PLAIN AND CIVIL: EXAMPLE 2.5.1. DEFORMATION DUE TO A LOAD

Tuesday 19 March 2019

EXAMPLE 2.5.1. DEFORMATION DUE TO A LOAD

A 1/4-inch diameter steel wire is subjected to 600-lb of tensile load. Knowing that $E=29x10^{6}psi$ and length of wire = 5ft, determine:

The elongation of the wire
The corresponding normal stress

The elongation of the wire is simply its deformation, which is determined from the given formula:

$\delta =\frac{PL}{AE}$

$\delta =\frac{600lb\left ( 5ft \right )\left ( \frac{12in}{1ft} \right )}{\frac{\pi }{4}\left ( 0.25in \right )^{2}\left ( 29000000\frac{lb}{in^{2}} \right )}$

${\color{Red} \mathbf{\delta =5.058x10^{-3}m\simeq 5.06mm}}$

The normal stress is computed as follows:

$\sigma =\frac{P}{A}=\frac{600lb}{\frac{\pi }{4}\left ( 0.25in \right )^{2}}$

${\color{Red} \mathbf{\sigma =12223.10psi}}$

PLAIN AND CIVIL

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Tuesday 19 March 2019

EXAMPLE 2.5.1. DEFORMATION DUE TO A LOAD

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