Composite Materials Design and Applications Part 12 doc

17.6.6 Interpretation in Terms of EnergyWe will limit ourselves to the surface energy density due to transverse shear stressesas: Substituting Equation 17.29, one obtains The first integ

The case of a multilayer plate such that for any two plies k and m one

has in the plane of the plate11:

Then Equations 17.19 and 17.20 reduce to

(17.23)

The preceding particular cases constitute a severe restriction among the variety

of practical laminations Nevertheless we will conserve in the following thesimplified forms of Equations 17.21, 17.22, and 17.23 because they well show thedirect connection between the warpings hx and hy and the transverse shear forces

Such a limiting case is rare in practice, because it imposes in particular:

1 cylindrical bending about y axis

G xy m

- akm

E ij k

E ij m

˝ÔÔ

- z d

h/2

–

h/2Ú

+

=

There appear two transverse shear coefficients k x and k y which require the

knowledge of the functions g(z) and p(z) for their calculations.

17.6.4 Warping Functions

Boundary conditions: We have assumed that the upper and lower faces

of the plate were free of any shear Then the transverse shear inEquations 17.17 and 17.18 leads to

then with Equation 17.25:

then with [17.26]:

Continuity at the interfaces: The continuity of the transverse shear at

the interfaces between layers results from the assumed perfect bonding

between two plies (see Paragraph 15.1.2) One then has at the interface between two consecutive plies k and k + 1:

- z d

h/2

–

h/2Ú

- z d

h/2

–

h/2Ú

- z d

h/2

–

h/2Ú

+ = 0 for z = ±h/2

+ = 0 for z = ±h/2

-txz k

txz

k+1; tyz k

tyz

k+1

then with Equations 17.17, 17.18 and Equations 17.25, 17.26:

Formulation of the warping functions: Let us substitute to g(z) and

p(z) the functions g0(z) and p0(z) such that:

g0(z) and p0(z) are called the warping functions Then the boundary conditions

and the interface conditions simplify, and Equations 17.23 allow one to formulate

the problems that permit a simple calculation of warping functions g0(z) and p0(z).

Form of the transverse shear stresses: Equations 17.17 and 17.18 then

take the simple forms:

(17.29)

G xz k

k x

dg k dz

+

=

G yz k

k y

dp k dz

+

¥–

=

dg0dz

- = 0 for z = ±h/2

G xz

k dg 0k dz

¥–

Transverse shear coefficients: One obtains these coefficients from the

Equation 17.5:

here is :

noting that :

One obtains :

(17.30)

here is :

leading to:

(17.31)

In summary, in the absence of body forces (inertia forces, example), thebending behavior uncoupled from the membrane behavior of a thick laminatedplate can be simplified in a few particular cases noted below The characteristicrelations are summarized in the following table

Bending Behavior (no in-plane stress resultants)

homogeneous orthotropic plate/orthotropic axes : x, y, z

or Laminated plate/midplane symmetry/orthotropic axes of plies: x,y,z/same Poisson ratios v xy and v yx for all plies/cylindrical bending about x or y axis.

or Laminated plate/midplane symmetry/orthotropic axes of plies: x,y,z/elastic

constants are proportional from one ply to another

z d

g0 (z) is the solution of the problem:

p0 (z) is the solution of the problem:

 Transverse shear coefficients k x and k y:

They are given by the formula:

- E11

k

EI11 - E12

k

EI12 -+

d2p0

dz2

- z·hG yzỊ

G yz k

- E22

k

EI22 - E12

k

EI12 -+

ĨƠƠƠƠÌƠƠƠỢ

=

17.6.6 Interpretation in Terms of Energy

We will limit ourselves to the surface energy density due to transverse shear stressesas:

Substituting Equation 17.29, one obtains

The first integral can be rewritten as:

or, taking into account Equation 17.27:

where one recognizes Equation 17.30 of the transverse shear coefficient k x, Thefirst integral is reduced to

Following a similar approach for the second integral and taking into account

Equations 17.28 and 17.31 for the transverse shear coefficient k y, the surface energydue to transverse shear takes the form:

17.7 EXAMPLES

Examples for plates in bending are shown in details in Part Four of this book, inChapter 18, “Applications.” We give here a few useful elements to treat theseexamples

2 - Q x

Ê ˆ2

z d

h/2

–

h/2Ú

=

12 - Q x

- g0d g0

z d

Ê ˆ g0d2g0

z2d

12 - Q x

-=

17.7.1 Homogeneous Orthotropic Plate

Warping functions: Equation 17.27 becomes12

Transverse shear stresses and shear coefficients: One deduces from

Remark: In Application 18.3.7 (Chapter 18), one treats the case of a thick

homogeneous orthotropic plate in cylindrical bending about the y axis The plate

supports a uniformly distributed load One can consider there the strong influence

of transverse shear in bending Two characteristics of the plate then apply directly

dg0dz

- = 0 for z = ±h/2

then12 dg0

dz

- 32 - 1 4z

2

h2

–

- 32 - 1 4z

2

h2

–

-z2

h2

–

Ê ˆ z2

z d

h/2

–

h/2Ú

=

k x

65 -

=

p0( )z = g0( )z

tyz

Q y h

- 32 - 1 4z

2

h2

–

=

The relative thickness h/a, where a is the length of the bent side of the plate

The ratio Ex/Gxz (for certain combinations of fiber/matrix, this ratio becomeslarge compared with unity; for example, for unidirectional)

17.7.2 Sandwich Plate

The plate consists of two orthotropic materials:

Material (1) for the skinsMaterial (2) for the core (see Figure 17.3)Assuming the proportionality of elastic coefficients for the two materials leads

to (see Section 17.6.3):

one deduces from there:

Figure 17.3 Sandwich Plate

H2 3

–12 -

➁ H2 3

12 -+

3

12 -

3

H2 3

–12 -+a12H2

3

12 -

-¥

17.7.2.1 Warping Functions

From the above one can write in Equation 17.2713

:

In addition :

Equation 17.27 then can be written as:

In Equation 17.28, one obtains an analogous formulation In effect, onecan write

The problem [17.28] is then written as :

Remark: These problems are identical to that which allows the calculation

of the warping function for the bending of a sandwich beam, and onecan consider it in Chapter 18, application 18.3.5 One can then carry outthe same steps of calculation The results obtained are shown below

H2 3

–

➁

+ H2 3

H2 3

–

➁

+ H2 3

-¥

¥–

ƠƠƠÌƠƠỢ

17.7.2.2 Transverse Shear Stresses

The relative thickness of the core is important (thin skins)

The relative thickness of the plate is large (thick plate)

H2

2

- z H2

2 -

£ £

E x➁ H2 2

4 - –z2

Ê ˆ E x➀ H12

4 - H2

2

4 - –

H2 3

–

➁

H2 3

+ -

2 -

£ £– ; tyz Q y 6 E y

H2 3

–

➁

H2 3

+ -

H2 3

–

➁

H2 3

+ -

H2 3

–

➁

H2 3

-H1

5 H2 5

5 - H1 2

H2 3

–+

H2 3

–

➁

H2 3

+ -+

H2 3

–

➁

H2 3

+ -

PART IV APPLICATIONS

We have grouped in this last part of the book exercises and examples for tions These have various objectives and different degrees of difficulties Leavingaside (except for special cases) the cases that are too academic, we will concernourselves with applications of concrete nature, with an emphasis on the numericalaspect of the results A few of these applications should be used as validationtests for numerical models

applica-TX846_Frame_C18a Page 341 Monday, November 18, 2002 12:40 PM

1 The following figure represents a beam made of duralumin that is supported

at two points It is subjected to a transverse load of F = 50 daN Calculatethe deflection—denoted as D—of the beam under the action of the force F

2 We separate the beam of duralumin into two parts with equal thickness

e p = 2.5 mm, by imaginarily cutting the beam at its midplane Each half isbonded to a parallel pipe made of polyurethane foam, making the skins

of a sandwich beam having essentially the same mass as the initial beam(in neglecting the mass of the foam and the glue) The beam is resting

on the same supports and is subjected to the same load F Calculate thedeflection caused by F, denoted by D¢ Compare with the value of D found

in Part 1 (Take the shear modulus of the foam to be: G c= 20 MPa.)

TX846_Frame_C18a Page 343 Monday, November 18, 2002 12:40 PM

For duralumin (see Section 1.6): E = 75,000 MPa One finds

2 Denoting by W the elastic energy due to flexure, one has1

· Ò

- x 1

2

- k GS

· Ò

- # 1

G c(e c+2e p ) b¥ -

2 -(
x– )
( - xº–2x)d

/2

Ú

+

d

0

/2Ú

¥

/2

Ú

+

0

/2Ú

· Ò -+

=

EI

· Ò # E p e p b (e c+e p)2

2 - E c

e c3b

12 -

¥+

one obtains for D¢:

Comparing with the deflection D found in Part 1 above:

Remarks:

The sandwich configuration has allowed us to divide the deflection by 14without significant augmentation of the mass: with adhesive film thickness0.2 mm and a specific mass of 40 kg/m3 for the foam, one obtains a totalmass of the sandwich:

m = 700 g (duralumin) + 50 g (foam) + 48 g (adhesive)This corresponds to an increase of 14% with respect to the case of the full beam

in Question 1

The deflection due to the shear energy term is close to 6 times moreimportant than that due to the bending moment only In the case of thefull beam in question 1, this term is negligible In effect one has:

k = 1.2 for a homogeneous beam of rectangular section, then:

(with G = 29,000 MPa, Section 1.6) The contribution to the deflection D of theshear force is then:

18.1.2 Poisson Coefficient of a Unidirectional Layer

D¢ = 1.22 mm

DD¢

- 141 -

=

k GS

- = 8.27¥108

k GS

- T T

F d

- x d

Ú = 0.02 mm << D

Show that two distinct Poisson coefficients n
t and nt
are necessary to terize the elastic behavior of this unidirectional layer Numerical application: a

charac-layer of glass/epoxy V f = 60% fiber volume fraction

Solution:

Let the plate be subjected to two steps of loading as follows:

1 A uniform stress s
along the
direction: the changes in lengths of thesides can be written as:

2 A uniform stress st along the t direction: for a relatively important elongation

of the resin, one can only observe a weak shortening of the fibers along

Using then another notation for the Poisson coefficient, the change inlength can be written as:

Now calculating the accumulated elastic energy under the two loadings above:

When s
is applied first, and then st is applied,

When st is applied first, and then s
is applied,

The final energy is the same:

12 -st¥b¥e¥Da2+s
¥a¥e¥Db2

with the values obtained above for Db2 and Da1:

Numerical application: n
t = 0.3, E
= 45,000 MPa, E t = 12,000 MPa (see Section 3.3.3):

Remark: The same reasoning can be applied to all balanced laminates having

midplane symmetry, by placing them in the symmetrical axes.3 However, depending

on the composition of the laminate, the Poisson coefficients in the two perpendiculardirections vary in more important ranges:

in absolute value and

one with respect to the other

One can find in Section 5.4.2 in Table 5.14 the domain of evolution of the globalPoisson coefficient nxy of the glass/epoxy laminate, from which one can deduce thePoisson coefficient nyx using a formula analogous to the one above, as:

18.1.3 Helicopter Blade

This study has the objective of bringing out some important particularities related

to the operating mode of the helicopter blade, notably the behavior due to normalload

Problem Statement:

Consider a helicopter blade mounted on the rotor mast as shown schematically

in the following figure

¥

=

nt
= 0.08

nyx /E y = nxy /E x

The characteristics of the rotor are as follows:

Rotor with three blades; rotational speed: 500 revolutions per minute

The mass per unit length of a blade at first approximation is assumed tohave a constant value of 3.5 kg /m

= 5 m; c = 0.3 m.

The elementary lift of a segment dx of the blade (see figure above) is

written as:

in which V is the relative velocity of air with respect to the profile of the blade.

In addition, C z (7∞) = 0.35 (lift coefficient)

r = 1.3 kg/m3

(specific mass of air in normal conditions)

We will not concern ourselves with the drag and its consequences Oneexamines the helicopter as immobile with respect to the ground (stationary flight

in immobile air) In neglecting the weight of the blade compared with the loadapplication and in assuming infinite rigidity, the relative equilibrium configuration

in uniform rotation is as follows:

2 -r(cdx )C z V2

=

1 Justify the presence of the angle called “flapping angle” q and calculate it.

2 Calculate the weight of the helicopter

3 Calculate the normal force in the cross section of the blade and at thefoot of the blade (attachment area)

The spar of the blade4 is made of unidirectional glass/epoxy with 60% fiber volumefraction “R” glass (s
rupture # 1700 MPa) The safety factor is 6 Calculate:

4 The longitudinal modulus of elasticity E
of the unidirectional

5 The cross section area for any x value of the spar, and its area at the foot

of the blade

6 The total mass of the spar of the blade

7 The elongation of the blade assuming that only the spar of the blade issubject to loads

8 The dimensions of the two axes to clamp the blade onto the rotor mast.Represent the attachment of the blade in a sketch

Solution:

1 The blade is subjected to two loads, in relative equilibrium:

Distributed loads due to inertia, or centrifugal action, radial (that means

in the horizontal plane in the figure, with supports that cut the rotor axis

Distributed loads due to lift, perpendicular to the direction of the blade

(Ax in the figure).

From this there is an intermediate equilibrium position characterized by the angle q

Joint A does not transmit any couple The moment of forces acting on the blade about the y axis is nil, then:

¥sin

¥

d

/10

Ú

=

dF z

12 -rc dx C z V2 1

2 -rc dx C z(xcosq w¥ )2

# 12 -rc dx C z x2w2

x q # m dx w2

x centrifugal load( )cos

=

12 -rcC zw2
4

=

q # 38

-rcC z m

-¥

or numerically:

Remarks:

One verifies that sinq = 0.073 # q and cosq = 0.997 # 1

When the helicopter is not immobile, but has a horizontal velocity, for

example v0, the relative velocity of air with respect to the blade varies

between v0 + wx for the blade that is forward, and –v0 + wx for a blade

that is backward If the incidence i does not vary, the lift varies in a cyclical

manner, and there is vertical “flapping motion” of the blade This is why

a mechanism for cyclic variation of the incidence is necessary

We have not taken into account the drag to simplify the calculations Thiscan be considered similarly to the case of the lift It then gives rise to anequilibrium position with a second small angle, called j, with respect tothe radial direction from top view, as in the following figure This is why

a supplementary joint, or a drag joint, is necessary

2 Weight of the helicopter: The lift and the weight balance themselves out.The lift of the blade is then:

then for the 3 rotor blades:

Mg = 3F z

Mg # 1

2 -rcC zw2

3

Mg = 2340 daN

3 Normal load: It is denoted as N(x):

at the foot of the blade (x = l/10):

4 Longitudinal modulus of elasticity:

Using the relation of Section 3.3.1:

with (Section 1.6): E f = 86,000 MPa; E m = 4,000 MPa

5 Section of the spar of the blade made of glass/epoxy:

The longitudinal rupture tensile stress of the unidirectional is

With a factor of safety of 6, the admissible stress at a section S(x) becomes

cos

d x

x x d x

Ú

N x( ) mw2

2 -
2

S x( ) = N x -( )s

S x( ) mw2

2s -
2

6 Mass of the spar (longeron) of the blade:

Specific mass of the unidirectional layer (see Section 3.2.3):

Then:

7 Elongation of the spar of the blade: The longitudinal constitutive relation

is written as (see Section 3.1):

8 Clamped axes: For 2 axes in 30 NCD16 steel (rupture shear strength trupt=

500 MPa; bearing strength sbearing= 1600 MPa); 4 sheared sections; factor ofsafety = 6:

=D
0.9
s

E

-=

D
= 2.4 cm