Graphical design for specified laminate
strain limits
• Strictly speaking, strain failure criteria should
be applied at ply level rather than at laminate
levels.
• However for in-plane loadings and several ply
angles present, it is not unreasonable to use
laminate strain limits.
• This greatly simplifies the search for the
optimum laminate.
Recall A* in terms of lamination
parameters
• Hooke’s law
• From Table 2.1
  x0 
 x 
 
* 0 


A
 y
 y 
 
 0 
 xy 
 xy 
U1 U 4 0  U 2
A*  U 4 U1 0    0
 0
0 U 5   0
0
U 2
0
0
 U5
0  V1*   U 3
 0
0 
U 3
U3
0
0 
0  V3*
U 3 
• What are the expressions for Poisson’s ratio and
shear modulus for a quasi-isotropic laminate?
• What other laminates will have the same
expression for the shear modulus?
Solution for laminate strains
• Inverting A* matrix analytically one obtains
 x0
U  U V  U V    U  U V  


U  U   U V  2U U  U V
*
2 1
1
2
1
2
4
*
3 3
x
2 *2
2 1
*
5 3
4
3
1
4
y
*
3
• Can have partial check by specializing to quasiisotropic laminate
 x0 
U1 x  U 4 y
U
2
1
 U 42 
• Does that check? Any other easy checks?
• How to change for 𝜀𝑦0 ?
Other two strain components
• Y-strain
U  U V  U V    U  U V  

 
U  U   U V  2U U  U V
0
y
*
2 1
1
2
1
2
4
*
3 3
y
*
5 3
4
2 *2
2 1
3
1
4
x
*
3
• Shear strain

 
U U V
• Checks?
• What property of the laminate is responsible
for zero coupling between shear strains and
normal stresses?
xy
0
xy
5
*
3 3
Example 7.1.1
• Design a graphite/epoxy laminate that will
withstand 𝜎𝑥 = 500𝑀𝑃𝑎, 𝜎𝑦 = 250𝑀𝑃𝑎, 𝜏𝑥𝑦 =
100𝑀𝑃𝑎 with largest possible 𝐸𝑥 . Use strain limits
for laminate limits.
• Material properties 𝐸1 = 181, 𝐸2 =
10.3,
𝐺12 = 7.17𝐺𝑃𝑎, 𝜈12 = 0.28, 𝜀1𝑡 =
𝑠
0.008, 𝜀2𝑡 = 0.004, 𝛾12
= 0.008
• From Example 2.4.1 𝑈1 = 76.37, 𝑈2 = 85.73, 𝑈3 =
19.71, 𝑈4 = 22.61, 𝑈5 = 26.88 𝐺𝑃𝑎
• Would a quasi isotropic laminate do?
Optimum point on Miki’s diagram
• Solving for when the strain limits are exactly
critical obtain
V3*  0.61  2.61V1*  3.58V1*2
V3*  16.4  26.0V1*  35.7V1*2
V3*  0.73
 xt constraint
 yt constraint
 xyt constraint
• Optimum at point A, 𝑉1∗ = 0.4180, 𝑉3∗ =
0.7296
Possible realization
Easiest is plies with fixed 𝑉1∗ or fixed 𝑉3∗
For example for 𝑉1∗ = 0.4180
Can do a combination of cross ply and
angle ply.
What percentage zeros in cross-plies?
To calculate angle of angle-ply laminate use cos2𝜃 =
04180, ⇒ 𝜃 = 32.65𝑜
• To calculate proportions need Eq. 4.2.7, 𝑉3∗ = 2𝑉1∗2 −
1 = −0.65
• Point A is at 𝑉3∗ =0.73, what proportion of angle ply
laminate?
•
•
•
•
•
Textbook alternative
• Use 0 ± 𝜃
•
•
•
•
𝑠
laminate. Solve
V1*  0.418   1 cos 0o   2 cos 2
V3*  0.73   1 cos 0o   2 cos 4
Two equations, three unknowns?
Get 𝜈1 = 0.6708, 𝜈2 = 0.3292, 𝜃 = ±70.07𝑜
More easily realizable. Example?
Textbook then checks that this laminate will
actually satisfy ply strain limits.
Design of laminates with two fiber
orientations
• With only two ply orientations, it is possible to
derive equations for strain limits and solve for
the equations of the curves defining the
constraints.
• Because quadratic equations are involved one
gets two possible solutions.
• With at least three constraints for each ply
direction, Miki’s diagram gets hairy
Example 7.1.2
• For graphite-epoxy in previous example, but
loading of 𝜎𝑥 = 500𝑀𝑃𝑎, 𝜎𝑦 = 250 𝑀𝑃𝑎, and
𝑠
𝜏𝑥𝑦 = 0, 𝜀1𝑡 = 0.00829, 𝜀2𝑡 = 0.00388, 𝛾12
=
0.00948
• Design a laminate ±𝜃1
𝑛1
±𝜃2
𝑛2
Two possible solutions spaces