Mechanics of composite _UTP

8/16/2019 Mechanics of composite _UTP

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Purdue University

EAS 557

Introduction to Seismology

Robert L. Nowac

Lecture !

Analysis o" Stress

The concept of stress involves the action of forces on a body. Twotypes are considered:

1) Body forces – . Which are generally forces per unit volume. For

gravitational forces these depend on the mass distribution withinthe body. These are generally non!contact forces.

") #urface forces – . These are forces per unit area acting along

surfaces of elemental volumes and are contact forces with ad$acent parts of the body. These forces give rise to the concept of traction

on e%ternal and internal surfaces. The traction on an e%ternalsurface of the body is related to the applied force by

where is the average force on the small surface centered on

the point &. is called the traction vector.


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'nits of traction and stress are in force(area. n cgs the units of stress

are dyne(cm". n # the units of stress are 1 *ewton(m" + 1 &ascal ,1 &ascal +1- dyne(cm"). ther units are

1 bar + 1-/ dyne(cm" + 1-0 &ascals

1 bar + 1.-1 2tmospheres 1 3ilobar + 10--- lb(in"

We now want to investigate tractions on internal surfaces within a

body. 4onsider a bar under tension

The e%ternal applied force results in tractions on the ends.

*ow cut the bar and apply forces on the two sides of the cut so that the

shape of either section of bar remains unchanged. 56uilibrium re6uires

that .

*ow define the tractions on the upper and lower

faces of the cut. From *ewton7s Third 8aw . Thus onthe arbitrary internal surface that we have constructed.

n !9 consider a cut of area at a point &. 8et be the force

applied to the side with normal


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2gain . n the other side of the cut .

From *ewton7s Third 8aw

Thus as in the case for the bar the tractions on either side of the cut are of the

same magnitude but opposite in sign.

f we made cuts in other directions say with normals and we

would obtain different values for the internal traction at &. *ote that for a

fluid the pressure is related to the traction by for any . But in a

solid the magnitude and direction of the traction depends on the orientation of

the surface element .

5%) 4onsider the walls of a house.


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For an elemental area of the wall at point . But at point

& will be non!ero and large.

t at first seems that there will be an infinite number of traction vectorsat a given point depending on the orientation of the small surface. But

4auchy proved that in fact all the various tractions at & can be derived from aset of si% numbers collectively grouped as the ;stress tensor


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where and are unit vectors in the x1 and x" directions. The vectors are

oriented in the conventional positive directions. Thus would be positive if they stretch the material. 2lso on the opposite faces

n 9 we will write

in which is the stress tensor at the point & where

are tensional or compressive stresses

are shear stresses

n inde% notation the stress tensor is where i is normal to coordinate

plane in which the traction acts and j indicates the component of the tractionvector.

4auchy showed that the stresses on any plane through an internal point

& can be written as a linear combination of the elements of the stresstensor. This is a fundamental theorem of solid mechanics. 4onsider the

triangle in "9 ,tetrahedron in !9)


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The traction on surface with normal is

. *ow we must find the condition of e6uilibrium of the

triangle. Balancing forces in the x1 and x" directions gives

n the x1 direction:

n the x" direction:

or

From geometry

We then find


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n inde% notation for !9 then

,with implied sum on j)

Thus the 3nowledge of the stress tensor specifies completely the state ofstress around the point &.

n order to prevent the body from spinning we also re6uire that the nettor6ue be ero. 5valuating the tor6ue for a reference point at the center of thes6uare we get

where is a force and is a moment arm. Then .

n the general !9 case . Because of this relation the stress

tensor is symmetrical where

with and . Thus only si% numbers areneeded to completely describe the state of stress of an internal point & in a

body.

t is important to separate between the isotropic and deviatoric part ofthe stress tensor . We define


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where is independent of the coordinate system and & is the pressure. Theminus sign is used to re6uire & to be positive when it is compressive. The

stress elements themselves are positive in tension. We rewrite as

or in inde% notation

where is the deviatoric part of the stress tensor.

For hydrostatic stress . For a hydrostatic increase ofstress with depth due to a uniform overburden then

where is the density of the overburden z is thic3ness and g is the

acceleration of gravity. 2s an e%ample the hydrostatic pressure at a depth of1- 3m underneath a thic3ness of roc3 with an average density of ---

3ilograms(m is

& + ,--- 3ilograms(m) ,1- % 1- m) ,=.> m(s")

+ ".=? % 1-> &ascals @ "=? A&a

+ ".=? 3bars

2t an average thic3ness of the crust of - 3m the hydrostatic pressure would be on the order of 1 &a or 1- 3bars.

#rans"ormations o" t$e Stress #ensor in Rotated %oordinate Systems


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The components of a vector in a rotated coordinate system can berelated to the components in the original system by a coordinate rotation

matri% . For e%ample the components of the vector above in twocoordinate systems can be written

where the components of are the direction cosines between the old and newcoordinate a%es. The rotation matri% is

2s an e%ample if then

8et then in the new coordinate system.

*ote for rotation matrices since these are orthogonal

matrices. *ow assume a general relationship between two vectors and

n a rotated coordinate system

Then the general relationship between and can be written


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or

Thus in the new coordinate system where

with . This general relationship has the same form in any rotated

coordinate system with the coefficients depending on the orientation of thecoordinate system.

*ow consider the traction vector on an internal plane with normal

#ince this can also be written

or

n a rotated coordinate system the traction vector is

with

*ow we choose so that is a diagonal matri%. n this specialcoordinate system all the shear stresses are ero and the normal stresses are

called the &rinci&le stresses acting with respect to the three rotated coordinatea%es. 8et

where are the rotated a%es with respect to the unrotated

a%es. Then can be written as


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or for each then

forms an eigenvalue problem. To find the eigenvalues the determinant

of must be ero. Thus

to solve for . *e%t solve for the eigenvectors and ma3e sure to ad$ust

to unit length. #tandard computer software such as Aatlab can be used to

find the principle stresses and the principle stress directions ,i + 1).

These can be ordered such that . #ince these are positive

for tensional stresses provides the ma%imum compressive stress. Thereare no shear or tangential stresses act on these new coordinate faces only

normal stresses.

n the "!9 case it is useful to derive these relations in detail


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8et where

and . Then

This can be multiplied out and written in terms of components of as

f we let then we can find the rotation angle to be

#ince

f then will be the ma%imum and minimum normal

stress. Thus

These would then be the eigenvalues in the above eigenvector analysis in

"9. From we find that


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2s an e%ample of the calculation of the principle stresses in "9 let

then

This gives for the principle stresses


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Thus

Then

and

These must then be normalied to unit length to get and . 'sing the "!9

formulas

#tress maps showing the orientation of ma%imum horiontal shearstress have been developed based on orientations of earth6ua3e faulting and in

situ stress measurements. ne techni6ue for finding this is from brea3outones in boreholes which align with the principle stress directions and are

often consistent over regional distances. A solid body is in static equilibrium when the resultant force and moment on each axis

is equal to zero. This can be expressed by the equilibrium equations. In this article we

will prove the equilibrium equations by calculating the resultant force and moment on

each axis. A more elegant solution may be derived by using Gauss’s theorem and

auchy’s formula. This approach may be found in international bibliography.

onsider a solid body in static equilibrium that neither moves nor rotates. !urface and

body forces act on this body. "e cut an infinitesimal parallelepiped inside the body andwe analyze the forces that act on it as shown in #ig. $. "e will assume that the stress

field is continuous and differentiable inside the whole body.


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Figure 1: Infinitesimal parallelepiped in static equilibrium

The stress components on each side is a function of the position since we have a non

uniform but continuous stress field. #or example on side % the normal stress

is . &n the opposite side ' the normal stress

is . (y ta)ing under consideration Taylor’s theorem we may write*

+$,

The higher order terms have been neglected because they are relatively small. "e

follow the same procedure for all the components as shown if #ig. $.

-quilibrium of the body demand that the resultant forces must vanish. (y summing up

the forces with direction parallel to axis we get*

+',

where and are the dimensions of the parallelepiped and is the

component of the body force parallel to . (y dividing with we get*

+/,

!imilarly we can obtain the equations for the other two directions. The final set of

equilibrium equations is*

+%,

(y using index notation we may write the three equilibrium equations in compact form*

+0,

The resultant moment on each axis must also vanish. (y ta)ing under consideration all

the forces that contribute to moment about axis we may write*

http://www.rockmechs.com/stress-strain/fundamentals/index-notation-tensor-vector/http://www.rockmechs.com/stress-strain/fundamentals/index-notation-tensor-vector/


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+1,

by dividing with and ta)ing the limit and we

derive*

+2,

#ollowing the same procedure for the other two axes lead to the conclusion that the

stress tensor is symmetric*

+3,

It should be noted that the above symmetry holds true only if no external body

moments proportional to volume exist. -lse the stress tensor should be considered

asymmetric. 4owever for the ma5ority of 6oc) 7echanics problems the stress tensor is

symmetric.

#or the case of two dimensional problems equilibrium equations simplify as follows*

+8,

Example

onsider a solid body which is sub5ect to the following stresses*

+$9,

alculate the body forces in order to achieve static equilibrium.

!olution

The resultant force on each axis must vanish. (y using equations +%, we get*

+$$,

+$',

+$/,

Suggested Bibliography

:.. #ung. A #irst ourse in ontinuum 7echanics. ;rentice 4all -nglewood liffs

.-. 7alvern. Introduction to the 7echanics of a ontinuous 7edium. ;rentice 4all

-nglewood liffs

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Mechanics of composite _UTP