A 6 × 19 wire rope with fibre core and tensile designation of 1570 is used to raise the load of 20 kN as shown in Fig. 23.7. The nominal diameter of the wire rope is 12 mm and the sheave has 500 mm pitch diameter. Determine the expected life of the rope assuming 500 bends per week.

Question

Accepted Answer

Given W = 20 kN
number of bends = 500 per week
For wire rope, construction = 6 × 19
[latex] d_{r} [/latex] = 12 mm tensile designation = 1570
D = 500 mm
Step I [latex] \left(p / S_{u} ight) [/latex] factor
It is observed from Fig. 23.7 that each side of the rope shares a load of (20/2) or 10 kN. Therefore, the wire rope is subjected to a maximum force of 10 kN.

Table 23.7 Wire-rope data

Sheave diameter (D) (mm)		Metallic area of rope (A) (mm²)	Diameter of wire [latex]\left(d_{w} ight)[/latex] (mm)	Modulus of elasticity of rope [latex]\left(E_{r} ight)[/latex] (N/mm²)	Type of Construction
Recommended	Minimum	Metallic area of rope (A) (mm²)	Diameter of wire [latex]\left(d_{w} ight)[/latex] (mm)		Type of Construction
[latex]72 d_{r}[/latex]	[latex]42 d_{r}[/latex]	[latex]0.38 d_{r}^{2}[/latex]	[latex]0.106 d_{r}[/latex]	97000	6 × 7
[latex]45 d_{r}[/latex]	[latex]30 d_{r}[/latex]	[latex]0.40 d_{r}^{2}[/latex]	[latex]0.063 d_{r}[/latex]	83000	6 ×19
[latex]27d_{r}[/latex]	[latex]18 d_{r}[/latex]	[latex]0.40 d_{r}^{2}[/latex]	[latex]0.045 d_{r}[/latex]	76000	6 × 37

From Eq. 23.4,

[latex] p=\frac{2 P}{d_{r} D} [/latex] (23.4).

[latex] p=\frac{2 P}{d_{r} D}=\frac{2\left(10 imes 10^{3} ight)}{(12)(500)}=\left(\frac{10}{3} ight) N / mm ^{2} [/latex] .

The ultimate tensile strength of the wires is 1570 N/mm². Therefore

[latex] \frac{p}{S_{u t}}=\frac{10}{3(1570)}=0.0021 [/latex].

Step II Life of wire rope
From Fig. 23.6, the life of the wire rope is [latex] 3.30 imes 10^{5} [/latex] bends before failure.

Table 23.6 Factor of safety for wire ropes in mining applications

Factor of safety	Application
	(a) Mining ropes For shafts of varying depths
10	Up to 300 m
9	300–500
8	500–700
7	700–1000
7	(b) Haulage ropes

[latex] herefore ext { life }=\frac{3.30 imes 10^{5}}{500} ext { weeks }=\frac{3.30 imes 10^{5}}{500(52)} ext { years } [/latex].

= 12.69 years.

Sheave diameter (D) (mm)		Metallic area of rope (A) (mm²)	Diameter of wire $\left(d_{w}\right)$ (mm)	Modulus of elasticity of rope $\left(E_{r}\right)$ (N/mm²)	Type of Construction
Recommended	Minimum	Metallic area of rope (A) (mm²)	Diameter of wire $\left(d_{w}\right)$ (mm)	Modulus of elasticity of rope $\left(E_{r}\right)$ (N/mm²)	Type of Construction
$72 d_{r}$	$42 d_{r}$	$0.38 d_{r}^{2}$	$0.106 d_{r}$	97000	6 × 7
$45 d_{r}$	$30 d_{r}$	$0.40 d_{r}^{2}$	$0.063 d_{r}$	83000	6 ×19
$27d_{r}$	$18 d_{r}$	$0.40 d_{r}^{2}$	$0.045 d_{r}$	76000	6 × 37

Question 23.3: A 6 × 19 wire rope with fibre core and tensile designation o...

Related Answered Questions