Plane Stress Transformations

Full text

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Plane Stress 6

Transformations

ASEN 3112 Lecture 6 – Slide 1

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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):

σ σ

σ τ τ

ττxx τxyyy τxz

zz yz zy zx

yx

σxx τx y

τx y

τyx

τyx

σyy

0 0

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 associated two-dimensional stress transformations

ASEN 3112 Lecture 6 – Slide 2

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Flat Plate in Plane Stress

x

y z

Inplane dimensions: in x,y plane Thickness dimension or transverse dimension

Top surface

ASEN 3112 Lecture 6 – Slide 3

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Mathematical Idealization as a Two Dimensional Problem

x y

Midplane

Plate

ASEN 3112 Lecture 6 – Slide 4

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x

y z

x

x x

x y In-plane stresses

σ σ

xx

τ = τxy yx yy

y In-plane strains h h

h

ε ε

xx

γ = γxy yxyy

y y

In-plane displacements

u u

x

y

h

y

y x

x In-plane internal forces

Thin plate in plane stress

pxx pxy pyy

+ sign conventions for internal forces, stresses and strains

dx

dx

dy

dy

dx dy

dx dy dx dy

h

In-plane body forces

b b

x

y

dx dy dx dy

Internal Forces, Stresses, Strains

ASEN 3112 Lecture 6 – Slide 5

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Stress Transformation in 2D

x y

z

x y

t

n z

P σxx

τxy

τyx

σyy

σnn

τnt

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

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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 θ

express σ , σ and τ in terms of the data

xx yy xy

nn tt nt

x y t

n z

σnn

τnt

τtn

σtt

θ

P

ASEN 3112 Lecture 6 – Slide 7

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Analytical Solution

2 2

2 2

2 2

σ = σ cos θ + σ sin θ + 2 τ sin θ cos θ σ = σ sin θ + σ cos θ − 2 τ sin θ cos θ τ = −(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 nn xx yy

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 geometric interpretation 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

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Double Angle Version

σ = cos 2θ + τ sin 2θ τ = − sin 2θ + τ cos 2θ

xy

xy nn

nt

2 2

σ + σ σ − σ xx yy xx yy

+ 2 σ − σ xx yy

Using double-angle trig relations such as cos 2θ = cos θ - sin θ and sin 2θ = 2 sin θ cos θ, the transformation equations may be rewritten as

2 2

Here σtt is omitted since it may be easily recovered as σ + σ − σ xx yy nn

ASEN 3112 Lecture 6 – Slide 9

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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 are called the principal angles.

nn

ASEN 3112 Lecture 6 – Slide 10

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Principal Angles

To find the principal angles, set the derivative of σ with respect to θ to zero. Using the double-angle version,

nn

nn

xx

xy xx yy

p yy

p1

p1 p2

p2

p

d σ

d θ = (σ − σ ) sin 2θ + 2τ cos 2θ = 0 This is satisfied for θ = θ if

tan 2θ = 2 τ σ − 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 the

two principal directions are orthogonal.

o o o

o

o

(*)

ASEN 3112 Lecture 6 – Slide 11

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Principal Stress Values

nn

Replacing the principal angles given by (*) of the previous slide into the expression for σ and using trig identities, we get

2

σ 1,2 = + τ 2 xy

1,2

2 σ + σ xx yy

2

σ − σ xx yy

in which σ 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

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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.

2 2

σ av = , R = + + τ xy

1 av 2 av

av

av

2 σ + σ xx

xx yy

yy

2

σ − σ xx yy

ASEN 3112 Lecture 6 – Slide 13

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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

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Numeric Example

x

x y

x t y

n P σ xx=100 psi

τ xy= τ yx=30 psi σ yy= 20 psi

θ

θ 1=18.44 θ 2= 108.44

x

|τ |max = R=50 psi 18.44 +45 = 63.44

principal planes principal directions

principal planes

principal planes

(a) (b) (c)

P

(d) σ1 =110 psi σ2 =10 psi

P

plane of max inplane shear

principal stress element

P 45

45

For computation details see Lecture notes

ASEN 3112 Lecture 6 – Slide 15

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Graphical Solution of Example Using Mohr's Circle

x y

P σ xx=100 psi τ xy= τ yx=30 psi σ yy= 20 psi

(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 = 10

2

coordinates of blue points are H: (20,30), V:(100,-30), C:(60,0)

τ = min −50 τ = 50max

Radius R = 50

2θ = 36.881

2θ 2= 36.88 +180 = 216.88

(b) Mohr's circle (a) Point in plane stress

ASEN 3112 Lecture 6 – Slide 16

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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

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General 3D Stress State

σ

σ , σ , σ

σ σ

τ τ

τ τ

xx

τ

xyyy

τ

xz

zz yz zx zy

yx

There are three (3) principal stresses, identified as

1 2 3

ASEN 3112 Lecture 6 – Slide 18

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Principal Stresses in 3D (2)

The σ turn out to be the eigenvalues of the stress matrix.

They are the roots of a cubic polynomial (the so-called characteristic 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 −σ

σ −σ

σ −σ τ τ

τ

τ

xx

τ τ

xy xz

yy

zz yz zy

zx

C (σ) = det

yx i

= −σ +

3

I

1

σ −

2

I σ +

2

I

3

= 0

ASEN 3112 Lecture 6 – Slide 19

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3D Mohr Circles

(Yes, There Is More Than One)

All possible stress states at the material point lie on the grey shaded area between the outer and inner circles

σ = normal stress τ = shear

stress

Principal stress σ 1 Outer Mohr's

circle Inner Mohr's

circles

σ Principal stress σ 2

3

Overall + max shear

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

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The Overall Maximum Shear is the Radius of the Outer Mohr's Circle

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 3

σ σ σ

1 1 3

σ − σ τ

overallmax

= R =

outer

2

2

σ − σ

1

2

3

σ − σ

2

2

3 1

σ − σ τ

overallmax

= max , , 2

ASEN 3112 Lecture 6 – Slide 21

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σ , σ , σ

1 2

=0

where σ and σ are the inplane principal stresses obtained as 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 ,

3

is zero:

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

1

σ

2

act in the x,y plane:

3

x

y z

σ

1

σ

2

σ

3

Thin plate

ASEN 3112 Lecture 6 – Slide 22

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Plane Stress in 3D: Overall Max Shear

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:

1 3

σ − σ 2

1 2 3

σ

1

2

(B) Inplane principal stresses have the same sign. Then If σ σ 0 and σ = 0, τ

overallmax

=

2

σ

3

If σ

3

σ 0 and σ = 0,

1

τ

overallmax

= − 2 τ

overallmax

= τ

inplanemax

=

2 3

σ σ σ

1

ASEN 3112 Lecture 6 – Slide 23

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Plane Stress in 3D: Example

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 = max 50

σ = 102 σ = 0

3

inplane

τ = maxoverall 55

Yellow-filled circle is the in-plane Mohr's circle

x y

P σ xx=100 psi τ xy= τ yx=30 psi σ yy= 20 psi

Plane stress example treated earlier:

ASEN 3112 Lecture 6 – Slide 24

Figure

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