basic concepts to start mechanics of...

Basic concepts to start Mechanics of Materials

Georges Cailletaud

Centre des MatériauxEcole des Mines de Paris/CNRS

A brief summary of continuum mechanics Notations

Notations (maths) (1/2)A vector v (element of a vectorial space) can be seen in a given frame as a column ofcomponents Vi . This is a tensor of order 1 (one index).

A second order tensor M∼ can be seen in a given frame as a matrix of components Mij

A fourth order tensor L≈

can be seen in a given frame as a four index table Lijkl

A scalar x is a tensor of order zero (no index)

• Einstein convention means “repeated index ≡ summation ≡ one order less”

(vector . vector) gives a scalar – (order_1 . order_1) gives order_0

vi vi =3

∑i=1

(vi )2 = x

(2nd order tensor . vector) gives a vector – (order_2 . order_1) gives order_1

Mij vj =3

∑j=1

Mij vj = wi

(2nd order tensor . 2nd order tensor) gives a 2nd order tensor – (order_2 . order_2) givesorder_2

Mij Njk =3

∑j=1

Mij Njk = Pik

Georges Cailletaud (Centre des Matériaux/UMR 7633 ) Basic concepts for MoM October 2013 2 / 24


Notations (maths) (2/2)

• Intrinsic notation means no component marked, one dot ≡ summation ≡ one order less

(vector . vector) gives a scalar – (order_1 . order_1) gives... order_0

v .v = x

(2nd order tensor . vector) gives a vector – (order_2 . order_1) gives... order_1

M∼ .v = w

(2nd order tensor . 2nd order tensor) gives a 2nd order tensor – (order_2 . order_2) gives...order_2

M∼ .N∼ = P∼

(2nd order tensor : 2nd order tensor) gives a scalar – (order_2 : order_2) gives... order_0

M∼ : N∼ = x

M∼ : L≈

: N∼ gives ?



Index expansion

• Tensorial product

Produce a 2nd order tensor from two vectors

index form : mij = ni lj intrinsic form : m∼ = n⊗ l

Produce a 4th order tensor from two 2nd order tensors

index form : Lijkl = Mij Nkl intrinsic form : L≈

= M∼ ⊗N∼



Notations (mechanics)

Displacement vector : ucomponents ui

Strain tensor (second order, symmetric) : ε∼components εij

Stress tensor (second order, symmetric) : σ∼components σij

Stress vector for a facet of normal n : T = σ∼ .ncomponents Ti = σij nj (sum on j)

Stress tensor from strain tensor (elastic constitutive equations) σ∼ = Λ≈

: ε∼

components σij = Λijkl εkl (sum on k and l)


A brief summary of continuum mechanics Kinematics and statics in small strains

Displacement-strainThe strain tensor is the symmetric part of the displacement gradient

εij =12

(ui,j + uj,i )

ε∼ =12

(∇u + ∇uT )

Kinematically admissible field : u = ud sur ∂Ωu

Compatibility equations (ex : 6 strain components derive from 3 displacementcomponents). For 3D cartesian coordinates

ε11,22 + ε22,11−2ε12,12 = 0 ε11,23 + ε23,11− ε12,13− ε13,12 = 0

. . .and circular permutations, that is :

εinmεljk εij,km = 0

with εijk = 0 if 2 indices are equalεijk = 1 for the case of an even permutation, =-1 for an odd permutation



Geometrical meaning of the terms in the strain tensor∆VV = Tr ε∼ = εii γ = 2ε12

dx

dxdxε22

dxε11

2

1

1

2

(1+ )

(1+ ) dx2

γ

dx1

The diagonal terms characte-rize the elongation of a unitsegment in the direction of theaxes

Non diagonal terms characte-rize the rotations with respectto the axes

Elongation in the direction defined by n

δ(n) = n.ε∼.n = εijninj



Stress

Volumetric forces : f d in the volume Ω

Surface forces : F d on the surface ∂ΩF

Statically admissible stress :in Ω

divσ∼

+ f d = 0 σij,j + f di = 0

on ∂ΩFσ∼.n = F d

σij nj = F di

Spherical part of the stress tensor :

S∼ =13

trace(σ∼ ) I∼ Sij =σll

3δij

Deviator associated to the stress tensor :

s∼ = σ∼−S∼ sij = σij −σll

3δij trace(s∼) = sll = 0



Physical meaning of the terms of the stress tensorx

x

σσ

σ

σ2 22

12

21

11

1

The diagonal terms characterizethe normal forces

The non diagonal termscharacterize the shear forces

Stress vector for a facet of direction n

T (n) = σ∼ .n Ti = σijnj

Normal stress for a facet of direction n

Tn(n) = n.T = n.σ∼ .n = σijninj

Shear on a facet of direction n

T t (n) = T −Tnn Tt =√

T 2−T 2n


A brief summary of continuum mechanics Internal/external forces

Work of internal/external forces

Stokes theorem for a scalar function f integrated on a volume Ω, n being thenormal to ∂Ω Z

Ωf,jdV =

Z∂Ω

f njdS

Work of internal forces (real stress field, kinematically admissible displacementfield)

−Wi =Z

Ωσijε′ijdV =

ZΩ

σiju′i,jdV

=Z

Ω

((σiju

′i ),j −σij,ju

′i

)dV =

Z∂Ω

σiju′i njdS−

ZΩ

σij,ju′i dV

Work of external forces

We =Z

Ωf di u′i dV +

Z∂ΩF

F di u′i dS


A brief summary of continuum mechanics Internal/external forces

Application of virtual work theorem

Total work (internal+external) should be zero for an isolated system

with Wi + We = 0, it comes :

−Z

∂Ωσiju

′i njdS +

ZΩ

σij,ju′i dV +

ZΩ

f di u′i dV +

Z∂ΩF

F di u′i dS = 0

equilibrium equation in Ω

σij,j + f di = 0 divσ∼ + f = 0

boundary condition on ∂ΩF

σijnj = F di σ∼ .n = F d

These relations do not depend on the material

The constitutive equations are the relations between stresses and strains


A brief summary of continuum mechanics Elastic potential, linear elasticity

Elastic potential

The behaviour, eventually non linear, is fully defined by a potential, given by its volumetricdensity. Its shape depends on the representative variable

Evolution between two equilibrium states, with σ∼∗ = σ∼ , and ε∼

′ = d ε∼, the elasticpotential W (ε∼) writes, in linear elasticity :

W (ε∼) =12

ε∼ : C≈

: ε∼ σ∼ =∂W∂ε

= C≈

: ε∼

Evolution between two equilibrium states, with ε∼′ = ε∼, and σ∼

∗ = dσ∼ , thecomplementary elastic potential W ∗(σ∼) writes, in linear elasticity :

W ∗(σ∼) =12

σ∼ : S≈

: σ∼ ε∼ =∂W ∗

∂σ= S≈

: σ∼

W and W ∗ convex and dW + dW ∗ = d(σijεij )

And :∂2W

∂εij∂εkl=

∂σij

∂εkl= Cijkl =

∂σkl

∂εij= Cklij



Linear elasticity

Linear elasticity (stiffness and compliance matrices) :

σ∼ = C≈

: ε σij = Cijklεkl

ε∼ = S≈

: σ εij = Sijklσkl

Symmetry relations :

Cijkl = Cijlk = Cjikl Sijkl = Sijlk = Sjikl

Energy related relations :

Cijkl = Cklij Sijkl = Sklij

General anisotropy = 21 coefficients ; orthotropy = 9 coeff ; cubic symmetry = 3coeff ; isotropic material = 2 coefficients

Isotropic material :s∼ = 2µε∼

devσll = 3κεll



Isotropic elasticity

Shear modulus µ such as τ = µγ

Compressibility modulus κ such as p =13

σll = κ∆VV

Young’s modulus E such as σ = E ε in tension

Poisson’s ratio ν such as εT =−νεL in tension (εT , transverse strain, εL,longitudinal strain)

Stress versus strain

σ∼ = λTr ε∼ I∼+ 2µε∼ σij = λεllδij + 2µεij

Strain versus stress

ε∼ =1 + ν

Eσ∼−

ν

ETrσ∼ I∼ εij =

1 + ν

Eσij −

ν

Eεllδij



Relations between the elastic coefficients

Expressions of λ, µ et κ

λ =Eν

(1 + ν)(1−2ν)2µ =

E1 + ν

3κ =E

1−2ν= 3λ + 2µ

Expressions of E et ν

E =µ(3λ + 2µ)

λ + µν =

λ

2(λ + µ)

Typical values :ν≈ 1/3 2µ≈ 3E/4 κ≈ E

Rubber :ν≈ 1/2 µ≈ E/3 κ E


A brief summary of continuum mechanics A few particular stress states

Pure tension

Onedimensional stress state σ∼ = σ0n⊗n ; for instance in a prism of axis x1, x1

being the tensile direction, and the lateral faces being free :

σ∼ :=

σ0 0 00 0 00 0 0

For a section S0, the force in direction x1 is : F = σ0S0

In the frame x1x2x3, the strain tensor writes :

ε∼ :=

σ0/E 0 00 −νσ0/E 00 0 −νσ0/E

If the length is L0, the elongation in direction x1 is : ∆L = εL0

The stiffness of the prism is R = F/∆L = ES0/L0



Pure shear

τ−τ

τ

τ

τ = σ12 = 2µε12

Deviatoric loading

Example of pure shear in theplane x1x2

σ∼ :=

0 τ 0τ 0 00 0 0

Rotation of π/4

σ∼ :=

τ 0 00 −τ 00 0 0



Circular bending

Only one component in the stress tensor, but non uniform in space :

σ∼ :=

σ11(x3) 0 00 0 00 0 0

For instance : σ11 =

Mx3

I, where M is the bending moment around x2, and

I =Z

Sx2

3 dS is the inertia of the section with respect to x2

A prism of axis x1 submitted to such a loading type presents a relative rotation of

its sections characterized by an angle θ such as θ,1 =MEI

The strain can be expressed as a function of the curvature u3,11 by :ε11 =−x3u3,11

For a rectangular section of height h along x3 and width b along x2 : I =bh3

12



Torsion

x 3

β

γ

Γ contour of the section

A line parallel to theprism axis becomeshelicoidal

Displacements

u1 =−αx3x2 u2 = αx3x1 u3 = αφ(x1,x2)

Stress

σ13 = µα(φ,1− x2) =µαθ,2 (1)

σ23 = µα(φ,2 + x1)=−µαθ,1 (2)

avec ∆φ = 0 ∆θ + 2 = 0 θ = 0 sur Γ

Torsion moment :

M =Z

S(x1σ23− x2σ13)dS

Torsion stiffness modulus :

D = 2µZ

Sθdx1dx2 = M/α



Torsion, circular section

For a circular prism of length L, and internal radius Re : β = αL

At the external surface γ = 2εθz = αR

Shear stress τ = µαr

θ can be expressed as θ =12

(R2− x21 − x2

2 )

A section perpendicular to x3 remains plane : φ = 0

Tube with an internal radius Ri and external Re : D = µπ(R4

e −R4i )

2Thin tube, with a radius R and a width e : D = 2µπeR3 = M/α thus

α =τ

µR=

M2πµeR3 , and τ =

M2eπR2



Cylindrical coordinates

Expressions limited to the classical examples where :- the displacement is colinear to er , u = ur er- the only non zero volumetric force is fr er .

Equilibrium

σrr ,r +σrr −σθθ

r+ fr = 0

Strainεrr = ur ,r εθθ =

ur

rthat is : rεθθ,r = εrr − εθθ

Assuming zero volume forces, internal radius a, external radius b

σrr = A− Br2 σθθ = A +

Br2

ur = Cr + D/r



Cylindre under pressure

Tube under pressure, internal pressure pi , external pressure pe

A =pia2−peb2

b2−a2 B =(pi −pe)a2b2

b2−a2

Solid cylinder (pi = 0, a = 0, pe = p),

σrr = σθθ = p

Internal pressure (pi = p, a, b),

σrr =a2

b2−a2

(1− b2

r2

)p σθθ =

a2

b2−a2

(1 +

b2

r2

)p

Thin tube under internal pressure p, radius R, thickness e,

σrr negligeable σθθ = pR/e



Spherical coordinates

Expressions limited to the classical examples where :- the displacement is colinear to er , u = ur er- the only non zero volumetric force is fr er .

Equilibrium

σrr ,r + 2σrr −σθθ

r+ fr = 0

Strainεrr = ur ,r εθθ =

ur

rthat is : rεθθ,r = εrr − εθθ

Zero volume force, internal radius a, external radius b

σrr = A− 2Br3 σθθ = σφφ = A +

Br3

ur = Cr + D/r2



Sphere under pressure

Sphere under pressure, internal pressure pi , external pressure pe

A =pia3−peb3

b3−a3 B =(pi −pe)a3b3

2(b3−a3)

Solid sphere (pi = 0, a = 0, pe = p),

σrr = σθθ = σφφ = p

Internal pressure (pi = p, a, b),

σrr =a3

b3−a3

(1− b3

r3

)p σθθ =

a3

b3−a3

(1 +

b3

2r3

)p

Thin tube mince under internal pressure p, radius R, thickness e,

σrr négligeable σθθ = pR/2e


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