Mechanics of Fiber Reinforced Polymer Composite Structures

Prof. Debabrata Chakraborty

Department of Mechanical Engineering
Indian Institute of Technology-Guwahati

Module-04
Macromechanics of Lamina-II
Lecture-08
Strength Failure Criteria-Part II

Hello and welcome to the second lecture of this fourth module.

(Refer Slide Time: 00:43)

In this module, we have been discussing the strength failure theories of lamina and in the last
lecture, we discussed that the philosophy of strength failure theories in lamina is same as those
for isotropic materials. But the fact that in orthotropic lamina the strength and stiffnesses are
direction dependent are actually taken into account in developing the failure theories in
orthotropic lamina. A major difference in applying failure theories in orthotropic lamina
compared to the failure theories in isotropic materials is that, in orthotropic lamina, instead of
finding out the principal stresses and maximum shear stress, the stresses with reference to the
principal material directions are determined since the strengths and stiffnesses of orthotropic
lamina are specified with reference to the principal material axis.
Therefore, for any state of stress the stresses with reference to the principal material axis are
determined and those are compared with the corresponding strengths to assess the safety or
failure of a lamina. There are two types of failure criteria viz. independent criteria and

153
interactive criteria. In the last lecture the maximum stress criteria was discussed. In today’s
lecture we shall discuss the maximum strain criteria and then few of the interactive criteria for
the strength failure theories of a lamina.

(Refer Slide Time: 02:43)

Maximum Strain Theory

This is also an independent or non-interactive theory. Analogous to the maximum stress
criteria, in maximum strain criteria, the material axes strains are determined and are compared
with the corresponding ultimate strains and the failure is said to have occurred if any of the
strains in the principal material axis exceeds the corresponding allowable strains.
Mathematically,

(σ ) ; ε (σ )
T C

=(ε ) = Ε
T
(
1 u) Ε
1 u C
1 u
1 u

 − ( ε1C ) < ε1 < ( ε1T )  1 1

 u u 

 − ( ε C ) < ε < ( ε T )  where

(σ =
T
); ε (σ ) C

 1 u 1 1 u

= (ε 2T )u 2 u

Ε
( ) Ε C
2 u
2 u

− (γ ) < γ < (γ ) 
2 2

 12 u 12 12 u 
(τ )
(γ )
12 u =
12 u

G12

154
Given the state of stress with respect to the x-y axes, σ x , σ y and τ xy , the state of stress with

respect to the material axes (1-2) σ 1 , σ 2 and τ 12 are determined as discussed earlier using the

stress transformation as

σ 1   c2 s2 2 sc  σ x 
   2  
σ 2 
= −2 sc  σ y 
2
s c
   − sc sc c 2 - s 2  τ xy 
τ 12    
And then corresponding material axes strains are determined by multiplying the material axes
stresses by the compliance matrix as

 ε1   S11 S12 0  σ 1 
    
 ε 2  =  S12 S 22 0  σ 2 
γ   0 S66  τ 12 
 12   0

These material axes strains are then compared with the corresponding allowable strains as the
conditions of safety

 − ( ε1C ) < ε1 < ( ε1T ) 

 u u 

− (ε C ) < ε < (ε T ) 
 1 u 1 1 u

− (γ ) < γ < (γ ) 
 12 u 12 12 u 

As shown in the Fig., from the stress strain curves of the lamina loaded in along its material
axes an noting the ultimate stresses (failure point) and the modulus (slope), the allowable
strains are obtained by dividing the strengths by the corresponding moduli as follows, with the
assumptions that the lamina is linearly elastic till its failure.

=

( ε1T )
(=
σ )
T

; (ε )
1 u (σ ) C
C
1 u

Ε Ε
1 u
 u
1 1

 T (=
σ )
T
(σ ) C

( ε 2 )u ; (ε )
2 u 2 u
= C

Ε Ε
2 u
 2 2

(τ )

 (γ )
12 u =
12 u

G12

Again like maximum stress theory, while comparing the strains, proper care should be taken to
decide the corresponding allowable strains depending upon the sign. For shear stress it is
independent of sign in the material axis.

155
A lamina is loaded in tension along 1 and the allowable longitudinal tensile strain (ε1T) u is

nothing but the ultimate longitudinal tensile stress

(σ ) T
1 u
divided by E1. Similarly, a lamina is

loaded in tension along 2 and the allowable longitudinal tensile strain (ε2T) u is nothing but the

ultimate longitudinal tensile stress

(σ )T
2 u
divided by E2. In the same way loading the lamina

in compression (ε1C) u and (ε2C) u could also be obtained from the stress strain plot. From the
shear loading of the lamina in the 1-2 plane from the stress strain plot, noting the failure point
and the slope, the corresponding allowable shear strain is (γ12) u could be obtained by dividing
the ultimate shear stress (τ12) u by G12
Again, like the maximum stress theory, maximum strain theory also consists of five different
criteria. To understand the maximum strain theory, we shall solve the same problem of
determination of off-axis tensile strength of a lamina maximum strain theory.

(Refer Slide Time: 09:37)

We have already determined the off-axis tensile strength using maximum stress theory, so we
would like to do the same exercise using maximum strain theory.

σ x 
 
0
 
As shown in the Fig., the state of stress with reference to x-y is  0  . Therefore, with reference

to the material axes the stresses are obtained using the transformation as

156
 σ 1   c2 s2 2 sc  σ x   σ xc2 
   2    2 
σ 2  = s c 2 −2 sc   0  =  σ xs 
   − sc sc c 2 - s 2   0  −σ sc 
τ 12     x 
Then the strains with reference to the material axes are determined as

 1 ν  1 2 
− 12  ( c −ν 12 s ) σ x 
2
 0 
E E1   σ x c 2   E1
 ε1   S11 S12 0  σ 1   1 
      ν     
 ( s −ν 21c ) σ x 
1 1
ε2  = S12 S 22 0  σ 2  = − 0   σ x s2  =
21 2 2

γ   0 E E2  −σ sc   E2
 12   0 S66  τ 12   2 
 1    
x
1
 0 0   − ( sc ) σ x 
 G12   G12 
So, once we have these strains in the material axis, now we can apply the maximum strain
theory. Using the maximum strain theory the conditions for safety are

 (σ 1C ) ( σ T
)   ( σ 1C )u ( σ 1T )u 
< ( c −ν 12 s ) σ x <  − 2 <σx < 2 
1
−
1

E1   ( c −ν 12 s ) ( c −ν 12 s 2 ) 
u 2 2 u
2
 − ( ε1C ) < ε1 < ( ε1T )   E1 E1
   
 (σ 2 )u ( σ 2T )u   ( σ 2C )u ( σ 2T )u 
 u u  C
− (ε C ) < ε < (ε T )  ⇒ − <
1 2
( s −ν 21c )σ x < E  ⇒ − s 2 −ν c 2 < σ x < s 2 −ν c 2 
2

  ( ) ( 21 ) 
 2 u 2 2 u
 E2 E2
 − (γ ) < γ < (γ )   2 21

 (τ 12 )  
 12 u 12 12 u 
1 (τ 12 )u 
 (τ 12 )u 
− u
<− ( sc ) σ x <  σ <
 G12 G12 G12   x ( sc ) 
   

(Refer Slide Time: 11:58)

157
Now in this case since we are considering tensile stress, therefore σx is positive. Therefore,

there are three conditions for safety instead of five (conditions corresponding to negative ε1

and ε 2 are omitted) So, the conditions are


σ x < 2
( σ 1T )u 

 ( c −ν 12 s 2 ) 
 
 ( σ 2T )u 
σ x < 2 
 ( s −ν 21c 2 ) 
 
 (τ 12 )u 
σ x < 
 ( sc ) 

Similar to what we did in maximum stress theory, here also we could plot σx with θ to
understand the influence of θ on the off-axis tensile strength of a lamina as shown in the Fig.

and we could see the different modes of failure (longitudinal tensile near θ = 0⁰, shear away

from θ = 0⁰ till certain angle and transverse tensile beyond 45⁰ and near 90⁰). Only difference
with maximum stress theory is due to the Poisson’s effect as could be seen from the plot. If we
put Poisson’s ratio as zero, there is no difference between the maximum stress theory and the
maximum strain theory.

(Refer Slide Time: 15:20)

158
After discussin the non-interactive theories, we shall now discuss some of the interactive
theories.
Tsai-Hill failure theory
Hill proposed a kind of distortion energy theory or the von Misses criterion for anisotropic
material and Tsai adopted that for an orthotropic lamina therefore it is Tsai-Hill theory. This
states that the condition for safety for an orthotropic material is given by this.

( G2 + G3 ) σ 12 + ( G1 + G3 ) σ 22 + ( G1 + G2 ) σ 32 − 2G3σ 1σ 2 − 2G1σ 2σ 3 − 2G2σ 3σ 1 + 2G4τ 232 + 2G5τ 132 + 2G6τ 122 < 1
This is a kind of extension of the von-Misses criterion. We know that von Misses criterion
actually represents the distortion energy in an isotropic material. Now, for any given state of
stress the state of stress and the corresponding strains are split into hydrostatic and deviatoric
parts. Hydrostatic is actually responsible for the volume change and the deviatoric part is
actually responsible for distortion and the energy corresponding to distortion is the distortion
energy. Using the distortion energy the von Misses equivalent stresses are obtained. Therefore,
this actually represents the distortion energy. But unlike isotropic materials, in orthotropic
materials, distortion cannot be separated from dilatation. While in an isotropic materials we
can actually separate the two stresses and the corresponding strains but in orthotropic it may
not be possible because of the existence of shear extension coupling. Since it is not possible to
separate the distortion energy from the dilatation energy and therefore, in true sense it is not
related to distortion even though it is adapted in the same line but this is not distortion energy
theory. Unlike the non-interactive criteria like maximum stress and maximum strain criteria
where there are five different sub criteria, here a single function predicts the strength. More
importantly it incorporates interactions between the strengths which were not there in
maximum stress and maximum strain criteria. However, the predicted strength is slightly lower
compared to that by maximum stress and strain criterion.
(Refer Slide Time: 18:32)

159
So, let us see that how these parameters G1, G2, G3, G4, G5, G6 in this are calculated? These are
kind of parameters related to the strengths. These are determined by loading the lamina along
the material axes till failure and then obtaining relations for determination of these parameters
as follows.

Case 1: σ 1 = (σ 1T ) (represents failure) 

u

⇒ ( G2 + G3 ) (σ 1 ) =
T 2
1 
u     
  1 2 1  
Case 2 : σ 2 = (σ 2T ) (represents failure) =
 
G1 −
2  (σ T ) 2 (σ T ) 2  
u
   2 u 1 u  
⇒ ( G1 + G3 ) (σ 2T ) =
2
1   1 1 
u
 ⇒
=  G = ; G 
Case 3 : σ 3 = (σ 2T ) (assuming strengths in 2 and 3 are same)   ( 1 )u ( 1 )u 
2 3
T 2 T 2 
2 σ 2 σ
 
u

⇒ ( G1 + G2 ) (σ 2T ) =  G = 1 
2
1
  
( )
u 6 2
 τ 
Case 4: τ 12 = (τ 12 ) (represents failure)   2 12 u 
u 
⇒ 2G6 (τ 12 ) = 1 
2

u 

For example, if we put only σ1 = (σ1T) u and all other stress are 0, and this represents a failure
condition and putting the expression

( G2 + G3 ) σ 12 + ( G1 + G3 ) σ 22 + ( G1 + G2 ) σ 32 − 2G3σ 1σ 2 − 2G1σ 2σ 3 − 2G2σ 3σ 1 + 2G4τ 232 + 2G5τ 132 + 2G6τ 122 < 1

( G2 + G3 ) (σ 1T )u =
2
1
equal to 1, we get . Similarly by putting other failure conditions, we get

three more relations as shown and we could solve for G1, G2, G3, G6 as shown.

(Refer Slide Time: 22:42)

160
Considering a lamina in plane stress (σ 3 = 0 = τ 13 0 ) and putting these values of G1,
τ 23 0=

G2, G3, G6 the Tsai-Hill Theory for a 2D lamina becomes-

Since the lamina is in the plane stress,
 
 σ 12 σ 1σ 2 σ 22 τ 122
(σ 3 = 0 τ 23 = 0) →
0 τ 13 = − + + < 1
 ( 1 )u ( 1 )u ( 2 )u ( 12 )u
 σT 2 σT 2 σT 2 τ 2 

However, since it does not consider the sign of the stresses, it actually underestimates the
strength. It could be clearly seen that since σ1 and σ2 are squared, the influence of +ve or –ve
stress is lost and we know that the strengths are different depending upon whether the normal
stress is +ve or –ve. This is modified to take care of the signs as follows..

=  X 1 (σ 1T ) if σ 1 > 0 
 u 
= 

(σ 1 )u
C
if σ 1 < 0 

=  X (σ T ) if σ 2 > 0 
 2 1 u

=
where 
 (σ 1C )u if σ 2 < 0 

 
= Y (σ 2T )u if σ 2 > 0 
 
  σ  2  σ  σ    σ  2 τ  2  =  (σ 2C )u if σ 2 < 0 
 1 −  1  2   +  2  +  12  < 1  
  X 1   X 2  X 2   Y   S    S = (τ 12 ) 
   u 
So, this is how the Tsai-Hill theory is slightly modified to take care of the sign of the stresses.
(Refer Slide Time: 25:14)

161
So, now suppose we want to do the same exercise what we have done for maximum stress and
maximum strain theory. That means we would like to determine the off-axis tensile strength of
a lamina using Tsai-Hill theory. As shown in the Fig., the lamina is subjected to tensile strength
σx, and we would like to know what is the σx at failure? From the given state of stress with
respect to the x-y axes, we determine the material axes stresses as follows.
σ x ≠ 0, σ y =
τ xy =0
σ 1   c2 s2 2 sc  σ x   σ xc2 
   2    2 
σ 2  = −2 sc   0  =  σ xs 
2
s c
τ   2  −σ sc 
 − sc sc c - s   0 
2
 12   x 
Here, c stands for cosθ, s stands for sinθ.
By putting these expressions of material axes stresses in the Tsai-Hill theory, the condition for
failure is

 
+   c2 s2 +
c4 1 1 s4 1
− 2
=
{(σ ) } {  (τ 12 )u } {(σ ) } {(σ ) } (σ )
2 2 2 2
T T T
1 u 1 u  2 u x

So, if we plot σx with θ say for a typical glass/epoxy and graphite/epoxy lamina we get the
plots as shown for the following properties

162
 (σ T
) = 1725MPa
(σ T
) = 500 MPa
1 u

(σ )
1 u
C
= 1350 MPa
(σ C
) = 350 MPa
1 u

(σ )
1 u
T
= 40 MPa
(σ T
) = 5 MPa
2 u

(σ )
2 u
C
= 275MPa
(σ )
2 u
C
= 75 MPa
(τ )
2 u
= 95MPa
(τ )
12 u
12 u = 35 MPa

As could be seen, it is a continuous curve and a single function predicts the strength. Unlike in
the case of maximum stress and maximum strain, there were actually three curves.
Here also, at θ = 0°, at failure σx = (σ1T) u, and at θ = 90°, at failure, σx = (σ2T) u, as expected
for the cases.
(Refer Slide Time: 30:52)

Here, the mode of failure cannot be determined. We can of course have a deeper look into all
the terms and make out but it is not visible looking at the expression straight away.
(Refer Slide Time: 31:50)

163
Now let us take the same example we have solved using maximum stress theory, just to have
a comparison of the strength prediction by maximum stress theory and by Tsai-Hill theory.
Example :
4p, σ y =
A 60° GR/E lamina subjected to σ x = − 6 p,τ xy = 8p ;
Using Tsai-Hill theory, determine max. value of p > 0,
so that the lamina is safe.  

(σ ) = 1725MPa, (σ ) = 1350MPa
T
1 u
C
1 u

Given, (σ ) 40
= T
=
2 uMPa, (σ ) 275MPa C
2 u

(τ ) = 95MPa
12 u

Solution:
We follow the same steps as discussed earlier that is we determine the material axes stresses
and then put those material axes stresses in the failure theory (in this case the Tsai-Hill theory)
to obtain the value of p at failure.

σ 1   c s 2 -2 sc   4 p  σ 1   3.4 
2

   2      
σ 2  =
s c2 2 sc  −6 p  → σ 2  =−5.4  p
τ   − sc sc c 2 - s 2   8 p     
 12     τ 12   −8.3

(Refer Slide Time: 33:22)

164
 Now given the strength properties as

=(σ 1T ) 1725
= MPa, (σ 1C ) 1350 MPa
u u

=(σ ) 40=
MPa, (σ ) 275MPa
T
2 u
C
2 u

(τ )
12 u = 95MPa

The material axes stresses as

σ 1   3.4 
   
σ 2 = −5.4  p
τ   −8.3
 12   

Depending upon the sign of the stresses, we get the following as the strength parameters

= X 1 (σ 1T ) if σ 1 > 0  
 u
 ⇒ X1 =
1725MPa 
= (σ 1 )
 C
if σ 1 < 0  
 u  
 
= X 2 (σ 1 )u if σ 2 > 0 
T

  ⇒ X2 =1350 MPa 
= (σ 1C )u if σ 2 < 0 
 
 
=Y (σ 2T )u if σ 2 > 0  
 ⇒Y = 275MPa 
= (σ 2C )u if σ 2 < 0 
 
 
S
=
 (τ=
12 )u 95MPa 

And putting in the Tsai-Hill theory as

165
   σ  2  σ   σ    σ  2 τ  2 
  1  −  1   2   +  2  +  12  < 1
  X 1   X 2   X 2    Y   S  
 
 3.4  2  3.4 × −5.4   −5.4  2  −8.3  2  2
  − +  +   p <1
 1725   1350 × 1350   275   95  
⇒ p 2 < 124.69
⇒ p < 11.16 × 106

(Refer Slide Time: 34:35)

We get that p should be less than 11.16 × 106 and using maximum stress theory we got
something like p = 11.4× 106. So, this predicts slightly less than that what is predicted by
maximum stress theory.

(Refer Slide Time: 36:17)

166
 Hoffman theory
One of the drawbacks of the Tsai-Hill theory in its general form is that it does not take care of
the sign of the stresses. So, to account for different strengths in tension and compression,
Hoffman actually added few linear terms in the Hill’s equation and put forward the Hoffman
theory as

C1 (σ 2 − σ 3 ) + C2 (σ 3 − σ 1 ) + C3 (σ 1 − σ 2 ) + C4σ 1 + C5σ 2 + C6σ 3 + C7τ 23 + C8τ 312 + C9τ 122 < 1

2 2 2 2

in plane 1 − 2 → (σ 3 = 0 τ=
23 0 τ=
13 0)
transverse isotropy in 2 − 3 → (σ 3T ) = (σ 2T ) , (σ 3C ) = (σ 2C ) , S31 = S12
u u u u

This takes care of the sign in addition to the interactions among different stresses. The
parameters C1, C2 , etc are determined by putting certain failure conditions as follows

 Case1: σ 1 (σ
= = 1 )u ; Case 2 : σ 1
T
(σ 1C )u 

 Case3 : σ (σ
= = 2 )u ; Case 4 : σ 2 (σ 2C )u 
T
 2

 Case5 : σ (σ
= = 3 )u ; Case6 : σ 3 (σ 3C )u 
T
 3

 Case7 : τ
=
 23 (τ=23 )u ; Case8 : τ 13 (τ13 )u ; 
 
Case9 : τ 12 = (τ 12 )
 u 
For plane stress in 1-2 plane ie. σ3 =τ23 =τ13 =0 and with 2-3 as the plane of transverse isotropy
i.e (σ 2T )
= (σ ) =
and (σ ) (σ
T
3 u =C
) , (τ ) (τ )
2 u
C
3 u 13 u 12 u
u

167
Now we obtain these C1, C2 all these are actually in terms of this 5 strength parameters of the
lamina and using these and putting these conditions we get the Hoffman's failure theory as the
condition for safety as

⇒ − C
σ 12
+
σ 1σ 2
−
σ 22
+
( σ 1C )u + (σ 1T )u
σ +
( σ 2C )u + (σ 2T )u
σ +
τ 122
<1
(σ 1 )u (σ 1T )u (σ 1C )u (σ 1T )u (σ 2C )u (σ 2T )u (σ 1C )u (σ 1T )u 1 (σ 2C )u (σ 2T )u 2 S122
So, here it is important to note that there are interactions between the stresses and it actually
takes care of the sign. Because there is a linear term it actually recognizes the sign of the stress
unlike in Tsai-Hill theory.
(Refer Slide Time: 40:01)

(Refer Slide Time: 41:31)

168
Every time a criterion is put forward, they are actually correlated with the experimental
observations and it was observed that all these are actually inadequate in representing the
experimental data. Therefore, to obtain a better correlation with the experimental data Tsai-Wu
failure theory was proposed with increasing number of terms in the equation so that it gives
better fit with the experimental data and there will be more interactions among the stress
parameters.
So, Tsai-Wu actually postulated the condition for safety as

Fiσ i + Fijσ iσ j < 1 i, j =

1, 2, 6
,

where i, j = 1, 2, 6 means that among the 6 strengths, σ1, σ2, σ3, τ12, τ23, τ13, all the possible
interactions are taken an on expanding this, will have 36 terms. However this is simplified for
a two-dimensional lamina with plane stresses 1-2 and transverse isotropy in 2-3 and
For a 2D lamina with plane stress in 1-2 and transverse isotropy in 2-3,
F1σ 1 + F2σ 2 + F6τ 12 + F11σ 12 + F22σ 22 + F66τ 122 + 2 F12σ 1σ 2 < 1

Still we have seven constants which need to be determined in terms of the strength parameters.

(Refer Slide Time: 43:43)

169
Putting different failure conditions of the lamina in the material axes 1-2, we get some relations
as follows from which these constants F1, F11, F2, F22, F6, F66 are determined as shown.

 1 1 
= F1 − C 
σ 1 (σ )=
= T
1 u , σ2 = τ 12 0 → F1 (σ ) + F11 (σ =
T
1 u ) 1  (σ 1 )u (σ 1 )u 
T 2
1 u
T

 
− (σ 1C )= 0 → − F1 (σ 1C ) + F11 (σ 1C ) =
2
σ1 = , σ 2 τ 12 = 1  F11 = 1

u u u

 ( σ C
)
1 u ( σ T
)
1 u


 1 1 
=  F2 − C 
σ 2 (σ 2 )u , σ 1 ==
= T
τ 12 0 → F2 (σ 2 )u + F22 (σ =
T
2 )u
T 2
1  (σ 2 )u (σ 2 )u 
T

 
σ 2 = − (σ 2C )u , σ 1 = τ 12 = 0 → − F2 (σ 2C )u + F22 (σ 2C )u = 1  F22 =
2
1


 ( σ C
2 u) ( σ T
)
2 u


F =0 
τ 12 (τ 12 )u , σ 1 ==
= σ 2 0 → F6 (τ 12 )u + F22 (τ 12= )u 1  6
2

 F66 = 1 
τ 12 =− (τ 12 =
)u , σ 1 σ 2 = 0 → − F6 (τ 12 )u + F22 (τ 12 )u =
2
1 
(τ 12 )u 
2


(Refer Slide Time: 45:33)

170
 1 1
For equal strength in tension and compression → F1= 0, F2= 0, F11= , F22=
(σ ) (σ )
T 2 T 2
1 u 2 u

σ 12 σ 22 τ 12 2
⇒ + 2 F12σ 1σ 2 + + =
1
(σ )T 2
(σ )
T 2
(τ 12 )u
2
1 u 2 u

Now still we did not get F12 which is associated with σ1σ2 and to determine F12 we need to
apply biaxial stress and need to observe where it fails. That means we take a lamina and apply
σ1 and σ2 and find out what is the combination of σ1 and σ2 at failure. And from that we can
get F12 but there may be large number of combinations. One condition is only σ1 and no σ2,
another condition is only σ2, and no σ1. So, in between there could be infinite combinations
between σ1 and σ2.

(Refer Slide Time: 47:03)

171
For that a biaxial test is conducted with equal stress σ1 = σ2 = σ and all other stresses are zero
and when putting this we get

( F1 + F2 ) σ + ( F11 + F22 + 2 F12 ) σ 2 =

1

And when we put the expressions for F1, F11, F2, F22, we get

     
1   1 1 1 1 σ +  1 1 σ 2 
F12 = 1− + + + +
  ( 1 )u ( 1 )u ( 2 )u ( 2 )u  ( 1 )u ( 1 )u ( 2 )u ( 2 )u
2 
2σ  σ T
σ C
σ T
σC   σT σC σ T
σC  
  

Here σ is the stress at failure for the conducted biaxial stress σ1 = σ2 = σ . So, once we know
this F12 then we can write the complete the Tsai-Wu theory. It is seen that especially for glass
epoxy, it correlates better compared to other interactive theories.
In summary like we have discussed the failure theories for orthotropic lamina. First, we
understood how they are different compared to the isotropic materials. The main difference is
that here the stresses in the material direction is more important as the strengths of orthotropic
lamina are defined with reference to material axis. Therefore, unlike isotropic material we do
not determine the principal stresses but we find out the material axis stresses.
Then those material axis stresses are compared to the corresponding strengths. In case of
independent non-interactive theories each of these strengths is compared independently and
then the failure or safety is assessed. In case of interactive theories, the combined effects of the
strengths are actually taken into account by the interaction terms and a single equation predicts
whether the failure will occur or not.

172