composite matls and strs 1

8/22/2019 Composite Matls and Strs 1

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Composite materials and

structures


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Poissons ratioIn general terms Poissons ratio, ij,

is defined as the ratio of the negative ofthe normal strain in the direction j to the

normal strain in the direction i, when the

only normal load applied is in direction i.


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Elastic constants for various epoxy matrix

composites.

(Fibres along x-axis)

Ma

teri

al

Exx

Gpa

Eyy

GPa

Gxy

GPa

xy yx Vf Sp.

gravi

ty.

Gra

phite

181 10.3 7.17 0.28 .015

94

0.7 1.6

Bor

on

204 18.5 5.59 0.23 0.5 2.0

Glass

38.6 8.27 4.14 0.26 0.45 1.8

Kev

lar

76 5.5 2.3 0.34 0.6 1.46


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Lamina stressstrain relations

referred to arbitrary axesor

Hookes law for a 2-D angle lamina


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A lamina is a thin layer of a composite material

which is generally of a thickness of the order ofo.125 mm. A laminate is constructed by stacking

a large number of such laminae in various

orientations in the direction of the thickness.

Structures are made of these laminates. These

structures are subjected to various loads such as

stretching, bending, twisting etc. The design

and analysis of such laminated structuresdemand the knowledge of the stresses and

strains in the laminate.


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The building blocks of a laminate are single

lamina. Understanding the mechanical analysis

of a lamina precedes that of a laminate. Alamina is not a homogeneous isotropic material,

even though it is made up of homogeneous

isotropic materials such as fibres and matrix..


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E.g., the stiffness of the lamina varies from

point to point, depending on whether the

point is on the fibre, the matrix, or the

fibre-matrix interface. This makes the

mechanical analysis of the lamina very

complicated. For this reason the macro-

mechanical analysis of a lamina is based on

average properties and considering the

lamina to be homogeneous. Micro

mechanics deals with the methods to find

the average properties


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As already mentioned, the laminates for the

construction of structural elements are generally

made by stacking the laminae in various

orientations as per the design requirements.

Hence to analyze a structure, the properties arerequired along arbitrary directions.

Now, that the properties of the lamina are

known in the directions along and transverseto the fibres, we need to workout the stress -

strain relations referred to arbitrary directions.


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The co-ordinate system used

to illustrate an angle lamina

is shown in the figure. Theaxes in the 1-2 direction is

called the local axes or material

axes or the natural axes. The

direction 1 is parallel to thefibres and direction 2 is perpendicular to the fibres.

Direction 1 is called as longitudinal or l direction

and 2 as transverse or t direction.

The x-y co-ordinate system are called the global axes

or off axes.

The angle between the two is angle .

2 1

x

y


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The stress strain relationship in the 1-2 co-

ordinate system or the local co-ordinate system

has already been established. The stress strainrelationship for the x-y co-ordinate system or

the global co-ordinate system need to be

developed now.


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The global and local stresses are related to each

other through the angle , as shown below:

x -1 1y = T 2

xy 12 ,

where, T is the transformation matrix given by,

C2 S2 2SC -1 C2 S2 -2SC

T = S2

C2

-2SC and T = S2

C2

2SC-SC SC C2-S2 SC -SC C2-S2

where C=cos and S= sin, C2-S2 = cos2


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Using the stress strain equation for the local

axes, that1 1

2 = Q 2

12 12

we can write the relation for global axes as,

x -1 1y = T Q 2

xy 12


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The global and local strains are related through

the transformation matrix,

1 x

2 = T y

(12)/2 (xy)/2

note:

x =u/x, y =v/y xy=(1/2)[u/y + v/x]xy= [u/y + v/x]


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The stress strain relation in the local co-

ordinate system can be written as,

1 Q11 Q12 0 1

2

= Q12

Q22

0 2

12 0 0 2Q66 12/2

where, Q11= E1/ (1-1221)

Q12 = 12 E2/ (1-1221) = 21 E1/(1-1221)Q22 = E2/(1-1221)

Q66 =G12


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Then the stressstrain relation in the global co-

ordinates can be written as,

{}global = [T]-1[Q][T]{}global

x x

y = [T]-1[Q][T] y

xy xy/2


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Where,

Q11 Q12 0

Q = Q12 Q22 00 0 2Q66


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Whereby,

x Q11 Q12 2Q16 xy = Q12 Q22 2Q26 y

xy

Q16

Q26

2Q66

xy

/2


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Q11 Q12 Q16 x= Q12 Q22 Q26 y

Q16 Q26 Q66 xy

Where Qij are called the elements of

transformed reduced stiffness matrix Q

and are given by,


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Q11 = Q11c4 + Q22S

4 + 2(Q12+2Q66)s2 c2

Q12 = (Q11+Q22-4Q66)s2 c2+Q12(c

4 +s4)

Q22 = Q11 s4 +Q22 c

4+2(Q12+2Q66)s2 c2

Q16 = (Q11-Q12-2Q66)c3s - (Q22-Q12-2Q66)s

3c

Q26 = (Q11-Q12-2Q66)cs3

-(Q22-Q12-2Q66)c3

s

Q66 = (Q11+Q22-2Q12-2Q66)s2c2+Q66(s

4+c4)


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Note that there are six elements in the[Q] matrix. However, they are

functions of four stiffness elements

Q11, Q12, Q22, Q66 and the angle of the

lamina .


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Inverting the above equation gives,

x S11 S12 S16 x

y = S12 S22 S26 y

xy S16 S26 S66 xy

Where Sij

are the elements of the

transformed reduced compliance

matrix and are given by,


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S11 = s11c4+(2s12+s66)s

2c2+s22s4

s12 = s12(s4+c4)+s11+s22-s66)s

2c2

s22 = s11s4

+(2s12+s66)s2

c2+s22c4

s16 = (2s11-2s12-s66)sc3-(2s22-s12-s66)s

3c

s26 = (2s11-2s12-s66)s3c-(2s22-2s12-s66)sc3s66 = 2(2s11-2s22-4s12-s66)s

2c2+s66(s4+c4)


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We know that,

s11 = 1/E1 s22=1/E2

s12=-12/E1 = -21/E2

s66= 1/G66=1/G12


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From these equations for the stress-

strain relation for a unidirectionallamina, it may be noted that when

they are loaded in the material axes

directions, there is no couplingbetween the normal and shearing

terms of strains and stresses.


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But, for loading other than the natural

axes, there is coupling between the normal

and shearing terms of stresses and strains.If normal stresses only are applied to

an angle lamina, the shear strains are

non-zero

and if shearing stresses only are applied

to an angle lamina the normal

stresses are non-zero.

----------00000000-----------


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Experimental characterisation of a Lamina.


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In order to analyze a composite structure,

the analyst needs five intrinsic macroscopicmaterial properties, El, Et, lt ( l and t being

longitudinal transverse directions to the

fibre axis) of a unidirectional lamina or ply.In addition we need at least 3

fundamental strengths of the lamina. (This

assumes that the tensile and compressivestrengths are equal. If not we need five

strengths.)


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1. X- axial or longitudinal strength

2.Y- transverse strength.

3. S- shear strength.These quantities are the ultimate stresses

of a lamina and have units of force per unit

area.


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To evaluate these values, four experiments

need to be conducted. Three tension tests

and one torsion test , as shown in the

figures below:

P

fig -1 fig-2

P

fig-3 fig-4

2

2

1P

P T

T

P

P


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1. Experiment-1.

Figure. I shows a tensile specimen with

fibres along the loading direction. Strain

gauges are fixed on the specimen to

measure the strain along and across the

fibre directions. (In order to average any

tendency to bending due to the grips of

the testing machine back to back biaxial

strain gauges should be used on both sides

of the specimen.)


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Since the fibre is oriented along the loading

direction, the l direction and 1 direction

coincide and =0.This experiment could be used to determine the

elastic constants El (or E1) and lt (or 12) and the

fundamental strength, X. By measuring the load P, thecross sectional area A, and the strains l (or 1) and t(or 2).

The youngs modulus, El (or E1) is given by,

El or E1 = P/(Al)lt (or 12) = - (t/l)

and X= Pultimate/A.


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2. Experiment-2.

Figure-2 shows a tensile specimen with fibres

along the transverse direction to the loadingdirection. Strain gauges are fixed as in

experiment -1.

This experiment can give the youngs modlus in

the transverse direction to the fibre or Et (or E2),

the Poisson's ratio tl, (or 21) and the strength

in the transverse direction Y.


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As in experiment-1, we have,

Et (or E2) = P/(At)

tl (or 21) = -(1/2)

Strength in the lateral direction of the

composite material, Y = Pultimate/A

A check on the accuracy of the measurement

can be made at this point, from the equationsthat, Eltl = Etlt


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3. Shear modulus or modulus of rigidity,

Glt

or G12

.

Figure-3 shows a tension specimen in which the

fibres are oriented at an angle with the

loading direction. If the test is conducted such

that =45 and the load and strain aremeasured, then Ex=P/Ax and Glt can be

computed using the formula,

(1/Ex) = (1/El +1/Et +1/Glt 2lt/El)/4 or

(1/Glt) = 4/Ex-1/El-1/Et+2lt/El .


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Proof:

When only x is applied x is given by,

x = s11x and Ex= (x / x ) = s11

where, s11 = s11c4+(2s12+s66)s

2c2 + s22s4

When = 45, cos =sin =(1/2)1/2,

then, s11

= ()[1/E1-2

12/E

1+1/G

12+1/E

2]

or, 1/Ex = ()[(1/E1)-2(12/E1)+(1/G12)+(1/E2)]


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4. Shear strength, S.

To determine the shear strength we do the

fourth test.A tube is constructed by circumferential

winding to only a few laminae thickness; it can

be considered so thin that the stresses can beasssumed to be constant through the thickness.

For torsional load, the state of stress at any

point in the material will be pure shear. If T is

the torsional load and r and t are the tubes

mean radius and thickness respectively, then ,

shear strength, S = ultimate = Tultimate/2r2t.


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property Boron/epoxyMPa Glass/epoxyMPa

El 280x103 56x103

Et 28x103 18.9x103

lt 0.25 0.25

Glt 10.5x103 8.75x103

X 1050 1050

Y 28 28

S 56 56


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Engineering constants of an angle lamina:

The engineering constants for a unidirectional lamina

were evaluated earlier. Using similar techniques , wecan evaluate the engineering constants of an angle

ply or lamina, Ex, Ey, xy etc in terms of the

transformed stiffness or compliance matrices. We

know the relation between strain and stress for an

angle lamina, as, 2 y 1

x S11 S12 S16 x

y = S12 S22 S26 yxy S16 S26 S66 xy

x

angle ply or angle lamina


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1. To find the youngs modulus in the direction x:

Apply a single stress in the direction x.

ie, x = 0.0 , y = xy= 0.0, we get,

x S11 S12 S16 x x x

y = S12 S22 S26 0xy S16 S26 S66 0

then,x= s11x , y = s12x and xy = s16x

the youngs modulus in the direction x is defined as,Ex= x/x = (1/ s11)xy = -(y/x) = -(s12/ s11)


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Shear coupling:

In an angular lamina, unlike unidirectional

lamina, interaction occurs also between theshear strain and the normal stresses. This is

called the shear coupling. The shear coupling

term which relates the normal stress in the x-

direction to the shear strain is denoted by mxand is defined as,

(1/mx) = -(x/E1xy) = - (1/ s16E1)

Note that mx is a non-dimensional parameter

like Poissons ratio.


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2. Youngs modulus in the y-direction:

Now apply the load in the y-direction alone.

(i.e.), x=0 y = 0 xy=0we can see that ,

Ey=1/s22, yx = - s12/s22

1/my =-1/(s26E1

y

y


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3. Reciprocal relationship:

from the above we can see the relation

between Ey, xy, and yx, as,( xy/ Ex ) = (yx/ Ey)

which is known as the reciprocal relationship of

the Poissons ratio.


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4. Shear modulus:

By applying stresses, x=y=0.0, and xy = 0.0, as

shown in figure, it is seen that,

1/mx = -1/s16E1

1/my = -1/s26E1 y xy

Gxy = 1/ s66 x

xy


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Hence the stress strain equation of an angle

lamina can also be written in terms of the

engineering constants of an angle lamina, as,

x 1/Ex -xy/Ex - mx/E1 x

y = -xy/Ex 1/Ey - my/E1 yxy -mx/E1 -my/E1 1/Gxy xy

The above six engineering consants of an angle

ply can also be written in terms of the

engineering constants of a unidirectional ply, as,


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(1/Ex) = S11 =s11c4+(2s12+s66)s

2c2+s22S4

= c4/E1 +(1/G12- 212/E1)s2c2 +s4/E2

xy = -Exs12 =-Ex s12(s4+c4) +(s11+s22-s66)s

2c2 +s22s4

=Ex (12/E1)(s4+c4)-((1/E1)+(1/E2)-1/G12)s

2c2

1/Ey = s22= s11s4+(2s12+s66)s

2c2+s22c4

=s4/E1+(-2 12/E1 +1/G12)s2c2 +c4/E2


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1/Gxy = s66 = 2(2s11+2s22-4s12-s66)s2c2 +s66(s

4+c4)

= 2(2/E1+2/E2+412/E1-1/G12)s

2

c

2

+1/G12( s

4

+c

4

)

mx = -s16E1

= -E1((2s11-2s12-s66 )sc3-(2s22-2s12-s66)s

3c

=E1(1/G12-2/E1-12/E1-1/G12)s3c

my = -s26E1

= -E1(2s11-2s12-s66)s3c-(2s22-2s12-s66)sc3)= -E1((1/G12-2/E1-212/E1)s

3c

+(2/E2+212/E1-1/G12) sc3)


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Example:

Find the following for a 60 angle lamina of a

graphite/epoxy,given : E1 =181 Gpa, E2= 10.3GPa

12=0.28, G12=7.17GPa

1. Transformed compliance matrix

2. Transformed reduced stiffness matrix3. Engineering constants, Ex, Ey,Gxy, mx, my,& xy.

if the applied stress is x=2.0 Mpa, y=-3.0MPa,

xy= 4MPa, calculate,

4. global strains, 5. local strains, 6. local stresses,

7.Principal stresses, 8. maximum shear stress,

9. Principal strains


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Q11 = E1/(1-1221)Q12= 12 E2/ (1- 1221)

= 21 E1 / (1- 1221)Q22 = E2/ (1-1221)

Q66

= G66

=G12

composite matls and strs 1

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