It is required to select a 6 ×19 wire rope with 1570 as tensile designation for a hoist on the basis of long life. The weight of the hoist along with the material is 5 kN. It is to be raised from a depth of 100 m. The maximum speed of 5 m/s is attained in 5 s. Determine the size of the wire rope

Question

It is required to select a 6 ×19 wire rope with 1570 as tensile designation for a hoist on the basis of long life. The weight of the hoist along with the material is 5 kN. It is to be raised from a depth of 100 m. The maximum speed of 5 m/s is attained in 5 s. Determine the size of the wire rope and the sheave diameter for long life on the basis of the fatigue as failure criterion. What is the factor of safety of this wire rope under static conditions?

Accepted Answer

[latex] ext { Given } \quad W=5 kN \quad h=100 m \quad v_{2}=5 m / s [/latex].

[latex] v_{1}=0 \quad t=5 s \quad ext { For wire rope, } [/latex].

construction = 6 × 19 tensile designation = 1570
Step I Total load on wire rope
The total force (P) acting on the wire rope consists of three factors—(i) the weight of the hoist; (ii) the weight of the wire rope; and (iii) the force due to acceleration. The weight of the hoist is given as

5 kN or 5000 N (i)
Referring to Table 23.4, the mass of the wire rope depends upon the nominal diameter [latex] \left(d_{r} ight) [/latex], which is unknown at this stage. As a trial value, the mass is assumed to be 40 kg per 100 m length. Since the material is to be raised from a depth of 100 m, the length of the wire rope is assumed as 100 m.

Table 23.4 Breaking load and mass for 6 × 19 (12/6/1) construction wire ropes with fibre core

Minimum breaking load corresponding to tensile designation of (kN)						Approximate mass (kg/100 m)		Nominal diameter [latex]\left(d_{r} ight)[/latex] (mm)
1960		1770		1570		Approximate mass (kg/100 m)
Steel core	Fibre core	Steel core	Fibre core	Steel core	Fibre core	Steel core	Fibre core
41.6	39	37.6	35	33	31	24.3	22.1	8
52.6	49	47.5	44	42	39	30.8	28.0	9
65	60	58.7	54	52	48	38.0	34.6	10
70.7	73	71.0	66	63	58	46	41.9	11
93.6	87	84.6	78	75	69	54	49.8	12
110	102	99	92	88	82	64.3	58.5	13
127	118	115	107	102	95	74.5	67.8	14
166	154	150	139	133	124	97.4	88.6	16
210	195	190	176	160	156	123.0	112	18
234	217	212	196	188	174	137	125	19
260	241	235	218	208	193	152.0	138	20
314	292	204	263	252	234	184.0	167	22
375	347	338	318	300	278	219.0	199	24
439	407	397	368	352	326	257	234	26

Therefore, the weight of the wire rope will be,

40 × 9.81 or 392.4 N (ii)
Total mass of hoist and wire rope

[latex] =\left[\frac{5000}{9.81}+40 ight] kg [/latex].

[latex] ext { Acceleration }=\frac{v_{2}-v_{1}}{t}=\frac{5-0}{5}=1 m / s ^{2} [/latex]

[latex] ext { Acceleration force }=\left[\frac{5000}{9.81}+40 ight](1) [/latex].

= 549.68 N (iii).

[latex] herefore \quad P=5000+392.4+549.68=5942.08 N [/latex].

Step II Value of (p) for long fatigue life
For 1570 tensile designation,

[latex] S_{u t}=1570 N / mm ^{2} [/latex].

The wire rope has a long fatigue life if the ratio [latex] \left(p / S_{u t} ight) [/latex] is equal to or less than 0.0015.

[latex] herefore \quad \frac{p}{S_{u t}}=0.0015 [/latex],

[latex] p=0.0015 S_{u t}=0.0015(1570) [/latex].

= 2.355 N/mm² (a)
Step II Nominal diameter of wire rope
From Table 23.7, the recommended sheave diameter for 6 × 19 wire rope is [latex] \left(45 d_{r} ight) [/latex].

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

[latex] herefore \quad D=45 d_{r} [/latex].

From Eq. 23.4,

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

[latex] p=\frac{2 P}{d_{r} D} \quad ext { or } \quad 2.355=\frac{2 P}{d_{r}\left(45 d_{r} ight)} [/latex] .

[latex] herefore \quad d_{r}^{2}=\frac{P}{52.99} mm ^{2} [/latex]. (b).

Substituting the value of P in the Eq. (b),

[latex] d_{r}^{2}=\frac{P}{52.99}=\frac{5942.08}{52.99} ext { or } d_{r}=10.59 mm [/latex].

The standard nominal diameter of the wire rope is 12 mm.

[latex] d_{r}=12 mm [/latex].

[latex] D=45 d_{r}=45(12)=540 mm [/latex].

Step IV Check for design
From Table 23.4, the mass of 100 m length wire rope of 12 mm diameter and with a fibre core is 49.8 kg.
Weight of wire = (49.8)(9.81) = 488.54 N

[latex] ext { Acceleration force }=\left[\frac{5000}{9.81}+49.8 ight](1) [/latex].

= 559.48 N
P = 5000 + 488.54 + 559.48 = 6048.02 N.

[latex] p=\frac{2 P}{d_{r} D}=\frac{2(6048.02)}{12(540)}=1.867 N / mm ^{2} [/latex].

[latex] \frac{p}{S_{u t}}=\frac{1.867}{1570}=0.00119 [/latex].

∴ [latex] p / S_{u t}<0.0015 [/latex].

Step V Static design
From Table 23.4, the breaking load of a 12-mm diameter wire rope is 69 kN.

[latex] (f s)=\frac{69 imes 10^{3}}{P}=\frac{69 imes 10^{3}}{6048.02}=11.4 [/latex].

Question 23.4: It is required to select a 6 ×19 wire rope with 1570 as tens...

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