## Q. 3.4

In the evening a contractor uses the wire rope on a large crane to lift a 1000-lb air compressor 10 ft above the ground, which will help prevent mischief occurring overnight. The wire rope has a diameter of d = 1/4 in.
and is composed of IPS wires in a 6 × 19 Seale IWRC construction; see Table 3.1.The wire rope modulus is $E_r$ = 12,000 ksi, and its compactness factor is CF = 0.40.The rope runs 10 ft between the crane winch at A and a pulley (sheave) at B, and it then runs L = 40 ft down to the attachment C at the compressor on the ground. The hook at the end of the wire rope, position C, has an attachment efficiency of 90%. (a) Calculate the elongation of the wire rope between the winch at A and the compressor at C, just as the compressor is lifted from the ground.
This is the length of wire rope which must be wound onto the winch drum before the compressor will move. (b) Calculate the factor of safety for the wire rope in this operation, assuming that failure occurs at the recommended maximum load of Table 3.1.

TABLE 3 . 1  Weight per Length and Recommended Allowable Loads (in tons of 2000 lb) Are Shown for 6 × 19 Seale IWRC (independent wire rope core) Construction Wire Ropes Using IPS (improved plough steel) Wires.

 Diameter (in.) Weight, lb/ft. Allowable load, tons 1/4 0.18 2.94 1/2 0.46 11.5 1 0.58 44.9 2 7.39 172

## Verified Solution

(a) It is necessary to first calculate the metallic area of the wire rope.
From Eq. 3.21, the metallic area of the rope is

$A_m$ = CF × d² = 0.40 × (0.25 in.)² = 0.025 in.²

The flexibility coefficient, from Eq. 3.19, is

$f_r = \frac{‏L}{A_mE_r} =\frac{(10 ft + 40 ft)} {(0.025 in^2)(12,000 ksi)} = 0.167 \frac{in.} {kip}$

Note that the entire length of the wire rope between the winch and the compressor (10 ft + 40 ft) elongates under the loading. You can verify this using free-body diagrams. The elongation of the wire rope is then

e = $f_rP = \left(0.167\frac{in.}{kip}\right)$ (1 kip) = 0.167 in.    Ans. (a)

(b) Calculation of the expected failure load, $P_F$, must account for the efficiency of the hook end attachment at C. Using the maximum load suggested for this wire rope in Table 3.1 of 5.88 kips, the allowable load after adjusting for the attachment efficiency is

$P_F$ = 0.90 × 5.88 kips = 5.29 kips

The greatest load on the wire rope occurs at B and is the 1000 lbs. (1 kip) of the compressor plus the weight of the wire rope. The total weight of wire rope hanging from the sheave at B to the attachment at C is,

W = (0.18 lb/ft)(40 ft) = 7.2 lb

This is, of course, a negligibly small load compared to that of the compressor.
Thus, we may reasonably neglect the weight of the wire rope for this relatively short length. The factor of safety is then

FS = $\frac{5.29 kips }{1 kip}$ = 5.29       Ans. (b)