A) 1.16mm
B) 2.32mm
C) 0.16mm
D) 1.35mm
Correct Answer: A
Solution :
[a] To find the minimum diameter, and hence minimum cross-sectional area, we assume that the force \[F=400\text{ }N\]brings us to the elastic limit. Then from the stress, \[F/A=379\times {{10}^{6}}Pa\], we get \[A=\frac{400N}{379\times {{10}^{6}}Pa}=1.0554\times {{10}^{-6}}{{m}^{2}}\] Then, \[A=\frac{\pi {{D}^{2}}}{4}\] \[{{D}^{2}}=\frac{4F}{\pi }=\frac{4(1.0554\times {{10}^{-6}}{{m}^{2}})}{\pi }\] \[=1.344\times {{10}^{-6}}\] and\[D=\sqrt{1.344\times {{10}^{-6}}{{m}^{2}}}\] \[=1.16\times {{10}^{-3}}m=1.16mm\]
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