Chapter 2 Macromechanical Analysis of a Lamina
Failure Theories
Dr. Autar Kaw
Department of Mechanical Engineering
University of South Florida, Tampa, FL 33620
H1σ1+H2σ2+H6τ12+H11
+H22
+H66
+2H12σ1σ2< 1

Tsai-Wu applied the failure theory to a lamina in plane stress. A lamina is
considered to be failed if
H 1 1  H 2 2  H 6 12  H 11 1  H 22 2  H 66 12  2 H 12  1 2  1
2
2
2
Is violated. This failure theory is more general than the Tsai-Hill failure
theory because it distinguishes between the compressive and tensile
strengths of a lamina.

The components H1 – H66 of the failure theory are found using the five
strength parameters of a unidirectional lamina.
a) Apply
  1
T
1

ult
,  2  0 ,  12  0
to a unidirectional lamina, the lamina will fail. Equation
(2.152) reduces to
H 1 σ 1 ult + H 11 σ 1 ult =1.
T 2
T
b) Apply
   1
C
1

ult
,  2  0 ,  12  0
to a unidirectional lamina, the lamina will fail. Equation
(2.152) reduces to
 H 1 σ 1
C

ult
+ H 11 σ 1C ult =1.
2
From Equations (2.153) and (2.154),
H 1=
1
σ 
H 11 =
T
1 ult

1
σ 
C
1 ult
1
σ  σ 
T
1 ult
C
1 ult
,
,
c) Apply
 0 ,  2   2

T
1
ult
,  12  0
to a unidirectional lamina, the lamina will fail. Equation
(2.152) reduces to
H 2 σ 2 ult + H 22 σ 2 ult =1.
T 2
T
d) Apply
 0 ,  2    2
C
1

ult
,  12  0
to a unidirectional lamina, the lamina will fail. Equation
(2.152) reduces to
 H 2 σ 2 ult + H 22 σ 2 ult =1.
C
C 2
From Equations (2.157) and (2.158),
H 2=
1
σ 
H 22 =
T
2 ult

1
σ 
1
σ  σ 
T
2 ult
,
C
2 ult
C
2 ult
,
e) Apply
1
 0 ,  2  0 ,  12   12 ult
to a unidirectional lamina, the lamina will fail. Equation
(2.152) reduces to
H
f) Apply 
1
6
 12 ult + H 66  12 ult2 =1.
 0 ,  2  0 ,  12    12 ult
to a unidirectional lamina, the lamina will fail. Equation
(2.152) reduces to
H
6
 12 ult + H 66  12 ult2 =1.
From Equations (2.157) and (2.158),
H 6=0,
H 66 =
1
 12 ult
2
,
Apply equal tensile loads along the two material axes in a unidirectional composite. If
σ x = σ y = σ τ xy = 0 , is the load at which the lamina fails, then
 H 1 + H 2 σ+  H 11 + H 22 + 2 H 12  σ 2 =1.
The solution of the Equation (2.165) gives
H 12 
1
2σ
2
1-(
2
H 1 + H 2 )σ  ( H 11 + H 22 ) σ .
H 1 1  H 2 2  H 6 12  H 11 1  H 22 2  H 66 12  2 H 12  1 2  1
2
2
2
Take a 450 lamina under uniaxial tension σ x . The stress σ x at failure is noted.
If this stress is σ x  0 then using Equation (2.94), the local stresses at failure are
1 
2 

2

2
 12  

2
Substituting the above local stresses in Equation (2.152),
H 1+ H 2 
σ
2

H 12 =
σ
2
2
+
σ
2
4
H 1
 H 11 + H 22 + H 66 + 2 H 12 =1.
+ H 2
σ

1
2
 H 11 + H 22 + H 66 .
2
H1σ1+H2σ2+H6τ12+H11  12 +H22  22 +H66  12
+2H12σ1σ2< 1
H 12  
H 12  
H 12  
1
as per Tsai-Hill failure theory8,
2
2 ( σ 1T )ult
1
2 ( σ 1T )ult ( σ 1C )ult
1
2
as per Hoffman criterion10,
1
T
1
( σ )ult ( σ
C
1
T
2
)ult ( σ )ult ( σ
C
2
)ult
as per Mises-Hencky criterion11.
= 4S
Find the maximum value of S  0 if a stress  x = 2S,  y = - 3S and  xy
are applied to a 600 lamina of Graphite/Epoxy. Use Tsai-Wu failure theory. Use the
properties of a unidirectional Graphite/Epoxy lamina from Table 2.1.

Using Equation (2.94), the stresses in the local axes are
 σ 1
 0 .2500
 

 σ 2  =  0 .7500
 

 τ 12 
 - 0 .4330
0 .7500
0 .2500
0 .4330
 0 .1714  10 1 


=  - 0 .2714  10 1  S.


 - 0 .4165  10 1 


0 .8660   2 S 


- 0 .8660   - 3 S 


- 0 .5000   4 S 
H12=
H1 
1
1500  10
H2 
1
40  10
6
6

1

1500  10
1
246  10
6
6
 0 Pa -1 ,
 2 .093  10 - 8 Pa -1 ,
H 6  0 Pa -1 ,
H 11 
H 22 
H 66 
1
( 1500  10 )( 1500  10 )
6
6
1
( 40  10 6 ) ( 246  10 6 )
1
( 68  10 6 )
2
 4 .4444  10 -19 Pa - 2 ,
 1.0162  10 -16 Pa - 2 ,
 2 .1626  10 -16 Pa - 2 ,
H 12  - 0 .5 4 .4444  10
-19
 1.0162

1
 10
-16
2
  3 .360  10 -18 Pa - 2 .
Substituting these values in Equation (2.152), we obtain
0 1.714 S + 2 .093  10 - 8 - 2 .714 S 
+ 0 - 4 .165 S + 4 .4444  10  19 1.714 S 
2
+ 1.0162  10 -16 - 2 .714 S  + 2 .1626  10  16  4 .165 S 
2
+ 2 - 3 .360  10 -18 1.714 S - 2 .714 S   1,
or
S< 22 .39 MPa
2
If one uses the other two empirical criteria for H12 as per Equation (2.171), one obtains
S< 22 .49 MPa for H 12  
S< 22 .49 MPa for H 12  
1
1
2
2 ( σ 1T )ult
,
1
2 ( σ 1T )ult ( σ 1C )ult
.
Summarizing the four failure theories for the same stress-state, the value of S obtained is
S
= 16.33 (Maximum Stress failure theory),
= 16.33 (Maximum Strain failure theory),
= 10.94 (Tsai-Hill failure theory),
= 16.06 (Modified Tsai-Hill failure theory),
= 22.39 (Tsai-Wu failure theory).