Material Science, Material Selection Question

In summary, the conversation revolves around selecting materials for a fuel tank for a rocket that will be used in space tourism. The materials need to have good fracture toughness at low temperatures and meet certain material index requirements. The conversation includes formulas and calculations used to determine the material indices and discuss the issue of the program not yielding the expected results.
  • #1
mathwurkz
41
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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.
[tex]\sigma = \frac{S_f pR}{2t}[/tex], [tex]\epsilon= \frac{\sigma}{E}[/tex], [tex]K_{IC}= C\sigma _f \sqrt{\pi a}[/tex], [tex]U= \frac{1 \sigma}{2 E}[/tex]
For part a), this is how I solved it.
[tex]\sigma = \frac{K_{IC}}{C \sqrt{\pi a)}}[/tex], divide both sides by E., we know, [tex]\frac{\sigma}{E} = \frac{\Delta l}{l}[/tex], so substituting gives me, [tex]\Delta l = \frac{l K_{IC}}{CE \sqrt{\pi a}}[/tex]
all in all...
[tex]\Delta l = \left[ \frac{1}{C \sqrt{\pi a}}\right] \left[l\right]\left[ \frac{K_{IC}}{E}\right][/tex] so...
[tex] M_1 = \frac{K_{IC}}{E}[/tex] to maximize the fracture toughness, under displacement-limited design.
b) [tex]U = \frac{1\sigma ^2}{2E} = \frac{1}{2E} \frac{K_{IC}^2}{C \sqrt{\pi a}}[/tex]
so i get...
[tex]M_2 = \frac{K_{IC}^2}{E}[/tex]
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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  • #2
Don't have CES available on this machine, but somehow for the 2nd index I would've used (intuitively before reading it completely) the integral of the stress-strain curve itself, but your question does state to go with fracture toughness on this as well (and if you've linear-elasticity going there would suggest even more so ... although fracture toughness will be maximized also if you just maximize the area under the stress-strain curve). So would've thought it reasonable to maximize fracture toughness and energy the structure can take without connecting the two.
Probably doesn't change anything but thought I'd mention if you want to try it as a criterion, but don't see anything 'wrong' in the way you've interpreted the question itself. Puzzling :confused:
 
  • #3


it is important to carefully analyze and interpret data to ensure accurate results. In this case, it seems that the material index you have calculated for part a (M1 = K_IC/E) may not be the most appropriate for the given problem. It is possible that using this index alone may not accurately reflect the fracture toughness at low temperatures. It may be helpful to also consider other factors such as the material's ductility and ability to withstand thermal stresses at high altitudes.

Additionally, for part b, the material index you have calculated (M2 = K_IC^2/E) may not accurately reflect the energy-limited failure under fracture at minimum mass. It may be beneficial to consider other factors such as the material's strength and weight to find a more suitable index for this situation.

In order to ensure the best material selection for the rocket fuel tank, it may be helpful to consult with other experts in the field and use multiple material selection tools to compare and validate results. It is also important to carefully review the input data and assumptions used in the material selection program to ensure accurate and reliable results.
 

1. What is material science?

Material science is an interdisciplinary field that combines principles from physics, chemistry, and engineering to study the properties, structure, and behavior of various materials. It involves understanding the relationship between the structure of a material and its resulting properties, and the development of new materials for specific applications.

2. How do scientists select materials for a specific application?

Scientists use a variety of criteria to select materials for specific applications. These may include physical and chemical properties such as strength, durability, and conductivity, as well as factors like cost, availability, and environmental impact. Material selection also involves considering the intended use and potential limitations or challenges the material may face.

3. What are some common types of materials used in material science?

Some common types of materials used in material science include metals, polymers, ceramics, and composites. Each of these materials has unique properties and characteristics that make them suitable for different applications. For example, metals are often used for their strength and conductivity, while polymers are known for their flexibility and versatility.

4. How does material science impact our daily lives?

Material science has a significant impact on our daily lives, as it is involved in the development and production of many of the products and technologies we use. From the clothes we wear to the electronic devices we use, material science plays a crucial role in shaping our modern world. It also helps improve existing products and create new ones that are more efficient, sustainable, and beneficial to society.

5. What are some emerging trends in material science?

Some emerging trends in material science include the development of new sustainable materials, such as bioplastics and carbon-neutral materials, to address environmental concerns. Nanotechnology is also a rapidly growing field in material science, with the potential to create new materials with unique properties and applications. Additionally, the integration of materials with technology, such as smart materials and wearable technology, is an emerging trend that has the potential to revolutionize industries like healthcare and transportation.

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