Question 17.1: Select a wire rope for a hoist required to lift 800 kg. The ...

Select a wire rope for a hoist required to lift 800 kg. The hoist design uses two lines to support the load attached to a vertically moving sheave, on which the swivel hook is mounted. The desired design life is 2 years, and the anticipated maximum use is 20 lifts per hour, 7.5 h a day for 220 days per year. A safety factor of 5 is required based on static ultimate strength and 1.5 based on fatigue.

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Load = 400 × 9.81 = 3924 N.

From Figure 17.4, going for a balance of resistance to abrasion and fatigue, a 6 × 31 WS or a 6 × 25 FW wire rope construction might be appropriate.

From Table 17.3,

Table 17.3: Material and construction data for selected wire rope classes.

Classification 6 × 7 6 × 19 6 × 37 8 × 19
Number of outer
strands
6 6 6 8
Number of wires per
strand
3  ̶  14 15  ̶  26 27  ̶  49 15  ̶  26
Maximum number of
outer wires
9 12 18 12
Approx diameter of
outer wires
d_{r} / 9 d_{r} / 13 to d_{r} / 16 d_{r} / 22 d_{r} / 15 to d_{r} / 19
Material available
(typically). Core: FC
IPS (200) I (80) IPS (200) I (80)
T (130) T (130)
IPS (200) IPS (200)
Material available
(typically). Core:
IWRC
IPS (190) I (190) EIPS (220) IPS (190)
T (220) EIPS (255) EIPS (200)
IPS (255)
Approx metallic crosssection
of rope.
Core: FC
0.384 d_{r}^{2} 0.404 d_{r}^{2} s^{4} 0.427 d_{r}^{2}( FW )^{4} 0.366 d_{r}^{2}( W )^{4}
Approx metallic crosssection
of rope.
Core: IWRC
0.451 d_{r}^{2} 0.470 d_{r}^{2} s^{4} 0.493 d_{r}^{2}( FW )^{4} 0.497 d_{r}^{2}( W )^{4}
Standard nominal ¼ \text { to } 5 / 8 \times 1 / 16^{\text {th }} s ¼\text { to } 5 / 8 \times 1 / 16^{\text {th }} s ¼ \text { to } 5 / 8 \times 1 / 16^{\text {th }} s ¼ \text { to } 5 / 8 \times 1 / 16^{\text {th }} s
rope diameters ¾ \text { to } 1 ½ \times 1 / 8^{\text {th }} s ¾ \text { to } 2 ¾\times 1 / 8^{\text {th }} s ¾ \text { to } 3¼ \times 1 / 8^{\text {th }} s ¾ \text { to } 1½ \times 1 / 8^{\text {th }} s
Unit weight of rope
(lb/ft)
1.5 d_{r}^{2} 1.6 d_{r}^{2} 1.55 d_{r}^{2} 1.45 d_{r}^{2}
Approx modulus of
elasticity of the rope
(psi) 0% to 20% of S_u
13 \times 10^{6}( FC ) 12 \times 10^{6}( FC ) 11.6 \times 10^{6}(F C) 9 \times 10^{6}
15 \times 10^{6}(\text { IWRC }) 14 \times 10^{6}(\text { IWRC })
Recommended
minimum sheave or
drum diameter
42 d_{r} 34 d_{r} 18 d_{r} 26 d_{r}

the nearest classifications compatible with 6 × 31 WS and 6 × 25 FW are 6 × 37 and 6 × 19, respectively. As a first iteration for the design, a 6 × 37 WS wire rope construction will be explored.

A safety factor of 5 based on ultimate tensile strength has been defined.
For a suddenly applied load,

\sigma_{\text {max suddenly applied }}=2 \sigma_{\text {max static }}

From Eqn (17.1),

\sigma_{t}=\frac{T}{A_{\text {rope }}}                                          (17.1)

 

\sigma_{\text {max suddenly applied }}=2 \sigma_{\text {max static }}=2 \frac{T}{A_{\text {rope }}}

Each wire rope carries 3924/2 = 1962 N. From Table 17.3, A_{\text {rope }}=0.427 d_{r}^{2}

 

\sigma_{\text {max suddenly applied }}=2 \frac{T}{A_{\text {rope }}}=2 \frac{1962}{0.427 d_{\text {rope }}^{2}}

From Table 17.3, the static ultimate tensile strength of improved plow steel (IPS) is 200,000 psi ≈ 1379 MPa.

With a static safety factor of 5, the design stress is given by

\sigma_{d}=1379 / 5=275.8 MPa

Equating the suddenly applied stress and the design stress, \sigma_{\text{max suddenly applied}}=\sigma_{\operatorname{design}}, and solving for the rope diameter gives

d_{\text {rope }}=\sqrt{\frac{3924}{275.8 \times 10^{6} \times 0.427}}=8.164 mm

From Table 17.3, the nearest larger standard rope diameter is 11/32 in or 12/32 in. Going for the larger of these, 12/32 in, d_{r \text { static }}=9.525 mm.

From Table 17.3 the minimum recommended sheave diameter is given by

d_{\text {sheave }}=18 d_{\text {rope }}=18 \times 9.525=171.5 mm

The bending stress in the outer wires can be estimated from

\sigma_{b}=\frac{d_{\text {wire }}}{d_{\text {sheave }}} E_{\text {rope }}

The approximate diameter of the outer wires, in this case, from Table 17.3 is d_{\text {wire }}=d_{\text {rope }} / 22.

So,

\sigma_{b}=\frac{d_{\text {wire }}}{d_{\text {sheave }}} E_{\text {rope }}=\frac{8.164 / 22}{171.5} 75.8 \times 10^{9}=191.5 MPa

The number of cycles is

N_{d}=20 \times 7.5 \times 220 \times 2=66,000

From Figure 17.5, the fatigue strength parameter, R_N = 0.0055.

The pressure corresponding to failure in 66,000 cycles is

p_{N_{f}}=R_{N} \sigma_{ uts }=0.0055 \times 1379=7.584 MPa

The fatigue safety facture is 1.5, so

p_{d} \text { fatigue }=\frac{p_{N_{f}}}{1.5}=\frac{7.584}{1.5}=5.056 MPa

The wire rope diameter based on fatigue is given by

d_{\text {rope fatigue }}=\frac{2 T}{p_{d \text { fatigue }} d_{\text {sheave }}}=\frac{2 \times 1962}{5.056 \times 10^{6} \times 0.1715}=9.054 mm

From Table 17.4,

Table 17.4: Maximum allowable bearing pressure between rope and sheave.

Sheave material
Rope Wood Cast iron Cast steel Manganese steel
Regular lay
6 × 7 1.0 2.1 3.8 10.1
6  ×  19 1.7 3.3 6.2 16.6
6  ×  37 2.1 4.0 7.4 20.7
8  ×  19 2.4 4.7 8.7 24.1
Lang lay
6 ×  7 1.1 2.4 4.1 11.4
6 ×  19 1.9 3.8 6.9 19.0
6 ×  37 2.3 4.6 8.1 22.8
8 ×  19 2.7 5.5 10.0 27.6

for a 6 × 37 wire rope on a cast-carbon steel sheave (BHN 160), the
allowable bearing pressure load based on wear is 8.1 MPa.

The wire rope diameter based on wear is given by

d_{\text {rope fatigue }}=\frac{2 T}{p_{d \text { fatigue }} d_{\text {sheave }}}=\frac{2 \times 1962}{5.056 \times 10^{6} \times 0.1715}=9.054 mm

In this case, examining the wire diameters based on static loading, fatigue and wear, the largest diameter is that for static loading.

In summary, the wire rope required is 3/8 in 6 × 37 WS IPS fiber core. The sheave is carbon steel (BHN-160) with a diameter of 171.45 mm.

17.4
17.5