Given a 6 × 19 monitor steel (Su = 240 kpsi) wire rope.(a) Develop the expressions for rope tension Ft , fatigue tension Ff , equivalent bending tensions Fb, and fatigue factor of safety nf for a 531.5-ft, 1-ton cage-and-load mine hoist with a starting acceleration of 2 ft/s^2 as depicted in

Question

Given a 6 × 19 monitor steel ([latex]S_{u}[/latex] = 240 kpsi) wire rope.
(a) Develop the expressions for rope tension [latex]F_{t}[/latex] , fatigue tension [latex]F_{f}[/latex] , equivalent bending tensions [latex]F_{b}[/latex], and fatigue factor of safety [latex]n_{f}[/latex] for a 531.5-ft, 1-ton cage-and-load mine hoist with a starting acceleration of [latex]2 ft/s^{2}[/latex] as depicted in Fig. 17–23. The sheave diameter is 72 in.
(b) Using the expressions developed in part (a), examine the variation in factor of safety nf for various wire rope diameters d and number of supporting ropes m.

Accepted Answer

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

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

[latex]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)[/latex]

[latex]=\frac{2124}{m} + 903d^{2}[/latex] lbf

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

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

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

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

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

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

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

Rope Young’s Modulus [latex]E_{r}[/latex], psi	Area of Metal [latex]A_{m}, in^{2}[/latex]	Diameter of Wires [latex]d_{w}[/latex], in	Better Sheave Diameter D, in	Minimum Sheave Diameter D, in	Weight per Foot Including Core w, lbf/ft	Weight per Foot w, lbf/ft	Wire Rope
13 × [latex]10^{6}[/latex]	0.38[latex]d^{2}[/latex]	0.111d	72d	42d		1.50[latex]d^{2}[/latex]	6 × 7
12 × [latex]10^{6}[/latex]	0.40[latex]d^{2}[/latex]	0.067d	45d	30d	1.76[latex]d^{2}[/latex]	1.60[latex]d^{2}[/latex]	6 × 19
12 × [latex]10^{6}[/latex]	0.40[latex]d^{2}[/latex]	0.048d	27d	18d	1.71[latex]d^{2}[/latex]	1.55[latex]d^{2}[/latex]	6 × 37

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

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

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

(b) Form a table as follows:

[latex]n_{f}[/latex]				d
m= 4	m= 3	m= 2	m= 1	d
5.029	3.865	2.641	1.355	0.25
6.536	5.150	3.617	1.910	0.375
7.254	5.879	4.263	2.336	0.500
7.331	6.099	4.573	2.612	0.625
6.918	5.911	4.578	2.731	0.750
6.210	5.425	4.330	2.696	0.875
5.320	4.736	3.882	2.520	1.000

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: [latex]\frac{5}{8}[/latex] -in ropes with three ropes or [latex]\frac{3}{8}[/latex] -in ropes with four ropes. The costs include not only the wires, but the grooved winch drums.

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

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