Transcript Document
Hygrothermal behavior of composite
laminates
• Composite materials can absorb moisture and
expand in the same way that they expand due to
temperature.
• We can use the same kind of analysis to deal with
stresses that arise from both.
• In this course we will focus on thermal behavior.
• The effect of temperatures on composite
laminates is more pronounced than on metals
because of large disparity in coefficient of
thermal expansion.
Effect of mismatch
• Graphite/epoxy typical thermal expansion coefficients are 𝛼1 =
0.02 × 10−6 ∘𝐶 , 𝛼2 = 22.5 × 10−6 ∘𝐶.
• A 200𝑜 𝐶 drop in temperature will produce a strain of -0.0045
in the transverse direction, and almost nothing in the fiber
direction.
• In a 0/90 laminate the fibers in one ply will not allow the matrix
in the other ply to shrink by more than a fraction of the -0.0045
strain.
• This will eat up a substantial part of the load carrying capacity.
• For cryogenic conditions (e.g. liquid hydrogen tanks) the
temperature drop is more than doubled, and the laminate can
fail just due to thermal loads.
• What can we do?
Thermal deformations
• If ply can expand or contract due to
hygrothermal effects, only strains, but no
stresses.
• Stress free expansion strain in a unidirectional
layer (Fig. 3.2)
Strains and stresses
• “Free” strains
1F 1
1
F
T
2 2
2 C
F 0
0
12
• Hooke’s law when total strains are different
from free strains.
1 Q11 Q12
2 Q12 Q22
0
0
12
0 1t 1F
0 2t 2F
Q66 12t
Interaction between plies
• 0/90/45 laminate
• Constraining effect of adjacent laminates
(0/90)s laminate
• Stress-free expansion of layers
• Constrained deformation
Residual strains
• Free strains in x-y system (Checks?)
xF m 21F n 2 2F m 21 n 2 2
F 2 F
2
2 F
2
n
m
n
m
y
1
2
1
2 T
F 2mn( F F ) 2mn( )
1
2
1
2
xy
m cos , n sin
• In vector form 𝜺𝐹𝑇 = 𝜶𝛥𝑇
• What is the coefficient of thermal expansion in
the 45-deg direction?
• The actual strains in the ply due to thermal loads
are denoted as 𝜺𝑁 for “non-mechanical”
• The “residual strains” or the mechanical strains of
non-mechanical origin are 𝜺𝑀 = 𝜺𝑁 − 𝜺𝐹
Residual stresses
• Residual strains in k-th layer
𝜺𝑅 = 𝜺0𝑁 + 𝑧
𝑘
𝑁
𝜿 −
𝐹
𝜺𝑘
• Residual stresses in k-th layer
xR Q11 Q12
R
y Q12 Q22
R Q Q
26
xy 16
xN xF
Q16 x0 N
Q26 y0 N z( k ) yN yF
N F
Q66 xy0 N
xy xy
• What terms will be the same in two layers,
under what conditions?
Thermal loads
• If a laminate is subjected to temperature change without any
loads then the stress and moment resultants should be zero.
N xR h /2 xR
R
R
N y y dz 0
N R h /2 R
xy
xy
M xR h /2 xR
R
R
M y y zdz 0
M R h /2 R
xy
xy
• Substituting from previous slide for in-plane equation for
symmetric laminate
0N
F
N
x h /2 x
Nx
0N
F
N
A y Q y dz N y
0 N h /2 F
N N
xy
xy
xy
• Loads needed to produce same strains without temperatures
Thermal loads for a single isotropic
layer
• For a single isotropic layer 𝜀1𝐹 = 𝜀2𝐹 = 𝛼𝛥𝑇
E
E
• The matrix A
0
1
1
2
E
A
1 2
0
• Thermal loads
2
E
1 2
0
0h
G
T
1
ETh
N
N A T
1
1
0
0
• Justify letter by letter!
With bending
• Thermal moment resultants
F
x
h /2
F
N
M Q y zdz
h /2
F
xy
• Calculating strains and curvatures under
thermal loads
𝐀𝜺0𝑁 + 𝐁𝜿𝑁 = 𝐍𝑁
𝐁𝜺0𝑁 + 𝐃𝜿𝑁 = 𝐌𝑁
With Tsai-Pagano Material constants
• Thermal loads
N xT
K K 2 cos 2
h /2 1
T
N
0.5
K
K
cos
2
y
1
Tdz
2
h /2
N T
K
sin
2
xy
3
• Thermal material constants
K1 U1 U 4 1 2 U 2 1 2
K 2 U 2 1 2 U1 2U 3 U 4 1 2
K 3 U 2 1 2 2 U 3 U 5 1 2
• Any checks that we can apply?