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A steel rod 2.0 m long has a cross-sectional area of `0.30cm^(2)`. The rod is now hung by one end from a support structure, and a 550 kg milling machine is hung from the strain, and the elongation of the rod. |
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Answer» `"Stress"=(F_(_|_))/(A)=((500kg)(9.8m//s^(2)))/(3.0xx10^(-5)m^(2))` `=1.8xx10^(8)Pa` `"Strain "=(Deltal)/(l_(0))=("Stress")/(Y)=(1.8xx10^(8)Pa)/(20xx10^(10)Pa)` `=9.0xx10^(-4)` Elongation = `Deltal` `="(strain)"xxI_(0)` `=(9.0xx10^(-4))(2.0m)` `=1.8 mm` |
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