ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 11 pps

Consider the class of invariant strength criteria which are formulated in a tensor-polynomial form as linear combinations of mixed invariants of the stress tensor σ ij and the strength t

For tension in the directions of axes 1and 2 in Fig 6.16 and for shear in plane 12, we

can write Eq (6.25) in the following forms similar to Eqs (6.10)

Here, σ45 and τ45 determine material strength in coordinates 1 and 2 (see Fig 6.16).

Then, Eq (6.25) can be reduced to

Comparing Eq (6.27) with Eq (6.23), we can see that Eq (6.27), in contrast to Eq (6.23),

includes a term with the product of stresses σ145and σ245 So, the strength criterion understudy changes its form with a transformation of the coordinate frame (from 1 and 2 to 1

and 2in Fig 6.16) which means that the approximation polynomial strength criterion in

Eq (6.23) and, hence, the original criterion in Eq (6.21) is not invariant with respect tothe rotation of the coordinate frame

Consider the class of invariant strength criteria which are formulated in a

tensor-polynomial form as linear combinations of mixed invariants of the stress tensor σ ij and

the strength tensors of different ranks S ij , S ij kl, etc., i.e.,

nota-materials referred to the principal material coordinates 1 and 2 (see Fig 6.16), we canpresent Eq (6.22) as

σ11= σ1, σ12= τ12, σ22= σ2 and S11= R1, S22= R2, S1111 = R0

11, S1122 = S2211= R0

12, S2222 = R0

22, S1212 = S2121= S1221= S2112= S0

12The superscript ‘0’ indicates that the components of the strength tensors are referred tothe principal material coordinates Applying the strength conditions in Eqs (6.14), we canreduce Eq (6.29) to the following form

any suitable value of coefficient R12 (in particular, we put R12 = 0), the criterion in

Eq (6.30) has an invariant tensor form, and coefficient R012 should be determined usingthis property of the criterion

Following Gol’denblat and Kopnov (1968) consider two cases of pure shear in nates 1 and 2 shown in Fig 6.17 and assume that τ+

coordi-45 = τ+45 and τ−

45 = τ−45, where theoverbar denotes, as earlier, the ultimate value of the corresponding stress In the general

Substituting Eq (6.34) into Eq (6.33), we arrive at the final form of the criterion underconsideration

F (σ1, σ2, τ12)=

"

σ12+ σ2 2

Let us transform stresses referred to axes (1, 2) into stresses corresponding to axes

(1 and 2)shown in Fig 6.16 Such a transformation can be performed with the aid of

Eqs (6.24) The matrix form of this transformation is

specifies the strength criterion for the same material but referred to coordinates (1,2).

The strength matrix has the following form

(see Fig 6.16) As can be seen, Eqs (6.35) and (6.43) have similar forms and follow fromeach other if we change the stresses in accordance with the following rule

If the actual material characteristics do not satisfy Eq (6.44), the tensor strength criterioncannot be applied to this material However, if this equation is consistent with experimentaldata, the tensor criterion offers considerable possibilities to study material strength Indeed,restricting ourselves to two terms presented in Eq (6.28) let us write this equation in

coordinates (1,2) shown in Fig 6.16 and suppose that φ= 45◦ Then

Here, l are directional cosines of axes 1and 2on the plane referred to coordinates 1 and 2

(see Fig 6.16), i.e., l11 = cos φ, l12= sin φ, l21 = − sin φ, and l22 = cos φ Substitution

of Eqs (6.46) in Eq (6.45) yields the strength criterion in coordinates (1,2)but written

in terms of strength components corresponding to coordinates (1, 2), i.e.,

Apply Eq (6.47) to the special orthotropic material studied above (see Fig 6.16) and forwhich, in accordance with Eq (6.22),

S pq = 0, S1111= S2222= R0

11= R0

22= 1

σ20S1122 = S2211= R0

Following Gol’denblat and Kopnov (1968), consider the material strength under tension

in the 1-direction and in shear in plane (1,2) Taking first σ φ

These equations allow us to calculate the material strength in any coordinate frame whose

axes make angle φ with the corresponding principal material axes Taking into account

Eqs (6.44) and (6.48), we can derive the following relationship from Eqs (6.49)1

transformation, i.e., composed invariants (such as Is )must be the same for all coordinateframes

6.1.4 Interlaminar strength

The failure of composite laminates can also be associated with interlaminar

frac-ture caused by transverse normal and shear stresses σ3 and τ13 , τ or σ and τ , τ

(see Fig 4.18) Since σ3 = σzand shear stresses in coordinates (1, 2, 3) are linked with

stresses in coordinates (x, y, z) by simple relationships in Eqs (4.67) and (4.68), the strength criterion is formulated here in terms of stresses σ z , τ xz , τ yz which can be founddirectly from Eqs (5.124) Since the laminate strength in tension and compression acrossthe layers is different, we can use the polynomial criterion similar to Eq (6.15) For thestress state under study, we get

As an example, Fig 6.18 displays the dependence of the normalized maximum deflection

w/R on the force P for a fiberglass–epoxy cross-ply cylindrical shell of radius R loaded with a radial concentrated force P (Vasiliev, 1970) The shell failure was caused by

delamination The shadowed interval shows the possible values of the ultimate force

0 0.4 0.8 1.2 1.6 2

P, kN

R w

Fig 6.18 Experimental dependence of the normalized maximum deflection of a fiberglass–epoxy cylindrical

shell on the radial concentrated force.

calculated with the aid of Eq (6.52) (this value is not unique because of the scatter ininterlaminar shear strength).

6.2 Practical recommendations

As follows from the foregoing analysis, for practical strength evaluation of fabric posites, we can use either the maximum stress criterion, Eqs (6.2) or second-orderpolynomial criterion in Eq (6.15) in conjunction with Eq (6.16) for the case of biax-ial compression For unidirectional composites with polymeric matrices, we can apply

com-Eqs (6.3) and (6.4) in which function F is specified by Eq (6.18) It should be

empha-sized that experimental data usually have rather high scatter, and the accuracy of morecomplicated and rigorous strength criteria can be more apparent than real

Comparing tensor-polynomial and approximation strength criteria, we can conclude thefollowing The tensor criteria should be used if our purpose is to develop a theory of mate-rial strength, because a consistent physical theory must be covariant, i.e., the constraintsthat are imposed on material properties within the framework of this theory should notdepend on a particular coordinate frame For practical applications, the approximation cri-teria are more suitable, but in the forms they are presented here they should be used onlyfor orthotropic unidirectional plies or fabric layers in coordinates whose axes coincidewith the fibers’ directions

To evaluate the laminate strength, we should first determine the stresses acting in theplies or layers (see Section 5.11), identify the layer that is expected to fail first andapply one of the foregoing strength criteria The fracture of the first ply or layer may notnecessarily result in failure of the whole laminate Then, simulating the failed element with

a suitable model (see, e.g., Section 4.4.2), the strength analysis is repeated and continued

up to failure of the last ply or layer

In principle, failure criteria can be constructed for the whole laminate as a homogeneous material This is not realistic for design problems, since it would benecessary to compare the solutions for numerous laminate structures which cannot prac-tically be tested However, this approach can be used successfully for structures that arewell developed and in mass production For example, the segments of two structures ofcomposite drive shafts – one made of fabric and the other of unidirectional composite,are shown in Fig 6.19 Testing these segments in tension, compression, and torsion, we

quasi-can plot the strength envelope on the plane (M, T ), where M is the bending moment and

T is the torque, and evaluate the shaft strength for different combinations of M and T

with high accuracy and reliability

Fig 6.19 Segments of composite drive shafts with test fixtures Courtesy of CRISM.

ν21 = 0.2, σ+1 = 510 MPa, σ−1 = 460 MPa, σ+2 = 280 MPa, σ−2 = 260 MPa,

τ12= 85 MPa The shear strain induced by torque T is (Vasiliev, 1993)

γ xy= T

2πR2B44

Here, T is the torque, R = 0.05 m is the shaft radius, and B44 is the shear stiffness of

the wall According to Eqs (5.39), B44 = A44h, where h= 5 mm is the wall thickness,

and A44 is specified by Eqs (4.72) and can be presented as (φ= 45◦)

4(1 − ν12ν21) (E1 + E2− 2E1ν12)

Using Eqs (5.122), we can determine strains in the principal material coordinates 1 and

2 of±45◦layers (see Fig 6.20)

The task is to determine the ultimate torque, Tu.

First, use the maximum stress criterion, Eqs (6.2), which gives the following fourvalues of the ultimate torque corresponding to tensile or compressive failure of ±45◦

Now apply the polynomial criterion in Eq (6.15), which has the form

As a second example, consider the cylindrical shell described in Section 5.12 (see

Fig 5.30) and loaded with internal pressure p Axial, N x , and circumferential, N y, stressresultants can be found as

N x=1

2pR, N y = pR

where R = 100 mm is the shell radius Applying constitutive equations, Eqs (5.125),

and neglecting the change in the cylinder curvature (κ y = 0), we arrive at the following

equations for strains

12, and the membrane stiffnesses B mnfor the shell under study are

presented in Section 5.12 Subscript ‘i’ in Eqs (6.54) indicates the helical plies for which

i = 1, φ1= φ = 36◦and circumferential plies for which i = 2 and φ2= 90◦.

The task that we consider is to find the ultimate pressure pu For this purpose, we usethe strength criteria in Eqs (6.3), (6.4), and (6.17), and the following material properties

σ+= 1300 MPa, σ+= 27 MPa, τ12 = 45 MPa

Calculation with the aid of Eqs (6.54) yields

Thus, we can conclude that failure occurs initially in the matrix of helical plies and takes

place at an applied pressure p (u1) = 1.1 MPa This pressure destroys only the matrix of the

helical plies, whereas the fibers are not damaged and continue to work According to themodel of a unidirectional layer with failed matrix discussed in Section 4.4.2, we should

take E2 = 0, G12= 0, and ν12 = 0 in the helical layer Then, the stiffness coefficients,

Eqs (4.72) for this layer, become

A (111) = E1cos4φ, A (121) = E1sin2φcos2φ, A (221) = E1sin4φ (6.55)

Calculating again the membrane stiffnesses B mn(see Section 5.12) and using Eqs (6.53),

u ,three modes of failure are possible The pressure causing failure

of the helical plies under longitudinal stress σ1( 1) can be calculated from the followingequation

which yields pu = 14.2 MPa The analogous value for the circumferential ply is

determined by the following condition

which gives pu = 9.84 MPa Finally, the matrix of the circumferential layer can fail under

tension across the fibers Since τ12( 2) = 0, we put

and find pu = 1.4 MPa

Thus, the second failure stage takes place at pu( 2) = 1.4 MPa and is associated with

cracks in the matrix of the circumferential layer (see Fig 4.36)

For p ≥ p ( 2)

u , we should put E2 = 0, G12 = 0, and ν12 = 0 in the circumferential

layer whose stiffness coefficients become

are the thicknesses of the helical and the circumferential layers Using again Eqs (6.54),

Thus, failure of the structure under study occurs at pu = 10 MPa as a result of fiber

fracture in the helical layer

0 2 4 6 8 10

(a)

0 2 4 6 8 10

The dependencies of strains, which can be calculated using Eqs (6.53), and the

appro-priate values of B mn are shown in Fig 6.21 (solid lines) As can be seen, the theoreticalprediction is in fair agreement with the experimental results The same conclusion can bedrawn for the burst pressure which is listed in Table 6.1 for two types of filament-woundfiberglass pressure vessels A typical example of the failure mode for the vessels presented

in Table 6.1 is shown in Fig 6.22

6.4 Allowable stresses for laminates consisting of unidirectional plies

It follows from Section 6.3 (see also Section 4.4.2) that a unidirectional fibrouscomposite ply can experience two modes of failure associated with

• fiber failure, and

• cracks in the matrix

Number of tested vessels

Experimental burst pressure

value (MPa)

Variation coefficient (%)

Fig 6.22 The failure mode of a composite pressure vessel.

The first mode can be identified using the strength criterion in Eqs (6.3), i.e.,

1 are the ultimate stresses of the ply under tension and compression

along the fibers, and i is the ply number For the second failure mode, we have the strength

criterion in Eq (6.18), i.e.,

2 are the ultimate stresses of the ply under tension and compression

across the fibers, and τ12is the ultimate in-plane shear stress

Consider a laminate loaded with normal, σ x and σ y , and shear, τ xy, stresses as

in Fig 6.23 Assume that the stresses are increased in proportion to some loading

parameter p Applying the strength criteria in Eqs (6.57) and (6.58), we can find two values of this parameter, i.e., p = pf which corresponds to fiber failure in at least one

of the plies and p = pm for which the loading causes matrix failure in one or more

plies Since the parameters pf and pmusually do not coincide with each other for moderncomposites, a question concerning the allowable level of stresses acting in the laminatenaturally arises Obviously, the stresses under which the fibers fail must not be treated as

allowable stresses Moreover, the allowable value pa of the loading parameter must be

less than pf by a certain safety factor sf, i.e.,

However, for matrix failure, the answer is not evident, and at least two different situationsmay take place

First, the failure of the matrix can result in failure of the laminate As an example,

we can take a ±φ angle-ply layer discussed in Section 4.5 whose moduli in the x- and

y-directions are specified by Eqs (4.147), i.e.,

E x = A11−A212

A22 , E y = A22−A212

A11 Ignoring the load-carrying capacity of the failed matrix, i.e., taking E2 = 0, G12 = 0,

and ν12 = ν21= 0 in Eqs (4.72) to get

A = E cos4φ, A = E sin2φcos2φ, A = E cos4φ

we arrive at E x = 0 and Ey= 0 which means that the layer under consideration cannot

work without the matrix For such a mode of failure, we should take the allowable loadingparameter as

where sm is the corresponding safety factor

Second, the matrix fracture does not result in laminate failure As an example of such

a structure, we can take the pressure vessel considered in Section 6.3 (see Figs 6.21and 6.22) Now we have another question as to whether the cracks in the matrix areallowed even if they do not affect the structure’s strength The answer to this questiondepends on the operational requirements for the structure For example, suppose that thepressure vessel in Fig 6.22 is a model of a filament-wound solid propellant rocket motorcase which works only once and for a short period of time Then, it is appropriate toignore the cracks appearing in the matrix and take the allowable stresses in accordancewith Eq (6.59) We can also suppose that the vessel may be a model of a pressurizedpassenger cabin in a commercial airplane for which no cracks in the material are allowed

in flight Then, in principle, we must take the allowable stresses in accordance with

Eq (6.60) However, it follows from the examples considered in Sections 4.4.2 and 6.3,

that for modern composites the loading parameter pmis reached at the initial stage of the

loading process As a result, the allowable loading parameter, pamin Eq (6.60), is so smallthat modern composite materials cannot demonstrate their high strength governed by thefibers and cannot compete with metal alloys A more realistic approach allows the cracks

in the matrix to occur but only if pm is higher than the operational loading parameter po.

Using Eq (6.60), we can presume that

and depends on the ratio pf /pm.

To be specific, consider a four-layered [0◦/45◦/−45◦/90◦] quasi-isotropic carbon–

epoxy laminate shown in Fig 6.23 which is widely used in aircraft composite structures.The mechanical properties of quasi-isotropic laminates are discussed in Section 5.7 Theconstitutive equations for such laminates have the form typical for isotropic materials, i.e.,