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mathwurkz
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I have a problem with this homework. Please have a look at my work and see if it checks out.
ROCKET FUEL TANK
A company has asked you to help their business in space tourism. They have designed a rocket that will be powered by nitrous oxide (reacted with rubber), and you are to select materials for the fuel tank (pressure vessel).
a) Considering the flight will be at high altitude (the edge of space), suggest materials that will have a good fracture toughness low temperature. THe M1 index should consider brittle failure under a displacement-limited design. Find materials with M1 > 10.
b) To increase passenger safety and spacecraft performance, consider M2 a second material index for energy-limited failure under fracture at minimum mass. Hint: Recall that toughness is the integral of the stress-strain curve (energy), and find an index that will maximize fracture toughness.
M2 > 0.25 required.
We are given 4 formulas.
\sigma = \frac{S_f pR}{2t}, \epsilon= \frac{\sigma}{E}, K_{IC}= C\sigma _f \sqrt{\pi a}, U= \frac{1 \sigma}{2 E}
For part a), this is how I solved it.
\sigma = \frac{K_{IC}}{C \sqrt{\pi a)}}, divide both sides by E., we know, \frac{\sigma}{E} = \frac{\Delta l}{l}, so substituting gives me, \Delta l = \frac{l K_{IC}}{CE \sqrt{\pi a}}
all in all...
\Delta l = \left[ \frac{1}{C \sqrt{\pi a}}\right] \left[l\right]\left[ \frac{K_{IC}}{E}\right] so...
M_1 = \frac{K_{IC}}{E} to maximize the fracture toughness, under displacement-limited design.
b) U = \frac{1\sigma ^2}{2E} = \frac{1}{2E} \frac{K_{IC}^2}{C \sqrt{\pi a}}
so i get...
M_2 = \frac{K_{IC}^2}{E}
The problem is when I go to the CES materials selection program. The index I have for part a ends up dominating and gives me a small list of materials. The problem is the next part of the problem tells us that parts a and b should have made our selections come down to 2 composites and 2 polymers. But what I end up getting from the program are just polymers. All the composites are below the slope line on the materials selection chart. So at this point, I am thinking that I must have misinterpreted the question wrong, got a wrong material index, or maybe the program is all buggy. If anyone can check what I am doing or even can tell me if my indicies do yield 2 composites and 2 polymers i'd appreciate it. THanks.
ROCKET FUEL TANK
A company has asked you to help their business in space tourism. They have designed a rocket that will be powered by nitrous oxide (reacted with rubber), and you are to select materials for the fuel tank (pressure vessel).
a) Considering the flight will be at high altitude (the edge of space), suggest materials that will have a good fracture toughness low temperature. THe M1 index should consider brittle failure under a displacement-limited design. Find materials with M1 > 10.
b) To increase passenger safety and spacecraft performance, consider M2 a second material index for energy-limited failure under fracture at minimum mass. Hint: Recall that toughness is the integral of the stress-strain curve (energy), and find an index that will maximize fracture toughness.
M2 > 0.25 required.
We are given 4 formulas.
\sigma = \frac{S_f pR}{2t}, \epsilon= \frac{\sigma}{E}, K_{IC}= C\sigma _f \sqrt{\pi a}, U= \frac{1 \sigma}{2 E}
For part a), this is how I solved it.
\sigma = \frac{K_{IC}}{C \sqrt{\pi a)}}, divide both sides by E., we know, \frac{\sigma}{E} = \frac{\Delta l}{l}, so substituting gives me, \Delta l = \frac{l K_{IC}}{CE \sqrt{\pi a}}
all in all...
\Delta l = \left[ \frac{1}{C \sqrt{\pi a}}\right] \left[l\right]\left[ \frac{K_{IC}}{E}\right] so...
M_1 = \frac{K_{IC}}{E} to maximize the fracture toughness, under displacement-limited design.
b) U = \frac{1\sigma ^2}{2E} = \frac{1}{2E} \frac{K_{IC}^2}{C \sqrt{\pi a}}
so i get...
M_2 = \frac{K_{IC}^2}{E}
The problem is when I go to the CES materials selection program. The index I have for part a ends up dominating and gives me a small list of materials. The problem is the next part of the problem tells us that parts a and b should have made our selections come down to 2 composites and 2 polymers. But what I end up getting from the program are just polymers. All the composites are below the slope line on the materials selection chart. So at this point, I am thinking that I must have misinterpreted the question wrong, got a wrong material index, or maybe the program is all buggy. If anyone can check what I am doing or even can tell me if my indicies do yield 2 composites and 2 polymers i'd appreciate it. THanks.
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