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Chapter 3
(f) The radius of the spring must be less than 2.122 in so it can be
mounted on the back of the disk.
Constraint 5
R _< 2.1211 in
(3.54)
Also to be fabricated the spring thickness should be more than 0.0015 in
Constraint 6
t _> 0.0015 in
(3.55)
The length must also be developed from some limited dimension (assume
0.100 in or stack up considerations.
Constraint 7
L > 0.100 in
(3.56)
The Eqs. (3.49)-(3.56) can be placed in a non linear program
values for R, t L found. Use ~ = 0.283 lbs/in 3 and E= 30 x 106 psi and compared to Example 3.13.
The spring problem submitted to a nonlinear optimization routine
found the following answers:
Ws = 0.00024 lbs
R = 0.900 in
t = 0.0015 in
L ---- 0.100 in
The answers indicate for a frequency greater than 200 Hz, the length
and thickness need a regional constraint from manufacturing considerationg.
When the frequency constraints
are less than 200 Hz the same
optimization routine found:
Ws = 0.0102 lbs
R = 0.897 in
t -- 0.0015 in
L = 4.281 in
The geometric programming solution in Example 3.13 using some of
the seven constraints in Example 3.9 results in the following values:
Ws = 0.00711 lbs
R = 0.900 in
t-- 0.0015 in
L = 2.96195 in