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

\text { Given } \quad W=5 kN \quad h=100 m \quad v_{2}=5 m / s .

v_{1}=0 \quad t=5 s \quad \text { For wire rope, } .

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 \left(d_{r}\right) , 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

Therefore, the weight of the wire rope will be,

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

=\left[\frac{5000}{9.81}+40\right] kg .

\text { Acceleration force }=\left[\frac{5000}{9.81}+40\right](1) .

= 549.68 N               (iii).

\therefore \quad P=5000+392.4+549.68=5942.08 N .

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

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

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

\therefore \quad \frac{p}{S_{u t}}=0.0015 ,

p=0.0015 S_{u t}=0.0015(1570) .

= 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 \left(45 d_{r}\right) .

Table 23.7 Wire-rope data

\therefore \quad D=45 d_{r} .

From Eq. 23.4,

p=\frac{2 P}{d_{r} D}                    (23.4).

p=\frac{2 P}{d_{r} D} \quad \text { or } \quad 2.355=\frac{2 P}{d_{r}\left(45 d_{r}\right)} .

\therefore \quad d_{r}^{2}=\frac{P}{52.99} mm ^{2} .              (b).

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

d_{r}^{2}=\frac{P}{52.99}=\frac{5942.08}{52.99} \text { or } d_{r}=10.59 mm .

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

d_{r}=12 mm .

D=45 d_{r}=45(12)=540 mm .

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

\text { Acceleration force }=\left[\frac{5000}{9.81}+49.8\right](1) .

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

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

\frac{p}{S_{u t}}=\frac{1.867}{1570}=0.00119 .

∴            p / S_{u t}<0.0015 .

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

(f s)=\frac{69 \times 10^{3}}{P}=\frac{69 \times 10^{3}}{6048.02}=11.4 .