Solid Mechanics2020/2021

Class 2

Cauchy’s formula

Stress tensor and its representation

Examples

September 29, 2020

Recap: Applied Forces in Continuum Mechanics

Body C with body and surface applied forces.

P is a part of C with surface Sp

Surface force density t(x, n), defined for each unit vector nand every point x in the body C.

t(x, n), the stress vector, is the force, per unit area, exerted across a surface S with unit normal n at x.

Body force, defined by a vector field b(x) giving the force, per unit volume, exerted by the environment on C.

Recap: Equilibrium conditions

Conditions for static equilibrium of part P of body C: the resultant applied forces on P and the resultant applied moment (about o) are zero.

f(P) = ∫Sp t(x, n) dS + ∫Vp b(x) dV = 0

m(P) = ∫Sp x x t(x, n) dS + ∫Vp x x b(x) dV = 0

Cauchy’s formula

Cauchy’s formula is one of the central results of continuum mechanics.

It asserts that the vector field stress vector t(x ,n) is linear in n.

In other words, there exists a tensor field σ(x) (or linear transformation) such that for each unit vector n

𝑡1𝑡2𝑡3

= σ

11σ

12σ13

σ21

σ22

σ23

σ31

σ32

σ33

𝑛1𝑛2𝑛3

, t(n) = σ n, ti = σij nj

Cauchy’s formulaNewton´s law of action and reaction

Consider surface Sp with unit normal n at x.

Consider a cylinder C centered at xwith small height h and bases normal to n with area ΔS.

Which forces act on the cylinder? Body and surface forces.

When h -> 0, ΔS remains small but finite, but the volume and lateral area tend to zero. Volume and lateral surface forces tend to zero.

Equilibrium of a small “coin” across a surface Sp

Cauchy’s formulaNewton´s law of action and reaction

Equilibrium of forces in the cylinder

f(C) = ∫Sc t(x, n) dS + ∫Vc b(x) dV = 0

becomes, when h -> 0,

t(x, n) ΔS + t(x, - n) ΔS = 0,

t(x, n) = - t(x, - n).

Equilibrium of a small “coin” across a surface Sp

Cauchy’s formulaEquilibrium of the tetrahedron

To obtain Cauchy’s formula: equilibrium of a small tetrahedron.

Vertex Q located at xQ. Three faces intersect at Q, are normal to e1, e2 e e3 and have areas ΔS1, ΔS2 e ΔS3.

Distance inclined face from Q is h, unit normal n = (n1, n2, n3) and area ΔS.

Tetrahedron volume is ΔV = 1/3 (h ΔS).

Surface and body forcesacting on a tetrahedron

Cauchy’s formulaEquilibrium of the tetrahedron

Equilibrium equation direction x1

(t1(xQ, - e1) + ε1) ΔS1 +

(t1(xQ, - e2) + ε2) ΔS2 +

(t1(xQ, - e3) + ε3) ΔS3 +

(t1(xQ, n) + ε) ΔS +

(b1(xQ) + εv) 1/3 (h ΔS) = 0

Surface and body forcesacting on a tetrahedron

Cauchy’s formulaTetrahedron face areas and their relationship

When w is a unit vector, the triple product

(a x b) · w, a x b parallel to n,

gives the projected area of the paralelogram formed by a and b in a plane normal to w.

Then:

A1 = A n1,

A2 = A n2,

A3 = A n3

Cauchy’s formulaEquilibrium of the tetrahedron

Equilibrium equation direction x1

(t1(xQ, - e1) + ε1) ΔS1 + (t1(xQ, - e2) + ε2) ΔS2 + (t1(xQ, - e3) + ε3) ΔS3 +

(t1(xQ, n) + ε) ΔS + (b1(xQ) + εv) 1/3 (h ΔS) = 0.

Using the tetrahedron areas relationship,

(t1(xQ, - e1) + ε1) n1 ΔS + (t1(xQ, - e2) + ε2) n2 ΔS + (t1(xQ, - e3) + ε3) n3 ΔS + (t1(xQ, n) + ε) ΔS + (b1(xQ) + εv) 1/3 (h ΔS) = 0.

Dividing by ΔS, when h -> 0 and noting that ε1, ε2, ε3, ε e εv tend to zero with h, one obtains

t1(xQ, - e1) n1 + t1(xQ, - e2) n2 + t1(xQ, - e3) n3 + t1(xQ, n) = 0

Cauchy’s formulaEquilibrium of the tetrahedron

t1(xQ, - e1) n1 + t1(xQ, - e2) n2 + t1(xQ, - e3) n3 + t1(xQ, n) = 0

Using Newton´s law of action and reaction in equilibrium equation in direction x1

- t1(xQ, e1) n1 - t1(xQ, e2) n2 - t1(xQ, e3) n3 + t1(xQ, n) = 0

and in directions x2 e x3

- t2(xQ, e1) n1 - t2(xQ, e2) n2 - t2(xQ, e3) n3 + t2(xQ, n) = 0

- t3(xQ, e1) n1 - t3(xQ, e2) n2 - t3(xQ, e3) n3 + t3(xQ, n) = 0.

For a more convenient notation, let σij(x) denote the j component of the stress vector t(x, n) that acts at point x, on a surface normal to ei

σij(x) = tj(x, ei)

Cauchy’s formulat1(x, n) = σ11(x) n1 + σ21(x) n2 + σ31(x) n3

t2(x, n) = σ12(x) n1 + σ22(x) n2 + σ32(x) n3

t3(x, n) = σ13(x) n1 + σ23(x) n2 + σ33(x) n3

σ(x) = σij(x) is the stress tensor

Cauchy’s formula

ti = σji nj, t(x, n) = σT(x) n,

𝑡1𝑡2𝑡3

= σ

11σ

21σ

31

σ12

σ22

σ32

σ13

σ23

σ33

𝑛1𝑛2𝑛3

Stress tensor and stress vector

By Cauchy’s formula t(x, n) = σT(x) n, the stress tensor

σ(x) = σij(x)

is a second order tensor with nine components.

When the nine components of the stress tensor are known, the state of stress is known at that point.

By Cauchy’s formula the stress vector t(x, n) that acts on any surface (defined by its normal) at that point is known.

Representing the stress tensor σ

The components of the stress tensor are represented in a matrix

[σij] = σ11

σ12

σ13

σ21

σ22

σ23

σ31

σ32

σ33

The first indice of σij indicates the face where the stress acts (normal to ei).

The second indice indicates the direction in which the stress acts.

Example: σ11, σ12 and σ13 are the components of the stress vector t(x, e1).

Representing the stress tensor σ

The components σij are represented as

surface forces acting on the faces of a cube

located at the point.

They describe the action of the exterior material

on the interior material.

σ11, σ22 and σ33 are normal stresses.

σ12, σ13, etc. are shear stresses.

Sign convention: for positiveness, exterior

normal and vector that represents the

component have the same sense.

2

3 x

1

32

12

13

31

33

11

23

3

x

1

x2

x'

21 22

x

31

11

x

33

13

12

3

1

32

2

x

x'x'

22

21

23

21

2313

12

31

21

31

33

2

12

x

1

22

113

32

33

13

x

23

22

32

11

xx2

x1

x3

Representing the stress tensor σ

In a cartesian coordinate system x y z the following notation is used:

[σij] = σ

xxσ

xyσ

xz

σyx

σ𝑦𝑦

σyz

σzx

σ𝑧𝑦

σzz

Example

Plate is a 30°parallelogram with side parallel to x1. Constant thickness. 2D problem.

Determine σ11, σ22, σ12 e σ21 in the plate. Other components zero.

[σij] = σ11

σ12

σ13

σ21

σ22

σ23

σ31

σ32

σ33

Plate with applied surface forces2D representation

Example

Surface forces applied in the horizontal side are known:

σ22 = - 10 sen (30°) = - 5 MPa, σ21 = - 10 cos (30°) = - 8.66 MPa

Surface forces applied in the inclined side are known.

Equilibrium equations for the generic triangular element:

σ11 sen(30°) t + 8.66 cos(30°) t -- 20 (1) t = 0

σ12 sen(30°) t + 5 cos(30°) t = 0

Generic triangular element

Resultσ11 = 25 MPa ,σ12 = - 8.66 MPa = σ21

Example

Alternative method using Cauchy’s formula ti = σji nj ,

n = (cos (120°), cos (30°), 0) is the exterior normal to the inclined face

- 20 = σ11 n1 + σ21 n2 = σ11 (- 0.5) +

(- 8.66) (0.866)

0 = σ12 n1 + σ22 n2 = σ12 (- 0.5) +

(- 5) (0.866),

and σ11 = 25 Mpa, σ12 = - 8.66 Mpa

Generic triangular element