Composite Materials Design and Applications Part 10 ppsx

14.2 ORTHOTROPIC MATERIAL: HILL–TSAI CRITERION14.2.1 Preliminary Remarks A parallel with the Von Mises criterion can be seen with the following remarks: For an orthotropic material, the

13.3 CASE OF A PLY

One can observe from Equation 13.7 that the stress–strain relations in the plane

x,y appear decoupled in the case when szz= 0 We suppose that this applies forthe plies making a thin laminate Each ply will be characterized in its plane bythe following relations which are extracted from relations 13.510 and 13.7:

In the orthotropic axes
,t:

Ê ˆ (c2–s2)2

G
t

+

4c2s2 E1

- 1

E t

2nt

E t

+ +

Ê ˆ (c2–s2) 2

G
t

+ -

Ï–

=

º c2

s2

–( ) nt

E t

- 1

2G
t

–

Ï–

of the Von Mises criterion.

14.1 ISOTROPIC MATERIAL: VON MISES CRITERION

The material is elastic and isotropic In Figure 14.1, one denotes by I,II,III theprincipal directions of the stress tensor S for a given point The corresponding

matrix is

The general form of the deformation energy dW for an elementary volume dV

surrounding the point considered can be written as:

which can be reduced to

=

One obtains then by replacing:

(14.1)

One can rewrite as following the quantity in brackets:

(14.2)

Remarks: If one denotes as the direction making the same angle with each

of the principal directions (following Figure 14.1), one observes on the face with thenormal , a stress such that: = S( )

- 1+n

E

- sI 2

sII 2

sIII 2

E

-(sI+sII+sIII)2º–

ÓÌ

sII 2

sIII 2

( ) (sI+sII+sIII)2

3 -–

sII 2

sIII 2

3(sIsII+sIIsIII+sIIIsI)–

dW dV

which can be compared with Equation 14.1 Thus,

The shear stress t also appears as the shear characteristic of the distortionenergy

One recognizes in Equation 14.2 the presence of the first and second scalarinvariants of the stress tensor independent of the coordinate system Incoordinate axes other than the principal directions, the second invariantcan be written as:

One then has for any coordinate system:

then:

The elastic domain (where the distortion energy is below a certain critical value)can then be characterized by the condition:

(14.3)

To determine the constant, a uniaxial test is sufficient; in effect if one denotes by

se the elastic limit obtained from a tension–compression test, one has:

then:

t2 13 - sI 2

sII 2

sIII

2 sI+sII+sIII

3 -

2

–( ) s33s11 t31

2

–( ) s33s11 t31

t23 2

t31 2

t31 2

14.2 ORTHOTROPIC MATERIAL: HILL–TSAI CRITERION

14.2.1 Preliminary Remarks

A parallel with the Von Mises criterion can be seen with the following remarks:

For an orthotropic material, the principal directions for the stresses do notcoincide with the orthotropic directions, unlike the isotropic case

A uniaxial test is not enough to determine all the terms of the equationfor the criterion because the mechanical behavior changes with the direc-tion of loading

For the fiber/resin composites, the elastic limit corresponds with the rupturelimit

The rupture strengths are very different when loading is applied along the

l direction or along the t direction.

The rupture strengths are different in tension as compared with in compression

One can then write in the orthotropic coordinates
,t,z — shown in Figure 14.2 —

an expression similar to Equation 14.3, as:

(14.4)

14.2.2 Case of a Transversely Isotropic Material

In the following, we will limit ourselves, for the purpose of simplification, to thecase of a transversely isotropic material.3 The constants a, b, c, d, e, f in Equation

14.4 above will be determined using the results of the following tests:

Test along the longitudinal direction
:

Figure 14.2 Principal Axes for an Orthotropic Ply

Test along the transverse direction t:

Test along the transverse direction z:

due to transverse isotropy:

and in developing4:

(14.5)

Remark: For the case of a “three-dimensional” orthotropic material, an analogous

reasoning to the previous presentation leads to a more general criterion, whichcan be written as:

14.2.3 Case of a Unidirectional Ply Under In-Plane Loading

When the stress state is plane-stress, in the plane defined by the axes
,t (see

Figure 14.2), one has

Equation 14.5 is simplified, and one obtains what is called “the Hill–Tsai criterion”for a ply subject to stresses within its plane:

––

-–

corresponding to the type of loadings in the numerators (tension orcompression).

Safety factor: Let a2

< 1 the Hill–Tsai expression found for a state ofstress s
, st, t
t. One can then increase the load by means of a multiplication

coefficient k to reach the limit as:

The margin of safety can then be defined as the expression:

which can also be written as:

14.3 VARIATION OF RESISTANCE OF A UNIDIRECTIONAL PLY

WITH RESPECT TO THE DIRECTION OF LOADING

14.3.1 Tension and Compression Resistance

We propose to evaluate the maximum stress sx that one can apply on a ply in

the direction x in Figure 14.3 The stresses s
, st, t
t in the orthotropic axes aregiven by Equation 11.4 as:

Figure 14.3 Direction of Loading Distinct from Orthotropic Axes

where one recalls that c = cosq and s = sinq Thus,

Replacing in the expression of the Hill–Tsai criterion of Equation 14.6, we have

The evolution of the sx rupture, when q varies, was discussed in Section 3.3.2

Figure 14.4 Pure Shear in x,y Axes

s
rupture will be the limit stress in compression, and st rupture the limit stress intension for 0∞ £ q £ 90∞.

-

-=

COMPOSITE BEAMS IN FLEXURE

Due to their slenderness, a number of composite elements (mechanical components

or structural pieces) can be considered as beams A few typical examples areshown schematically in Figure 15.1 The study of the behavior under loading ofthese elements (evaluation of stresses and displacements) becomes a very complexproblem when one gets into three-dimensional aspects In this chapter, we propose

a monodimensional approach to the problem in an original method It consists ofthe definition of displacements corresponding to the traditional stress and momentresultants for the applied loads This leads to a homogenized formulation for theflexure—and for torsion This means that the equilibrium and behavior relationsare formally identical to those that characterize the behavior of classical homogeneousbeams Utilization of these relations for the calculation of stresses and displacementsthen leads to expressions that are analogous to the common beams

We will limit ourselves to the composite beams with constant characteristics(geometry, materials) in any cross section, made of different materials—which

we call phases—that are assumed to be perfectly bonded to each other

To clarify the procedure and for better simplicity in the calculations, we willlimit ourselves in this chapter to the case of composite beams with isotropic phases.The extension to the transversely isotropic materials is immediate When thephases are orthotropic, with eventually orthotropic directions that are changingfrom one point to another in the section, the study will be analogous, with amuch more involved formulation.1

15.1 FLEXURE OF SYMMETRIC BEAMS WITH ISOTROPIC PHASES

In the following, D symbolizes the domain occupied by the cross section, in the

y,z plane The external frontier is denoted as ∂D One distinguishes also (see

Figure 15.2) the internal frontiers which limit the phases, denoted by l ij for twocontiguous phases i and j The area of the phase i is denoted as S i; its moduli

of elasticity are denoted by E i and G i The elastic displacement at any point ofthe beam has the components: u x (x,y,z); u y (x,y,z); u z (x,y,z)

The beam is bending in the plane of symmetry x,y under external loads whichare also symmetric with respect to this plane

1

The only restrictive condition lies in the fact that one of the orthotropic directions is supposed

15.1.1.2 Longitudinal Displacement

By definition, longitudinal displacement is denoted by u(x) and written as:

which consists of a mean displacement u(x) and an incremental displacement

Du x as:

where one notes that:

(15.2)

15.1.1.3 Rotations of the Sections

By definition, this is the fictitious rotation—or equivalent—given by the followingexpression:

Or, with the above:

15.1.1.4 Elastic Center

Origin 0 of the coordinate y is chosen such that the following integral is zero:

We call elastic center the corresponding point 0 that is located as in the expressionabove Then Du x takes the form:

15.1.1.5 Transverse Displacement along y Direction

By definition, this is v(x) that is given by the following expression:

It follows from this definition that:

where one notes that:

15.1.1.6 Transverse Displacement along z Direction

By definition, this is w(x) given by

It follows from this definition and from the existence of the plane of symmetry

x,y of the beam a zero average transverse displacement, as: w(x) = 0

In summary, we obtain the following elastic displacement field:

hx represents the longitudinal distortion of a cross section, that is, the

quantity that this section displaces out of the plane which characterizes

it if it moves truly as a rigid plane body

hy and hz represent the displacements that characterize the variations ofthe form of the cross section in its initial plane

15.1.2 Perfect Bonding between the Phases

For the phases i and j in Figure 15.3, in the plane of an elemental interface with

a normal vector of , the relations between the strain tensors e are:

which can also be written as:

in which ny and nz are the cosines of the outward normal , and dG representselement of frontier ∂D If one assumes the absence of shear stresses applied over

the lateral surface of the beam, then txy ny + txz nz =0 along the external frontier

∂D Then for longitudinal equilibrium we have5

where one recognizes the shear stress resultant:

Then transforming the second integral into an integral over the external frontier

∂D of the domain D of the cross section6

:

if one remarks that:

which is the transverse density of loading on the lateral surface of the beam,transverse equilibrium can be written as:

5

We have neglected the body forces which appear in the local equation of equilibrium in the

form of a function f x If these exist (inertia forces, centrifugal forces, or vibration inertia, for example), one obtains for the equilibrium: in which represents the longitudinal load density.

-+p y = 0

d dx

where appears the moment resultant:

Then transforming the second integral7:

where one notes that:

which can be called a density moment on the beam Then one obtains theequilibrium relation:

The case where a density moment could exist in statics is practically nil, wetherefore assume that mz = 0 In summary, one obtains for the equations ofequilibrium:

(15.7)

15.1.4 Constitutive Relations

Taking into account the isotropic nature of the different phases, the constitutive

relation can be written in tensor form for Phase i as:

One deduces, in integrating over the domain occupied by the cross section ofthe beam:

Taking into account the form of the displacements in Equation 15.3, one can write

which leads, with the notation in Equation 15.1, to the relation:

(15.8)

Taking into account the form of the displacements in Equation 15.3 one can write

This leads, with the notation in Equation 15.1, to the relation:

(15.9)

Taking into account the form of the displacements in Equation 15.3, one can write

which gives with the notation in Equation 15.1, the relation:

15.1.5 Technical Formulation

15.1.5.1 Simplifications

We extend to the composite beams the simplifications made for the homogeneousbeams as:

1 syy and szz << sxx at almost all points of the cross section.8

2 We neglect the variation of warping {hx, hy, hz} between two neighboring

infinitely near sections in order to calculate the flexure stresses, that are

sxx, txy, and txz.9

15.1.5.2 Expression for the Normal Stresses

With the previous simplifications, one extracts from the constitutive relation:

the following simplified form:

· Ò -+

=bending extension

which shows the discontinuity of normal stresses due to the difference in the

longitudinal moduli, as illustrated in Figure 15.4

15.1.5.3 Shear Stress Expression

15.1.5.3.1 Characterization of Warping

Starting from the local equilibrium described by the relation:

we study the flexure shear stress in the plane of the cross section, noted as:

Taking into consideration Equations 15.11 and 15.7, and the simplification 2, atthe beginning of Section 15.1.5.1, one can write

and with the displacement field in Equation 15.3:

Putting hx in the form:

Figure 15.4 Normal and Bending Stresses

-y E i

ES

· Ò

-dN x dx

- E i

EI z

· Ò -– T y¥y

=–

EI z

· Ò -¥y

–

=

In the above relation appears a k coefficient which is analogous to the shear

coefficient for homogeneous beams

15.1.5.3.2 External Limit Condition

We have supposed that the lateral surface of the beam is free from shear Thisgives, along the external contour ∂D of the cross section, the relation:

and, with the displacement field in Equation 15.3 and the simplifications describedabove:

Introducing the function g(y,z) (Equation 15.12) and with 15.14, one obtains

Substituting the function g(y,z) with the function g o (y,z) such that:

—2

G i

– · ÒGS

-EI z

· Ò -¥y

∂y

-dS D

∂y

-dS D

one verifies that go is solution of the problem:

We denote go(y,z) as the longitudinal warping function for the cross section.

15.1.5.3.3 Conditions at the Interfaces

The continuity conditions already described in Section 15.1.2 lead for the warping

function, at the interface between two phases i and j,

then:

15.1.5.3.4 Uniqueness of the Solution

This is given by Equation 15.5 which is interpreted here as:

15.1.5.3.5 Form of the Shear Stresses

One can easily verify the following expressions:

then again:

15.1.5.3.6 Shear Coefficient for the Section

The shear coefficient for the section is obtained starting from the Equation 15.5:

-EI z

· Ò -¥y in domain D

∂z

-t=G i

T y GS

· Ò

-grad g o

h

This necessitates the knowledge of the warping function go Then the above

relation can be rewritten as:

which leads to:

In summary, in the absence of body forces (inertia forces for example), thebending of a composite beam in its plane of symmetry can be characterized by

a homogenized formulation — equivalent to a classical homogeneous beamsolution—in the following manner:

• Elastic center 0: it is such that

15.1.6 Energy Interpretation

Denoting by dWs as the deformation energy of an elementary portion of a beam

with length dx due to the application of normal stresses sxx, one has

(Contd.)

Stresses:

• longitudinal warping function g o (y, z): it is the

solution to the problem

with internal continuity:

and the uniqueness condition:

=

shear stresses txy =G i

T y GS

· Ị -∂g o

∂z

- ˛Ơ

˝Ơ

-EI z

· Ị

-y in domain D of the section.

=+

∂g o

∂n

- = 0 on the boundary ∂D

ĨƠƠÌƠỢ

˝Ơ

¸ along internal boundaries
ij

Taking into account Equation 15.11 for the normal stresses:

(the above expression simplifies due to the definition of the elastic center 0 in15.16); therefore:

15.1.6.2 Energy Due to Shear Stresses

Denoting dWt as the deformation energy of an elementary portion of a beam

with length dx due to the application of shear stresses , one has:

then, taking into account the form of the shear stresses in 15.16:

with the value from 15.16 of the Laplacian of the warping function go11:

dWsdx

Ú

+

dWsdx

- 1

2 - M z

2

ES

· Ò -+

- 1

2 - G i

- 1

2 - T y

- 1

2

- T y2GS