Question

A helicopter is paying out a spool of 1/8" diameter wire that is supporting a probe that has a weight of 100 lb. The wire is made from steel (E = 30E6 psi, yield strength = 30,000 psi, specific weight 0.29 lb/in'). a) Ignoring the weight of the wire, determine the factor of safety against a yield failure due to 3. the 100 lb load. Determine the length of wire (in feet) that can be deployed (ignore any stretching) before the wire begins to yield due to a combination of the probe's weight and the wire's weight. Based on the un-stretched length calculated above, determine amount the wire stretches (axial deformation) due to the probe's weight and the wire's weight. b) c)

Answer 1

Given Diameter of wire = 1/8 in = dw weight of proble = w_p = 1001b E = 30 times 10^5 psi 6y = 30 times 10^3 psi r = Specific weight of wire = 0.291 b/in ^3 (a) Fos against yield for only weight of probe Therefore 6y/Fos = W_p/A_w (Aw Rightarrow c/s area opf wire) (Fos rightarrow factor of safety) 30 times times 10^3/Fos = 100/pi/4 dw^2 = 100/pi/4 (1/8)^2 Therefore Fos = 3.682 (b) Length opf wire upto yielding (L): The yield strain epsilon_y = 6y/E = 30 times 10^3 /30 times 10^6 = 1 times 10^-3 For maximum stretch without permanent deformation yield strain must be equal to (strain due to self weight of wire + strain due to weight of prope) (epsilon_sw Rightarrow strain due to self weight epsilon_w Rightarrow strain due to weight) Therefore epsilon_y = epsilon_sw + epsilon_w epsilon_y = gamma L/26 + W_p /AE Therefore 1 times ^-3 = 0.29 times L/2 times 30 times 10^6 + 100/pi/4 (1/8)^2 times 30 times 10^6 Therefore L = 150698.39 in or 12558.2 ft (e) stretch in wire due to probes weight R .self weight of L = 150698.39 in strecth.