Transcript Flexural stiffness design using Miki’s diagram

Flexural stiffness design using Miki’s
diagram
• Flexural lamination parameters
I
W   sk cos 2 k
*
1
k 1
I
W   sk cos 4 k
*
3
k 1
sk  (2 / h)3 ( zk3  zk31 )
• Boundaries of the domain
W3*  2W1*2  1
1  W1*  1
• What laminates have the same position on the
Miki in-plane diagram as on the Miki flexural
diagram?
Examples
• (0/90)s : 𝑧0 = −2𝑡, 𝑧1 = −𝑡, 𝑧2 = 0,h=4t
𝑠0 = 2 4 𝑡 3 −𝑡 3 + 8𝑡 3 = 0.875,
𝑠90 = 2 4 𝑡 3 0 + 𝑡 3 = 0.125
𝑊1∗ = 0.875cos0𝑜 + 0.125cos180𝑜 = 0.75
𝑊3∗ = 0.875cos0𝑜 + 0.125cos360𝑜 = 1
• 02 ± 45 :
– 𝑊1∗ = 0.875cos0𝑜 + 0.125cos90𝑜 = 0.875,
𝑊3∗ = 0.875cos0𝑜 + 0.125cos180𝑜 = 0.75
Stiffest laminate under lateral loads
• Recall displacement under sine load
w
a 4 q0 sin  x / a  sin  y / b 
4
 4  D11  2  D12  2 D66  (a / b)2  D22  a / b  
• To find stiffest laminate we need to maximize
S= 𝐷11 +2 𝐷12 + 2𝐷66 𝑎 𝑏 2 + 𝐷22 𝑎 𝑏 4
• From Table 2.1
12D11
*
 U1  W1*U 2  W3*U 3
D22
 U1  W1*U 2  W3*U 3
3
h
*
D12*  U 4  W3*U 3
D66
 U 5  W3*U 3
D11* 
• This implies that S is a linear function of the
lamination parameters, and the stiffest laminate
is an angle ply. Why?
Example 8.2.1a
• Design a 16-layer 20x15” laminated graphite
epoxy plate to maximize its fundamental
frequency. Material properties are:
𝐸1 = 18.5, 𝐸2 = 1.89, 𝐺12 = 0.93𝑀𝑠𝑖, 𝜈12 =
0.3, 𝑡 = 0.005", 𝜌 = 0.057𝑙 𝑏 𝑖 𝑛3
• Tsai-Pagano material properties (in Msi) are
𝑈1 = 8.3252, 𝑈2 = 8.3821, 𝑈3 =
1.9643, 𝑈4 = 2.5366, 𝑈5 = 2.8943
Normalized fundamental frequency
• Normalized frequency
 mn
2
12  a 4mn
*
4
*
*
2
*
4


D
m

2(
D

2
D
)(
mna
/
b
)

D
(
na
/
b
)
11
12
66
22
 4h2
• For our data
107 11  6.42  1.81W1*  1.27W3*
108  21  2.78  1.07W1*  0.462W3*
• For maximum frequency we want negative 𝑊1∗
and negative 𝑊3∗ , so angles near 60-deg.
• Why?
Maximization of frequency
• Iso-frequency contours on Diagram.
• Maximum where iso-frequency line is tangent
to diagram
Should be omega21 in
figure
10 11  6.42  1.81W  1.27W
7
*
1
*
3
W3*  2W1*2  1
dW3*
*

4
W
1  1.81/1.27
*
dW1
• Get
• Text suggests ±55.4
• Can we do better?
W1*opt  0.3542, W3opt  0.749   55.4o
4𝑠