1. 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.

2. Recall A* in terms of lamination parameters
Hooke’s law
From Table 2.1
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?

3. Solution for laminate strains
Inverting A* matrix analytically one obtains
Can have partial check by specializing to quasi-isotropic laminate
Does that check? Any other easy checks?
How to change for 𝜀 𝑦 0 ?

4. Other two strain components
Y-strain
Shear strain
Checks?
What property of the laminate is responsible for zero coupling between shear strains and normal stresses?

5. Example 7.1.1
Design a graphite/epoxy laminate that will withstand 𝜎 𝑥 =500𝑀𝑃𝑎,  𝜎 𝑦 =250𝑀𝑃𝑎,  𝜏 𝑥𝑦 =100𝑀𝑃𝑎 with largest possible 𝐸 𝑥 . Use strain limits for laminate limits.
Material properties 𝐸 1 =181,  𝐸 2 =10.3,   𝐺 12 =7.17𝐺𝑃𝑎,  𝜈 12 =0.28, 𝜀 1 𝑡 =0.008,   𝜀 2 𝑡 =0.004,  𝛾 12 𝑠 =0.008
From Example 𝑈 1 =76.37,  𝑈 2 =85.73,  𝑈 3 =19.71,  𝑈 4 =22.61,  𝑈 5 =26.88 𝐺𝑃𝑎
Would a quasi isotropic laminate do?

6. Optimum point on Miki’s diagram
Solving for when the strain limits are exactly critical obtain
Optimum at point A, 𝑉 1 ∗ =0.4180,  𝑉 3 ∗ =0.7296

7. Possible realization Easiest is plies with fixed 𝑉 1 ∗ or fixed 𝑉 3 ∗
For example for 𝑉 1 ∗ =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= 𝑜
To calculate proportions need Eq , 𝑉 3 ∗ =2 𝑉 1 ∗2 −1=−0.65
Point A is at 𝑉 3 ∗ =0.73, what proportion of angle ply laminate?

8. Textbook alternative Use 0 ± 𝜃 𝑠 laminate. Solve
Two equations, three unknowns?
Get 𝜈 1 =0.6708,  𝜈 2 =0.3292, 𝜃=± 𝑜
More easily realizable. Example?
Textbook then checks that this laminate will actually satisfy ply strain limits.

9. 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

10. Example 7.1.2
For graphite-epoxy in previous example, but loading of 𝜎 𝑥 =500𝑀𝑃𝑎,  𝜎 𝑦 =250 𝑀𝑃𝑎, and 𝜏 𝑥𝑦 =0, 𝜀 1 𝑡 = ,   𝜀 2 𝑡 = ,  𝛾 12 𝑠 =
Design a laminate ± 𝜃 1 𝑛1 ± 𝜃 2 𝑛2

11. Two possible solutions spaces