A 15-passenger elevator in a 15-story building is raised by using a one-inch-diameter 1^'' steel rope. Assuming that the city code requires a factor of safety of 11 against the ultimate strength of the rope, check the adequacy of the rope and its elongation.

Question

A 15-passenger elevator in a 15-story building is raised by using a one-inch-diameter 1^{\prime } steel rope. Assuming that the city code requires a factor of safety of 11 against the ultimate strength of the rope, check the adequacy of the rope and its elongation.

Accepted Answer

One-inch [latex]1^{\prime \prime }[/latex] -diameter rope:

Net resisting area [latex]=0.523 \space in.^2[/latex]

Ultimate strength = 27 tons [latex]=54 \space k[/latex]

Rope wt. [latex]=2.0 \space \# /ft.[/latex]

Rope length = 14 stories of rope plus [latex]10 ^{\prime }[/latex] to pulley

Loads on rope:

15 passengers [latex]@150\# ea. [/latex] [latex]=2,250\# [/latex]

15-passenger elevator cab [latex]=1,250\# [/latex]

Rope length [latex]=14 imes 10^\prime [/latex] per story [latex]+ 10^\prime [/latex] (to pulley) [latex]=150^\prime imes 2.0 \space \# /ft. [/latex] = [latex] \underline{300\# } [/latex]

[latex]\space [/latex] [latex]P=3,800\# [/latex]

Ultimate strength [latex]=54,000 \#[/latex]

[latex]\underset{(working) }{ safe \space strength } = \frac{ultimate \space strength }{safety \space factor} =\frac{54,000\# }{11}= 4,909\# [/latex]

[latex]P_{actual }= 3,800\# \lt 4,909\# \space allowable \space OK[/latex]

[latex]\delta =\frac{PL}{AE} =\frac{(3.8 \space k )(150^\prime imes 12 \space in./ft.)}{(0.523 \space in.^2)(29,000 \space k/in.^2)} =0.45 ^{\prime \prime }[/latex]

Question 5.15: A 15-passenger elevator in a 15-story building is raised by ...

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