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− | We will now investigate two important questions regarding the feasibility of the project, which will help us demonstrate that our structure will work for its intended use. Is our material usable as a building material? | + | We will now investigate two important questions regarding the feasibility of the project, which will help us demonstrate that our structure will work for its intended use. Is our material usable as a building material? And how much further does development need to be focused on structural properties? To answer these questions, we have run structural dynamics simulations with material properties derived from our <a href="https://2018.igem.org/Team:DTU-Denmark/ModellingTheDesign" target="_blank">compressive strength experiments</a>, performed in the labs at DTU Civil Engineering. |
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It is very normal to test a material based on pressure along one axis, which we also did. The question is now how to compare this characterization to a more complex case, where pressure exists on one axis, and shearing (dragging the surface parallel with itself) along another. For this, the so-called von Mises yield criterion (1) was utilized. The basic assumption is that it is not volumetric change that breaks an object, but uneven stress. An example is a balloon; if you put it under water, there is pressure on all sides and it is pushed into a smaller volume but does not fracture. But if you squeeze a balloon between your hands, it will eventually pop. In more technical terms, the cause of this phenomena is the distortion energy which only increases based on differences in stress.<br><br> | It is very normal to test a material based on pressure along one axis, which we also did. The question is now how to compare this characterization to a more complex case, where pressure exists on one axis, and shearing (dragging the surface parallel with itself) along another. For this, the so-called von Mises yield criterion (1) was utilized. The basic assumption is that it is not volumetric change that breaks an object, but uneven stress. An example is a balloon; if you put it under water, there is pressure on all sides and it is pushed into a smaller volume but does not fracture. But if you squeeze a balloon between your hands, it will eventually pop. In more technical terms, the cause of this phenomena is the distortion energy which only increases based on differences in stress.<br><br> | ||
− | A complex solid modeled as an infinite combination of infinitesimally small boxes. Stress only needs to be described on one of the faces in the pair because of the infinitesimally small size and Newtons third law(3). The stress on a face can be split into two types. If you have a paper tissue, and you are either pulling the paper apart or pushing (crumbling), it is one type ($\sigma$), but if you start to shear the paper that will be a different type of stress($\tau$). The same view can be applied in 3D, where the shearing can happen in two directions on each facepair per axis, but the pull and push forces can occur on all three axes. The stress on each of the sides (faces) will be a combination of the forces described above. <br> | + | A complex solid can be modeled as an infinite combination of infinitesimally small boxes. Stress only needs to be described on one of the faces in the pair because of the infinitesimally small size and Newtons third law (3). The stress on a face can be split into two types. If you have a paper tissue, and you are either pulling the paper apart or pushing (crumbling), it is one type ($\sigma$), but if you start to shear the paper, that will be a different type of stress ($\tau$). The same view can be applied in 3D, where the shearing can happen in two directions on each facepair per axis, but the pull and push forces can occur on all three axes. The stress on each of the sides (faces) will be a combination of the forces described above. <br> |
− | The von Mises stress is an attempt | + | The von Mises stress is an attempt to calculate the size of the combined stress differences, analogous to vector length, which in the 2D push/pull scenario reduces to the pull/push stress. This is exactly what we measured in our characterisation tests, that is, how high stress can be before cracking. Since it is an infinitesimally small box, if you shear on one side, the other side in that plane will also be sheared, e.g. shear on the x face in y-direction also gives shear on the y face. Generally these are the shears with the same subscript but different order. This means some of the stresses are identical giving a simpler formula. |
− | The formula for the von Mises stress is (1) (for the notation fig. 1): | + | The formula for the von Mises stress is (1) (for the notation, see fig. 1): |
$$\sigma_{vM}=\sqrt{\frac{(\sigma_{xx}-\sigma_{yy})^2+(\sigma_{yy}-\sigma_{zz})^2+(\sigma_{zz}-\sigma_{xx})^2+6(\tau_{xy}^2-\tau_{yz}^2+\tau_{zx}^2)}{2}}$$ | $$\sigma_{vM}=\sqrt{\frac{(\sigma_{xx}-\sigma_{yy})^2+(\sigma_{yy}-\sigma_{zz})^2+(\sigma_{zz}-\sigma_{xx})^2+6(\tau_{xy}^2-\tau_{yz}^2+\tau_{zx}^2)}{2}}$$ | ||
<p style="text-align:center;"> <img src="https://static.igem.org/mediawiki/2018/d/d6/T--DTU-Denmark--cauchi-stress-tensor.png | <p style="text-align:center;"> <img src="https://static.igem.org/mediawiki/2018/d/d6/T--DTU-Denmark--cauchi-stress-tensor.png | ||
− | " style="max-width: 100%;" > <figcaption><p style="text-align:center; font-size:14px;"><b>fig. 1: </b> The stress on each surface is here called $\sigma$ | + | " style="max-width: 100%;" > <figcaption><p style="text-align:center; font-size:14px;"><b>fig. 1: </b> The stress on each surface is here called $\sigma$. The first index indicates the surface that is affected, and the second indicating direction with respect to surface, pull/push will have identical subscripts and shear is the other. |
</p></figcaption> | </p></figcaption> | ||
</p> | </p> | ||
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− | Using COMSOL®,we can run a simulation and get the von Mises stress out over the entire structure and ensure that it is below the max value measured at DTU Civil Engineering. | + | Using COMSOL®, a type of modeling software combining different physics in one simulation, we can run a simulation and get the von Mises stress out over the entire structure and ensure that it is below the max value measured at DTU Civil Engineering. |
− | The simulation is based on the FEM (Finite Element Method), where you first construct a 3D representation of the structure. Then you split the structure into many smaller elements (see fig. 2), which can theoretically be of any shape. You can then either do a temporal solution or a stationary one. If you want to do a temporal solution you take the governing physics (in our case mechanical physics) and calculate the time derivative from the corresponding equations. Technically | + | The simulation is based on the FEM (Finite Element Method), where you first construct a 3D representation of the structure. Then you split the structure into many smaller elements (see fig. 2), which can theoretically be of any shape. You can then either do a temporal solution or a stationary one. If you want to do a temporal solution you take the governing physics (in our case mechanical physics) and calculate the time derivative from the corresponding equations. Technically discretizing would force us to redefine gradients as the difference between the value at the evaluation element and the values of the neighboring elements, and the same principle applies to other vector operators. Now knowing the temporal change of the physical properties of any element, take a step forward in time, thus changing the values by the just calculated temporal change times the amount of time. This discretization of time and space is not a complete representation and leads to an error, but the smaller the elements and time step, the better the approximation. If you want a stationary solution, you can do the same, but continue until the temporal change is zero, and then the steps do not matter anymore. A stationary solution only depends on the boundary conditions, which are the values at the boundary. For instance, the velocity being zero on a face means that the face is stationary all the time. |
</p> | </p> | ||
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The real structure will be made of the triangular shaped pieces, meaning it will not be a perfect dome, but as a rough approximation, it can be modeled as one. Since the variability of the strength of the bricks is large, it also makes sense to calculate the order of the magnitude of the stress. This includes an assumption that the binding of the bricks is not going to be the weakest point, but the model tells us whether the material itself is strong enough to be used.<br> | The real structure will be made of the triangular shaped pieces, meaning it will not be a perfect dome, but as a rough approximation, it can be modeled as one. Since the variability of the strength of the bricks is large, it also makes sense to calculate the order of the magnitude of the stress. This includes an assumption that the binding of the bricks is not going to be the weakest point, but the model tells us whether the material itself is strong enough to be used.<br> | ||
The structure is modeled as a dome with a certain thickness connected to a base with the same thickness, where all of the structure made of <i>Aspergillus</i> and the material properties are measured with equipment from DTU Civil Engineering. <br><br> | The structure is modeled as a dome with a certain thickness connected to a base with the same thickness, where all of the structure made of <i>Aspergillus</i> and the material properties are measured with equipment from DTU Civil Engineering. <br><br> | ||
− | We have chosen to test the stress caused by the application of more pressure than gravity | + | We have chosen to test the stress caused by the application of more pressure internally than outside, and the gravity of Mars. Since the forces do not depend on time, we can expect the physical behaviour to also not depend on time, and we can look for the stationary solution which is faster.<br><br> |
The boundary conditions at the bottom of the base are fixed to “the ground” (which is not included). The martian pressure of 0.006 bars is set to push on the outside of the structure, and 1 atm is set to push from the inside.<br><br> | The boundary conditions at the bottom of the base are fixed to “the ground” (which is not included). The martian pressure of 0.006 bars is set to push on the outside of the structure, and 1 atm is set to push from the inside.<br><br> | ||
− | The radius is set after having a height of 5 meters at 45 degrees, which is where the sphere starts going down more than outwards. Unless otherwise noted a thickness of 1 meter is used to make it easy to relate other values and so we have a large safety margin. | + | The radius is set after having a height of 5 meters at 45 degrees, which is where the sphere starts going down more than outwards. Unless otherwise noted, a thickness of 1 meter is used to make it easy to relate other values, and so we have a large safety margin. |
The material parameters have been obtained experimentally and the highest load bearing bricks have been chosen ($\sigma_{max}$). | The material parameters have been obtained experimentally and the highest load bearing bricks have been chosen ($\sigma_{max}$). | ||
Unless otherwise noted the parameters in table 1 have been used. | Unless otherwise noted the parameters in table 1 have been used. | ||
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<tr> | <tr> | ||
<td class="tg-0pky">V[m^3]</td> | <td class="tg-0pky">V[m^3]</td> | ||
− | <td class="tg-0pky">$(5 | + | <td class="tg-0pky">$(5\cdot \sqrt{2})^3$</td> |
<td class="tg-0pky">353.6</td> | <td class="tg-0pky">353.6</td> | ||
</tr> | </tr> | ||
<tr> | <tr> | ||
<td class="tg-0lax">R[m]</td> | <td class="tg-0lax">R[m]</td> | ||
− | <td class="tg-0lax">(V | + | <td class="tg-0lax">$(V\cdot 6/(4\cdot \pi))^{1/3}$</td> |
<td class="tg-0lax">5.5</td> | <td class="tg-0lax">5.5</td> | ||
</tr> | </tr> | ||
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<p style="text-align:justified"> | <p style="text-align:justified"> | ||
− | Changing Young’s modulus does not change the maximum stress significantly when looking at the range of Young's modulus our best bricks | + | Changing Young’s modulus does not change the maximum stress significantly when looking at the range of Young's modulus of our best bricks; it changes only how much the walls are displaced. To see how these are related, see fig. 3. |
<p style="text-align:center;"> <img src="https://static.igem.org/mediawiki/2018/2/2d/T--DTU-Denmark--Youngs_modulus.gif" style="max-width: 100%;" > <figcaption><p style="text-align:center; font-size:14px;"><b>fig. 3: </b> Here you see young’s modulus change within a range of our hardest bricks. The arrows at the legend is the minimum and maximum value. The black line indicate the original placement of the walls. The black lines indicate the original dome shape.</p></figcaption> | <p style="text-align:center;"> <img src="https://static.igem.org/mediawiki/2018/2/2d/T--DTU-Denmark--Youngs_modulus.gif" style="max-width: 100%;" > <figcaption><p style="text-align:center; font-size:14px;"><b>fig. 3: </b> Here you see young’s modulus change within a range of our hardest bricks. The arrows at the legend is the minimum and maximum value. The black line indicate the original placement of the walls. The black lines indicate the original dome shape.</p></figcaption> | ||
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− | Increasing the thickness of the shell gives more material to spread the stress over. The largest stress is on the inside. Since fungi are very light and filled with air, they insulate quite well, and we might need rather thick walls for protection from the cold Martian environment. But neglecting this structurally we only need a thickness of 0.34 m. | + | Increasing the thickness of the shell gives more material to spread the stress over. The largest stress is on the inside. Since fungi are very light and filled with air, they insulate quite well, and we might need rather thick walls for protection from the cold Martian environment. But neglecting this structurally, we only need a thickness of 0.34 m. |
</p> | </p> | ||
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− | Since the maximum stress is on the inside it decreases slower as thickness increases since the added thickness gets further and further away from the stress zone. This is shown in fig. 5. | + | Since the maximum stress is on the inside, it decreases slower as thickness increases since the added thickness gets further and further away from the stress zone. This is shown in fig. 5. |
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</p> | </p> | ||
<p style="text-align:justified"> | <p style="text-align:justified"> | ||
− | As <a target="_blank" href="https://2018.igem.org/Team:DTU-Denmark/Design">mentioned</a> 1 atm pressure inside makes it easy for plants to grow and does not alter the operation of vital machinery. But it does require a thicker wall | + | As <a target="_blank" href="https://2018.igem.org/Team:DTU-Denmark/Design">mentioned</a>, 1 atm pressure inside makes it easy for plants to grow and does not alter the operation of vital machinery. But it does require a thicker wall – according to our simulations, approximately double the thickness. Fungi have the advantage of growing in a couple of weeks, so the extra thickness can quickly be grown. But in scenarios where only humans need to be there and resources (e.g. heat) are scarce, a lower internal pressure could be preferable to lessen stress and strain on the structure. See fig. 6 for the stress distribution for these cases. |
Latest revision as of 03:35, 18 October 2018
Structural Integrity
We will now investigate two important questions regarding the feasibility of the project, which will help us demonstrate that our structure will work for its intended use. Is our material usable as a building material? And how much further does development need to be focused on structural properties? To answer these questions, we have run structural dynamics simulations with material properties derived from our compressive strength experiments, performed in the labs at DTU Civil Engineering.
Here you see how varying the thickness (th) of the walls changes the stress in the material. The arrows over and under the legend are minimal and maximal value. For more info see the corresponding section
Theory
It is very normal to test a material based on pressure along one axis, which we also did. The question is now how to compare this characterization to a more complex case, where pressure exists on one axis, and shearing (dragging the surface parallel with itself) along another. For this, the so-called von Mises yield criterion (1) was utilized. The basic assumption is that it is not volumetric change that breaks an object, but uneven stress. An example is a balloon; if you put it under water, there is pressure on all sides and it is pushed into a smaller volume but does not fracture. But if you squeeze a balloon between your hands, it will eventually pop. In more technical terms, the cause of this phenomena is the distortion energy which only increases based on differences in stress.
A complex solid can be modeled as an infinite combination of infinitesimally small boxes. Stress only needs to be described on one of the faces in the pair because of the infinitesimally small size and Newtons third law (3). The stress on a face can be split into two types. If you have a paper tissue, and you are either pulling the paper apart or pushing (crumbling), it is one type ($\sigma$), but if you start to shear the paper, that will be a different type of stress ($\tau$). The same view can be applied in 3D, where the shearing can happen in two directions on each facepair per axis, but the pull and push forces can occur on all three axes. The stress on each of the sides (faces) will be a combination of the forces described above.
The von Mises stress is an attempt to calculate the size of the combined stress differences, analogous to vector length, which in the 2D push/pull scenario reduces to the pull/push stress. This is exactly what we measured in our characterisation tests, that is, how high stress can be before cracking. Since it is an infinitesimally small box, if you shear on one side, the other side in that plane will also be sheared, e.g. shear on the x face in y-direction also gives shear on the y face. Generally these are the shears with the same subscript but different order. This means some of the stresses are identical giving a simpler formula.
The formula for the von Mises stress is (1) (for the notation, see fig. 1):
$$\sigma_{vM}=\sqrt{\frac{(\sigma_{xx}-\sigma_{yy})^2+(\sigma_{yy}-\sigma_{zz})^2+(\sigma_{zz}-\sigma_{xx})^2+6(\tau_{xy}^2-\tau_{yz}^2+\tau_{zx}^2)}{2}}$$
fig. 1: The stress on each surface is here called $\sigma$. The first index indicates the surface that is affected, and the second indicating direction with respect to surface, pull/push will have identical subscripts and shear is the other.
Using COMSOL®, a type of modeling software combining different physics in one simulation, we can run a simulation and get the von Mises stress out over the entire structure and ensure that it is below the max value measured at DTU Civil Engineering. The simulation is based on the FEM (Finite Element Method), where you first construct a 3D representation of the structure. Then you split the structure into many smaller elements (see fig. 2), which can theoretically be of any shape. You can then either do a temporal solution or a stationary one. If you want to do a temporal solution you take the governing physics (in our case mechanical physics) and calculate the time derivative from the corresponding equations. Technically discretizing would force us to redefine gradients as the difference between the value at the evaluation element and the values of the neighboring elements, and the same principle applies to other vector operators. Now knowing the temporal change of the physical properties of any element, take a step forward in time, thus changing the values by the just calculated temporal change times the amount of time. This discretization of time and space is not a complete representation and leads to an error, but the smaller the elements and time step, the better the approximation. If you want a stationary solution, you can do the same, but continue until the temporal change is zero, and then the steps do not matter anymore. A stationary solution only depends on the boundary conditions, which are the values at the boundary. For instance, the velocity being zero on a face means that the face is stationary all the time.
fig. 2: Splitting of structure into small elements
The Case
The real structure will be made of the triangular shaped pieces, meaning it will not be a perfect dome, but as a rough approximation, it can be modeled as one. Since the variability of the strength of the bricks is large, it also makes sense to calculate the order of the magnitude of the stress. This includes an assumption that the binding of the bricks is not going to be the weakest point, but the model tells us whether the material itself is strong enough to be used.
The structure is modeled as a dome with a certain thickness connected to a base with the same thickness, where all of the structure made of Aspergillus and the material properties are measured with equipment from DTU Civil Engineering.
We have chosen to test the stress caused by the application of more pressure internally than outside, and the gravity of Mars. Since the forces do not depend on time, we can expect the physical behaviour to also not depend on time, and we can look for the stationary solution which is faster.
The boundary conditions at the bottom of the base are fixed to “the ground” (which is not included). The martian pressure of 0.006 bars is set to push on the outside of the structure, and 1 atm is set to push from the inside.
The radius is set after having a height of 5 meters at 45 degrees, which is where the sphere starts going down more than outwards. Unless otherwise noted, a thickness of 1 meter is used to make it easy to relate other values, and so we have a large safety margin.
The material parameters have been obtained experimentally and the highest load bearing bricks have been chosen ($\sigma_{max}$).
Unless otherwise noted the parameters in table 1 have been used.
Parameter | calculation | value |
---|---|---|
Thickness[m] | 1 | |
V[m^3] | $(5\cdot \sqrt{2})^3$ | 353.6 |
R[m] | $(V\cdot 6/(4\cdot \pi))^{1/3}$ | 5.5 |
Young[kPa] | 6000 | |
$\sigma_{max}$[kPa] | 1500 | |
Pressure inside [kPa] |
101 | |
Pressure outside[kPa] |
0.6 | |
g[m/s] (2) | 3.7 |
Table 1: These are the parameters used and for derived ones, how they were calculated
Results
Changing Young’s modulus does not change the maximum stress significantly when looking at the range of Young's modulus of our best bricks; it changes only how much the walls are displaced. To see how these are related, see fig. 3.
fig. 3: Here you see young’s modulus change within a range of our hardest bricks. The arrows at the legend is the minimum and maximum value. The black line indicate the original placement of the walls. The black lines indicate the original dome shape.
Increasing the thickness of the shell gives more material to spread the stress over. The largest stress is on the inside. Since fungi are very light and filled with air, they insulate quite well, and we might need rather thick walls for protection from the cold Martian environment. But neglecting this structurally, we only need a thickness of 0.34 m.
fig. 4: Here you see the stress distribution with a thickness resulting in our material stressed a little less than breaking at the most stressed spot. Meaning the maximum stress seen next to the arrow is smaller than $\sigma_{max}$. The black lines indicate the original dome shape
Since the maximum stress is on the inside, it decreases slower as thickness increases since the added thickness gets further and further away from the stress zone. This is shown in fig. 5.
fig. 5: Simulation with varying thickness. Notice how the change of the maximum value written over the legend gets slower and slower. The black lines indicate the original dome shape.
As mentioned, 1 atm pressure inside makes it easy for plants to grow and does not alter the operation of vital machinery. But it does require a thicker wall – according to our simulations, approximately double the thickness. Fungi have the advantage of growing in a couple of weeks, so the extra thickness can quickly be grown. But in scenarios where only humans need to be there and resources (e.g. heat) are scarce, a lower internal pressure could be preferable to lessen stress and strain on the structure. See fig. 6 for the stress distribution for these cases.
fig. 6: Here we see the same simulation, above with 1 atm, and below with 0.0006 atm. The thickness of the shell on a is 1 m and b is 0.5 m. Showing the compromise between material need and pressure inside. Both maxima is around one third of the maximum breaking stress measured in lab, giving a safety margin. The arrows at the legend is the minimum and maximum value. The black lines indicate the original dome shape.
Another way to minimise the stress is by adding weight on top to counteract the outwards force. This could either be just mass eg. regolith or useful mass in the form of solar cells, or a protective layer against micrometeorites. This effect is less pronounced than the others, but also limits the displacement. The effect of this on stress and displacement is shown in fig. 7.
Fig. 7a: (left for mobile screens)
Fig. 7b: (right for mobile screens)
Fig. 7c: (left for mobile screens).
Fig. 7d: (right for mobile screens)
Fig. 7: Left side is without added mass, right is with 2000 kg/m^2. the black lines indicate where the shell started. The number next to the arrows are minima and maxima values.
Future Work
Here the structure has been simulated as a cohesive dome, and it could be very enlightening to also research how a structure with not as strong binding holds together, and how much the not completely spherical structure we have planned to make of bricks results in higher max stress. In addition to this, the wind load caused by the martian winds would also be an interesting addition to secure that the structure is actually able to withstand the conditions.
Results
In conclusion using FEM we have showed that choosing Aspergillus oryzae is strong enough for our use case and a wall thickness of a 3.4 m is enough to ensure structural integrity. This gives us grounds for choosing this over Ganoderma resinaceum which is used by Ecovative, though it can still be used for more demanding parts of a colony. The cold on mars can also be a factor requiring a thicker wall for isolation. Which will also give more secure walls for holding up to winds.
(1) McGinty B. Von Mises Stress. Accessed(17-10-2018) at http://www.continuummechanics.org/vonmisesstress.html
(2) William M. 2016. How Strong is the Gravity on Mars?. Accessed(17-10-2018) at https://www.universetoday.com/14859/gravity-on-mars/
(2) University of Auckland. The Cauchy Stress Tensor. Accessed(17-10-2018) at http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/Chapter_3_Stress_Mass_Momentum/Stress_Balance_Principles_03_The_Cauchy_Stress_Tensor.pdf