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Tutor profile: Aaron A.

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Aaron A.
Licensed Mechanical Engineer
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Questions

Subject: Mechanical Engineering

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Question:

A tank holding carbon dioxide at 6,000 psi is made of a steel cylinder with caps welded to the ends. The cylinder's inner dimensions must be 13 inches diameter by 5 feet length to ensure the proper volume. What is the factor of safety if the wall thickness of the vessel is 1.5 inches? Assume the steel's yield strength is 45 ksi. Use thick walled theory and the Distortion Energy Theory.

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Aaron A.
Answer:

Because it is not otherwise stated, it is assumed that the external pressure is atmospheric. Therefore $$p_o=0$$. The following information can be pulled from the problem statement: $$p_i=6,000$$ psi $$r_i=13\times0.5=6.5$$ in $$t=1.5$$ in $$r_o=6.5+1.5=8$$ in Because we know that the caps are welded to the ends of the vessel, we know that we will have tangential stress ($$\sigma_t$$), radial stress $$(\sigma_r)$$, and longitudinal stress ($$\sigma_l$$). $$\sigma_t=\frac{r_i^{2}p_i}{r_o^{2}-r_i^{2}}(1+\frac{r_o^{2}}{r_i^{2}})=\frac{(6.5^{2})(6,000)}{8^{2}-6.5^{2}}(1+\frac{8^{2}}{6.5^{2}})=29,310$$ psi $$\sigma_t=\frac{r_i^{2}p_i}{r_o^{2}-r_i^{2}}(1-\frac{r_o^{2}}{r_i^{2}})=\frac{(6.5^{2})(6,000)}{8^{2}-6.5^{2}}(1-\frac{8^{2}}{6.5^{2}})=-6000$$ psi $$\sigma_l=\frac{r_i^{2}p_i}{r_o^{2}-r_i^{2}}=\frac{(6.5^{2})(6,000)}{8^{2}-6.5^{2}}=11,655$$ psi Since $$\sigma_t, \sigma_r, \sigma_l $$ are principle stresses, we can use these to directly find the von Mises stress. $$\sigma'=\sqrt{\frac{(\sigma_t-\sigma_r)^{2}+(\sigma_r-\sigma_l)^{2}+(\sigma_l-\sigma_t)^{2}}{2}}$$ $$=\sqrt{\frac{(29,310+6000)^{2}+(-6000-11,655)^{2}+(11,655-29,310)^{2}}{2}}$$ $$=30,579$$ psi $$=30.6$$ ksi Using the distortion energy theory, the factor of safety can be found by: $$n=\frac{S_y}{\sigma'}=\frac{45}{30.6}=1.47$$ $$\longleftarrow$$ ANSWER

Subject: Algebra

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Question:

What is the quadratic equation? Use it to solve for solutions of $$3x^{2}+5x-16=0$$

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Aaron A.
Answer:

$$x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ when $$ax^{2}+bx+c=0$$ $$x=\frac{-5\pm\sqrt{5^{2}-4(3)(-16)}}{2(3)}$$ $$x=\frac{-5\pm\sqrt{217}}{6}$$ $$x=1.622$$ & $$x=-3.288$$

Subject: Physics

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Question:

What is Newton's Third Law of Motion? Please give an example of the application of this law.

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Aaron A.
Answer:

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. An example of this is the recoil of a cannon when fired. The forces generated that cause the action of the projectile in the form of forward motion also cause the cannon to react in the form of recoil.

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