Figure shows a guide wire used to deliver catheters in the h

Figure shows a guide wire, used to deliver catheters in the human body during angioplasty
operations. The compression spring at the end is used as a flexible support for the nose as it opens
the artery and prevents its puncture by the guide wire. The spring has an outside diameter of 0.026
inch, a wire diameter of 0.003 inch, a free length of 0.11 inch and 30 total coils, and has squared and ground ends. Shear Module(G) is 11 Mpsi. The tensile strength Sut is 131.7 ksi.
Determine the following:
a) The mean diameter
b) The spring index
c) The number of active coils
d) The solid length
e) The rate of the spring
f) The force needed to compress the spring to its solid length
g) The safety factor against static overload if the spring is compressed to its solid length, when Ssy= 0.4 Sut. Use the Wahl Factor.

Solution

Mean Diameter of Spring = Outer Diameter - Wire Diameter

= 0.026 - 0.003

= 0.023 inch

Sprin Index = Mean Coil diameter / Wire diameter

= 0.023 / 0.003

= 7.67

As this is a Compression Spring with no of colis 30, Then the active coils will be 29

Wire Lengh = 3.14 * Mean Diameter * No of Coils

= 3.14 * 0.023 * 30

= 2.16 inch

Rate of Spring = (Shear Modulus of Material)⁴ / (8 (D)³ * Active Coils)

= (11)⁴ / (8 * (0.026)³ * 29)

= 366 mpsi / inch

Force needed = 366 * 0.11 = 40.26 lbs

Waht Shear factor = (4c-1 / 4c-4) + 0.615 / c

= 1.12