Download - 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 stress- strain 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.
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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