A wire of length 4.5 m has a percentage strain of 0.050% when loaded with a tensile force. Determine the extension in the wire.

Original length of wire = 4.5 m = 4500 mm and strain = \(\cfrac{0.050}{100}\) = 0.00050

Strain \(\varepsilon\) = \(\cfrac{extension \,x}{original\,length \,L}\) hence, extension x = \(\varepsilon\)L = (0.00050)(4500) = 2.25 mm

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