Introduction to Elasticity Part 6 pps

The kinematic equations relate strains to displacement gradients,and the equilibrium equations relate stress to the applied tractions on loaded boundaries and alsogovern the relations am

Constitutive Equations

David RoylanceDepartment of Materials Science and EngineeringMassachusetts Institute of Technology

Cambridge, MA 02139October 4, 2000

Introduction

The modules on kinematics (Module 8), equilibrium (Module 9), and tensor transformations(Module 10) contain concepts vital to Mechanics of Materials, but they do not provide insight onthe role of the material itself The kinematic equations relate strains to displacement gradients,and the equilibrium equations relate stress to the applied tractions on loaded boundaries and alsogovern the relations among stress gradients within the material In three dimensions there aresix kinematic equations and three equilibrum equations, for a total of nine However, there arefifteen variables: three displacements, six strains, and six stresses We need six more equations,and these are provided by the material’s consitutive relations: six expressions relating the stresses

to the strains These are a sort of mechanical equation of state, and describe how the material

is constituted mechanically

With these constitutive relations, the vital role of the material is reasserted: The elasticconstants that appear in this module are material properties, subject to control by processingand microstructural modification as outlined in Module 2 This is an important tool for theengineer, and points up the necessity of considering design of the material as well as with thematerial

Isotropic elastic materials

In the general case of a linear relation between components of the strain and stress tensors, wemight propose a statement of the form

ij =Sijklσklwhere theSijkl is a fourth-rank tensor This constitutes a sequence of nine equations, since eachcomponent ofij is a linear combination of all the components ofσij For instance:

23=S2311σ11+S2312σ12+· · · + S233333

Based on each of the indices ofSijkl taking on values from 1 to 3, we might expect a total of 81independent components inS However, both ij and σij are symmetric, with six rather thannine independent components each This reduces the number ofS components to 36, as can beseen from a linear relation between the pseudovector forms of the strain and stress:

It can be shown1 that the S matrix in this form is also symmetric It therefore it contains only

21 independent elements, as can be seen by counting the elements in the upper right triangle ofthe matrix, including the diagonal elements (1 + 2 + 3 + 4 + 5 + 6 = 21)

If the material exhibits symmetry in its elastic response, the number of independent elements

in theS matrix can be reduced still further In the simplest case of an isotropic material, whosestiffnesses are the same in all directions, only two elements are independent We have earliershown that in two dimensions the relations between strains and stresses in isotropic materialscan be written as

The quantity in brackets is called the compliance matrix of the material, denoted S or Sij It

is important to grasp the physical significance of its various terms Directly from the rules ofmatrix multiplication, the element in theithrow andjth column ofSij is the contribution of the

jth stress to the ith strain For instance the component in the 1,2 position is the contribution

of the y-direction stress to the x-direction strain: multiplying σy by 1/E gives the y-directionstrain generated by σy, and then multiplying this by −ν gives the Poisson strain induced inthex direction The zero elements show the lack of coupling between the normal and shearingcomponents

The isotropic constitutive law can also be written using index notation as (see Prob 1)

If we wish to write the stresses in terms of the strains, Eqns 3 can be inverted In cases ofplane stress (σz =τxz =τyz = 0), this yields

whereD = S−1 is the stiffness matrix

Hydrostatic and distortional components

Figure 1: Hydrostatic compression

A state of hydrostatic compression, depicted in Fig 1, is one in which no shear stresses existand where all the normal stresses are equal to the hydrostatic pressure:

σx=σy =σz =−pwhere the minus sign indicates that compression is conventionally positive for pressure butnegative for stress For this stress state it is obviously true that

In many cases other than direct hydrostatic compression, it is still convenient to “dissociate”the hydrostatic (or “dilatational”) component from the stress tensor:

σij = 1

Here Σij is what is left over from σij after the hydrostatic component is subtracted The Σij

tensor can be shown to represent a state of pure shear, i.e there exists an axis transformationsuch that all normal stresses vanish (see Prob 5) The Σijis called the distortional, or deviatoric,

component of the stress Hence all stress states can be thought of as having two components asshown in Fig 2, one purely extensional and one purely distortional This concept is convenientbecause the material responds to these stress components is very different ways For instance,plastic and viscous flow is driven dominantly by distortional components, with the hydrostaticcomponent causing only elastic deformation.

Figure 2: Dilatational and deviatoric components of the stress tensor

Example 1 Consider the stress state

where the factor 2 is needed because tensor descriptions of strain are half the classical strainsfor which values of G have been tabulated Writing the constitutive equations in the form ofEqns 8 and 9 produces a simple form without the coupling terms in the conventionalE-ν form.

Example 2 Using the stress state of the previous example along with the elastic constants for steel (E = 207 GPa, ν = 0.3, K = E/3(1 − 2ν) = 173 GPa, G = E/2(1 + ν) = 79.6 Gpa), the dilatational and distortional components of strain are



 The total strain is then

ij =1

3 kkδij+ eij=



 0.00960 0.03780.0378 0.0285 0.04410.05670.0441 0.0567 −0.00930

Finite strain model

When deformations become large, geometrical as well as material nonlinearities can arise thatare important in many practical problems In these cases the analyst must employ not only adifferent strain measure, such as the Lagrangian strain described in Module 8, but also differentstress measures (the “Second Piola-Kirchoff stress” replaces the Cauchy stress when Lagrangianstrain is used) and different stress-strain constitutive laws as well A treatment of these for-mulations is beyond the scope of these modules, but a simple nonlinear stress-strain modelfor rubbery materials will be outlined here to illustrate some aspects of finite strain analysis.The text by Bathe2 provides a more extensive discussion of this area, including finite element

For the large-strain case, the following analogous stress-strain relation has been proposed:

λ2

x = 1 + 2x = 3

where here x is the Lagrangian strain and σ∗

m is a parameter not necessarily equal to σm.The σ∗

m parameter can be found for the case of uniaxial tension by considering the transverse

contractionsλy =λz:

λ2y = 3

E(σy− σ∗m)Since for rubberλxλyλz= 1, λ2

y = 1/λx Making this substitution and solving forσ∗

λ2x= 3E

whereG = E/2(1 + ν) = E/3 for ν = 1/2 This result is the same as that obtained in Module

2 by considering the force arising from the reduced entropy as molecular segments spanningcrosslink sites are extended It appears here from a simple hypothesis of stress-strain response,using a suitable measure of finite strain

Anisotropic materials

Figure 3: An orthotropic material

If the material has a texture like wood or unidirectionally-reinforced fiber composites asshown in Fig 3, the modulusE1 in the fiber direction will typically be larger than those in thetransverse directions (E2 and E3) WhenE1 6= E26= E3, the material is said to be orthotropic

It is common, however, for the properties in the plane transverse to the fiber direction to beisotropic to a good approximation (E2 = E3); such a material is called transversely isotropic.The elastic constitutive laws must be modified to account for this anisotropy, and the followingform is an extension of Eqn 3 for transversely isotropic materials:

a strain applied in the 2-direction Since the 2-direction (transverse to the fibers) usually hasmuch less stiffness than the 1-direction, it should be clear that a given strain in the 1-directionwill usually develop a much larger strain in the 2-direction than will the same strain in the2-direction induce a strain in the 1-direction Hence we will usually have ν12> ν21 There arefive constants in the above equation (E1,E2,ν12,ν21 andG12) However, only four of them areindependent; since theS matrix is symmetric, ν21/E2 =ν12/E1

A table of elastic constants and other properties for widely used anisotropic materials can

be found in the Module on Composite Ply Properties

The simple form of Eqn 11, with zeroes in the terms representing coupling between normaland shearing components, is obtained only when the axes are aligned along the principal materialdirections; i.e along and transverse to the fiber axes If the axes are oriented along some otherdirection, all terms of the compliance matrix will be populated, and the symmetry of the materialwill not be evident If for instance the fiber direction is off-axis from the loading direction, thematerial will develop shear strain as the fibers try to orient along the loading direction as shown

in Fig 4 There will therefore be a coupling between a normal stress and a shearing strain,which never occurs in an isotropic material

Figure 4: Shear-normal coupling

The transformation law for compliance can be developed from the transformation laws forstrains and stresses, using the procedures described in Module 10 (Transformations) By suc-cessive transformations, the pseudovector form for strain in an arbitraryx-y direction shown inFig 5 is related to strain in the 1-2 (principal material) directions, then to the stresses in the 1-2directions, and finally to the stresses in thex-y directions The final grouping of transformationmatrices relating thex-y strains to the x-y stresses is then the transformed compliance matrix

Figure 5: Axis transformation for constitutive equations.

in thex-y direction:

Example 3 Consider a ply of Kevlar-epoxy composite with a stiffnesses E 1= 82,E 2 = 4,G 12= 2.8 (all GPa) and

ν 12= 0.25 The compliance matrix S in the 1-2 (material) direction is:

The apparent engineering constants that would be observed if the ply were tested in the x-y rather than 1-2 directions can be found directly from the trasnformed S matrix For instance, the apparent elastic modulus in the x direction is E x= 1/S 1,1= 1/(.8830 × 10 −10) = 11.33 GPa.

Problems

1 Expand the indicial forms of the governing equations for solid elasticity in three dimensions:

equilibrium : σij,j = 0kinematic : ij = (ui,j+uj,i)/2constitutive : ij = 1 +ν

E σij−

ν

Eδijσkk+αδij∆Twhereα is the coefficient of linear thermal expansion and ∆T is a temperature change

2 (a) Write out the compliance matrix S of Eqn 3 for polycarbonate using data in theModule on Material Properties

(b) Use matrix inversion to obtain the stiffness matrixD

(c) Use matrix multiplication to obtain the stresses needed to induce the strains

(b) Use matrix inversion to obtan the stiffness matrix D

(c) Use matrix multiplication to obtain the stresses needed to induce the strains

(b) GivenG = 357 MPa and K = 1.67 GPa, obtain the deviatoric and dilatational straintensorseij and (1/3)kkδij.

(c) Add the deviatoric and dilatational strain components obtained above to obtain thetotal strain tensor ij

(d) Compute the strain tensor ij using the alternate form of the elastic constitutive lawfor isotropic elastic solids:

ij = 1 +ν

E σij −

ν

EδijσkkCompare the result with that obtained in (c)

5 Provide an argument that any stress matrix having a zero trace can be transformed to onehaving only zeroes on its diagonal; i.e the deviatoric stress tensor Σij represents a state

Statics of Bending: Shear and Bending Moment Diagrams

David RoylanceDepartment of Materials Science and EngineeringMassachusetts Institute of Technology

Cambridge, MA 02139November 15, 2000

Introduction

Beams are long and slender structural elements, differing from truss elements in that they arecalled on to support transverse as well as axial loads Their attachment points can also bemore complicated than those of truss elements: they may be bolted or welded together, so theattachments can transmit bending moments or transverse forces into the beam Beams areamong the most common of all structural elements, being the supporting frames of airplanes,buildings, cars, people, and much else

The nomenclature of beams is rather standard: as shown in Fig 1, L is the length, or span;

b is the width, and h is the height (also called the depth) The cross-sectional shape need not

be rectangular, and often consists of a vertical web separating horizontal flanges at the top andbottom of the beam1.

Figure 1: Beam nomenclature

As will be seen in Modules 13 and 14, the stresses and deflections induced in a beam underbending loads vary along the beam’s length and height The first step in calculating these quan-tities and their spatial variation consists of constructing shear and bending moment diagrams,

V (x) and M (x), which are the internal shearing forces and bending moments induced in thebeam, plotted along the beam’s length The following sections will describe how these diagramsare made

Figure 2: A cantilevered beam.

Free-body diagrams

As a simple starting example, consider a beam clamped (“cantilevered”) at one end and jected to a load P at the free end as shown in Fig 2 A free body diagram of a section cuttransversely at position x shows that a shear force V and a moment M must exist on the cutsection to maintain equilibrium We will show in Module 13 that these are the resultants of shearand normal stresses that are set up on internal planes by the bending loads As usual, we willconsider section areas whose normals point in the +x direction to be positive; then shear forcespointing in the +y direction on +x faces will be considered positive Moments whose vectordirection as given by the right-hand rule are in the +z direction (vector out of the plane of thepaper, or tending to cause counterclockwise rotation in the plane of the paper) will be positivewhen acting on +x faces Another way to recognize positive bending moments is that they causethe bending shape to be concave upward For this example beam, the statics equations give:

Figure 3: Shear and bending moment diagrams

1There is a standardized protocol for denoting structural steel beams; for instance W 8× 40 indicates a wide-flange beam with a nominal depth of 800and weighing 40 lb/ft of length

As stated earlier, the stresses and deflections will be shown to be functions of V and M , so it

is important to be able to compute how these quantities vary along the beam’s length Plots of

V (x) and M (x) are known as shear and bending moment diagrams, and it is necessary to obtainthem before the stresses can be determined For the end-loaded cantilever, the diagrams shown

in Fig 3 are obvious from Eqns 1 and 2

Figure 4: Wall reactions for the cantilevered beam

It was easiest to analyze the cantilevered beam by beginning at the free end, but the choice

of origin is arbitrary It is not always possible to guess the easiest way to proceed, so considerwhat would have happened if the origin were placed at the wall as in Fig 4 Now when a freebody diagram is constructed, forces must be placed at the origin to replace the reactions thatwere imposed by the wall to keep the beam in equilibrium with the applied load These reactionscan be determined from free-body diagrams of the beam as a whole (if the beam is staticallydeterminate), and must be found before the problem can proceed For the beam of Fig 4:

X

Fy = 0 =−VR+ P ⇒ VR= P

X

Mo= 0 = MR− P L ⇒ MR= P LThe shear and bending moment at x are then

V (x) = VR= P = constant

M (x) = MR− VRx = P L − P xThis choice of origin produces some extra algebra, but the V (x) and M (x) diagrams shown inFig 5 are the same as before (except for changes of sign): V is constant and equal to P , and Mvaries linearly from zero at the free end to P L at the wall

Distributed loads

Transverse loads may be applied to beams in a distributed rather than at-a-point manner asdepicted in Fig 6, which might be visualized as sand piled on the beam It is convenient todescribe these distributed loads in terms of force per unit length, so that q(x) dx would be theload applied to a small section of length dx by a distributed load q(x) The shear force V (x) set

up in reaction to such a load is

V (x) = −

Z x