ADVANCED MECHANICS OF COMPOSITE MATERIALS Episode 9 potx

Consider a layer whose material stiffness coefficients Amn do not depend on if we take e = h/2, i.e., if the reference plane coincides with the middle-plane of the layer shown in Fig.. 5

Consider a layer whose material stiffness coefficients Amn do not depend on

if we take e = h/2, i.e., if the reference plane coincides with the middle-plane of the layer

shown in Fig 5.9 In this case, Eqs (5.5) and (5.15) take the following de-coupled form

y

Fig 5.9 Middle-plane of a laminate.

The stiffness coefficients, Eqs (5.34), become

Finally, for an isotropic layer, we have

E x = Ey = E, νxy = νyx = ν, Gxy = Gxz = Gyz = G = E

Consider the general case, i.e., a laminate consisting of an arbitrary number of layers

with different thicknesses hi and stiffnesses A (i) mn (i = 1, 2, 3, , k) The location of an arbitrary ith layer of the laminate is specified by the coordinate ti, which is the distance

from the bottom plane of the laminate to the top plane of the ith layer (see Fig 5.10).

Assuming that the material stiffness coefficients do not change within thickness of the

layer, and using piece-wise integration, we can write parameter Imn in Eqs (5.29) and(5.32) as

where r = 0, 1, 2 and t0= 0, tk = h (see Fig 5.10) For thin layers, Eqs (5.41) can be

reduced to the following form, which is more suitable for calculations

in which hi = ti − ti−1is the thickness of the ith layer.

The membrane, coupling, and bending stiffness coefficients of the laminate are specified

2

i k

Fig 5.10 Structure of the laminate.

Consider transverse shear stiffnesses that have two different forms determined byEqs (5.30) and (5.31) in which

(5.46)

Laminates composed of unidirectional plies have special stacking-sequence notations Forexample, notation [0◦2/+45◦/−45◦/90◦

2] means that the laminate consists of 0◦ layer

having two plies, ±45◦ angle-ply layer, and 90◦ layer also having two plies Notation

[0◦/90◦] means that the laminate has five cross-ply layers

Symmetric laminates are composed of layers that are symmetrically arranged withrespect to the laminate’s middle plane as shown in Fig 5.11 Introduce the layer

coordinate zi , (see Fig 5.11) Since for any layer which is above the middle surface

z = 0 and has the coordinate zi there is a similar layer which is located under the middlesurface and has the coordinate (−zi), the integration over the laminate thickness can be

performed from z = 0 to z = h/2 (see Fig 5.11) Then, the integrals for Bmn and Dmn similar to Eqs (5.6) and (5.7) must be doubled, whereas the integral for Cmn similar toEqs (5.8) is equal to zero Thus, the stiffness coefficients entering Eqs (5.5) become

A (i)55A (i)66 − A (i)56

!2 (5.49)

k / 2

k / 2 i

To indicate symmetric laminates, a contracted stacking-sequence notation is used, e.g.,[0◦/90◦/45◦]s instead of[0◦/90◦/45◦/45◦/90◦/0◦] Symmetric laminates are character-ized by a specific feature – their bending stiffness is higher than the bending stiffness ofany asymmetric laminate composed of the same layers To show this property of sym-metric laminates, consider Eqs (5.28) and (5.29) and apply them to calculate stiffness

coefficients with some combination of subscripts, e.g., m = 1 and n = 1 Since the coordinate of the reference plane, e, is an arbitrary parameter, we can find it from the condition C11= 0 Then,

Introduce a new coordinate for an arbitrary point A in Fig 5.12 as z = t −(h/2) Changing

t to z, we can present Eq (5.29) in the form

z A

Fig 5.12 Coordinate of point A referred to the middle plane.

11 In this case, Eq (5.52) gives e = h/2.

Thus, symmetric laminates provide the maximum bending stiffness for a given ber and mechanical properties of layers and, being referred to the middle-plane, donot have membrane–bending coupling effects This essentially simplifies the behavior

num-of the laminate under loading and constitutive equations which have the form specified

by Eqs (5.35)

5.5 Engineering stiffness coefficients of orthotropic laminates

It follows from Eqs (5.28) that the laminate stiffness coefficients depend, in the general

case, on the coordinate of the reference surface e By changing e, we can change the bending stiffness coefficient Dmn Naturally, the result of the laminate analysis undertaken

with the aid of the constitutive equations, Eqs (5.5) does not depend on the particular

pre-assigned value of the coordinate e because of the coupling coefficients Cmn which

also depend on e To demonstrate this, consider an orthotropic laminated element loaded with axial forces N and bending moments M uniformly distributed over the element width

as in Fig 5.13 Suppose that the element displacement does not depend on coordinate y Then, taking Nx = N, Mx = M, ε0

y = 0 and κy= 0 in Eqs (5.44), we get

N = B11ε0x + C11κ x , M = C11ε x0+ D11κ x (5.55)where, in accordance with Eqs (5.28),

B11= I ( 0)

, C11= I ( 1) − eI ( 0)

, D11= I ( 2) − 2eI ( 1) + e2I ( 0) (5.56)

M N

h

N e

x y z

Fig 5.13 Laminated element under tension and bending.

Here, as follows from Eqs (5.41)

(r = 0, 1, 2) are coefficients which do not depend on the coordinate of the reference

plane e It is important to emphasize that forces N in Fig 5.13 act in the reference plane

z = 0, and the strain ε0in Eqs (5.55) is the strain of the reference plane

Solving Eqs (5.55) for ε0and κx, we have

As can be seen, the parameter D1does not depend on e.

Consider now the same element but loaded with forces P applied to the middle plane of

the element as in Fig 5.14 As follows from Fig 5.15 showing the element cross section,the forces and the moments in Fig 5.13 induced by the forces in Fig 5.14 are

e t

A

h / 2

h / 2

Fig 5.15 Cross section of the element.

Substitution of Eqs (5.60) into Eqs (5.58) yields

of the element However, Eq (5.61) includes e which is also expected because ε0is the

strain in the plane z = 0 located at the distance e from the lower plane of the element (see Fig 5.15) Let us find the strain ε t x at some arbitrary point A of the cross section for which z = t − e (see Fig 5.15) Using the first equation of Eqs (5.3), we have

This equation includes the coordinate of point A and does not depend on e Thus, taking

an arbitrary coordinate of the reference plane, and applying Eqs (5.56) for the stiffness

coefficients, we arrive at values of C11 and D11, the combination of which provides the

final result that does not depend on e However, the derived stiffness coefficient D11 is

not the actual bending stiffness of the laminate which cannot depend on e.

To determine the actual stiffness of the laminate, return to Eqs (5.58) for ε x0and κx

Suppose that C11= 0, which means that the laminate has no bending–stretching coupling

effects Then, Eq (5.59) yields D = B11D11and Eqs (5.58) become

ε0x= N

B11 , κ x= M

It is obvious that now B11 is the actual axial stiffness and D11 is the actual bending

stiffness of the laminate However, Eqs (5.63) are valid only if C11 = 0 Using thesecond equation of Eqs (5.56), we get

Substituting this result into Eqs (5.56) and introducing new notations Bx = B11 and

D x = D11for the actual axial and bending stiffness of the laminate in the x-direction, we

where coordinates zi and zi−1are shown in Fig 5.11 Note, that if the number of layers

k is not even, the central layer is divided by the plane z = 0 into two identical layers,

so k becomes even.

To find the shear stiffness, consider the element in Fig 5.13 but loaded with shear

forces, S, and twisting moments H , uniformly distributed along the element edges as

shown in Fig 5.16 It should be recalled that forces and moments are applied to the

element reference plane z = 0 (see Fig 5.13) Taking Nxy = S and Mxy = H in the

corresponding Eqs (5.44), we get

x y

y

q y

q x

w A

Fig 5.17 Deformation of the element under torsion.

where κxyis given in notations to Eqs (5.3), i.e.,

κ xy= ∂θ x

∂θ y

The deformed state of the laminated element (see Fig 5.16) loaded with twisting moments

only is shown in Fig 5.17 Consider the deflection of point A with coordinates x and y.

It follows from Fig 5.17 that w = xθx or w = yθy Introduce the gradient of the torsional

Since θdoes not depend on x and y, θx = yθ, θy = xθand w = xyθ Using Eq (5.75),

we have κxy = 2θ Then, Eq (5.74) yields

also transverse shear forces V (see Fig 5.6), we must first determine the actual transverse

(through-the-thickness) stiffnesses of a laminate

h

x y

Fig 5.18 Laminated element loaded with transverse shear forces.

Consider an orthotropic laminated element loaded with transverse shear forces Vx = V

uniformly distributed over the element edge as in Fig 5.18 From Eqs (5.20), we havetwo possible constitutive equations, i.e.,

For the orthotropic material, A (i)55 = G (i)

xz , where G (i) xz is the transverse shear modulus of

the ith layer Thus, Eqs (5.79) take the form

(5.80)

As shown in Section 5.1, S55 gives the upper bound and S55 gives the lower bound ofthe actual transverse shear stiffness of the laminate For a laminate consisting of identical

layers, i.e., for the case G (i) xz = Gxz for all the layers, both equations of Eqs (5.80) give

the same result S55 = S55 = Gxz h However, in some cases, following from Eqs (5.80)the results can be dramatically different, whereas for engineering applications we musthave instead of Eqs (5.78) a unique constitutive equation, i.e.,

and the question arises whether S55or S55should be taken as Sxin this equation Since for

a homogeneous material there is no difference between S55 and S55, we can expect thatthis difference shows itself in the laminates consisting of layers with different transverseshear moduli

Consider, for example, sandwich structures composed of high-stiffness thin facinglayers (facings) and low-stiffness light foam core (Fig 5.19a) The facings are made

Fig 5.19 Three-layered (sandwich) and two-layered laminates.

of aluminum alloy with modulus Ef = 70 GPa and shear modulus Gf = 26.9 GPa The foam core has Ec = 0.077 GPa and Gc = 0.0385 GPa The geometric and stiffness

parameters of two sandwich beams studied experimentally (Aleksandrov et al., 1960) are

presented in Table 5.1 The beams with length l= 280 mm have been tested under

trans-verse bending The coefficient Sa in the table corresponds to the actual shear stiffnessfound from experimental results Actually, experimental study allows us to determine theshear parameter (Vasiliev, 1993)

k G= D

which is presented in the third column of the table and depends on the bending stiffness, D, and the beam length, l Since the sandwich structure is symmetric, we can use Eq (5.67) for Dx in which 2k= 2 (the core is divided into two identical layers as in Fig 5.19a)

The results of the calculation are listed in the last column of Table 5.1 The shear stiffness

coefficients S55and S55can be found from Eqs (5.80) which for the structure in Fig 5.19a

Table 5.1

Parameters of sandwich structures.

hf (mm) hc (mm) kG Shear stiffness (GPa × mm) Bending stiffness

(GPa × mm 3 )

The results of the calculation are presented in Table 5.1 As can be seen, coefficients S55

are in good agreement with the corresponding experimental data, whereas coefficients S55

are higher by an order of magnitude Note, that S55, providing the lower boundary for

the exact shear stiffness, is higher than the actual stiffness Sa The reason for this effect

has been discussed in Section 5.1 Coefficient S55 specifies the lower boundary for thetheoretical exact stiffness corresponding to the applied model of the laminate, but not forthe actual stiffness following from experiment For example, the actual shear stiffness

of the sandwich beams described above can be affected by the compliance of adhesivelayers which bond the facings and the core and are not allowed for in the laminate model

So, it can be concluded that the shear stiffness coefficient S55 specified by the sponding equation of Eqs (5.79) can be used to describe the transverse shear stiffness

corre-of composite laminates However, there are special structures for which coefficient S55

provides a better approximation of shear stiffness than coefficient S55 Consider, for ple, a two-layered structure shown in Fig 5.19b and composed of a high-stiffness facingand a low-stiffness core Assume, as for the sandwich structure considered above, that

exam-Gf = 26.9 GPa and Gc = 0.0385 GPa, so that Gf/Gc = 699, and take hc = 9.9 mm, and hf = 2.4 mm It is obvious that the core, having such a low shear modulus, does not

work, and the transverse shear stiffness of the laminate is governed by the facing layer.For this layer only, we get

As can be seen, coefficient S55is very far from the value that would be expected However,

structures of type for which the stiffness coefficient S55 is more appropriate than the

coefficient S55 are not typical in composite technology and, being used, they usually donot require the calculation of transverse shear stiffnesses For laminated composites it can

be recommended to use the coefficient S (Chen and Tsai, 1996) Thus, the transverse

shear stiffness coefficient in Eq (5.81) can be taken in the following form

in the analysis of composite structures To evaluate the possibility of neglecting transverse

shear deformation, we can use parameter kG specified by Eq (5.82) and compare it withunity The effect of the transverse shear deformation is demonstrated in Table 5.2 forthe problem of transverse bending of simply supported sandwich beams with various

parameters kG listed in the table The right hand column of the table shows the ratio

of the maximum deflections of the beam, w, found with allowance for transverse shear deformation (wG) and corresponding to the classical beam theory (w∞) As can be seen,

for beams number 4 and 5, having parameter kG which is negligible in comparison withunity, the shear deformation practically does not affect the beams deflections

Returning to the problem of torsion, we consider an orthotropic laminated strip with

width b loaded with a torque M as in Fig 5.20 In contrast to the laminate shown

The effect of transverse shear deformation on the deflection of sandwich beams.

z

Mt

Fig 5.20 Torsion of a laminated strip.

Mtz

e

M xy

V x

Fig 5.21 Forces and moments acting in the strip cross section.

in Fig 5.6, the strip in Fig 5.20 is loaded only at the transverse edges, whereas the

longitudinal edges y = ±b/2 are free The shear stresses τxz and τyz induced by torsion

give rise to the shear forces Nxy, twisting moment Mxy and transverse shear force Vx

shown in Fig 5.21 Applying the corresponding constitutive equations, Eqs (5.44) and(5.81), we get

N xy = B44γxy0 + C44κxy , M xy = C44γ xy0 + D44κxy (5.85)

where the stiffness coefficients B, C, D, and S are specified by Eqs (5.69) and (5.82) Pre-assign the coordinate of the reference plane e in accordance with Eq (5.71).

Then, C44 = 0 and Eqs (5.85) reduce to

where Bxy and Dxy are given by Eqs (5.73) Since the strip is loaded with a torque Mt

only (see Fig 5.20), Nxy = 0, and as follows from Eq (5.87), γ0

xy = 0 So, we have

only two constitutive equations, i.e., Eqs (5.86) and (5.88) for Vx and Mxy which are

expressed in terms of the transverse shear strain γx and the twisting deformation κxy.

Applying Eqs (5.14) and (5.75), we have

Consider the deformation of the strip Assume that the strip cross section rotates around

the longitudinal axis x through an angle θ which depends only on x (Fig 5.22) Then, as

follows from Fig 5.22,

q y

q

Fig 5.22 Rotation of the strip cross section.