Pamela Meyer

2022-01-10

Consider a cylindrical specimen of a steel alloy 8.5 mm (0.33 in.) in diameter and 80 mm (3,15 in) long that is pulled in tension. Determine its elongation when a load of 65,250 N $\left(14,500l{b}_{f}\right)$ is applied.

Donald Cheek

The equation for stress is:
$\sigma =\frac{F}{A}$
The area if a cylinder is given by:
$A=\pi {\left(\frac{d}{2}\right)}^{2}$
Which results in:
$\sigma =\frac{F}{\pi {\left(\frac{d}{2}\right)}^{2}}=\frac{65250}{\pi {\left(\frac{3.5×{10}^{-3}}{2}\right)}^{2}}=678MPA$
Refer to Fig 6.22 on page 208 we can see the associated stress is 0.00125. We can find the amount elongated from:
$\mathrm{△}L=ϵ{L}_{0}=0.00125\left(80\right)=0.1mm$

sonSnubsreose6v

This problemasks that we calculate the elongation $\mathrm{△}l$ of a specimen of steel the stress-strain behavior of which is shown in Figure 6.21. First it becomes necessary to compute the stress when a load of 65,250 N is applied using Equation 6.1 as
$\sigma =\frac{F}{{A}_{0}}=\frac{F}{\pi {\left(\frac{{d}_{0}}{2}\right)}^{2}}=\frac{65.250N}{\pi {\left(\frac{8.5×{10}^{-3}m}{2}\right)}^{2}}=1150MPa\left(170.000\psi \right)$
Referring to Figure 6.21, at this stress level we are in the elastic region on the stress-strain curve, which corresponds to a strain of 0.0054. Now, utilization of Equation 6.2 to compute the value of $\mathrm{△}l$
$\mathrm{△}l=ϵ{l}_{0}=\left(0.0054\right)\left(80mm\right)=0.43mm\left(0.017in.\right)$

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