Unit2 Notes

7/25/2019 Unit2 Notes

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16.21 Techniques of Structural Analysis and

Design

Spring 2005

Unit #2 - Stress and Momentum balance

Raul Radovitzky

February 6, 2005

Stress at a point

We are going to consider the forces exerted on a material. These can beexternal or internal. External forces come in two flavors: body forces (givenper unit mass or volume) and surface forces (given per unit area). If we cut abody of material in equilibrium under a set of external forces along a plane,as shown in Fig.1, and consider one side of it, we draw two conclusions: 1)

the equilibrium provided by the loads from the side taken out is provided bya set of forces that are distributed among the material particles adjacent tothe cut plane and that should provide an equivalent set of forces to the onesloading the part taken out, 2) these forces can now be considered as externalsurface forces acting on the part of material under consideration.

The stress vector at a point on S is defined as:

ft = lim (1)

S 0 S

If the cut had gone through the same point under consideration but along aplane with a different normal, the stress vector t would have been different.Lets consider the three stress vectors t(i) acting on the planes normal to thecoordinate axes. Lets also decompose each t(i) in its three components inthe coordinate system ei (this can be done for any vector) as (see Fig.2):

t(i) = ijej (2)

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s f

surface forces

body forces

n

body forces

n

Figure 1: Surface force f on area S of the cross section by plane whosenormal is n

ij is the component of the stress vector t(i) along the ej direction.

Stress tensor

We could keep analyzing different planes passing through the point withdifferent normals and, therefore, different stress vectors t(n) and one mightwonder if there is any relation among them or if they are all independent. The

answer to this question is given by invoking equilibrium on the (shrinking)tetrahedron of material of Fig.3.The area of the faces of the tetrahedron areS1, S2, S3 and S. The stress vectors on planes with reversed normalst(ei) have been replaced with t

(i) using Newtons third law of actionand reaction (which is in fact derivable from equilibrium): t(n) = t(n).Enforcing equilibrium we have:

t(n)S t(1)S1 t(2)S2 t

(3)S3 = 0 (3)

where V is the volume of the tetrahedron and f is the body force perunit volume. The following relation: Sni = Si derived in the following

mathematical aside:

By virtue of Greens Theorem:

dV = ndSV S

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11

12

13

21

22

23

31

32

33

t

t

t

(1)

(2)

(3)

x1

x2

x3

e

e

e1 2

3

Figure 2: Stress components

t

(3)

t(2)

t(1)

e

e

e

1

2

3

n

t(n)

Figure 3: Cauchys tetrahedron representing the equilibrium of a tetrahedronshrinking to a point

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applied to the function = 1, we get

0 = ndSS

which applied to our tetrahedron gives:

0 = Sn S1e1 S2e2 S3e3

If we take the scalar product of this equation with ei, we obtain:

S(n ei) = Si

or

Si = Sni

can then be replaced in equation 3 to obtain:

e3)t(3)S t(n) (n e1)t

(1) + (n e2)t(2) + (n = 0

ore1t

(1) + e2t(2) + e3t

(3)t(n) = n (4)

The factor in parenthesis is the definition of the Cauchy stress tensor :

e1t(1) + e2t(2) + e3t(3) = eit(i) = (5)

Note it is a tensorial expression (independent of the vector and tensor com-ponents in a particular coordinate system). To obtain the tensorial com-ponenents in our rectangular system we replace the expressions of t(i) fromEqn.2

= eiijej (6)

Replacing in Eqn.4:t(n) = n (7)

or:

t(n) = n ijeiej = ij

n ei

ej =

ijni

ej (8)

tj = ijni (9)

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Transformation of stress components

Consider a different system of cartesian coordinates ei. We can express ourtensor in either one:

= klekel = e e (10)mn m n

We would like to relate the stress components in the two systems. To thisend, we take the scalar product of (10) with e and ej:i

ei e

j = kl e

i el =

mnimnj =

mn e

i em e ej =

ijn ek e

j

or (11)ij = kl e

i elek e

j

The factors in parenthesis are the cosine directors of the angles between theoriginal and primed coordinate axes.

Principal stresses and directions

Given the components of the stress tensor in a given coordinate system, thedetermination of the maximum normal and shear stresses is critical for thedesign of structures. The normal and shear stress components on a planewith normal n are given by:

tN = t(n)

n

= kinkni

tS = t(n)2 t2N

It is obvious from these equations that the normal component achieves itsmaximum tN = t

(n)when the shear components are zero. In this case:

t(n) = n = n = In

or in components:

kink = ni

kink = kink (12)

ki ki nk = 0

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which means that the principal stresses are obtained by solving the previous

eigenvalue problem, the principal directions are the eigenvectors of the prob-lem. The eigenvalues are obtained by noticing that the last identity can besatisfied for non-trivial n only if the factor is singular, i.e., if its determinantvanishes:

= 011 12 13

21 22 233231 33

which leads to the characteristic equation:

3 + I12 I2 + I3 = 0

where:

I1 = ii = 11 + 22 + 33 (13)

1212 +

23123 + 2I2 = iijj ijji = 1122 + 2233 + 3311

2(14)

I3 = det[] = ij (15)

are called the stress invariants because they do not depend on the coordinatesystem of choice.

Linear and angular momentum balance

We are going to derive the equations of momentum balance in integral form,since this is the formulation that is more aligned with our integral approachin this course. We start from the definition of linear and angular momentum.For an element of material at position x of volume dV , density , mass dVwhich remains constant, moving at a velocity v, the linear momentum isvdV and the angular momentum x (vdV ). The total momenta of thebody are obtained by integration over the volume as:

vdV and x vdVV V

respectively. The principle of conservation of linear momentum states thatthe rate of change of linear momentum is equal to the sum of all the externalforces acting on the body:

D

Dt VvdV =

V

fdV +S

tdS (16)

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where D is the total derivative. The lhs can be expanded as:Dt

D D vDt V

vdV =V Dt

(dV )v +V

t

dV

DbutDt

(dV ) = 0 from conservation of mass, so the principle reads:

v dV = fdV + tdS (17)

t VV S

Now, using what weve learned about the tractions and their relation to thestress tensor:

v dV = fdV + n dS (18)

t V

V S

This is the linear momentum balance equation in integral form. We canreplace the surface integral with a volume integral with the aid of the diver-gence theorem:

n dS = dVS V

and then (18) becomes:

v dV = 0

t f

V

Since this principle applies to an arbitrary volume of material, the integrand

must vanish: v = 0 (19)

t f

This is the linear momentum balance equation in differential form. In com-ponents:

viji,j + fi =

t

Angular momentum balance and the symmetry of the

stress tensor

The principle of conservation of angular momentum states that the rate ofchange of angular momentum is equal to the sum of the moment of all theexternal forces acting on the body:

Dx vdV = x fdV + x tdS (20)

Dt V V S

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It can be conveniently written as

vj vi xitj xjti dS + xifj xjfi dV = xi dV

txj

tS V V

Using ti = kink, the divergence theorem and (19), this expression leads to(see homework problem):

(ij ji)dV = 0V

which applies to an arbitrary volume V , and therefore, can only be satisfiedif the integrand vanishes. This implies:

(21)ij = ji

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Unit2 Notes