Question 17.6: Given a 6 × 19 monitor steel (Su = 240 kpsi) wire rope.(a) D...

(a) Rope tension Ft from Eq. (17–46) is given by

F_{t} =\left(\frac{W}{m} + wl\right) \left(1 +\frac{a}{g}\right)                     (17-46)

F_{t} =\left(\frac{W}{m} + wl\right) \left(1 +\frac{a}{g}\right)=\left[ \frac{2000}{m} + 1.60d^{2}(531.5)\right] \left( 1 +\frac{2}{32.2}\right)

=\frac{2124}{m} + 903d^{2} lbf

From Fig. 17–21, use p/S_{u} = 0.0014. Fatigue tension F_{f} from Eq. (17–47) is given by

F_{f} =\frac{(p/S_{u})S_{u}Dd}{2}           (17-47)

F_{f} =\frac{(p/S_{u})S_{u}Dd}{2} =\frac{0.0014(240 000)72d}{2} = 12 096d lbf

Equivalent bending tension F_{b} from Eq. (17–48) and Table 17–27 is given by

F_{b} =\frac{E_{r}d_{w} A_{m}}{D}                   (17-48)

F_{b} =\frac{E_{r}d_{w} A_{m}}{D} =\frac{12(10^{6})0.067d(0.40d^{2})}{72} = 4467d^{3} lbf

Table 17–27
Some Useful Properties of 6 × 7, 6 × 19, and 6 × 37 Wire Ropes

Factor of safety n_{f} in fatigue from Eq. (17–50) is given by

n_{f} =\frac{F_{f} − F_{b}}{F_{t}}              (17-50)

n_{f} =\frac{F_{f} − F_{b}}{F_{t}} =\frac{12 096d − 4467d^{3}}{2124/m + 903d^{2}}

(b) Form a table as follows:

Wire rope sizes are discrete, as is the number of supporting ropes. Note that for each m the factor of safety exhibits a maximum. Predictably the largest factor of safety increases with m. If the required factor of safety were to be 6, only three or four ropes could meet the requirement. The sizes are different: \frac{5}{8} -in ropes with three ropes or \frac{3}{8} -in ropes with four ropes. The costs include not only the wires, but the grooved winch drums.