Chapter2_15_failuretheories_tsaiwu
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Transcript Chapter2_15_failuretheories_tsaiwu
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).