COMPOSITE MATERIALS

PROF. R. VELMURUGAN

Lecture 38
Module IV - Failure Theory of Composites

Failure Theory of Composites

For isotropic material, the load carrying capacity of a structure can be determined from the
principal stresses and ultimate tensile, compressive and shear strengths since the elastic constants
of an isotropic material are direction independent. However, in the case of orthotropic materials,
the strengths and elastic constants are direction dependent. Hence, different strength values can be
obtained for an orthotropic material depending upon the direction of the application of load. More
over uniaxial stress applied in any direction other than the principal material axes produces
multiaxial sstrains along the principal material axes of orthotropic material. Therefore, the
strengths of orthotropic materials must be predicted through an appropriate failure criterion.
Many failure theories are not general but are applicable only to some specific types of
composites. In this module, the theories used for fiber composites will be discussed. To use these
theories, applied stresses/strains are transformed into the corresponding stresses/strains along the
principal material directions.
Maximum-Stress Failure Theory:
It states that failure will occur if any one of the stresses (induced by the applied loads) in
the principal material axes exceed the corresponding allowable stress. Therefore, to avoid failure
all the following inequalities must be satisfied.

L < LU

(5.9)

T < TU

(5.10)

LT < LTU

(5.11)

where, L , T , and LT are the stresses produced by the applied loads and

LU , TU , and LTU are the allowable stresses.

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Here, it is to be noted that when the applied normal stress is compressive then the
appropriate allowable compressive stress must be used. Further, there is no interaction between the
failure modes; that is why all the inequalities should be satisfied simultaneously to avoid failure.
While solving problem, the applied stresses (i.e. x , y etc) are to be resolved into the
stresses along the principal material directions (i.e. L , T etc.) using stress transformation law.
If an orthotropic lamina is subjected to a stress x making an angle with the longitudinal
direction, the stresses in the principal material direction are (using the stress transformation)

L = x cos 2

(5.12)

T = x sin 2

(5.13)

LT = x sin cos

(5.14)

It can be concluded from the equations (5.1) and (5.2) that the applied stress x should be
the least of the following stress values to avoid failure:

x <

LU
cos 2

(5.15)

x <

TU
sin 2

(5.16)

x <

LTU
sin cos

(5.17)

On the other hand, if the applied stress x does not satisfy any of the above conditions then
the following failures will occur.
if,

LU
cos 2

then, longitudinal fiber failure

(5.18)

if,

TU
sin 2

then, transverse failure

(5.19)

if,

LTU
then, shear failure
sin cos

(5.20)

Thus, the safe value of x depends on the fiber orientation angle .

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At small values of , the longitudinal tensile failure will occur and in this case the lamina
strength is calculated from

LUt
.
cos 2

At high values of , the transverse tensile failure will occur and the lamina strength is
calculated from

TUt
.
sin 2

At intermediate values of , the in-plane shear failure will occur and the lamina strength is
calculated from the expression

LTU
.
cos sin

Maximum-Strain Failure Theory:

It states that failure will occur if any one of the strains (due to the applied loads) in the
principal material direction exceeds the corresponding allowable strain. Therefore, to avoid failure
the following inequalities must be satisfied.

L < LU

(5.21)

T < TU

(5.22)

LT < LTU

(5.23)

where, L , T , and LT
are the strains, due to the applied loads and

LU , TU , and LTU are the allowable strains

As already stated in the maximum stress theory, when the normal strain due to the applied
load is compressive (shortening) then the appropriate allowable compressive strain must be used.
Again, as in the case of previous theory here also there is no interaction between the failure modes,
that is why all the inequalities should be satisfied simultaneously to avoid failure.
If the material is linearly elastic up to the ultimate failure, the allowable strains can be
replaced by the corresponding strength values, as given below:

LU =

LU
EL

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(5.24)

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TU =

TU

LTU =

(5.25)

ET

LTU
GLT

(5.26)

If an orthotropic lamina, with fiber orientation angle of with the longitudinal direction, is
subjected only to stress x, the strains in the principal material directions are (using the strain-stress
relations and the equation (5.2))
=
L

1
(cos 2 LT sin 2 ) x
EL

=
T

1
(sin 2 TL cos 2 ) x
ET

LT =

1
(sin cos ) x
GLT

(5.27)

(5.28)

(5.29)

It can be concluded from the equations (5.5) and (5.6) that the applied stress x should be
the least of the followings to avoid failure:

x <

LU
(cos LT sin 2 )

(5.30)

x <

TU
(sin TL cos 2 )

(5.31)

x <

LTU
sin cos

(5.32)

As the material is assumed to be elastically linear up to the ultimate failure, both the
maximum-stress theory and the maximum-strain theory will lead to almost an identical result, but
due to the Poisson ratio involved in the latter theory, there may be a slight difference. When the
material is not linearly elastic up to failure, both the theories will predict differently.

Tsai-Hill Failure Theory:

This theory provides a single criterion to predict the failure of a lamina. It states that under
plane stress condition the failure will occur when the following inequality is satisfied.

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11 11 22 22 12

2 +
+
1

LU LU TU LTU

(5.33)

The above relation is valid if the material is transversely isotropic. If the material is orthotropic and
the state of stress is plane stress condition the criterion is
The point to remember here is that this theory will not provide the information about the mode of
failure as in the case of maximum stress/strain theory because the interactions of the modes of
failure are taken into account. This theory considers the interaction of strength values. The
limitation of this theory is that it does not consider the compressive strength values of the lamina.
If an off axis load x is applied on the laamina, the criterion becomes
Tsai-Wu Failure Theory:
This theory provides a single criterion to predict the failure of lamina. It states that under
plane stress condition the failure will occur when the following inequality is not satisfied.
2
A1 L + A11 L2 + A2 T + A22 T2 + 2 A12 L T + A66 LT
<1

where, A1

1 1
'

LU LU

A11

1
=

'
LU LU

A2

1
=
TU

A22

1
=

'
TU TU

A12

A66

(5.34)

(5.35)

1
'
TU

1
A11 A22
2

2
LTU

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The general expression of Tsai-Wu criterion is obtained by substituting the values of Egn. ( 5.35)
in Egn. ( 5.34).

Failure prediction for delamination initiation

The delamination or ply separation is due to inter-laminar stresses, which can reduce the
failure strength of laminas.
In order to predict the initiation of determination at the free edges where inter-laminar
stresses are vulnerable, the following quadratic failure criterion is suggested.
2
2
z2 x z y z
+ 2 + 2 =
1
2
zU
x zU y zU
t

(5.36)

where, the average inter-laminar stresses are defined by

, x z , y z

1
=
xc

xc

, x z , y z ) dx

(5.37)

xc = 2*t,

(5.38)

critical distance over which the inter-laminar stresses are averaged.

Consequence of Lamina failure:

After a lamina fails, the stresses and strains in the remaining laminae increase and the
laminate stiffness is reduced. For the analysis purpose the following methods can be considered.
(i)

Total discount method In this method, once a lamina fails then its stiffness and
strength in all directions are assigned to zero.

(ii)

Limited discount method Zero stiffness and strength are assigned to failed lamina
for the transverse and shear modes if the lamina failure is in the matrix material. If the
lamina fails by fiber rupture, the total discount method is adopted.

(iii)

Residual property method In this method, residual strength and stiffness are assigned
to the failed lamina.

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COMPOSITE MATERIALS

PROF. R. VELMURUGAN

Problems:
The material properties are, E1= 147 GPa, E2= 15 GPa, G12= 12 GPa and 12 = 0.3 . For the lamina
with orientation [45o], find the lamina stresses due to the load of NXX = 100 kN/m. Verify for
failure through the different failure criteria, if the strength values are

LU = 1200 MPa
TU = 60 MPa
LTU = 90 MPa

References:
"Mechanics of Composite Structural Elements", H Altenbach, J Altenbach and W Kissing,
Springer publications.
B D Agarwal, L J Broughtman and K Chandrashekhara, "Analysis and Performance of Fiber
Composites", John Wiley and Sons, Inc

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COMPOSITE MATERIALS

PROF. R. VELMURUGAN

Lectures 39 and 40
Problems (Module IV):

Problem 5.1: A unidirectional composite lamina is subjected to stresses as shown in Fig. 5.1. It
has the allowable tensile stress of 750 MPa in the fiber direction and 50 MPa in the fiber transverse
direction and the allowable compressive stress of 400 MPa in the fiber direction and 100 MPa in
the fiber transverse direction. The allowable shear stress is 50 MPa. Determine whether, the lamina
will fail under the applied stresses using the maximum-stress theory.

Case 1 : x = 50 MPa,(Tension),y = 25 MPa (Comp), xy = 50 MPa (+ve)

Case 2 : x = 100 MPa,(Comp), y = 25 MPa (Comp), xy = 50 MPa (+ve)
Case 3 : x = 50 MPa,(Comp), y = 150 MPa (Comp), xy = 50 MPa (+ve)

xy
L

T
30o

y
Solution:
Case 1:
Given data:The applied stresses,
x = 50 MPa, (Tension)
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y = -25 MPa (Compression)

xy = 50 MPa (Positive shear)
The allowable stresses,
LU = 750 MPa, (Tension)
LU = 400 MPa (Compression)
TU = 50 MPa, (Tension)
TU = 100 MPa (Compression)
xy = 50 MPa (Shear)
Fiber Orientation angle, = 30o
Since, the fibers are oriented at 30o to the x axis, stress-transformation matrix is used to get the
stresses along the principal material directions.
L
T

cos230 o

sin230o

2 cos30 sin30o

50

sin230 o

cos230 o

-2 cos30 sin30 o

-25

-sin30 ocos30 o sin30 o cos30 o cos230-sin230 o

LT

L

=
T

LT

cos 2 30

sin 2 30

sin 30 cos 30

L
T
LT

sin 2 30
cos 2 30
sin 30 cos 30

2 cos 30 sin 30 50

2 cos 30 sin 30 25
cos 2 30 sin 2 30 50

74.55
=

-49.55

50

750
MPa

<

-7.48

-100 (C)

MPa

50
(5.39)

Since, all the induced stresses are within the permissible limits, as per the maximum stress
theory the lamina will not fail under this applied loading condition.

Case 2:
Given data:-

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The applied stresses,

x = -100 MPa, (Compression)
y = -25 MPa (Compression)
xy = 50 MPa (Positive shear)
L

T

LT

cos 2 30

sin 2 30

sin 30 cos 30

sin 2 30
cos 2 30
sin 30 cos 30

2 cos 30 sin 30 100

2 cos 30 sin 30 25

cos 2 30 sin 2 30 50

-400(C)

-37.95

LT

-87.05

MPa

<

-100(C)

MPa

50

57.48

(5.40)

Since, the induced shear stress exceeds the permissible limit, as per the maximum stress
theory the lamina will fail in shear under the applied loading condition.
Case 3:
Given data:The applied stresses,
x = -50 MPa, (Compression)
y = -150 MPa (Compression)
xy = 50 MPa

L

=
T

LT

cos 2 30

sin 2 30

sin 30 cos 30

(Positive shear)

sin 2 30
cos 2 30
sin 30 cos 30

2 cos 30 sin 30

2 cos 30 sin 30
cos 2 30 sin 2 30

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50

150
50

10

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-400(C)

-31.7

-168.3

LT

MPa

<

-100(C)

MPa

50

-18.3

Since, the induced transverse compressive stress exceeds the permissible limit, as per the
maximum stress theory the lamina will experience transverse failure under the applied loading
condition.

Problem 5.2: Determine whether the lamina explained in the problem 5.1 will fail under the same
loading conditions using the maximum-strain theory if :
EL = 40 GPa,

ET = 10 GPa,

GLT = 4.5 GPa,

LT = 0.22

Solution:
The allowable strains are first calculated using equations (1.5).

LU = LU / EL = 750 / 40 x 103 = 0.0188 [stretching]

(5.41)

LU = LU / E L = 400 / 40103 = 0.010 [shortening]

(5.42)

TU= TU / ET= 50 /10 103= 0.005 [stretching ]

(5.43)

3
= TU
/ E=
100 /10 10=
0.010 [shortening]
TU
T

(5.44)

3
LTU= LTU / GLT= 50 / 4.5 10=
0.011 [shearing]

(5.45)

Case 1:
From the problem 5.1,
L

T =

LT

74.55

49.55 MPa

7.48

From the stress-strain relationship,

TL

LT ET
EL

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(5.46)

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= 0.22 x 10 / 40
= 0.055

L TL T
EL

(5.47)

ET

= 74.55 / 40 x 103 0.055 x (- 49.55) / 10 x 103

= 0.0021

T
ET

LT L

(5.48)

EL

= (- 49.55) / 10 x 103 0.22 x 74.55 / 40 x 103

= - 0.0054

LT

LT
GLT

(5.49)

= (-7.48) / 4.5 x 103

= -0.0017
On comparison of strains calculated and allowable strains,

L < LU ,

,
T < TU

LT < LTU

(5.50)

Therefore, according to the maximum-strain theory, the lamina will not fail under the case-1
loading condition.

Case 2:
From the problem 5.1,
L

T =

LT

37.95

87.05 MPa

57.48

From the stress-strain relationship,

= (-37.95) / 40 x 103 0.055 x (-87.05) / 10 x 103

= -0.0005

= (-87.05) / 10 x 103 0.22 x (-37.95) / 40 x 103

= - 0.0085

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LT

PROF. R. VELMURUGAN

= 57.48 / 4.5 x 103

= 0.0128

On comparison of strains calculated and the allowable strains,

L < LU ,

LT <

T < TU ,

and

LTU

(5.51)

Therefore, according to the maximum-strain theory, the lamina will fail under the case-2
loading condition.

Case 3:
From the problem 5.1,

L
T

-31.7
=

LT

-168.3

MPa

-18.3

From the stress-strain relationship,

L

= (-31.7) / 40 x 103 0.055 x (-168.3) / 10 x 103

= 0.0001

= (-168.3) / 10 x 103 0.22 x (-31.7) / 40 x 103

= - 0.0167

LT

= (-18.3) / 4.5 x 103

= -0.0041

On comparison of strains calculated and the allowable strains,

L < LU ,
, LT < LTU
T < TU
Therefore, according to the maximum-strain theory, the lamina will fail under the case-3
loading condition.

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Problem 5.3: Redo the problem 5.1 using Tsai-Hill theory.

Solution:
Case 1:
From the problem 5.1,
L
T

74.55
=

-49.55

LT

-7.48

LU

750

TU

LTU

-100(C)

MPa

MPa

50

74.55 74.55*( 49.55) 49.55 7.48

=
+
+

7502
100
750
50

2

= 0.2843 < 1 OK.

As it is less than unity, the lamina will not fail under this load.
Case 2:
From the problem 5.1,

L

T =

LT

37.95

87.05 MPa

57.48

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400 ( C )
LU

= 100 ( C ) MPa
TU
50

LTU

37.95 (37.95) *(87.05) 87.05 57.48

=
+

+

4002
100
400

50

= 2.109 > 1. NOT OK.

As it is greater, than unity the lamina will fail under this load.

Case 3:
From the problem 5.1,
L
T

-31.7
=

-168.3

LT

-18.3

'LU

-400(C)

TU

LTU

-100(C)

MPa

MPa

50

31.7 (31.7) *(168.3) 168.3 18.3

=
+

+

4002
400
100 50
2

= 3.006 > 1. NOT OK.

As, it is greater than unity the lamina will fail under this load.

Problem 5.4: Redo the problem 5.1 using Tsai-Wu theory.

Solution:
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From the problem 5.1,

LU 750

=
400
LU
MPa

TU 50

Case 1:

LT
A1

1
=
LU

74.55

49.55

7.48

MPa

1
'
LU

(5.52)

= ( 1 / 750) ( 1 / 400)
= -0.0012
A11

1
=

'
LU LU

(5.53)

= 1 / ( 750 x 400)
= 3.33e-006
A2

1
=
TU

1
'
TU

(5.54)

= ( 1 / 50) ( 1 / 100)
= 0.010
A22

1
=

'
TU TU

(5.55)

= 1 / (50 * 100)
= 2.00e-004
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A12

PROF. R. VELMURUGAN

1
A11 A22
2

(5.56)

= - 1/2 ( )
= -1.290 e-005
A66

(5.57)

2
LTU

= 1 / ( 50 )2
= 4.00 e-004
2
A1 L + A11 L2 + A2 T + A22 T2 + 2 A12 L T + A66 LT
= 0.0448 < 1

Therefore, the lamina will not fail under this loading condition according to Tsai-Wu theory.

Case 2:
L
T

- 37.95
=

LT

- 87.05

MPa

57.48

The values of A1, A11, A2, A22, A12, and A66 are irrespective of the loading condition;
therefore, the values calculated in the case 1 will be used here.
2
A1 L + A11 L2 + A2 T + A22 T2 + 2 A12 L T + A66 LT
= 1.93

< 1

Therefore, the lamina will fail under this loading condition according to Tsai-Wu theory.

Case 3:
L
T
LT

-31.7
=

-168.3

MPa

-18.3

2
A1 L + A11 L2 + A2 T + A22 2T + 2 A12 L T + A66 LT
= 4.019

< 1

Therefore, the lamina will fail under this loading condition according to Tsai-Wu theory.

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17