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ASEN 3112 - Structures 6 Plane Stress Transformations ASEN 3112 Lecture 6 – Slide 1

Transcript of Plane Stress Transformations - Home | University of · PDF file · 2015-09-09Plane...

ASEN 3112 - Structures

6Plane 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 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