Material Science in Space Missions
The concepts of rocketry were laid out long ago, and over the past 20–30 years, there has been enormous development in rocketry science. We have all read about the sending of rockets and space shuttles into space. Whenever someone tells us about space rockets, we mostly remember the physics applied to them. But apart from physics, we need a whole branch of chemistry like Material Science engineering for designing the rockets. Material Science and engineering play an important role in the selection of materials for rocket building. Without proper material usage, a lot of space disasters can happen, which can cause huge losses to us. Over the past few years, we have made so many advances in this field that can make space projects more efficient. In this post, we will get a glimpse of how Material Science and Metallurgical Engineering play a vital role in space missions.
What is a Structure and what are its types?
Anything that has a specific and regular arrangement followed throughout is said to have a feature called structure. This Structure is further classified as:
- Nuclear structure
- Atomic structure
- Crystal structure
- Micro structure
- Macro structure
Nuclear structures, as defined by their name, arise due to differences in the number of electrons, protons, and neutrons. Both the atomic and nuclear structures do not affect the mechanical properties of the material. Crystal structures (for example, simple cubic (SC), body-centered cubic (BCC), etc.), micro-structures (like grain structures), and macro-structures affect the mechanical properties of the materials.
Mechanical Properties
Some of the most classical examples of mechanical properties possessed by the materials are Tensile strength, compressive strength, ductility, malleability, hardness, fatigue, and toughness. Though the terms sound similar when heard, they are actually different.
In childhood, we learned that ductility is the property of materials to be drawn into wires and malleability is the property of materials to be beaten into sheets. In more engineering fashion, the definition can now be modified as follows: Ductility is the ability of the material to deform without rupturing under tensile stress. Malleability is the ability of the material to deform without tearing apart under compressive stress.
Q. Experiments report that gold has a Youngs Modulus of 80 GPa and steel made of a particular composition has a Youngs Modulus of 200 GPa. Find which metal is most ductile.
Ans: We know that Young's modulus (Y) equals stress or strain. For a given amount of stress, say 10e9 Pa, the gold has a strain of 0.0125 (1/80), stating that when gold of 1 m length is subjected to 10e9 Pa of stress, it extends 1.25 cm. When 1 m of steel is subjected to the same condition, it elongates only 5 mm. So the ability of gold to be drawn into a long wire is greater than that of steel.
The next property is hardness! Not about the hardness of water but about material. Here, we determine the hardness of a material as the measure of the depth of penetration of a substance on application of pressure. So this is our mathematical interpretation of the hardness of a material. On a physical level, we describe it as the resistance of the material to scratching or wearing.
Q. A needle of edge diameter 0.5 mm applies a constant pressure of 150 Pa on two different materials A and B for a period of 100 sec. The needle pierced a distance of 2mm in material A and 5 mm in material B. Compare the hardness of the material.
Ans: Hardness of Material A >Hardness of Material B As hardness is inversely proportional to the depth of penetration.
What Next ? Toughness ! The ability of an object to absorb energy prior to fracture is called Toughness. Mathematically, it is defined as the area between the stress-strain graph, which gives you the energy per unit volume stored in the material. Now try the following question:
Q. Given the stress-strain graph of two hypothetical materials, A and B, compare the toughness levels of A and B qualitatively.
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| graph MS.01 |
Ans: On calculating the approximate area of the graph using the given info, we get the area of the graph under A = (0.5 * 1600 * 0.05) + (0.5 * (0.52-0.05) * 200) + ( (0.52-0.05) * 1600 )
area under A = 839 MJ/m^3
area of graph under B (approx.) = (0.099 * 800 * 0.5) + (800 *(0.5-0.099))
area under B (approx.) = 360.4 MJ/m^3
So we can qualitatively tell that toughness of material A is approximately 2.32 times tougher than B or more approximately 2 times tougher than B. Interesting !!
This number states that Material A can survive an impact of 839 J given over 1 m^3 of the the object. Any further increase in energy for the given volume will fracture or rupture the material.
Finally, we are left with fatigue and creep. Fatigue is the weaking of material due to the repetitive exposure to tensile and compressive stress. Creep is defined as the failure of a material due to subjecting material to constant load usually at high temperature.
It is qualitatively stated that the strength of a material is inversely proportional to the ductility or the toughness of the material. Whenever we say that the toughness of material is too low, the material is so brittle.
Why Titanic Ship (made with qualitied steel of those time) failed, on hitting a ice berg even after sustaining for long time in the ice cold water?
The answer to the question relies on the toughness of the material. Consider the given graph below where material A has a BCC crystal structure and B has a FCC crystal structure.
If we consider at a temperature around 5 degree Celsius the material A made of BCC crystal structure has toughness of 200 J/m^3 where the B has around 150 J/m^3. When the temperature of the material is subjected say -5 degree Celsius, the toughness of material A drastically reduced to order of 40 J/m^3 whereas B has still some toughness of range 100-120 J/m^3, Now you could have guessed the answer for the question. Yes, the titanium ship was made of best quality steel of that time but still it had a very sharp Ductile to Brittle Transition Temperature (DBTT) at nearly 0 C. So when the ship was sailing in the ice cold water it's toughness barrier was very less. So as soon as it crashed with an ice berg, the steel failed and the ship ruptured. If the material B had been used, this would not have happened since there is smooth change between ductile to brittle transition. Even at -5 C, it could retain considerable amount of impact energy.
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| graph MS.02 |
If we consider at a temperature around 5 degree Celsius the material A made of BCC crystal structure has toughness of 200 J/m^3 where the B has around 150 J/m^3. When the temperature of the material is subjected say -5 degree Celsius, the toughness of material A drastically reduced to order of 40 J/m^3 whereas B has still some toughness of range 100-120 J/m^3, Now you could have guessed the answer for the question. Yes, the titanium ship was made of best quality steel of that time but still it had a very sharp Ductile to Brittle Transition Temperature (DBTT) at nearly 0 C. So when the ship was sailing in the ice cold water it's toughness barrier was very less. So as soon as it crashed with an ice berg, the steel failed and the ship ruptured. If the material B had been used, this would not have happened since there is smooth change between ductile to brittle transition. Even at -5 C, it could retain considerable amount of impact energy.
Q. Considering the knowledge of toughness only, which material from the above graph would you prefer for making a rocket that would travel in outer region of space where temperature ranges between 258 K to 290 K. (There are several other factors to be taken care, this question is to make you understand the dependence of toughness on temperature and crystal structure of material)
More on corrosion, metallurgical aspects of Material Science Engineering will be dealt in upcoming posts. Stay curious !
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