Khufu Boat museum Egypt
Content
AR0133 (2020/21 Q3) Technoledge Stuctural Design Final report
Jens Slagter Shreyas Vadodaria Vasilka Espinosa
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Design
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Components
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Assembly order
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Safety analysis
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Connections
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Sustainability
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Structural verification
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Reflection
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Appendix
4599780 5268613 5383803 2
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Design
During our design concept, some important considerations were written down to give us some direction. We divided them into structural and architectural consideration. Keywords like heat gain, vision to the pyramid, types of glass, sandstorm, sustainability, dry assembly, and minimize the number of different components was mentioned. We had five different designs, each of them focusing on one or more of the keywords. All of them were doable, except for one. That one was an organic vault brick roof, a doable design with regular brick, cement, and steel, but a much more complex project with a load-bearing glass structure.
https://youtu.be/vgyLAKZnRZk (animated video)
Appendix A (a few sketches of our fist designs and brainstorms)
We didn’t know much about how to calculate a catenary arch. We only knew the basics concept, as its ability to withstand the weight of the material from which it is constructed, without collapsing, and that it is a strong arch because the vertical forces of gravity are redirected into compression forces pressing along the arch’s curve. Besides that, we were too concern about the heat gaining of the glass, which made us decide to add a second arch, a bigger one, that would work as a second skin façade, a decision that would bring us many complications in the future.
To comprehend better the catenary arch, a physical model was made to see how the arch works, where the additional shape needed to be attached and to discover a mathematical derivation of such an arch.
Despite the recommendation of not choosing this design due to lack of time and its complexity, we decided to stick with it and simplify to a catenary vault structure made entirely of glass bricks. Until this point, orthomorphic brick was the main load-bearing glass component to be considered, which is one of the most promising brick for dry-assembled interlocking, besides its curved geometry prevent residual stress concentration. We wanted to have a museum that could be entirely disassembled, and its bricks could be reused, trying to keep a circular structural component, that could expand its lifetime by allowing it to be demountable. We wanted as well to minimize the number of different components.
At this point was recommended to us develop our glass brick. We tried different shapes and sizes, seeing what would suit our purpose best. We wanted to use standard brick as much as possible, to reduce the number of different types of blocks. When we started our calculation, it soon becacame clear that the full brick glass vault was way too heavy. To reduce the weight of the structure, the massive vault was replaced by a smaller arch rib. The additional arch (the bigger one) was another challenge. It would have been best to start that bigger curve at the bottom, but the transition period would be very long, resulting in many different blocks with a lot of mass. That is why it was placed on top with a steeper angle. This angle was adapted to reduce structural problem, which will be described better in the Structural verification chapter.
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Design
The final design consists of a structural grid that is divided longitudinally every 3 meters. This results in more arches, but less weight per arch. The bigger arch works as a second skin façade, protecting the South faced from the sun. Furthermore, the space in between the arches works as a foyer, that could be further designed.
In the exposition hall, where the Khufu boat is located, we created a route with ramps that surrounds the boat, giving the visitor a different perspective of the boat while walking in.
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Design In this short section, we can observe how the two arches work together with their height difference. The small arch is 10.30 meters tall, while the second one is almost 15 meters. In the drawing some details are pointed out, representing the keystone brick, the joint between the two arches, the standard connection and the bottom connection, which will be further represented. The arches are the load-bearing structure of this design, which is made of dry-assembled glass bricks. Those bricks have a standard size and in between each one, an interlayer will be placed. For each set of bricks, a wedge is used, allowing the bricks to follow the arch curvature. By standardizing and optimizing the components according to the curvature, the number of different elements is minimized. There is an inner and outer curved glass panel that follows the arch curvature until the top, with an exception of the top part of the smaller arch.
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Design
South View
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Design
The east view shows the translucent insulated tempered glass facade, and the south east view shows the application of ceramic fritted fully tempered hot bend glass on the top of the structure in order to reflect harsh sun light and to maintain indoor thermal comfort.
South East View
East View
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Components
16.15m long; 2.59m wide; 2.89m high
There are many glass plates on the outside of the building. The curvature is between flat and 4 m. However, because the radius enlarges the panels can be stacked next to each other and transported. The east and west facade have some bigger panels that would require special transportation (53 foot high cube container, see image on left).
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Components
At the bottom we have 4 alternative designs for the “splitting brick (nr 4)”. The original design is made out of cast glass. Alternative A is based upon hollow-core slab floor as pieces of the inside core are removed to increase cooldown surface and reduce mass. Alternative B is based upon making smaller subsection areas since the subsection of the normal brick is overdimensioned. C is a titanium joint. Placed over the top of two regular bricks. It has the same thermal expansion coefficient and can therefore be used. D is made up out of natural stone. There is no annealing time, but it weighs more than the glass. The other glass bricks (section 1 - 3) are used as keystone, regular stacking brick or base brick. The section on top (5 - 17) describes the various wedges used throughout our design. Not every wedge is connected to an outside panel, not every wedge has a glass panel connected to the side and not every wedge is just used to tilt the arc. This results in many wedges and caps. Section 18 - 28 shows the additional smaller items used within the building. 19 is an important one since it is placed in between every brick and wedge. 18 is used within the key-stone wedge (to be able to slide it in). 22 24 are clamps for the east and west facade(could be replaced by strip running along the outside of the facade). Number 25 is the base that the rib is placed in. 26 is the U profile to allow for thermal expansion of the whole structure. 27 and 28 are similar to 25 and 26, but only for the east and west facade.
For the exact quantities and materials see Appendix B
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Assembly order
South view
Stage 1 : Concrete Foundation and Plinth Construction
Stage 2 : Erecting scaffolding for base panel and erecting catenary form scaffolding for catenary arch ribs
Stage 3 : Erecting scaffolding for the panels above, which will support the catenary form
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Assembly order
Stage 4 : Starting to assemble the base of the catenary arch and connecting two ribs with panels simultaneously
Stage 5 : The process of constructing the catenary ribs and connecting glass panels
Stage 7 : Later step by step couple of arches are constructed to finish the whole building.
Stage 6 : Finally first two arch are completed
Stage 8 : At last east and west facade will be constructed after sliding in the boat.
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Safety analysis
The risk analysis involves eight steps, that are: identifying the hazards; what will be the consequence and the damage to the constructive element; the probability of happening; how long the structural elements would be exposed; how serious will be the consequence at complete failure; the risk value; protective measure to mitigate the risk; and the final risk value after protective measures being taken, when necessary. Our analyses consider the different condition of structural failure and as well, risk related to the use of glass in a hot desert climate. Since glass is normally not recommended in hot climates due to its overheating, and extra precaution was taken about it. Heatwave and high indoor temperature were considered risk related to the user of the space, but not to the structure. The RD values for both are high, in the heatwave, this is thanks to the high probability of happening and its consequence, while in the high indoor temperature the exposure has a higher value.
culation is made taking into account the protection measure, showing that we could decrease the probability or the exposure or the consequence of a certain scenario. The RD values lower than 70, were not necessary to be taken as a solution, either because there was not a solution for the problem, like a terrorist attack, or because the RD values were too small to be considered a risk. Even having values lower than 70, for the sake of the exercise of finding protective measures, we described some possible solution for it.
In 2015, one of the worst heat waves in Egypt happened, killing 61 people of heat stroke in three days, and another 581 people had been admitted to the hospital with heat exhaustion. Therefore, the heat was one of our concerns during this project, being considered since the design concept. That is why we choose to add a second arch, which would work as a second skin façade on the south elevation. Besides that, we design a double-glazing façade on the north and south by inserting curved glass panels inside and outside of the arch structure. Structurally speaking the temperature of the desert would not impact the glass elements, since Borosilicate has a maximum service temperature between 200°C and 400°C, and the curved glass panels, which is a hot bent curved float glass, have it maximum service temperature between 110°C and 460°C (FROM GRANTA EDUPACK). The risk scenario with RD value higher than 70 were the ones related to dropping the glass element during the assembly process, the heatwave, and the high indoor temperature, somehow proving that our speculation about the heat impact could have in the building was plausible. In general, all RD values higher than 70 have a protection measure described apart. Consequently, a new RD cal-
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Safety analysis
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Design of connections
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Design of connections
The detail here shows the key brick of the arches. And the exploded view shows the assembly of wedges and key brick.
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Design of connections
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Design of connections The detail shown here is the important part of the structure, where two arches meets. The key connection plays vital role structurally. And the titanium wedges makes sure that the branching of arche takes place at definite angle.
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Design of connections
The connection shown here is, joinery between exterior east and west facade to the titanium wedge of the key connection brick.
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Design of connections
The connection shown here is the connection of arches in intermediate ribs, where the exterior and interior glass panels are secured by titanium wedges.
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Design of connections
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Design of connections
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Design of connections
The connection of first glass brick laid on the base of concrete is very important to set an angle of the arches. The connection has SS ‘U” profile with interlayer of rubber bearing for thermal expansion movement along the length of the structure.
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Design of connections
The connection showing the joinery at the bottom of the east and west facade, where insulated glass panel in placed into SS “U” profile with flexible rubber gasket for the movement due to thermal expansion of the structure.
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Design of connections
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Sustainability Reduce glass material needed Dry assembly ( Polyurethane PU interlayer) Less types of components Addition of 2nd skin Natural ventilation via earth duct Ventilated cavity to reduce thermal heat gain
Wind Catcher
Cavity between glass panels
Earthducts
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Structural verification
5) Normal force in base R = normal force in base [kN] = 58 [kN] Rv = vertical force in base [kN] = 56 [kN] Θ = (tan)^(-1) (sin(L/2a)) = 1.33 [rad]
Preliminary hand calculation The base design started off as a perfect catenary arc. The first step was to define the geometry and to describe it mathematically. This way we are able to know directions and length. This section of the calculation is based upon Gohnert and Bradley (2020). It shows our first calculation with a rectangular brick vault (no arc).
1)
General catenary curve formula
2)
In order to calculate the stresses we only have to divide this by the area it is working on and we have σn of 5,52 [MPa]. This is far within compression strength of borosilicate (268 MPa). The highest moment is at the top.
catenary curve for specific height
H = height of arc [m] L = length of base [m] a = scale factor
The minimal width of the structure is determined to be 17,1 [m]. Proportions of the curve are roughly based upon the proportions of the pyramid. The pyramid is 220 [m] x 139 [m] that gives us a +/- 10,5 [m]. The length of the catenary arc is then the derivative of the catenary function (1).
6) Maximum moment Mtop = 352 [kNm]
To check the bending stresses at the top we need to include the second moment of area (in this case still a brick). Then calculate the stress around that is working around the N.A. (see figure below) 7) Second moment of area b = width of brick [m] = 0,05 m h = height of brick [m] = 0,21 m A = surface area [m^2] = 0,01 m^2 d = distance to side [m] = 0,025 m Izz = second moment of area [m^4] = 3,86E-5
3) Length of catenary arc S = length of arc [m] = 34.06 m L = length of base [m] = 17,5 m a = scale factor 4.11
8) stresses due to bending My = maximum moment around y axis [kNm] = 352 kNm h = height of section [m] = 0,025 m Izz = second moment of area [m^4] = 3,86E-5 m^4 σ = stress [ kN/m^2] = 105.280,2 kN/m^2 = 105 [MPa}
Now that we know the length of the full curve, we can calculate the weight and thus the vertical reaction forces at the base of the arc. 4) weight of the curve W = weight [kg/] = 11546,3 kg = 113,3 kN Ɣ = specific weight [kg/m^3] = 2260 kg/m^3 S = length of arc [m] = 34,06 m b = width of arc [m] = 0,05 m h = thickness of arc[m] = 3 m
The preliminary calculation show that this arc is able to deal with the vertical force at the bottom, but does not withstand the moment in the top section (this does not include normal force at the top, this is equal to the horizontal force in the base). There is tension due to bending. The values are higher than the yield strength of borosilicate 26.8 [MPa] (even without safety factors). This is only discovered at the final stage of the process when the calculation was correct
So that means that the vertical reaction force equals 56,63 [kN]
(Bending Moment Diagram, n.d.).
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The full weight of the curve can be summarized into a point load at a distance. When applying this to the 2nd skin facade we find the Fv of the pointload. We know the angle because it is still a catenary shape produced by karamba. After measurement this angle appeared to be 60 °.
Structural verification Structural considerations and evolution
Izz (formula 7) = 0,0025 + 0,041 x 0,004^2 Izz = 0,00106 [m^4] (see appendix H, final hand calculation of Izz )
As we could see the structure will fail when the forces are to high we needed a lighter structure. This is why the full brick vault has been changed to vault-ribs made out of borosilicate glass bricks. It is for the calculation important to know what kind of brick we are using. In the initial calculation, a rectangular brick of 0,21 x 0,05 x 0,1 [m] is initially selected. However due to sustainability and assembly reasons we changed the brick design (see image on left).
Forces in junction (see appendix E): Fv (panels) = 5,72 [kN] Fv (bricks) = 6,46 [kN] 9) Reaction force in B e = eccentricity of point load [m] Fpoint = normal force in junction L = width of total arc [m]
The new construction is made up out of a glass brick vault rib and one 3 meter wide glass panels (2 x 3 [m] x 7,5 [kg/m^2]) on either side of it. This results in a new weight of the structure (formula 4). Reaction force at bottom = 11 [kN]. A quick check shows that compression stresses at the bottom are within range, but still the bending moment in the top are around 51,91 [MPa]. (See appendix D & E for more detailed preliminary calculation.) We discovered how to calculate the bending moment at the very end of our proces, due to the time it took to understand the shape and architectural considerations and the overlooked bending moment check in the first calculations. For architectural considerations, we wanted a 2nd skin facade over the primary catenary arc (see image on right). The place where it joins is very important as a high point load is hitting the arc. The best would be to start at the base, but that requires a big transition period to separate the two arches. We went for a joint somewhere around ⅓rd of the length. In a later stage this is defined. The angle of inclination decides how much horizontal force is transferred into the primary arc. For the simplicity of the calculation we summarise the 2nd skin into a point load on an angle (Θ). The next part of the calculation will go about the 2nd skin We need to calculate the length of the 2nd skin. Some parts of this calculation are based upon dimensions of a rhino model. Since we want to maintain the relationship with the pyramid and in order to walk outside the primary arc, but still inside the 2nd skin, we enlarge the width by 5,6 [m]. This gives us a height of 14,69 [m]. Now we have the width and height. With some help of karamba and kangaroo, we determined the length to be 12.95 [m] and area centroid is located at 15,7 : 10,2 [m] (this has been done by rhino, because taking the primitive of a hyperbolic cosine function is outside of our knowledge).
F
10) eccentricity of point load e = eccentricity of point load [m] = 0,52 [m] Θ = angle of inclination [°] = 60 degrees x = horizontal distance to junction [m] = 6.31 [m] y = vertical distance to junction [m] = 9,87 [m] Rvb = 8,6 [kN] Rva = 10,24 [kN]
The point load is already causing a lot of deformation of the primary arc. The best thing to do is minimize the effect it is having on the arc. This meaning reducing the eccentricity of the point load. This means the perfect catenary shape needs to be transformed into a parabolic shape in order to adjust the angle of inclination. To do so we went through a few steps:
27 1) find the correct angle
2) Draw arc that follows angle
By drawing a line from junction to bottom.
The 2nd skin should start according to angle.
3) find centroid and calculate reaction forces See image on top of page.
4) Calculate vertical and horizontal components
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Define load combinations: (we used sand as a load in the first draft, but this has been eliminated when we discovered that the load calculation became more complex, see appendix W note 5)
Structural verification
The calculations took 5 iteration before the final (at least compression strength) number was trusted to be right. We did however foresee that the stresses are going to be a problem. The bending moment in the structure is hard to determine. To keep the project going we made a few alternative brick designs in order to increase the surface area, height, width and/or the relationship between (see appendix F & G).
The compressive stress test at the base is conducted first. Since there is no bending moment, the stresses are pure compression. Therefore the check will be based upon the compressive strength divided by a safety factor of 4. **see note 6 of appendix H
Final calculation The dimensions are final (see appendix K). The properties of the glass bricks and glass plates are final and brick number 2 of appendix F is chosen as a check for the final calculation (lot of surface, relative low weight). To see full calculation and steps see appendix H until O. Checking exact location of maximal bending moment (located in beam AI):
First calculating the weight of the second skin:
Formula 11) see appendix H
Calculate weight of primary arc (focus on left bottom): Checking maximum bending moment in AI: The first value was wrongfully calculate. There was a value of -0.7 kNm while the value on top was 6.36. When simulating this in matrix frame there was a big discrepancy. That is why in the last week a re-do of the moment calculation was performed (for schematic diagram, see appendix H & I): Formula 12) see appendix I
Combining all forces and additional effects (focus on left bottom):
Calculating reaction forces of live loads (See appendix I & J)
** see note 3 appendix H
This number seems trustworthy, however when performing the same calculation on top (also done twice) the first calculation provided a bending moment of 6.35 kNm and the second one of 24.46 kNm. Here the 2nd calculation is a lot less trustworthy. (see appendix H & O). An alternative method is provided, however this is beyond the scope of the course.
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Structural verification
Load case 1
1.2 x dead weight + 1.5 x (maintenance)
Digital (matrix frame) model (see appendix P → U for all models)
Compression check bottom: N/A 492 [kN/m^2] of the 268 Mpa /4 = 67000 kN/m^2
1.01
Bending check: (My * ½ z) / Izz Bending moment max=-6.58 kNm 10043.5 [kN/m^2]
4.34
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-8.14 -11.72
27.6 -6.58 28.2
4.34
2.63
Maximum tensile stress: 9551.49 [kN/m^2] this is within the parameters of the yield strength, but outside when you include the safety factor of 4 (143% of allowed).
2.35
40.1 -20.2
Bending moment [kNm]
Deformation [mm]
-11.25
Normal forces [kN]
-16.24
Load case 2 1.2 x dead weight + 1.5 x (pressure + suction + maintenance)
Compression check bottom: N/A 431 [kN/m^2] of the 268 Mpa /4 =67000 kN/m^2
-3.66
-5.29 66.7 84.1 -2.62 -5.79
-13.45 11.31
-6.01 -6.49
5.52
112
128.5
Bending check: (My * ½ z) / Izz Bending moment max= 15.72 kNm 27796 [kN/m^2]
-7.75
15.72
-6.65
-17.72 -12.36
Bending moment [kNm]
Deformation [mm]
Maximum tensile stress: 27364 [kN/m^2] this is barely outside the parameters of the yield strength (by 2 %), but outside when you include the safety factor of 4 (408% of allowed).
-3.55
Normal forces [kN]
Load case 3 1.2 x dead weight + 1.5 x (pressure + suction)
-5.83 58 -5.99 -13.20 80.1
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Compression check bottom: N/A 419 [kN/m^2] of the 268 Mpa /4 =67000 kN/m^2
-2.08 -7.49
10.92
Bending check: (My * ½ z) / Izz Bending moment max= 16.6 kNm 29352 [kN/m^2]
4.93 125
16.6
-11.73 -5.54
Deformation [mm]
Bending moment [kNm]
-17.2
Normal forces [kN]
Maximum tensile stress: 29 28932 [kN/m^2] this is outside the yield strength parameters (by 8% and 432% outside incl. safety.)
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Structural verification
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Structural verification
Fins outer facade global dimensioning The maximum height that the glass panels span is 14 [m]. This gives us fins of 10 to 12 [cm]. In order to quickly check if these are usable, a buckling test with Euler’s formula was advised. Formula 12) Eurler’s formula E = Youngs modulus = 69.000 GPa I = second moment of area L = length = 14 [m] n = ‘buckling number’ Fcr = critical axial force However, these fins are used to make sure that the big glass panels won’t bend to much and break due to wind. This means the fins will not resist a lot carry axial load of the brick structure, more transversal load to prevent bending. So instead we will check the dimensions according to the maximum allowable deformation of 14.04[m]/250 = 56.16 [mm]. If the construction is hinged on 2 places (see appendix V for more information): Formula 13) deflection of fixed and roller hinge: q= wind pressure = 3 [kN/m] L = Length of panel = 14.04 [m] E = young’s modulus = 6.9E+07 [kN/m^2] I = second moment of area [m^4] W = deformation = 0.05615 [m] This means when we rewrite the equation we can derive the specific height related to a certain width. When we take 6 [mm] thick panels we need at least 92.1 [cm] long fins. (30 [cm] if fixed bottom is allowed (see appendix V). However the fins require to much space to be implemented between the last arc and the outer facade. The tolerance is 12 mm (excluding thickness of 2 glass panels). This means the fins need to be on another location (outside) or a stiffer material needs to be selected.
Lateral stability The panels on the outside provide the lateral stability. The wind is then directed down alongside the length of the building. The panels are fixed together, this is allowed due to the fact that all materials have the same thermal expansion coëfficient. When the building expands, it does so in every direction. Because of the length of the building the expansion could pose as a problem. A system of roller and hinge supports are provided to allow for expansion
Further advise and solutions. The structure fails when the variable loads are applied and therefore some answers to solve this are provided. The building can be ‘dig in’. This will greatly reduce the outside forces on the building. When we lower it the moment that is generated by the wind will be higher and the total force lower. Another option is to choose a bigger brick. This will probably not solve the high tensions. An increased area does help a lot, but will also increase its own weight and annealing time. More vault-ribs. This could be a very possible solution but it requires more bricks and wedges. The bending moments generated by wind are greatly reduced. This will result in lower maximal stresses. The necessary size of fins on the east and west facade cannot be implement into our current design. They can not be placed on the outside of the building since this will generate tensile stress in the finns instead of compression. A solution could be to place the fins right through the 2 layers of glass (part inside, part outside). Another solution could be a different material like steel (bends less easy) or aluminium (able to withstand tension). Another solution for the fins can be that we connect the east and west facade panels to the glass panels that provide lateral stability. This requires the panels on the interior to also be able to transfer some loads. Then the very long glass panels are not just only connected by the top and bottom, but also have a support somewhere around the middle. This reduces the deformation and therefore reduces the fin-size. For critical notes about structural numbers and values, see notes at end of appendix.
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Reflection
Jens: The project went the way I hoped it would go. I think we did something different and really learned something about modern techniques, problem-solving, digit group work in Covid-19 times and a lot about glass structures. I have a small background in civil engineering. That is why the part of the course that drew my interest, was about combining new structural materials into a very complex geometry. It was a stressful but very informative journey to understand and mathematically derive all the forces and stresses in a structure by hand. I learned a lot about how to go about a project. However the workload was enormous due to the fact that the structure was so complex and new to us. It is hard to understand everything inside the geometry. This resulted in a clear and unfortunately unbreakable hierarchy in the groups in terms of work division. I stumbled a lot of times when I went through the calculations and I could only work in Excel since there are so many angles. This caused my overview to be blurred from time to time. Fortunate enough my roommates also took structural mechanics at various faculties. By sparring with them about how to derive the moments of the full structure, I was still able to understand the structure. The group work during this course was tough. The absence of physical meetings resulted in long explanations and discussions through ZOOM and very often we misunderstood things. Resulting in wrongful calculations, drawing but most of all we lost a lot of time re-doing things In conclusion, I really enjoyed the lectures and the consultations and I am thrilled about the subject of glass.
Shreyas The course of Structural Design was intriguing to me, as the brief we choose was to design a glass structure in harsh climate of Egypt. Having said that, to design a structure beside Pyramid was the best context ever. Having an architecture background, it was interesting challenge for me to take cast glass brick as a structural material to design a local traditional form of catenary arch. Though structurally the geometry was quite challenging as it was completely new for me and for the team as well. Ultimately we all took the challenge to resolve the structure as much as we can under the available time and circumstances of online education. Personally it was quite challenging to grasp all the structural calculation for the form we were designing, and there were tough times to get one calculation for the geometry. At the end it was a huge learning experience, and that learning for me will continue after this course, as the field of structure design has lots to offer all together. It would strengthen my skill sets for sure. There were time when we as a team struggled a bit to communicate with each other due to distant learning and lack of time spent in person with each other. And it was tough to manage all the courses completely online, which created imbalance in devotion of time per course for me. Inclusive of all ups and downs along the journey, the learning from the course was really great and thanks to all mentors for making the journey so smooth as it was, and for pushing us to stretch our limits and perform at the best than ever.
Vasilka When I chose this course, I was a bit reluctant. Sincerely it was not a course that was my first choice. However, after the first lectures, the challenges were cheering me up. It was very interesting how the previous courses of Bucky Lab Engineering (Material Science and Structural Mechanics) were being put into practice. I enjoyed it and liked to focus mainly on the structure, I could see from a different perspective and understand better the structure, the material, and even the architectural decision. One of the thing that I regret is not taking much responsibility for the calculations. Online communication (and education) it’s hard, especially when you don’t know much about your peers. We manage some obstacles, but others remain, creating perhaps some negative atmosphere within the group. I hope that in the future we can put some of the water under the bridge and maybe grab a beer or two, and enjoy a time together personally. Some suggestion for the next Structural Design course, not to put a submission in a non-commercials hour, have the final presentation in week 9 and the submission report in week 10. That way we can at least rest during the weekend. Starting tomorrow some of us are having a week full of lectures from 9h to 17h, and it doesn't seem fear to the next professor either, because some of us will be already burned out. I had a break down during this quarter and still, I’m not recovered. I stopped working for one week, but I forced myself to continue working because I didn’t want my group to have to assume all of the workloads. Besides the difficulties, it was a very enriching course, the lectures instigating, giving me food for thoughts.
References: bending moment diagram. (n.d.). [Illustration]. http://www.abuildersengineer.com/2014/06/beams-section-modulus.html Gohnert, M., & Bradley, R. (2020). Catenary solutions for arches and vaults. Journal of Architectural Engineering, 26(2), 04020006.
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Appendix
A
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Appendix
B
Types of material, their function and quantities
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Appendix
C
Location of catenary arc brick rotation points and length and location of glass panels
** note 2
(see end of appendix)
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Appendix
D
Deformation and moment check preliminary calculation (only primary arc)
Matrix frame simulation of primary arc deformation. No secondary arc added.
Matrix frame simulation of primary arc moments. No secondary arc added. The panels are simulated as point loads on their respective wedge.
Hand calculation of primary arc deformation due to normal force of its own dead weight. The colored sections show where the plates are connected to the structure. The weight of the panels are added and the normal component inside each brick section is calculated. This gives the change in length, initially intended to be used to create one big ‘williot diagram’, this turned out to be to much work for the course, but the individual change in length is calculated. The forces of the primary arc are not integrated inside this deformation check due to the complexity of the distribution of load.
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Appendix
E
First calculation forces of 2nd skin
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Appendix
F
Brick design alternatives
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Appendix
G
Brick design alternatives
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Appendix
H
Properties of used materials and geometry
** note 6 (see end of appendix) ** note 4 (see end of appendix)
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Appendix
I
Load cases 60%
Maintenance Load
40%
In order to understand the distribution of load that is descending from the 2nd skin we had to make a small model in matrix frame. We assume that the force distribution is the same for all load cases. We identified that from all the incoming force, 40% is directed towards the left base and 60% to the right imposition. The load cases are based upon this.
Vertical point load on top of construction (maintenance): F vertical joint = 0.4 x 0.5 x maintenance load (assuming equal distribution on top section) F vertical in left bottom = 0.2 [kN] This calculation would be more trustworthy when the sum of moments is used instead of adding up the downward forces. However, this way there is understanding of the changes of forces and their directions.
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Windload pressure
Appendix Load cases
J ** note 5 (see end of appendix)
Wind force workin on top part: Fwh top = wind force x grid width x height Windforce = 1 kN/m^2 Grid width = 3 [m] Height = 4,78 [m] (from junction to top) Fwh top = 14.34 [kN] Wind force in base left: 0.4 x Fwh top x height to junction x windforce x grid width Additional windforce horizontal in left base = 35.50 [kN] The normal forces are calculated by taking the tangent ( ) of the angle related to the position of the segment. ** note that this is still not the correct distribution. The calculation is done by adding up the wind working on the structure to discover the normal forces in the brick section. The calculation could be improved by doing the sum of moments around the base points.
Windload suction (roof) Wind force workin on top part: Fwv top = wind force x grid width x working width Windforce = 1 kN/m^2 Grid width = 3 [m] width= 9.4 [m] (from junction horizontal position of centroid) Fwv top = 1.32 [kN] Wind force in base left: 0.4 x Fwv top x width to junction x windforce x grid width Additional windforce vertical in left base = 3.84 [kN] The normal forces are calculated by taking the tangent ( ) of the angle related to the position of the segment. ** note that this is still not the correct distribution. The calculation is done by adding up the wind working on the structure to discover the normal forces in the brick section. The calculation could be improved by doing the sum of moments around the base points.
41
Appendix
K
** note 1 (see end of appendix)
42
Appendix
L
** note 3 (see end of appendix)
See live load section
** note 4 (see end of appendix)
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Appendix
M
This calculation is in order to locate where the highest bending moment will occur. The perpendicular distance from the curve to the to the extension of the normal force is calculated. When cross referenced with the list of segment dimensions the bending moment is located in beam AI.
The distance is calculated as follows: A = angle at bottom [rad] C = angle of junction [rad] B = A - C [rad] h1 = perpendicular distance to first segment = sin (B) * length segment [m] D = angle of next segment = arctan (∆h/∆x) [rad] E=B F=B-A+D h2 = sin (F) * segment length + h1
M = F x distance [kNm] “Huilie” is defined as a moment rotating counter clockwise. The ‘smiley’ is defined clockwise. At this point in time it was hard to keep track what forces needed to be included 2 calculations were made in order to check the bending moment. The distance the point load is working on is calculated by :
∆H = height difference joint & top ∆X = Horizontal distance from joint to top
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Appendix
N The bending moment due to inclined forces are calculated by multiplying the normal force times the eccentricity. However, it is safer to divide the force into its horizontal and vertical components. Then multiplying by the difference in horizontal and vertical position. This is done in the 2nd calculation (shown below). The result is trustworthier.
** note 3 (see end of appendix)
After the calculation is redone, we derive to a more trustworthy value for the maximum bending moment. However, because the values differ so much a digital model is produced in matrix frame. By comparing both results we concluded that the maximum occurring bending moment is close to 6 kNm. The value for the top bending moment is missing some forces that need to be included. The hand calculation is becoming less trustworthy. That is why we transferred to a digital model after calculating the stresses. However one more way of calculating the final bending moment is sketched. This includes superposition of the separate forces. This way the deformation can be calculated as well. All forces from wind to maintenance to the point load (from 2nd skin) are then looked at individually and the calculation is done for every part. (see image on left). 1) Only own weight of bricks 2) Only own weight of panels 3) Point load 2nd skin dead weight 4) Wind pressure on primary 5) Point load 2nd skin wind pressure 6) Wind suction on primary 7) Point load 2nd skin wind suction 8) Point load maintenance 2nd skin
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Appendix
O
Hand calculation of bending moment load cases (as told before, not very trustworthy, for final values look at matrix frame model)
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Appendix
P
Forces due to dead weight
-4.49
0.07 -7.6
-0.42
-2.288
-10.64
4.02
-3.5
0.77
-15.61 -1.64
Reaction forces [kN]
-19.09
-0.58
-10.76
Normal forces [kN]
Shear forces [kN]
1.8 32.5
18.3 36.2
5.72 27.9
Deformation [mm]
4.13
0.38
3.75
2.97 2.75
Bending moment [kNm]
47
Appendix
Q
Forces due to wind pressure
2.70
3.27
2.43
-5.67
-2.14
-7.16
8.74
6.67 0.43
2.85
Reaction forces [kN]
-6.91
-0.65
Normal forces [kN]
Shear forces [kN] 2 -9.61
1
12.93 3.32
16.55
4 7
Deformation [mm]
21.31
Bending moment [kNm]
48
Appendix
R
Forces due to wind suction
-0.29
2.81 0.49
2.55
-0.99
4.53
4.82 0.32
Reaction forces [kN]
-0.14
0.47
-0.17
Normal forces [kN]
Shear forces [kN]
-0.51 -2.14
3.2
-1.62
2.2 3.9
0.94
-0.66
1.2
Deformation [mm]
Bending moment [kNm]
49
Appendix
S
Load combination 1
-0.44 4.15
-8.14 -11.72
1.16
-20.2
-1.59 -0.48
Reaction forces [kN]
-11.25
-0.17
Normal forces [kN]
Shear forces [kN]
-16.24
1.01 4.34
24 27.6
40.1
-6.58 28.2
Deformation [mm]
4.34
2.63
2.35
Bending moment [kNm]
50
Appendix
T
Load combination 2 -3.66
2.85
-2.62
-7.75
-6.01 -6.49 -2.31 -5.82
5.48
-6.65
2.23
-12.36
Normal forces [kN]
Shear forces [kN]
Reaction forces [kN]
-17.72
-5.29 66.7 -5.79 84.1 112 128.5
Deformation [mm]
-13.45
11.31 5.52
15.72
Bending moment [kNm]
51
Appendix
U
Load combination 3 -3.55
2.87
1.1
-2.08 -7.49
-2.89 -5.9
5.71
-11.73 -5.54
2.13
Normal forces [kN]
Shear forces [kN]
Reaction forces [kN]
-17.2
-5.83 58 -5.99 80.1
-13.20 10.92
110
4.93 125
Deformation [mm]
16.6
Bending moment [kNm]
52
Appendix
V
Deflection check of fins east or west facade
Loadcase due to wind pressure. We consider a wind pressure of 1 kN and our grid size is 3 [m]. This give the wind =3 kn/m The length of the system is dimensioned on the longest panel → 14.04 [m] and the young’s modulus of structural glass is 69 GPa or 6.9E+07 {kN/m^2] When we decide we want glass fins we can solve the deflection equation according to a different second moment of area. Iyy = 5 x q x l^4 / 384 x E x W
Formula 13) deflection of fixed and roller hinge: q= wind pressure = 3 [kN/m] L = Length of panel = 14.04 [m] E = young’s modulus = 6.9E+07 [kN/m^2] I = second moment of area [m^4] W = deformation = 0.05615 [m]
If we do apply the same strategy as before, but now with a hinge system (U-profile at the bottom is fixed). Iyy necessary = 0.000156679532 [m^4] This results in a working height of 315 [mm]
Iyy = 1/12 x b x h^3 We use 6 [mm] glass that mean h = (60 x q x l^4 / 384 x E x W x b) ^⅓ h = 0.921 [m]
This has to be checked if it is possible. For the course there is no more time, but the fixed connection results in a momentum. This generates tension within the construction element. (M x ½ h / Iyy). This should be below the yield strength (26.6) in order to be safely used.
53
Appendix
W
Critical structural notes: ** notes 1 Take a look at the weight division within the deadweight of the primary arc. It needs to be as close to symetrical as possible. Since the glass panels also weight something. The panels in the interior do not need 3 layers and therefore do not weight the same. We can suggest a different material that weighs more in order to maintain equal load distribution and an more even displacement. This can also benefit the transparency from entrance to the boat area. 2
the keystone and the part before do not have lateral stability due to the absence of the windows (they are the colors that are highlighted. Lateral stability needs to be solved with different connection system in this part, otherwise the bricks will shift. This can be done with steel thin crosses.
3 The exact centroid of the structure shifts as soon as the second arc is added. This results in a non-symmetrical distribution of its own weight. This is not taken into account in this calculation. To do so, there should be a division of the structure as a whole. then the area * the mass * distance of a fixed point should be calculated for each division of the structure. Then when dividing this number by the total area, you get the distance (either in x or y direction) from the fixed point to the centroid. When the centroid is known the calculation can be more precise. The prognosis is that more force will shift towards the left bottom of the structure.
when computing with rhinoceros the centremark shifts with a value of x= -0,73 [m] this can be used to recalculate the dead weight reaction forces at the bottom
4 the compressive strength test can be used in the bottom. The moment is 0 thus the tensile stresses due to bending moment will be 0 as well. The compressive strength can be used because there are only normal forces going into the foundation. Because we are unsure about the strength of the connection between the titanium wedge and the borosilicate element we enlarge that factor to 4. This means we use 268/4= 67 [MPa] 5 The loadcases for the Egypte site are less complex as the ones in Iceland (we already had a lot of trouble). The absence of snow load is evident, but it was surprising to see that sandload was not one of the factors. The specific weight of sand is relatively high. It is not uncommon that sand falls out of the air when no wind is blown in specific areas. Therefore it should be included in the loadcases. when sand reaches and average depth of 1 cm it is already heavier than the distributed maintenance load hen sand is blown into a corner, depths higher of 1 cm can easily be reached. I
6 The yield strength is the exact strength a material yields. In order to include safety we use different factor. The safety value of yield strength normally is found by dividing it by 3. However, because we are unsure about the strength of the connection between the titanium wedge and the borosilicate element we enlarge that factor to 4. This means we use 26.8/4= 6.7 [MPa]. We also include this safety factor for compressive strength check.
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