unit m2 - massachusetts institute of...

Paul A. Lagace © 2006

MIT - 16.001/16.002 Fall, 2006

16.001/002 -- “Unified Engineering”Department of Aeronautics and Astronautics

Massachusetts Institute of Technology

Unit M2.4Stress and Strain Transformations

Readings:CDL 4.5, 4.6, 4.7, 4.11, 4.12, 4.13

CDL 4.14, 4.15

Unit M2.4 - p. 2Paul A. Lagace © 2006

MIT - 16.001/16.002 Fall, 2006

LEARNING OBJECTIVES FOR UNIT M2.4Through participation in the lectures, recitations, and workassociated with Unit M2.4, it is intended that you will beable to………

• ….explain the bases for the transformations of thestates of stress, strain, and deformation

• ….cite the equations for 3-D transformations of stress,strain, and deformation

• ….transform the states of stress, strain, anddeformation for any 2-D configuration

• ….apply the concepts associated with principalstress/strain/axes


MIT - 16.001/16.002 Fall, 2006

Earlier, we learnt how to transform coordinates and forces.We may want to do the same with stresses and strains.Why?…..Recall the

Motivation

--> We may want to describe the behavior (stress and strain) of astructure with reference to more than one set of axes:Figure M2.4.1 Example of loading axes on airplane

fuselage axes

wing axes


MIT - 16.001/16.002 Fall, 2006

Different “loading axes” for wing and fuselage.Definition: Loading axes are axes along which loading is applied/oriented

Also, wing axes not oriented with fluid flow axes (transformation needed here)

Figure M2.4.2 Example of a fibrous composite and different axes

fiber direction

s11

s12

s22s21 x2

x1

x1~

x2~

Know stresses (or strains) along loading axes but want to knowstresses in axis system referenced to fibers.


MIT - 16.001/16.002 Fall, 2006

Let’s first consider the mathematical form of the transformation:

Tensorial FormWe learnt in Unit U4 that to transform from one axis system to another, weneed the direction cosines:Figure M2.4.3 Two different axis systems

x2

x1

x1~

x2~

q

(direction) cosine of angle from xn to xm lnm =

xn = lnm xm

--> Axes and forces are first-order tensors (1 subscript) and require 1 direction cosine for transformation.

--> Stresses and strains are second-order tensors (2 subscripts) and require 2 direction cosines for transformation.

~ ~

~~


MIT - 16.001/16.002 Fall, 2006

˜ s mn = l ˜ m p l ˜ n q s pq

˜ e mn = l ˜ m p l ˜ n q e pq

˜ x m = l ˜ m p xp

˜ u m = l ˜ m p up

˜ E mnpq = l ˜ m r l ˜ n s l ˜ p t l ˜ q u Erstu



˜ x m = l ˜ m p xp

˜ u m = l ˜ m p up


Thus:



˜ x m = l ˜ m p xp

˜ u m = l ˜ m p up


and for displacement:

These are the tensor equations to transform stress and strain fromthe xm - system to the xn - system.~

(we won’t write this out in full until we go to two dimensions)**Remember from before the

IMPORTANT CONCEPT: The axis system in which we describe a quantity (or set thereof) does not change the quantity (or set thereof), only its description.

So, the stress state and the strain state do not change, we just describethem differently. In order to see this let’s consider the physical bases forthese transformations.


MIT - 16.001/16.002 Fall, 2006

Physical Bases--> Stress

We are looking at the same stress state as referenced to two sets ofcoordinate axes/systems.

Figure M2.4.4 Infinitesimal cubes of stress in two axis systems

x2

x3

x1 x 1~ x 2

~x 3

~transform

fi

s11

s33

s22

s 11~

s 22~s 33

~infinitesimal cubein equilibrium

rotated infinitesimalcube still in equilibrium

So the transformation of stresses is based on equilibrium(we’ll prove it when we go to 2-D)


MIT - 16.001/16.002 Fall, 2006

--> StrainHere we are looking at the same physical deformation as referencedto two different sets of coordinate axis/systems.Thus, the geometry stays the same

andThe transformation of strain is based on geometry

The transformation laws are generally in 3-D and their bases can beproven in 3-D. However, it is easier to consider these items by looking attheir….

Two-Dimensional Forms

And there are numerous problems which we deal with in 2-D (e.g., planestress)

Since we have written the tensorial forms of the transformationequations in 3-D, we can do the same in 2-D (just use Greeksubscripts!):


MIT - 16.001/16.002 Fall, 2006

2-D forms for stress and strain

Writing the first out in full:--> Stress

where q is angle from xm to xm axes (q +CCW)

~ ~

~ ~

~

~

~

~

s1 1 = cos2 q s1 1 + sin2 q s 2 2 + 2 cosq sinq s1 2

s2 2 = sin2 q s1 1 + cos2 q s 2 2 - 2 cosq sinq s1 2

s1 2 = - sinq cosq s1 1 + cosq sinq s 2 2

+ cos2 q - sin2 q( ) s1 2

˜ s ab = laq lbt sq t

˜ e ab = laq lbt eb tq


MIT - 16.001/16.002 Fall, 2006

Figure M2.4.5 Illustration of axis transformation

x1~

x2~

x3 , x3~

x1

x2

q

Let’s look at the physical basis for this and show where one of theseequations comes from

Consider a….


MIT - 16.001/16.002 Fall, 2006

Figure M2.4.6 Unit square (of unit depth dx3) in equilibriumx2

x1

s11s11

s22

s22

s21

s21

s12

s12

Now cut this diagonally or at some angle q (more generally):Figure M2.4.7 Cut of unit square and axes acting along each face

x2

x1

x1~x2

~

q

s22

s21

s11

s12

s12~

s11~

dx2

x2~

d

p2 - q

q

A


MIT - 16.001/16.002 Fall, 2006

• Align x1 perpendicular to cut face; x2 parallel to cut face• This is still of unit depth dx3• This must still be in equilibrium

Take :form = (stress) (length) (depth)and can see dx2 = dx2 cosq

A = dx2 sinq~~

s1 1( ) dx2( ) dx3 - s1 1 cosq( ) dx2 cosq( ) dx3 - s2 2 sinq( ) dx2 sinq( ) dx3

- s1 2 sinq( ) dx2 cosq( ) dx3 - s2 1 cosq( ) dx2 sinq( ) dx3 = 0

~ ~ ~ ~

~ ~

Use this and cancel out dx3:s1 1 dx2 = s1 1 cosq dx2 cosq + s 2 2 sinq dx2 sinq~ ~ ~

~ ~

Canceling out dx2 and combining:

same as via tensorial form!

~ ~

~

~~

+ s1 2 sinq dx2 cosq + s 2 1 cosq dx2 sinq

s1 1 = cos2 q s1 1 + sin2 q s 2 2 + 2 cosq sinq s1 2

F1 = 0Â :~


MIT - 16.001/16.002 Fall, 2006

This can be done for s22 and s12 and could be done in 3-D as well!Note: Transformation equations for engineering notation form of stresses are the same (but can’t write it tensorially)

Now let’s consider…--> Strain

For tensor notation, the 2-D (and 3-D) transformation equations havethe exact same form:

where q is angle from xn to xm axes (q +CCW)~

Using geometry, these equations can be shown physically.

For a “baseline feeling,” consider….

~ ~

~

~

~

e1 1 = cos2 q e1 1 + sin2 q e2 2 + 2cosq sinq e1 2

e2 2 = sin2 q e1 1 + cos2 q e2 2 - 2cosq sinq e1 2

e1 2 = - sinq cosq e1 1 + cosq sinq e2 2

+ cos2 q - sin2 q( ) e1 2


MIT - 16.001/16.002 Fall, 2006

Figure M2.4.8 Two squares drawn on a bar, one at 45° relative angle to the other

undeformedx2

x1

x1~x2

~

deformed

One appears stretched/elongated, yet they are the samedeformation field!

Use geometry!

Note (on engineering notation): the transformation equations change due to the factor of 2 in the shear strain. BE CAREFUL!


MIT - 16.001/16.002 Fall, 2006

There are some important aspects associated with stress/straintransformations. The most important of these is:

Principal Stresses/Strains/Axes

There is a set of axes into which any state of stress (or strain) can beresolved such that there are no shear stresses (or strains). These areknown as the principal axes of stress (or strain) and the resolved set ofstresses (or strains) are known as the principal stresses (or strains).

Can see this readily in 2-D by considering the equation:s1 2 or e1 2( )

Set s12 to 0 and solve for qp . Then use qp in equations for s11 and s22.

~ ~

~

~ ~ ~

s1 2 = - sinq cosq s1 1 + sinq cosq s2 2

+ cos2 q - sin2 q( ) s1 2

principal


MIT - 16.001/16.002 Fall, 2006

More generally in 3-D can determine the principal stresses as the rootof the equations:

This has three roots (for t) which we label (in decreasing numericalorder):

s1 1 - t( ) s1 2 s1 3

s1 2 s 2 2 - t( ) s 2 3

s1 3 s 2 3 s3 3 - t( ) = 0

sI, sII, sIII -- THE PRINCIPAL STRESSES

The three directions (axes) along which these principal stresses actcan be found via:

s II - s I( ) lI1 + s1 2 lI2 + s1 3 l I3 = 0s1 2 l I1 + s2 2 - s I( ) lI2 + s2 3 lI3 = 0l I1

2 + lI2 2 + lI3

2 = 1where are the direction cosines between the axis along which sIacts and the original axes xn

~ ~ ~

~ ~~

~ ~ ~

~ l In

2


MIT - 16.001/16.002 Fall, 2006

--> Similar equations can be written for the case of sII and sIII.The resulting equations have…

• eigenvalues (roots) -- these are the principal stresses• eigenvectors -- these are the principal axes

--> write this out explicitly for 2-D: s1 1 - t s1 2

s1 2 s2 2 - t = 0

fi s1 1 s 2 2 - s1 1 + s2 2( ) t + t 2 - s1 22 = 0

fi t 2 - t s1 1 + s2 2( ) + s1 1s2 2 - s1 22( ) = 0

Solve via quadratic formula. Roots are sI and sII

To get qp use:s I = cos2 qp s1 1 + sin2 q p s 2 2 + 2 cosqp sinq p s1 2

and solve for qp

or use:

fi t 2 - t s1 1 + s2 2( ) + s1 1 s 2 2 - s1 22( ) = 0


MIT - 16.001/16.002 Fall, 2006

s1 2 = 0 = - sinq cosq s1 1 + sinq cosq s2 2 + cos2 q - sin2 q( ) s1 2 s1 2 = 0 = - sinq cosq s1 1 + sinq cosq s2 2 + cos2 q - sin2 q( ) s1 2

(no shear stress in principal axes)

~

use the following trigonometric identities

cos2 q = 12

1 + cos2q( )

sin2 q = 12

1 - cos2q( )

sinq cosq = 12

sin2q

fi 0 = -s1 1 12

sin2q + s 2 2 12

sin2q

+ s1 2 12

+ 12

cos2q - 12

+ 12

cos2qÊ Ë Á

ˆ ¯ ˜

gives:


MIT - 16.001/16.002 Fall, 2006

12

sin2q s1 1 - s 2 2( ) = s1 2 cos2q( )

Could use this result in transformation equations for s11 and s22 toget sI and sII (optionally)

~ ~

--> There are ways to check work when doing transformation because of….

Invariants(Invariant --> doesn’t vary [with axis system] )

• (s11 + s22 + s33) is a constant!Thus, sum of normal stresses is invariant

fi tan 2q = 2s1 2

s1 1 - s2 2

fi qp = 12

tan -1 2s1 2

s1 1 - s 2 2

Ê

Ë Á Á

ˆ

¯ ˜ ˜


MIT - 16.001/16.002 Fall, 2006

--> Prove in 2-D:s1 1 + s2 2 = cos2 q s1 1 + sin2 q s2 2 + 2 cosq sinq s1 2

+ sin2 q s1 1 + cos2 q s 2 2 + 2cosq sinq s1 2

= s1 1 cos2 q + sin2 q( ) + s2 2 sin2 q + cos2 q( )+ s1 2 2 cosq sinq - 2 cosq sinq( )

s1 1 + s2 2 = s1 1 + s 2 2

s1 1 + s2 2 = cos2 q s1 1 + sin2 q s2 2 + 2 cosq sinq s1 2



s1 1 + s2 2 = s1 1 + s 2 2




s1 1 + s2 2 = s1 1 + s 2 2




s1 1 + s2 2 = s1 1 + s 2 2 So:

= 1 = 1

= 0

Q. E. D.(doesn’t vary with q!)

Note: all these same concepts can be done for strain:- principal strains- principal axes- invariants

(use same equations)

So: (sI + sII + sIII) = (s11 + s22 + s33)

~ ~

~ ~




s1 1 + s2 2 = s1 1 + s 2 2

-


MIT - 16.001/16.002 Fall, 2006

Associated with the concept of principal stresses/strains/axes, is theconcept of

Maximum (Extreme) Shear Stresses/Strains/Planes

These are planes along which the value(s) of the shear stresses/strainsare maximized.

(important in failure considerations)--> Can see what direction this is relative to the principal stresses/strains by considering:

s1 2 = - sinq cosq s I + sinq cosq s II

(recall s12 = 0 in principal axes)Maximize, take derivative:

∂s1 2

∂q = 0 = s I sin2 q - cos2 q( ) + - sin2 q + cos2 q( ) s II

fi 0 = s I - s II( ) sin2 q - cos2 q( )fi sin2 q = cos2 q fi q = 45°

~


MIT - 16.001/16.002 Fall, 2006

Can generalize to 3-D and show that maximum shearstresses/strains occur on planes oriented 45° to the principal axes

Values are:s I - s II

2, s II - s III

2, s I - s III

2 (use s1 2 equation)

There is one final concept to look at with regard to transformations.Before calculators/computers, use of geometry was made to do stressand strain calculations. Done via:

Mohr’s Circle(see handout M-3 for description)

• Based on invariants (radius of circle is sum of normalstresses/strains)

• Easy to see principal stresses/strains, important angles,maximum shear stresses/strains

• Relatively simple for 2-D, can be extended (not easily to3-D)

~

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