Transcript Composite Design - Plymouth University

Composite Strength
and Failure Criteria
Micromechanics of failure in a
unidirectional ply
In the fibre direction (‘1’), we assume
equal strain in fibre and matrix. The
applied stress is shared:
s1 = sf Vf + sm Vm
Failure of the composite depends on
whether the fibre or the matrix reaches its
failure strain first.
Failure in longitudinal tension

1T

f
s  s Vf
Failure in longitudinal compression
• Failure is difficult to model, as it may be
associated with different modes of failure,
including fibre buckling and matrix shear.
• Composite strength depends not only on
fibre properties, but also on the ability of the
matrix to support the fibres.
• Measurement of compressive strength is
particularly difficult - results depend heavily
on method and specimen geometry.
Failure in longitudinal compression

1C
s

Em 
 2 f Vf  1  Vf 

E
f 

Microbuckling
Shear failure mode
Failure in transverse tension
High stress/strain
concentrations
occur around fibre,
leading to interface
failure. Individual
microcracks
eventually
coalesce...
Failure in transverse compression
May be due to one or
more of:
• compressive
failure/crushing of
matrix
• compressive
failure/crushing of
fibre
• matrix shear
• fibre/matrix
debonding
Failure by in-plane shear
Due to stress concentration
at fibre-matrix interface:
Five numbers are needed to characterise
the strength of a composite lamina:
s1T* longitudinal tensile strength
s1C* longitudinal compressive strength
s2T* transverse tensile strength
s2C* transverse compressive strength
12* in-plane shear strength
‘1’ and ‘2’ denote the principal material
directions; * indicates a failure value of stress.
Typical composite strengths (MPa)
UD CFRP
UD GRP
woven GRP SiC/Al
s1T*
2280
1080
367
1462
s1C*
1440
620
549
2990
s2T*
57
39
367
86
s2C*
228
128
549
285
12*
71
89
97
113
The use of Failure Criteria
• It is clear that the mode of failure and hence
the apparent strength of a lamina depends on
the direction of the applied load, as well as
the properties of the material.
• Failure criteria seek to predict the apparent
strength of a composite and its failure mode
in terms of the basic strength data for the
lamina.
• It is usually necessary to calculate the
stresses in the material axes (1-2) before
criteria can be applied.
Maximum stress failure criterion
Failure will occur when any one of the stress
components in the principal material axes
(s1, s2, 12) exceeds the corresponding
strength in that direction.
s 1T * (s 1  0) 
s 1   C*

 s 1 (s 1  0)
Formally, failure occurs if:
s 2T * (s 2  0) 
s 2   C*

 s 2 (s 2  0)
*
 12   12
Maximum stress failure criterion
All stresses are independent. If the lamina
experiences biaxial stresses, the failure
envelope is a rectangle - the existence of
stresses in one direction doesn’t make the
lamina weaker when stresses are added in the
other...
Maximum stress failure envelope
s2
s2T*
s1
s1T*
s1C*
s2C*
Orientation dependence of strength
The maximum stress criterion can be
used to show how apparent strength and
failure mode depend on orientation:
s 1  s x cos 2 q
s 2  s x sin2 q
 12  s x sinq cosq
s2
q
s1
12
sx
Orientation dependence of strength
At failure, the applied stress (sx) must be
large enough for one of the principal
stresses (s1, s2 or 12) to have reached
its failure value.
Observed failure will occur when the
minimum such stress is applied:
s 1* cos2 q

 *

*
2
s x  mins 2 sin q

 *


sin
q
cos
q
12


Orientation dependence of strength
s 1* cos2 q
Off-axis tensile strength (E-glass/epoxy)
 12* sinq cosq
1500
strength (MPa)
1250
1000
long tension
750
in-plane shear
trans tension
500
250
0
0
10 20 30 40 50 60 70 80 90
reinforcement angle
s 2* sin2 q
Daniel & Ishai (1994)
Maximum stress failure criterion
• Indicates likely failure mode.
• Requires separate comparison of
resolved stresses with failure stresses.
• Allows for no interaction in situations of
non-uniaxial stresses.
Maximum strain failure criterion
Failure occurs when at least one of the
strain components (in the principal material
axes) exceeds the ultimate strain.
1T * (1  0) 
1   C *

 1 (1  0)
 2T * ( 2  0) 
 2   C*

  2 ( 2  0)
*
 12   12
Maximum strain failure criterion
The criterion allows for interaction of
stresses through Poisson’s effect.
For a lamina subjected to stresses s1, s2,
12, the failure criterion is:
s 1T * , 1  0
s 1   12s 2   C *
 s 1 , 1  0
s 2T * ,  2  0
s 2   21s 1   C *
 s 2 ,  2  0
 12   12*
Maximum strain failure envelope
For biaxial stresses (12 = 0), the failure
envelope is a parallelogram:
s2
s1
Maximum strain failure envelope
In the positive quadrant, the maximum
stress criterion is more conservative than
maximum strain.
max strain
s2
The longitudinal tensile
stress s1 produces a
compressive strain 2.
This allows a higher value
of s2 before the failure
strain is reached.
max stress
s1
Tsai-Hill Failure Criterion
• This is one example of many criteria
which attempt to take account of
interactions in a multi-axial stress state.
• Based on von Mises yield criterion,
‘failure’ occurs if:
2
2
2
 s 1  s 1s 2  s 2    12 
 *  
  *    *   1
2
*
s
s1
 1
 s 2    12 
Tsai-Hill Failure Criterion
• A single calculation is required to determine
failure.
• The appropriate failure stress is used, depending
on whether s is +ve or -ve.
• The mode of failure is not given (although inspect
the size of each term).
• A stress reserve factor (R) can be calculated by
setting
2
2
2
 s 1  s 1s 2  s 2    12 
1
 *  
  *    *   2
2
*
s
R
s1
 1
 s 2    12 
Orientation dependence of strength
The Tsai-Hill criterion can be used to
show how apparent strength depends on
orientation:
s 1  s x cos 2 q
s 2  s x sin2 q
 12  s x sinq cosq
s2
q
s1
12
sx
apparent strength (MPa)
UD E-glass/epoxy
Orientation dependence of strength
1200
1000
long tension
800
trans tension
600
shear
400
Tsai-Hill
200
0
0
10
20
30
40
50
angle (o )
60
70
80
90
Tsai-Hill Failure Envelope
• For all ‘quadratic’ failure criteria, the
biaxial envelope is elliptical.
• The size of the ellipse depends on the
value of the shear stress:
s2
s1
12 = 0
12 > 0
Comparison of failure theories
• Different theories are reasonably close
under positive stresses.
• Big differences occur when
compressive stresses are present.
A conservative
approach is to
consider all
available
theories: