ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 14 pps

8.16 with the corresponding expression for the thickness of a metal pressure vessel, which is hm = pR/σ , we can see that the thickness of an optimal composite vessel is 1.5 times more t

Fig 7.51 For this type of loading,

N x=1

2pR, N y = pR, Nxy= 0

where Nx and Ny are the circumferential and axial stress resultants, respectively, p the internal pressure and R is the cylinder radius Thus, we have ny = 2 and λ = 1/3 Since

N xy = 0, the structure of the laminate is symmetric with respect to the cylinder meridian,

and Eqs (8.12)–(8.14) can be reduced to

Comparing Eq (8.16) with the corresponding expression for the thickness of a metal

pressure vessel, which is hm = pR/σ , we can see that the thickness of an optimal composite vessel is 1.5 times more than hm Nevertheless, because of their higher strength

and lower density, composite pressure vessels are significantly lighter than metal ones

To show this, consider pressure vessels with radius R = 100 mm made of different

materials and designed for a burst pressure p= 20 MPa The results are listed in Table 8.1

As can be seen, the thickness of a glass–epoxy vessel is the same as that for the thickness

of a steel vessel, because the factor 1.5 in Eq (8.16) is compensated by the composite’sstrength which is 1.5 times greater than the strength of steel However, the density of aglass–epoxy composite is much lower than the density of steel, and as a result, the mass

of unit surface area of the composite vessel is only 27% of the corresponding value for

a steel vessel The most promising materials for pressure vessels are aramid and carboncomposites, which have the highest specific tensile strength (see Table 8.1)

Consider Eq (8.17) which shows that there can exist an infinite number of optimallaminates with one and the same thickness specified by Eq (8.16)

The simplest is a cross-ply laminate having k = 2, φ1= 0◦, h

1= h0, and φ2 = 90◦,

h2 = h90 For this structure, Eq (8.17) yields h90 = 2h0. This result seems obvious

because Ny /N x = 2 For symmetric ±φ angle-ply laminate, we should take k = 2,

h1= h2= hφ / 2, φ1= +φ, φ2= −φ Then,

cos2φ=1

3, φ = φ0= 54.44◦

As a rule, helical plies are combined with circumferential plies as in Fig 7.51 For this

case, k = 3, h1 = h2 = hφ / 2, φ1 = −φ2 = φ, h3 = h90, φ3 = 90◦, and Eq (8.17)gives

h90

Since the thickness cannot be negative, this equation is valid for 0 ≤ φ ≤ φ0 For

φ0 ≤ φ ≤ 90◦, the helical layers should be combined with an axial one, i.e., we should

put k = 3, h1= h2= hφ / 2, φ1= −φ2= φ and h3= h0, φ3= 0◦ Then,

h0

h φ = 1

The dependencies corresponding to Eqs (8.18) and (8.19) are presented in Fig 8.3

As an example, consider a filament wound pressure vessel whose parameters are listed

in Table 6.1 The cylindrical part of the vessel shown in Figs 4.14 and 6.22 consists of

a ±36◦ angle-ply helical layer and a circumferential layer whose thicknesses h

1 = hφ and h2= h90 are presented in Table 6.1 The ratio h90/ h φ for two experimental vessels

is 0.97 and 1.01, whereas Eq (8.18) gives for this case h90/ h φ = 0.96 which shows

that both vessels are close to optimal structures Laminates reinforced with uniformly

stressed fibers can exist under some restrictions imposed on the acting forces Nx , N y ,

and Nxy Such restrictions follow from Eqs (8.13) and (8.14) under the conditions that

h i ≥ 0, 0 ≤ sin2φ i ,cos2φ i ≤ 1 and have the form

0≤ λ ≤ 1, −1

2 ≤ λnxy ≤ 1

2

In particular, Eqs (8.13) and (8.14) do not describe the case of pure shear for which

only the shear stress resultant, Nxy, is not zero This is quite natural because the strength condition σ1(i) = σ1 under which Eqs (8.12)–(8.14) are derived is not valid for shearinducing tension and compression in angle-ply layers

To study in-plane shear of the laminate, we should use both solutions of Eq (8.7) and

assume that for some layers, e.g., with i = 1, 2, 3, , n − 1, σ (i)

1 = σ1whereas for the

other layers (i = n, n + 1, n + 2, , k), σ (i) = −σ1 Then, Eqs (8.1) can be reduced to

0 0.4 0.8 1.2 1.6 2

Fig 8.3 Optimal thickness ratios for a cylindrical pressure vessel consisting of±φ helical plies combined with

circumferential (90 ◦) or axial (0◦) plies.

the following form

For the case of pure shear (Nx = Ny = 0), Eqs (8.20) and (8.21) yield h+ = h−

and φi = ±45◦ Then, assuming that φi = +45◦for the layers with hi = h+i , whereas

φ i = −45◦for the layers with hi = h−i , we get from Eq (8.22)

h = h++ h−=2Nxy

σ

The optimal laminate, as follows from the foregoing derivation, corresponds to a±45◦angle-ply structure as shown in Fig 8.2b.

8.2 Composite laminates of uniform strength

Consider again the panel in Fig 8.1 and suppose that unidirectional plies or fabriclayers, that form the panel are orthotropic, i.e., in contrast to the previous section, we

do not now neglect stresses σ2and τ12 in comparison with σ1 (see Fig 3.29) Then, theconstitutive equations for the panel in a plane stress state are specified by the first threeequations in Eqs (5.35), i.e.,

we suppose that the fibers in each layer are directed along the lines of principal strains,

or principal stresses because τ12(i) = G12γ12(i) for an orthotropic layer and the condition

γ12(i) = 0 is equivalent to the condition τ (i)

12 = 0 (see Section 2.4) Using the third equation

in Eqs (4.69), we can write these conditions as

This equation can be satisfied for all the layers if we take

sin φi cos φi

−NxyE (i)1 1+ ν (i)

To determine the stresses that act in the optimal laminate, we use Eqs (4.69) and (8.26)

that specify the strains in the principal material coordinates of the layers as ε = ε = ε,

γ12= 0 Applying constitutive equations, Eqs (4.56), substituting ε from Eq (8.27) and

writing the result in explicit form with the aid of Eqs (8.24), we arrive at

is the laminate stiffness coefficient

If all the layers are made from the same material, Eqs (8.28) and (8.29) are simplified as

m= 1, and Eqs (8.30) reduce to Eqs (8.9) and (8.10)

To determine the thickness of the optimal laminate, we should use Eqs (8.31) in junction with one of the strength criteria discussed in Chapter 6 For the simplest case,

con-using the maximum stress criterion in Eqs (6.2), the thickness of the laminate can be

found from the following conditions σ1= σ1or σ2= σ2, so that

fibers and transverse characteristics are determined by the matrix properties, σ2/σ1 n.

In accordance with Eqs (8.32), this means that h1 h2, and the ratio h2/ h1varies from12.7 for glass–epoxy to 2.04 for boron–epoxy composites Now, return to the discussionpresented in Section 4.4.2 from which it follows that in laminated composites, transverse

stresses σ2 reaching their ultimate value, σ2, cause cracks in the matrix, which do notresult in failure of the laminate whose strength is controlled by the fibers To describethe laminate with cracks in the matrix (naturally, if cracks are allowable for the structureunder design), we can use the monotropic model of the ply and, hence, the results ofoptimization are presented in Section 8.1

Consider again the optimality condition Eq (8.25) As can be seen, this equation can

be satisfied not only by strains in Eqs (8.26), but also if we take

Parameters of typical advanced composites.

Parameter Fabric–epoxy composites Unidirectional-epoxy composites Boron–A1

Glass Carbon Aramid Glass Carbon Aramid Boron

Thus, the optimal laminate consists of two layers, and the fibers in both layers are directedalong the lines of principal stresses Suppose that the layers are made of the same com-

posite material and have the same thickness, i.e., h1= h2= h/2, where h is the thickness

of the laminate Then, using Eqs (8.24) and (8.35), we can show that B11 = B22 and

B24 = −B14 for this laminate After some transformation involving elimination of γ xy0

from the first two equations of Eqs (8.23) with the aid of Eq (8.34) and similar

transfor-mation of the third equation from which ε0and ε0are eliminated using again Eq (8.34),

"

1− ν2 12

#

+ E2 2

"

1− ν2 21

Elastic modulus, E (GPa) 36.9 75.9 50.3 114.8 201.1 95.2 205.4 Poisson’s ratio, ν 0.053 0.039 0.035 0.035 0.21 0.06 0.176

elastic constants of the corresponding isotropic material For typical advanced composites,these constants are listed in Table 8.3 (the properties of unidirectional plies are takenfrom Table 3.5) Comparing the elastic moduli of the optimal laminates with those forquasi-isotropic materials (see Table 5.1), we can see that for polymeric composites thecharacteristics of the first group of materials are about 40% higher than those for thesecond group However, it should be emphasized that whereas the properties of quasi-isotropic laminates are universal material constants, the optimal laminates demonstratecharacteristics shown in Table 8.3 only if the orientation angles of the fibers are found

from Eqs (8.35) or (8.38) and correspond to a particular distribution of stresses σx , σ y, and τxy.

As follows from Table 8.3, the modulus of a carbon–epoxy laminate is close to themodulus of aluminum, whereas the density of the composite material is lower by a factor

of 1.7 This is the theoretical weight saving factor that can be expected if we change fromaluminum to carbon–epoxy composite in a thin-walled structure Since the stiffness ofboth materials is approximately the same, to find the optimal orientation angles of thestructure elements, we can substitute in Eq (8.38) the stresses acting in the aluminumprototype structure A composite structure designed in this way will have approximatelythe same stiffness as the prototype structure and, as a rule, higher strength because carboncomposites are stronger than aluminum alloys

To evaluate the strength of the optimal laminate, we should substitute strains fromEqs (8.36) into Eqs (4.69) and thence these strains in the principal material coordinates

of the layers – into constitutive equations, Eqs (4.56), that specify the stresses σ1 and

σ2(τ12 = 0) acting in the layers Applying the appropriate failure criterion (see Chapter 6),

we can evaluate the laminate strength

Comparing Tables 1.1 and 8.3, we can see that boron–epoxy optimal laminates haveapproximately the same stiffness as titanium (but is lighter by a factor of about 2) Boron–aluminum can be used to replace steel with a weight saving factor of about 3

For preliminary evaluation, we can use a monotropic model of unidirectional pliesneglecting the stiffness and load-carrying capacity of the matrix Then, Eqs (8.37) takethe following simple form

E= E1

As an example, consider an aluminum shear web with thickness h = 2 mm, elastic

constants E = 72 GPa, ν = 0.3 and density ρ = 2.7 g/cm3 This panel is loaded with

shear stress τ Its shear stiffness is B44a = 57.6 GPa · mm and the mass of a unit surface

is ma= 5.4 kg/m2 For the composite panel, taking σx = σy= 0 in Eq (8.38) we have

φ= 45◦ Thus, the composite panel consists of+45◦and−45◦ unidirectional layers of

the same thickness The total thickness of the laminate is h= 2 mm, i.e., the same as for an

aluminum panel Substituting E1= 140 GPa and taking into account that ρ = 1.55 g/cm3

for a carbon–epoxy composite which is chosen to substitute for aluminum we get B44c =

70 GPa · mm and mc = 3.1 kg/m3 The stresses acting in the fiber directions of the

composite plies are σ1c= ±2τ Thus, the composite panel has a 21.5% higher stiffness and

its mass is only 57.4% of the mass of a metal panel The composite panel also has higherstrength because the longitudinal strength of unidirectional carbon–epoxy composite undertension and compression is more than twice the shear strength of aluminum

The potential performance of the composite structure under discussion can be enhanced

if we use different materials in the layers with angles φ1and φ2specified by Eqs (8.35).According to the derivation of Obraztsov and Vasiliev (1989), the ratio of the layers’thicknesses is

Superscripts 1 and 2 correspond to layers with orientation angles φ1and φ2, respectively

8.3 Application to optimal composite structures

As stated in the introduction to this chapter, there exists special composite structures forwhich the combination of the specific properties of modern composites with the appropriatedesign concepts and potential of composite technology provide a major improvement ofthese structures in comparison with the corresponding metal prototypes Three such specialstructures, i.e., geodesic filament-wound pressure vessels, composite flywheels, and ananisogrid lattice structure are described in this section

8.3.1 Composite pressure vessels

As the first example of the application of the foregoing results, consider filament-woundmembrane shells of revolution, that are widely used as pressure vessels, solid propellant

+f

Fig 8.4 Axisymmetrically loaded membrane shell of revolution.

rocket motor cases, tanks for gases and liquids, etc (see Figs 4.14 and 7.51) The shell is

loaded with uniform internal pressure p and axial forces T uniformly distributed along the contour of the shell cross section r = r0as in Fig 8.4 Meridional, Nα, and circumferential,

N β ,stress resultants acting in the shell follow from the corresponding free body diagrams

of the shell element and can be written as (see e.g., Vasiliev, 1993)

N α = h"σ1cos2φ + σ2sin2φ − τ12sin 2φ#

in which h is the shell thickness Stresses σ1, σ2, and τ12are related to the correspondingstrains by Hooke’s law, Eqs (4.55), as

ε1= 1

E1(σ1 − ν21σ2), ε2= 1

E2(σ2 − ν12σ1), γ12= τ12

whereas strains ε1, ε2, and γ12 can be expressed in terms of the meridional, εα, and

circumferential, εβ, strains of the shell using Eqs (4.69), i.e.,

ε1 = εαcos2φ + εβsin2φ

ε2 = εαsin2φ + εβcos2φ

γ12= (εβ − εα ) sin 2φ

(8.44)

Since the right-hand parts of these three equations include only two strains, εα and εβ,

there exists a compatibility equation linking ε1, ε2, and γ12 This equation is

Now, assume that in accordance with the results presented in the previous section theoptimal shell is reinforced along the lines of principal stresses, i.e., in such a way that

τ12= 0 In accordance with the last equation of Eqs (8.43), for such a shell γ12 = 0 and,

as follows from Eqs (8.44), εα = εβ = ε1= ε2

Putting τ12= 0 in the last equation of Eqs (8.45), we can conclude that for the optimalshell

The first two equations of Eqs (8.45) yield the following expressions for stresses acting

in the tape of the optimal shell

Substituting Nα from the first equation of Eqs (8.40) into Eq (8.49), we have

Assume that the optimal shell is a structure of uniform stress Differentiating Eq (8.50)

with respect to r and taking into account that according to the foregoing assumption

σ1= constant, we arrive at the following equation in which zis eliminated with the aid

This equation specifies either the thickness or the orientation angle of the optimal shell

Consider two particular cases First, consider a fabric tape of variable width w(r) being

laid up on the surface of the mandrel along the meridians of the shell of revolution to be

fabricated Then, φ= 0, and Eq (8.51) takes the form

2,1− λ



λ= 1

2(2 − n)

Here, Bx is the β-function (or the Euler integral of the first type) The constant of gration is found from the condition z(r = R) = 0 Meridians corresponding to various

inte-n -numbers are presented in Fig 8.5 For n= 1 the optimal shell is a sphere, whereas for

n = 2 it is a cylinder As follows from Eq (8.52), the thickness of the spherical (n = 1) and cylindrical (n = 2, r = R) shells is a constant Substituting Eqs (8.52) and (8.55) in

Eq (8.54) and taking into account Eq (8.49), we have

σ1=σ2

n = pR

2hR This equation allows us to determine the shell thickness at the equator (r = R), hR,

matching σ1or σ2with material strength characteristics

As has been noted already, the shells under study can be made by laying up fabric tapes

of variable width, w(r), along the shell meridians The tape width can be related to the shell thickness, h(r), as

does not depend on r, and its thickness is δ Then, the relevant equation similar to

Eq (8.56) can be written as

where φR = φ (r = R) It should be noted that this equation is not valid for the shell part

in which the tapes are completely overlapped close to the polar opening

Substituting h(r) from Eq (8.58) in Eq (8.51), we arrive at the following equation for

the tape orientation angle

is the monotropic model, which ignores the stiffness of the matrix For this model, we

should take n= 0 in the foregoing equations Particularly, Eq (8.59) yields in this case

This is the equation of a geodesic line on the surface of revolution Thus, in the optimalfilament-wound shells the fibers are directed along the geodesic trajectories This substan-tially simplifies the winding process because the tape placed on the surface under tensionautomatically takes the form of the geodesic line, provided there is no friction between

the tape and the surface As follows from Eq (8.60), for φ = 90◦, the tape touches theshell parallel to radius

and a polar opening of this radius is formed in the shell (see Fig 8.4)

Transforming Eq (8.48) with the aid of Eqs (8.60) and (8.61) and taking n = 0,

we arrive at the following equation which specifies the meridian of the optimal filamentwound shell

Integrating Eq (8.62) with due regard for the condition 1/z(R) = 0 which, as above,

requires that for r = R the tangent to the meridian be parallel to z-axis, we have

"

r2− η2#$

R2− r2 0