ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 6 ppt

As an example, consider a boron–aluminum unidirectional composite whose mental stress–strain diagrams Herakovich, 1998 are shown in Fig.. Unidirectional anisotropic layer Consider now a

As an example, consider a boron–aluminum unidirectional composite whose mental stress–strain diagrams (Herakovich, 1998) are shown in Fig 4.17 (circles) alongwith the corresponding approximations (solid lines) plotted with the aid of Eqs (4.65).

experi-4.3 Unidirectional anisotropic layer

Consider now a unidirectional layer studied in the previous section and assume that its

principal material axis 1 makes some angle φ with the x-axis of the global coordinate

frame (see Fig 4.18) An example of such a layer is shown in Fig 4.19

4.3.1 Linear elastic model

Constitutive equations of the layer under study referred to the principal material dinates are given by Eqs (4.55) and (4.56) We need now to derive such equations for

coor-the global coordinate frame x, y, and z (see Fig 4.18) To do this, we should transfer stresses σ1, σ2, τ12, τ13, τ23 acting in the layer and the corresponding strains ε1, ε2, γ12,

γ13, γ23 into stress and strain components σ x , σ y , τ xy , τ xz , τ yz and ε x , ε y , γ xy , γ xz , γ yz

using Eqs (2.8), (2.9) and (2.21), (2.27) for coordinate transformation of stresses andstrains According to Fig 4.18, the directional cosines, Eqs (2.1), for this transformation

are (we take x= 1, y= 2, z= 3)

t12, MPa

g12, %

e2, %

s2, MPa

Fig 4.17 Calculated (solid lines) and experimental (circles) stress–strain diagrams for a boron–aluminum

composite under transverse loading (a) and in-plane shear (b).

y

x

Fig 4.18 A composite layer consisting of a system of unidirectional plies with the same orientation.

Fig 4.19 An anisotropic outer layer of a composite pressure vessel Courtesy of CRISM.

equations to the following form (Verchery, 1999)

where A0n are stiffness coefficients corresponding to φ = 0 It follows from Eqs (4.72)that,

It can be directly checked that Eqs (4.73) provide three invariant stiffness characteristics

whose forms do not depend on φ, i.e.,

A11(φ) + A22(φ) + 2A12(φ) = A0

11+ A0

22+ 2A0 12

A44(φ) − A12(φ) = A0

44− A0 12

A55(φ) + A66(φ) = A0

55+ A0 66

D1= A11A22A44− A11A224− A22A214− A44A212+ 2A12A14A24

D2= A55A66− A2

56and

As can be seen in Eqs (4.71) and (4.75), the layer under study is anisotropic in plane

xy because the constitutive equations include shear–extension and shear–shear coupling

coefficients η and λ For φ= 0, the foregoing equations degenerate into Eqs (4.55) and(4.56) for an orthotropic layer

The dependencies of stiffness coefficients on the orientation angle for a carbon–epoxycomposite with properties listed in Table 3.5 are presented in Figs 4.20 and 4.21.Uniaxial tension of the anisotropic layer (the so-called off-axis test of a unidirectionalcomposite) is often used to determine material characteristics that cannot be found intests with orthotropic specimens or to evaluate constitutive and failure theories Such a

0

20 40 60 80 100 120

Fig 4.20 Dependencies of tensile (A11, A22) and shear (A44 ) stiffnesses of a unidirectional carbon–epoxy layer

on the orientation angle.

10 20 30 40

Fig 4.21 Dependencies of coupling stiffnesses of a unidirectional carbon–epoxy layer on the orientation angle.

test is shown in Fig 4.22 To study this loading case, we should take σ y = τxy = 0 inEqs (4.75) Then,

ε x= σ x

E x , ε y = −νxy σ x

E x , γ xy = ηxy, x σ x

As can be seen in these equations, tension in the x-direction is accompanied not only

with transverse contraction, as in orthotropic materials, but also with shear This results

in the deformed shape of the sample shown in Fig 4.23 This shape is natural becausethe material stiffness in the fiber direction is much higher than that across the fibers.Such an experiment, in cases where it can be performed, allows us to determine the

in-plane shear modulus, G12 in principle material coordinates using a simple tensile testrather than the much more complicated tests described in Section 3.4.3 and shown in

Figs 3.54 and 3.55 Indeed, if we know E x from the tensile test in Fig 4.23 and find E1,

E2, and ν21 from tensile tests along and across the fibers (see Sections 3.4.1 and 3.4.2),

we can use the first equation of Eqs (4.76) to determine

ε and ε2, so that material deformation is associated mainly with shear An off-axis test

Fig 4.22 An off-axis test.

g g

Fig 4.23 Deformation of a unidirectional layer loaded at an angle to fiber orientation.

0.5 1 1.5 2 2.5

σ1= σxcos2φ, σ2= σxsin2φ, τ12= −σx sin φ cos φ (4.79)

Thus, applying stress σ x and changing φ we can induce proportional loading with different combinations of stresses σ1, σ2, and τ12to evaluate putative constitutive or failure theoriesfor a material under study

However, the test shown in Fig 4.23 can hardly be performed because the test fixture(see Fig 4.22) restrains the shear deformation of the specimen and induces a correspondingshear stress The constitutive equations for the specimen loaded with uniaxial tension as

in Fig 4.23 and fixed as in Fig 4.22 follow from Eqs (4.75) if we take σ y= 0, i.e.,

in which elastic constants are specified by Eqs (4.76) The shear stress, being of a reactive

nature, can be found from Eq (4.81) if we put γ xy= 0 Then,

is the apparent elastic modulus that can be found from the test shown in Fig 4.22 As

follows from Eq (4.83), Eax , in general, does not coincide with E x as used in Eq (4.78)

for G12.

Thus, measuring σ x and ε x we can determine E xfrom Eq (4.82) only under the

condi-tion E xa = Ex , which means that the shear–extension coupling coefficient η must be zero Applying Eqs (4.76) and assuming that φ = 0 and φ = 90◦, we arrive at the following

Consider the first set of inequalities in Eqs (4.85) and assume that the first of them, whichhas the following explicit form

So, the set of conditions in Eqs (4.85) can be reduced to one inequality in Eq (4.88),which can be written in a final form as

Since the first condition in Eqs (4.86) can be presented as

1+ ν21

2G12

we can conclude that it is satisfied

Consider the second inequality in Eqs (4.86) and write it in an explicit form, i.e.,

which means that the condition in Eq (4.93) is satisfied

So, the set of conditions in Eqs (4.86) is reduced to one inequality in Eq (4.92), whichcan be written in the following final form

Thus, Eq (4.84) determines the angle φ0for the orthotropic materials whose mechanicalcharacteristics satisfy the conditions in Eqs (4.91) or (4.94) Such materials, being loaded

at an angle φ = φ0, do not experience shear–stretching coupling The shear modulus can

be found from Eq (4.78) in which E x = σx /ε x , where σ x and ε x are the stress and thestrain determined in the off-axis tension test shown in Fig 4.22

Consider as examples unidirectional composites with typical properties (Table 3.5)

(1) For fiberglass–epoxy composite, we have E1 = 60 GPa, E2 = 13 GPa,

(2) For aramid–epoxy composite, E1 = 95 GPa, E2 = 5.1 GPa, G12= 1.8 GPa,

ν12 = 0.018, ν21= 0.34

E1

2(1 + ν21) = 36.45 GPa, E2

2(1 + ν12) = 2.5 GPa, and φ0= 61.45◦(3) For carbon–epoxy composite with E1 = 140 GPa, E2 = 11 GPa, G12 = 5.5 GPa,

ν12 = 0.021, ν21= 0.27, we have

E1

2(1 + ν21) = 55.12 GPa, E2

2(1 + ν12) = 5.39 GPa

As can be seen, the conditions in Eqs (4.91) and (4.94) are not satisfied, and angle φ0

does not exist for this material

As can be directly checked with the aid of Eqs (4.76), there exists the following tionship between the elastic constants of anisotropic materials (Verchery and Gong, 1999)

This equation means that η x, xy = 0 for materials whose modulus Exreaches the extremum

value in the interval 0 < φ < 90◦ The dependencies of E

x /E1on φ for the materials

considered above as examples, are shown in Fig 4.25

As can be seen, curves 1 and 2 corresponding to glass and aramid composites reach

the minimum value at φ0 = 54.31◦ and φ0 = 61.45◦, respectively, whereas curve 3 for

carbon composite does not have a minimum at 0 < φ < 90◦.

The dependence E x (φ) with the minimum value of E x reached at φ = φ0, where

0 < φ < 90◦, is typical for composites reinforced in two orthogonal directions For

example, for a fabric composite having E1 = E2 and ν12 = ν21, Eq (4.84) yields the

well-known result φ0= 45◦ For a typical fiberglass fabric composite with E1= 26 GPa,

E2= 22 GPa, G12= 7.2 GPa, ν12= 0.11, ν21= 0.13, we have

In conclusion, it should be noted that the actual application of Eq (4.78) is hindered by

the fact that the angle φ0 specified by Eq (4.84) depends on G12, which is not known and needs to be determined from Eq (4.78) To find G12, we actually need to perform several tests for several values of G12in the vicinity of the expected value and the corresponding

values of φ0 following from Eq (4.84) and to select the correct value of G12, which satisfies in conjunction with the corresponding value of φ0, both equations – Eqs (4.78)

and (4.84) (Morozov and Vasiliev, 2003)

Consider the general case of an off-axis test (see Fig 4.22) for a composite specimen

with an arbitrary fiber orientation angle φ (see Fig 4.26) To describe this test, we need

to study the coupled problem for an anisotropic strip in which shear is induced by tensionbut is restricted at the strip ends by the jaws of a test frame as in Figs 4.22 and 4.26

As follows from Fig 4.26, the action of the grip can be simulated if we apply a bending

moment M and a transverse force V such that the rotation of the strip ends (γ in Fig 4.23)

will become zero As a result, bending normal and shear stresses appear in the strip thatcan be analyzed with the aid of composite beam theory (Vasiliev, 1993)

To derive the corresponding equations, introduce the conventional assumptions of beam

theory according to which axial, u x , and transverse, u y, displacements can be presented as

u x = u(x) + yθ, uy = v(x)

where u and θ are the axial displacement and the angle of rotation of the strip cross section

x = constant and v is the strip deflection in the xy-plane (see Fig 4.26) The strains

corresponding to these displacements follow from Eqs (2.22), i.e.,

where ( )= d ( ) /dx and ε is the elongation of the strip axis These strains are related

to stresses by Eqs (4.75) which reduce to

l

V M

Fig 4.26 Off-axis tension of a strip fixed at the ends.

The inverse form of these equations is

σ x = B11ε x + B14γ xy , τ xy = B41ε x + B44γ xy (4.97)where

E x

(4.99)

Here, ε1 = u

1 and σ x ( 1) = σ = F/ah, where F is the axial force applied to the strip,

a the strip width, and h is its thickness Since σ x ( 1)= constant, Eqs (4.99) give

These stresses are statically equivalent to the axial force P , the bending moment M, and the transverse force V , which can be introduced as

The second equation of Eqs (4.105) shows that V = C1, where C1 is a constant of

integration Then, substituting this result into Eq (4.104) and eliminating ε2with the aid

Taking in the third equation of Eqs (4.105) V = C1 and substituting M from

Eq (4.103), we arrive at the following equation for θ2

θ

2 = 12C1

a3hB11

P M V

Integration yields

θ2= 6C1

a3hB11 + C2x + C3

The total angle of rotation θ = θ1+ θ2, where θ1is specified by Eqs (4.100), should be

zero at the ends of the strip, i.e., θ (x = ±l/2) = 0 Satisfying these conditions, we have

θ2= 3C1

a3hB11



2x2−l22

−0.5 −0.4 −0.3 −0.2 −0.1 0.1 0.2 0.3 0.4 0.5

vE x ls

x l

Fig 4.28 Normalized deflection of a carbon–epoxy strip (φ= 45 ◦, l = 10).

Now, we can write the relationship between modulus E xcorresponding to the ideal test

shown in Fig 4.23 and apparent modulus E xa that can be found from the real test shown

in Figs 4.22 and 4.26 Using Eqs (4.98), (4.100), (4.106), and (4.110), we finally get

infinitely short strip (l → 0), taking into account Eqs (4.98), we get

1− ηx, xy η xy, x = B11

In accordance with Eq (4.97), this result corresponds to a restricted shear deformation

(γ xy = 0) For a strip with finite length, Ex < E xa< B11 The dependence of the normalizedapparent modulus on the length-to-width ratio for a 45◦carbon–epoxy layer is shown in

Fig 4.29 As can be seen, the difference between E xa and E x becomes less than 5%

for l > 3a.

1 1.1 1.2 1.3 1.4

4.3.2 Nonlinear models

Nonlinear deformation of an anisotropic unidirectional layer can be studied rather

straightforwardly because stresses σ1, σ2, and τ12 in the principal material coordinates(see Fig 4.18) are statically determinate and can be found using Eqs (4.67) Substitutingthese stresses into the nonlinear constitutive equations, Eqs (4.60) or Eqs (4.64), we can

express strains ε1, ε2, and γ12 in terms of stresses σ x , σ y , and τ xy Further substitution of

the strains ε1, ε2, and γ12 into Eqs (4.70) yields constitutive equations that link strains

ε x , ε y , and γ xy with stresses σ x , σ y , and τ xythus allowing us to find strains in the global

coordinates x, y, and z if we know the corresponding stresses.

As an example of the application of a nonlinear elastic material model described byEqs (4.60), consider a two-matrix fiberglass composite (see Section 4.4.3) whose stress–strain curves in the principal material coordinates are presented in Fig 4.16 These curves

allowed us to determine coefficients ‘b’ and ‘c’ in Eqs (4.60) To find the coupling coefficients ‘m,’ we use a 45◦ off-axis test Experimental results (circles) and the corre-sponding approximation (solid line) are shown in Fig 4.30 Thus, the constructed modelcan be used now to predict material behavior under tension at any other (different from 0,

45, and 90◦) angle (the corresponding results are given in Fig 4.31 for 60◦) or to studymore complicated material structures and loading cases (see Section 4.5)

As an example of the application of the elastic–plastic material model specified by

Eq (4.64), consider a boron–aluminum composite whose stress–strain diagrams in cipal material coordinates are shown in Fig 4.17 Theoretical and experimental curves(Herakovich, 1998) for 30 and 45◦ off-axis tension of this material are presented inFig 4.32

prin-0 4 8 12 16

s x, MPa

e x, %

Fig 4.30 Calculated (solid line) and experimental (circles) stress–strain diagram for 45 ◦off-axis tension of a

two-matrix unidirectional composite.

0 4 8 12 16

s x, MPa

e x, %

Fig 4.31 Theoretical (solid line) and experimental (dashed line) stress–strain diagrams for 60 ◦off-axis tension

of a two-matrix unidirectional composite.

0 40

Fig 4.32 Theoretical (solid lines) and experimental (dashed lines) stress–strain diagrams for 30 ◦ and 45◦

off-axis tension of a boron–aluminum composite.

4.4 Orthogonally reinforced orthotropic layer

The simplest layer reinforced in two directions is the so-called cross-ply layer thatconsists of alternating plies with 0 and 90◦orientations with respect to the global coordi-

nate frame x, y, and z as in Fig 4.33 Actually, this is a laminated structure, but being

formed with a number of plies, it can be treated as a homogeneous orthotropic layer (seeSection 5.4.2)

1 23

Fig 4.33 A cross-ply layer.

4.4.1 Linear elastic model

Let the layer consist of m longitudinal (0◦) plies with thicknesses h (i)

0 (i = 1,

2, 3, , m) and n transverse (90◦) plies with thicknesses h (j )

90 (j = 1, 2, 3, , n) made from one and the same composite material Then, stresses σ x , σ y , and τ xythat comprisethe plane stress state in the global coordinate frame can be expressed in terms of stresses

in the principal material coordinates of the plies as