- The smallest diameter which should be used
- The corresponding stress due to the load
With the given allowed deformation, we can draw out the diameter of the rod from its area:
The stress uses the same formula:
Tuesday, 19 March 2019
EXAMPLE 2.5.2. LEAST DIAMETER TO COUNTER DEFORMATION
A 2.5m long steel rod must not stretch more than 1.2mm when a 9kN load is applied to it. Knowing that E=200 GPa, determine the following:
=\frac{9kN\left&space;(\frac{1000N}{1kN}&space;\right&space;)2.50m}{\frac{\pi&space;}{4}d^{2}200x10^{9}\frac{N}{m^{2}}} "1.20mm\left ( \frac{1m}{1000mm} \right )=\frac{9kN\left (\frac{1000N}{1kN} \right )2.50m}{\frac{\pi }{4}d^{2}200x10^{9}\frac{N}{m^{2}}}")
The stress uses the same formula:
} "\sigma =\frac{P}{A}=\frac{9000N}{\frac{\pi }{4}\left (0.01093^{2} \right )}")
The stress uses the same formula:
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