CONTINUUM MECHANICS

Strain at a point

Group 3Eduard Hutapea (221140 )

Micky kololu (22114028)

At any instant of time t, a continuum having a volume and a bounding surface will occupy a certain region in physical space. The position of this continuum with respect to a set of co-ordinate axes is said to be the configuration of the continuum at the particular time, t.

deformation occurs as a result of relative displacements between points in a body. As such, the state of deformation in a continuous medium can be defined on either a purely mathematical sense or on a purely geometrical sense. Thus the problem is that given the positions of points in a body before and after deformation, it is required to determine the change in the distance between two arbitrary, infinitesimally close points in the body irrespective of either the causes of deformation or the low according to which the body resists this deformation.

Deformations

o o\ A

l A'

B

B’

p p

figure a figure b

• Deformation of continuum medium

𝜇𝐴

𝜇𝐵

Look figure b :

Because there is force P thus : A become A’ B become B’

If we choose segment AB : called is unit deformation/relative/unitair

presentation of displacement

zOA =

AA’=

OA’= y

x

• General case 3-D

A

A’

O

𝜉

𝜇

𝑥

If we have :

A = operator of transformation

= matrix cooficient constant A= (Aij)

= a11 x + a12 y + a13 z

= a21 x + a22 y + a23 z

= a31 x + a32 y + a33 z

aij constant those given at the point

A = A1 x A2

M = M1x M2

A = I = if i = j ; i ≠ j

O

M M1

M2A

A1

A2

As discussed in the previous section, deformation occurs as a result of relative displacements between points in a body. If the shape and size of the body remains unchanged then this type of deformation is known as a rigid body motion. However, if the shape and size of the body are altered, then a state of strain has been set up. As the body (or continuum) is transformed from one configuration to another, the matter in the neighbourhood of each point is translated and rotated and hance strained.

Concept of strain

Strain may be regarded as normalized displacement. If a structure is subjected to a stress state, it will deform. However, the magnitude of the deformation is dependent on the size of the structure as well as the magnitude of the applied stresses. In order to render the deformation as a scale-independent parameter, the concept of strain is utilized.

Strain in its simplest form, strain is the ratio of displacement to the undeformed length

• Finite strain

It should be noted that strain is a 3-D phenomenon that requires reference to all three cartesian coordinate axes. However, it is instructive to start with 2-D strain, and then once the basic concepts have been introduced,3-D strain follows as a natural progression.

Figure of normal strain and shear strain

Referring to fig. beside, let B and C be two

Infinitesimally close neighbouring points

In a body before deformation and their new

Positions B’ and C’ after deformation.

The cartesian coordinates of these four points

Can be written as follows : z

B : x, y ,z

C : x + dx, y + dy, z + dz

B’: x’, y’, z’ or x + ux, y + uy, z + uz …. (1)

C’: x’ + dx, y’+ dy, z’ + dz

Or [x+dx+ux+dux], [y+dy+uy+duy], [z+dz+uz+duz]

In the original configuration, the square

Of the distance between the points B and y

c is given by :

…. (2)

Similarlly in the deformed state,

…. (3)

Expanding (x’+dx’) and employing x

taylor’s theorem and neglecting terms

Involving higher powers of

C’ B’

C

B

…. (4)

Thus , …. (5)Or , {da’} = [F] {da} ….(5a)Where [F] = matrix containing spatial deformtion gradients.Subtituting eq. (5) in eq. (3) and expressing in terms of matrix [F],

…. (6)

Infinitesimal strain is homogeneous strain over a vanishingly small element of a finite strained body. To find the components of the strain matrix, we need to consider the variation in coordinates of ends of an imaginary line inside a body as the body is strained.

The point P with coordinates (x,y,z) moves when the body Is strained, to a point P* with Coordinates (x+ux, y+uy, z+uz).The components of movement Ux, uy and uz, may vary with location within the body, andso are regarded as functionsOf x, y and z.

• Infinitesimal strain

If ux = ux* , uy = uy* and uz = uz* , thus displacement relative between points P and Q are zero.Line element wich joint P and Q will experience length change because there are load, and body said experience of strain. Components of displacement (ux, uy, uz) from point P and (ux*, uy* , uz*) from point Q.

Because :

Thus : displacement inkremental can be express as :

or

{dr} express initial length element line PQ{express displacement relative from points line element because deformed from not condition experience of strain to condition experience of strain.

Element with length dx assumed stretch Component normal strain :

Components displacement relative because normal strain : y

Element at the plane x-y experienced distorsion.Because small, element experienced pure shear, thus it’s component displacement : y magnitude shear strain defined as :

Thus :

With way its same, for elementWhich experience pure shearAt the plane y-z and z-x we get :

And

z

Thus component of total displacement become : y

Or in the matrix :

With {) = matrix of strain

or

Because we ignore rotation , then : {d}={d, and we get :

Thus

= =

with the way at same we get matrix T2 (rotation). Where :

Where :

or

= = aij xj – xi = aij xj - = (aij- )xjIf: A = aij I = A= H+I H = aij-

Remember : multiply from 2 transformation which make displacement point M to M’ A = A2 – A1

If we have 2 transformation A1 and A2 (very small) :

A1 = H1 + IA2 = H2 + IMultiply A1A2 = ( H2 + I)-(H1 +I) = H2H1 +H1 +H2 +I H1 H2 very small, can be ignoreA = A2A1 = H1 + H2 + IA = H + I

Multiply from 2 transformation

So : multiply 2 transformation matrix A = quantity matrix displacement + IIn our case :

We can write : pure strain (small) transformation of rotation (small) = D+R+IWhere :H = D + R

So: matrix T = quantity matrix pure strain and matrix rotation.

Example : original volume = v volume after transformation = v’

aij = determinant from matrix transformation aij = Hij +

Where :Hij =

Or : So :

= 1 + (deformation of volumic)

Change of volume result a transformation

Principal axes

here there is no distortion (because outher components = 0)

called is principal axes

direction from this axes called is invariant

If : …….(1)

If there is a direction “invariant” from this transformation, mean that if there is a point m in this direction, then in this transformation m m’.

m’ m‘ also in the same direction

0 m

sb2

sb1

sb3

……….. (2)We search direction from initial medium which verification relations second equation, which remained on the direction of the invariant verification first equation.

Or (

so there are 3 equations with 3 unknown parameters (x, y, z) such condition is called: the homogeneous system

we have demonstrated the system with a single solution :X = 0Y = 0 Z = 0 ………. (3)

,

we find the value of is annulled determinant : ……………(4)we have shown that the is symmetry, equation 3 will always have real roots (may be the same).

when the 1, 2, 3 is a root of the equation 3, then: 1, 2, 3 annul each determinant and give direction "invariant“

, and

Is a principal directiion and to three axes mutually perpendicular

1, 2, and 3 principal deformationI1, I2, an I3 principal axes

Pure deformation have the matrixWhere Quantity from deformation in the diagonal row = deformation volumic (.

If , decompotition then :

+ Pure isotrop deviationDeformation deformation deformation matrix 1 matrix 2Matrix 1 :- All direction is the principal direction- That occurred = volume change ( not form)

Matrix 2 :-’ +=0- That occurred = form change (not volume).

Deviatoric Strains

the meaning of the matrix :

Where : = 1/3 kk + meaning 1/3 here is k to 3 major axis direction as large, so it must be divided by 3.