEN 3112 Lecture 6 – Slide 23
Plane Stress in 3D: ExampleASEN 3112 - Structures
σ = normal stress
τ = shear stress
0 20 40 60 80 100
50
40
30
20
10
0
−10
−20
−30
−40
−50
σ = 1101
τ = 50max
σ = 102
σ = 03
inplane
τ = 55maxoverall
Yellow-filled circle is the in-plane Mohr's circle
x
y
P σ =100 psixx
xyτ = τ =30 psiyx
σ = 20 psiyy
Plane stress example treated earlier:
ASEN 3112 Lecture 6 – Slide 24