A 600-foot-long roof cable cannot be permitted to stretch more than three feet when loaded, or the roof geometry will be affected too drastically. If Es = 29 × 10^3 ksi and the load is 1,500 k, determine the required cable diameter needed to avoid excessive elongation or overstress. Ft = 100 ksi (a

Question

A 600-foot-long roof cable cannot be permitted to stretch more than three feet when loaded, or the roof
geometry will be affected too drastically. If [latex]E_{s} =29 imes 10^{3} \space ksi[/latex] and the load is 1,500 k, determine the required cable diameter needed to avoid excessive elongation or overstress. [latex]F_{t}=100 \space k[/latex] (allowable tensile stress).

Accepted Answer

[latex]\delta =\frac{PL}{AE} \space ; A=\frac{PL}{\delta E}=\frac{1,500 \space k (600^{\prime })}{3^{\prime }(29 imes 10^{3} \space k/in.^{2} )} =10.34 \space in.^{2}[/latex]

[latex]A=\frac{\pi D^{2} }{4} \space ; D=\sqrt{\frac{4(10.34 \space in.^{2} )}{\pi }} =3.63^{\prime \prime }[/latex]

(diameter based on the deformation requirement)

[latex]f=\frac{P}{A} \space ; A=\frac{P}{f_{t}} =\frac{1,500 \space k}{100 \space k/in.^{2}}=15 \space in.^{2}\Longleftarrow [/latex] governs

[latex]D=\sqrt{\frac{4(15 \space in.^{2} )}{\pi }} =4.37^{\prime \prime }\Longleftarrow [/latex] governs

(diameter based on tensile stress)

Question 5.13: A 600-foot-long roof cable cannot be permitted to stretch mo...

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