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WHY: COMPOSITE FAILURE THEORY
WHY: COMPOSITE FAILURE THEORY
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 criterionFailure will occur when any one of the stress components in the principal material axes (1, 2, 12) exceeds the corresponding strength in that direction.
*1212
2*
2
2*
22
1*
1
1*
11
)0()0(
)0()0(
C
T
C
T
Formally, failure occurs if:
)( RankimTheoryStressMaximum directionthatinStrengthapisinges Pr)(
00
11
111
whenFwhenF
c
t
00
22
222
whenFwhenF
c
t
06 For
CosSinFFxc
Sin
cFxcFCos
cFxcF
CosSinFFxt
Sin
tFxtF
Cos
tFxtF
CosSinTSinxCosx
atfiberConsider
62
22
1
62
22
1
226
22
21
''
Maximum stress failure criterionFailure will occur when any one of the stress components in the principal material axes (1, 2, 12) exceeds the corresponding yield strength in tension/ compression in that direction.
*1212
2*
2
2*
22
1*
1
1*
11
)0()0(
)0()0(
C
T
C
T
Formally, failure occurs if:
Maximum stress failure criterionAll 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
1
2
2T*
1T*
2C*
1C*
Orientation dependence of strengthThe maximum stress criterion can be used to show how apparent strength and failure mode depend on orientation:
21
12
x
cossinsincos
12
22
21
x
x
x
Orientation dependence of strengthAt failure, the applied stress (x) must be large enough for one of the principal stresses (1, 2 or 12) to have reached its failure value.
Observed failure will occur when the minimum such stress is applied:
cossinsincos
min*12
2*2
2*1
*x
Orientation dependence of strength
Off-axis tensile strength (E-glass/epoxy)
0
250
500
750
1000
1250
1500
0 10 20 30 40 50 60 70 80 90
reinforcement angle
stre
ngth
(MPa
)
long tensionin-plane sheartrans tension
2*1 cos
2*2 sin
cossin*12
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.
*1212
2*
2
2*
22
1*
1
1*
11
)0()0(
)0()0(
C
T
C
T
Maximum strain failure criterion
The criterion allows for interaction of stresses through Poisson’s effect.For a lamina subjected to stresses 1, 2, 12, the failure criterion is:
*1212
2*
2
2*
21212
1*
1
1*
12121
0,0,
0,0,
C
T
C
T
Maximum strain failure envelope
For biaxial stresses (12 = 0), the failure envelope is a parallelogram:
1
2
Maximum strain failure envelope
Maximum strain failure envelope
In the positive quadrant, the maximum stress criterion is more conservative than maximum strain.
1
2
The longitudinal tensile stress 1 produces a compressive strain 2. This allows a higher value of 2 before the failure strain is reached.
max strain
max stress
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:
12
*12
12
2
*2
22*
1
21
2
*1
1
Tsai-Hill Failure Criterion• A single calculation is required to determine
failure.• The appropriate failure stress is used, depending
on whether 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
*12
12
2
*2
22*
1
21
2
*1
1 1R
Orientation dependence of strength
The Tsai-Hill criterion can be used to show how apparent strength depends on orientation:
21
12
x
cossinsincos
12
22
21
x
x
x
UD E-glass/epoxy Orientation dependence of strength
0
200
400
600
800
1000
1200
0 10 20 30 40 50 60 70 80 90
angle (o)
appa
rent
stre
ngth
(MPa
)
long tensiontrans tensionshearTsai-Hill
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:
1
2
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:
Maximum Strain Energy Theory/ Distortion Energy TheoryThe total energy absorbed by a body may be divided into two parts, one part associated with the change in volume (change in size) and the other associated with the distortion (change in shape) of the body. This theory assumes that yielding begins when the distortion energy equals the distortion energy at yield in uniaxial tension. This theory usually fits (but not always) the experimental data better than the other theories. It is expressed mathematically as follows:
σx2 – σxσy + σy2 + 3τxy2 = σys2
‐ + + =1
The quantities σx, σy and τxy are the state of stress at a point and σys is the yield stress of the material in simple tension. Yielding will occur when the left hand side of Equation (2) reaches the value of 1. For composite materials, we need a failure criteria for individual plies. The strength of a composite laminate is based on the strength of the individual plies within the laminate. As the load is applied to the laminate, there will be a first ply failure followed by other ply failures until the last ply fails, which would be the ultimate failure.
2
2
ys
x
2ys
yx
2
2
ys
y
2
23ys
xy
Unidirectional composite laminae have highly directionally dependent strengths. Furthermore, the longitudinal strength can be twenty times the transverse and shear strengths. For ant state of applied stress, we cannot say which component of stress is responsible for failure. In fact, all components contribute to failure. Thus we shall use a failure criterion that includes interaction among all the stress or strain components, the failure criterion used here for orthotropic materials is one that is analogous to the distortion energy theory for isotropic materials. Without derivation, it is expressed mathematically as follows:
where X = Longitudinal Tensile StrengthX' = Longitudinal Compressive StrengthY = Transverse Tensile StrengthY' = Transverse Compressive StrengthS = Shear Strength associated with x,y coordinate system
Yield will occur when the left hand side of Equation (3) reaches a value of 1. For simplicity, we can write Equation (3) as follows:
Fxx σx2 + 2Fxy σx σy + Fyy σy2 + Fss σs2 + Fx σx + Fy σy = 1Where
We can also write Equation (4) in terms of Strain as follows:
Gxx εx2 + 2Gxy εx εy + Gyy εy2 + Gss εs2 + Gx εx + Gy εy = 1where
Gxx = Fxx Qxx2 + 2Fxy Qxx Qxy + Fyy Qxy
2
Gyy = Fxx Qxy2 + 2Fxy Qxy Qyy + Fyy Qyy
2
Gxy = Fxx Qxx Qxy + Fxy [Qxx Qyy + Qxy2] + Fyy Qxy Qyy
Gss = Fss Qss2
Gx = Fx Qxx + Fy Qxy
Gy = Fx Qxy + Fy Qyy
B. Strength Ratio The strength ratio for a lamina is defined as the ratio of the allowable strain to actual strain or allowable stress to actual stress:
When the left hand sides of Equation (4) and (6) are less than 1, the stresses and strains are defined as actual stresses and strains. When the left hand sides are equal to 1, the stresses and strains are defined as allowable values. Since εx = Rε and σa = Rσ, we write Equations (4) and (6) as follows:
Fxx σxa2 + 2Fxy σxa σya + Fyy σya2 + Fss σsa2 + Fx σxa + Fy σya = 1or Fxx σx2 R2 + 2Fxy σxR σyR + Fyy σy2R2 + Fss σs2R2 + Fx σxR + Fy σyR = 1or [Fxx σx2 + 2Fxy σx σy + Fyy σy2 + Fss σs2]R2 + [Fx σx + Fy σy ]R = 1 (10)and [Gxx εx2 + 2Gxy εx εy + Gyy εy2 + Gss εs2]R2 + [Gx εx + Gy εy]R = 1 (11)
C. Laminate Strength The strength of laminate is determined by examining the strength ratios of each ply. The ply with the lowest strength ratio will fail first and it is assumed to fail when it reaches a value of 1. The state of stress resultant when this ply failure occurs is called the first‐ply‐failure state. The plies with higher strength ratios will fail later, when the externally applied load is increased. This successive ply failure progresses until the last ply or ultimate failure occurs.
To determine the allowable stress resultants, proceed as follows:
Allow the stress resultants to increase until the strength ratio for one of the plies is reduced to 1. This ply is assumed to fail. Eliminate the failed ply by reducing its elastic constants to small values, but allowing it to remain in the laminate in its original position. Allow the stress resultants to increase further until the strength ratio of another ply is reduced to 1, then eliminate it the same way. Repeat this process until ultimate failure occurs.
We solve either Equation (10) or (11) for the two roots of R. one root is the Strength Ratio corresponding to a particular set of stresses, while the other root corresponds to the same set of stresses but reversed in sign.
• Example Problem A laminate of [0/45/–45/90]s is subjected to the stress resultants shown. The 0° and 90° plies are Glass/Epoxy 1002 and the 45° and –45° plies are Graphite/Epoxy T300/5208. The thickness per ply is 0.000125m. Determine the strength ratio of the 45° lamina. N1 = 100x106(8)(0.000125) = 100,000 N/mN2 = 0N6 = 40x106(8)(0.000125) = 40,000 N/mFrom Section III‐D, Figure 1b, we see how to obtain σx, σy, σs, from N1, N2, N6. Then from Figure 2, Section VIII‐B, we see how to obtain R:From Equation 8, Section III‐A, we have
and ε1 = 3.59x10–3 , ε2 = –1.99x10–3 , ε6 = 1.576x10–3
From equation (9), Section II‐D, we have
and σ1 = 186.5 MPa , σ2 = 106.9 MPa, σ6 = 142.0 Mpa
From equation (4), Section II‐C, we have
and σx = 288.7 MPa , σy = 4.68 MPa, σs = –39.84 MPa
From Table 7, Section II‐B, we have X = 1500 MPa X' = 1500 MPaY = 40 MPa Y' = 240 MPaS = 68 Mpa
From Equations (5), Section VIII‐B, we have Fxx = 0.444x10–18 Fyy = 0.444x10–18
Fxy = 0.444x10–18 Fss = 0.444x10–18
Fx = 0.444x10–18 Fy = 0.444x10–18
From Equation 10, Section VIII‐B, we have[0.444(288.7)2x10–6 + 2(–3.36)(288.7)(4.68)x10–6 + 101.6(4.68)2 x10–6 + 216.2 (–39.84) 2 x 106 ] R2 + [0 (288.7) x 10–3 + 20.93 (4.68) x10–3] R–1 = 00.3733R2 + 0.0980 R –1 = 0R = 1.51, –1.77If we reverse the stress resultants, we obtain R = 1.77, –1.51. The answer to our problem is R = 1.51.
39
GLOBAL/ LOCAL ANALYSIS FOR PREDICTING STRUCTURAL BEHAVIOR
Structural Design and Analysis
Structural Design and Analysis
Problem (Chapter VIII)1. A laminate consists of Graphite/Epoxy, T300/5208, layers as follows: [0, 45, –45, 90]s. The applied stress resultants are M1 = 25 in‐lb/in, M2 = 0, M6= 0, N1 = 1500 lb/in, N2 = 0, N6 = 0. Strength ratios for the upper and lower surfaces of each lamina are shown in the figure.Verify the value of R = 0.610 for the upper surface of layer number 7. What does this value imply? Based on the values of the strength ratios for this laminate, what action(s) would you suggest?
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