1.

A circular section cable has a tensile force of 1 kN applied to it and the force produces a stress of 7.8 MPa in the cable. Calculate the diameter of the cable.

Answer»

Stress \(\sigma\) = \(\cfrac{forcce F}{area A}\) hence, cross-sectional area, A = \(\cfrac{forcce F}{stress\,\sigma}\)

\(\cfrac{1\times10^3}{7.8\times10^6}\) = 128.2 x 10-6 m2

Circular area = \(\pi\)r2 = 128.2 x 10-6 m2

from which, r2\(\cfrac{128.2 \times10^{-6}}{\pi}\) and radius r = \(\sqrt{\cfrac{128.2 \times10^{-6}}{\pi}}\)

= 6.388 10-3 m = 6.388 mm

and diameter d = 2 x r = 2 x 6.388 = 12.78 mm



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