Assume the data of Example 23.1 and determine the true factor of safety taking into account the bending stresses. The sheave diameter can be taken as (45dr).

Question

Assume the data of Example 23.1 and determine the true factor of safety taking into account the bending stresses. The sheave diameter can be taken as [latex] \left(45 d_{v}\right) [/latex].

Accepted Answer

[latex] ext { Given } D=45 d_{r} [/latex].

Step I Bending load
From Eq. 23.3 and Table 23.7,

Table 23.3 Breaking load and mass for 6 × 7 (6/1) construction wire ropes

Minimum breaking load corresponding to tensile designation of wires 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
45	42	41	38	36	33	25.2	22.9	8
57	53	51	48	46	42	31.8	28.9	9
70	65	64	59	56	52	39.1	35.7	10
85	79	77	71	68	63	47.6	43.2	11
101	94	91	85	81	75	56.6	51.5	12

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] P_{b}=\frac{A E_{r} d_{w}}{D} [/latex].

[latex] =\frac{\left(0.40 d_{r}^{2} ight)(83000)\left(0.063 d_{r} ight)}{\left(45 d_{r} ight)} [/latex].

[latex] =46.48 d_{r}^{2}=46.48(10)^{2}=4648 N [/latex].

Step II Total load on wire rope
The total force acting on the wire rope consists of three factors discussed in the previous example, plus the bending load. The total force is given by,

[latex] \left[\frac{5000}{2}+67.89+\frac{509.68}{2}+6.92 ight]+4648 ext { or } 7477.65 N [/latex].

Step III Factor of safety

[latex] (f s)=\frac{48000}{7477.65}=6.42 [/latex].

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 23.2: Assume the data of Example 23.1 and determine the true facto...

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