180
Now
Chapter 3
Ul
1.77814RtL
a~- -- -1 _
v(8)
0.00589
~ = U[ = 1 = 15.9253 x 10
.3
8" l= UI, = 1 = 5.37663 x 10
-3
_L = U~"= 1 = 600(t/R)
(3.146)
(3.147)
(3.148)
(3.149)
from orthogonality Eq. (3.67) and Eq. (3.85)
Eat = 1
(3.138)
61 = 1 and 811 = 1~1
(3.139)
all = ]/’2
(3.140)
a’i’ =/*3
(3.141)
The powers on R, t, and L are zero for a minimum
at -- 3a’1/2 - 38’~’/2 - al ’tt = 0
(3.142)
61 -- a’l/2 - 8’1’/2 + al" = 0
(3.143)
a~ + a’1/2 + 38’[/2 + 0 = 0
(3.144)
Solving using Eq. (3.138) and Eqs. (3.142)-(3.144)
al =1 611=5/2
a’[=-3/2
8~"=-1/2
(3.145)
Now to substitute a values Eq. (3.145) into Eq. (3.137) yields
V(8) = 0.00589 lbs
This will be the weight of the spring go Eq. (3.129) and (3.133) if optimum
values for R, t and L can be found.
The equations to size the dimensions are developed from Eqs.
(3.129)-(3.132).
Now
Chapter 3
Ul
1.77814RtL
a~- -- -1 _
v(8)
0.00589
~ = U[ = 1 = 15.9253 x 10
.3
8" l= UI, = 1 = 5.37663 x 10
-3
_L = U~"= 1 = 600(t/R)
(3.146)
(3.147)
(3.148)
(3.149)
from orthogonality Eq. (3.67) and Eq. (3.85)
Eat = 1
(3.138)
61 = 1 and 811 = 1~1
(3.139)
all = ]/’2
(3.140)
a’i’ =/*3
(3.141)
The powers on R, t, and L are zero for a minimum
at -- 3a’1/2 - 38’~’/2 - al ’tt = 0
(3.142)
61 -- a’l/2 - 8’1’/2 + al" = 0
(3.143)
a~ + a’1/2 + 38’[/2 + 0 = 0
(3.144)
Solving using Eq. (3.138) and Eqs. (3.142)-(3.144)
al =1 611=5/2
a’[=-3/2
8~"=-1/2
(3.145)
Now to substitute a values Eq. (3.145) into Eq. (3.137) yields
V(8) = 0.00589 lbs
This will be the weight of the spring go Eq. (3.129) and (3.133) if optimum
values for R, t and L can be found.
The equations to size the dimensions are developed from Eqs.
(3.129)-(3.132).