Answer:
a) 0.00996 m
b) 109090909 Pa
Explanation:
Unit conversions:
[tex]E = 200GPa = 200\times10^9 Pa[/tex]
1.2 mm = 0.0012 m
8.5 kN = 8500 N
If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is
[tex]\epsilon = \frac{\Delta L}{L} = \frac{0.0012}{2.2} = 0.000545455[/tex]
With elastic modulus being E = 200 GPa, then its maximum stress must be
[tex]\sigma = E\epsilon = 200\times10^9*0.000545455 = 109090909 Pa[/tex]
Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area
[tex]A = \frac{F}{\sigma} = \frac{8500}{109090909} = 7.79\times10^{-5}m^2[/tex]
And its corresponding diameter is
[tex]A = \pi d^2/4[/tex]
[tex]7.79\times10^{-5} = \pi d^2/4[/tex]
[tex]d^2 = \frac{4*7.79\times10^{-5}}{\pi} = 9.92\times10^{-5}[/tex]
[tex]d = \sqrt{9.92\times10^{-5}} = 0.00996 m \approx 1 cm [/tex]
(a) The rod must have a minimum diameter of 1 cm
(b) The normal stress in the rod corresponding to the smallest diameter is [tex]10.9\times 10^{7}\,Pa[/tex].
Given that the Elastic modulus is,
(b) [tex]E= 200\,GPa= 200\times 10^9 \,Pa= 2\times 10^{11} \,Pa[/tex]
We know that the longitudinal strain is given by,
[tex]\epsilon =\frac{\Delta L}{L} =\frac{1.2\times 10^{-3}m}{2.2\,m} =5.45 \times 10^{-4}[/tex]
We know that elastic modulus is given as the normal stress per strain;
[tex]E = \frac{\sigma}{\epsilon}[/tex]
So, the normal stress is;
[tex]\implies \sigma = E \,\epsilon= (2\times 10^{11}\,Pa) \times (5.45\times 10^{-4})= 10.9 \times 10^7\,Pa[/tex]
(a) We know that the stress is force per unit area.
[tex]\sigma = \frac{F}{A}[/tex]
[tex]\implies A = \frac{F}{\sigma} = \frac{8500\,N}{10.9 \times 10^{7}} =7.79\times 10^{-5}\,m^2[/tex]
But the area of cross-section is;
[tex]A= \pi r^2 =7.79\times 10^{-5}\,m^2[/tex]
[tex]\implies \pi r^2 =\sqrt{\frac{7.79\times 10^{-5}\,m^2}{\pi} } = 0.0049\,m[/tex]
So diameter, [tex]d = 2r = 2\times 0.0049=0.0099\,m\approx 1\,cm[/tex]
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