|Feb27-10, 11:46 AM||#1|
building a cardboard bridge
I am building a cardboard bridge that is supposed to span 7-9 feet and made of cardboard. I am having trouble however designing the abutments. The bridge will simply be placed on a table (no suspension designs btw) and will have a maximum 10% grade.
What will be the best way to design the edges of the bridge?
|Feb27-10, 04:21 PM||#2|
The design of abutments depends on the support reactions.
These, in turn, depend upon the type of bridge chosen.
Some more information about your ideas would help.
|Feb27-10, 04:28 PM||#3|
Here is a diagram of my concern.
Is this something I should worry about? How do I adequately prevent that part from folding in?
|Feb27-10, 06:16 PM||#4|
building a cardboard bridge
Is it supposed to be free standing on the table or to span between two tables or what?
Have you considered the horizontal and vertical reactions at the supports?
I think they have given you a clue here as you state the max grade is 10%
A simply supported beam has only vertical reactions.
No horizontal abutment reactions or moments.
An arch has no abutment moments but does have both horizontal and vertical abutment reactions.
How are you to fabricate this bridge?
Is it to be made from sheets of cardboard, cut to shape?
Glue? Pin joints?
I know you said you cannot use a suspension design, but can you use string to prestress the beam?
If you want to use a truss, and the above fabrication methods are available I would suggest cutting the cardboard into rectangles and rolling these into longish cylinders and gluing the joins. Also change the truss design so the compression members are much shorter and/or fatter than the tension ones, otherwise the cardboard will buckle.
We can look again at the abutment question when you clarify what the bridge is to stand on, as your current idea will not work well.
|Feb27-10, 08:02 PM||#5|
First, it is to span between two tables.
Secondly, I have not considered the reactions at the supports. I am most likely not imagining the flow of pressure throughout the system. Now, the arch abutments in West Point Bridge Designer 2010 are vertical, but provide horizontal reactions?
Here is a reference picture:
However, seeing as how I will simple place the bridge onto the two tables, like so,
How should I design the abutments (highlighted in blue),
I don't understand if this would suffice for 160lbs.
This bridge will be made from cardboard, and can be joined by any means possible. I believe the beam can be prestressed, but to be honest, I have no idea what you mean by using a string to prestress the beam.
Lastly, thank you for the idea. To roll up the cardboard, assuming the length of the card board is how long the member should be, how much would you recommend the width /the diameter of the member?
|Feb28-10, 08:59 AM||#6|
Is this load to be a point load in the middle or evenly distributed over the deck?
An indication of what you are studying and at what level, would certainly help.
Do you understand what the term 'abutment reactions' means?
You have correctly identified one weakness of your underslung design. The truss needs to continue beyond the abutment line.
Of course you have produced a planar drawing. To support 160 lbs worth of loading you will need more than one truss, may be several. Depending upon how the load is distributed you will need cross members to transfer the load evenly between the supporting trusses.
Incidentally for the purposes here a truss is a skeletal form of a beam.
I would not advise pursuing an arch. Arches are compression structures and sheet cardboard is weak in compression.
Most concrete bridges with curver undersides (soffits) are actually curved beams, not arches.
|Feb28-10, 11:40 AM||#7|
Also, this is a high school engineering class, we have gone over reactions (each joint has to evened out at 0N, and each member has to "cancel out" the force of another member), I am possible confused as to why a standard abutment does not provide a horizontal reaction.
Thanks for confirming my problem about the weakness of an underslung design, I started the thread to try and solve this problem. I thought about continuing the truss under the table, but I could not think of a way to attach it under the table, and still have the bridge be "portable." Also, since the span of the bridge will be 3 1/2 feet long, I thought it would be okay if I had 2 trusses, but I haven't gotten to crossing the loads, my textbook did not explain it.
The arch approach was mainly an aesthetic one, but I can see how problems could arise especially towards the middle. On that note, do you have any suggested readings that can may be explain the designs of different bridges vs. the requirements of their members?
|Feb28-10, 12:23 PM||#8|
Firstly you do not need to reproduce my entire post in quotes every time. Just any points or emphasis you wish to discuss.
OK I get the picture about where you are at.
This is a considerable task for high school. It will be a real feather if and when you achieve it.
You did say at the outset that the span was to be 7 - 9 feet, now you say 3.5 feet???
I will post some sketches shortly to help matters. In order to understand what is going on you will need to understand
Resolution of forces at a point
Moments of a force about a point
Friction Force as equal to [tex]\mu [/tex] times the normal reaction
Please confirm you understand these?
We will do what is commonly known as a line beam analysis in a simple form. I do not intend to try to calculate individual member forces - that is beyond high school - just to do enough simple mechanics to get a handle on things.
Do you realise the difference between a point load and a distributed load? A teacher in the middle of the bridge constitutes a near point load. A real difficulty will be creating a walkway that a human can stand on without falling through, whether the bridge spans anything or not.
|Feb28-10, 01:17 PM||#9|
Sorry for requoting the whole post, I wasn't aware I saw doing so.
I don't think the tasks will be incredibly difficult considering we made a paper bridge that held 200 times its own weight (the difference was were given the blueprints.) As for the specs of the bridge I meant to say it is going to 7 feet long and 3.5 feet wide.
I understand the three points you posted, and once I am able to finally grasp "why it works", I will be able to calculate individual member forces (I don't think this is beyond high school, and also provides a benchmark for how sturdy we should make members).
|Feb28-10, 04:44 PM||#10|
I've been thinking a bit more about the issue.
7 ft x 3ft 6ish is about the size of a door.
Modern doors are not solid but of cellular construction, very often with a honeycomb like cardboard core and two very thin skins of veneer to form the faces.
I think that such a door, laid across two tables, would support a teacher who had had too much Jack Daniels.
So how does it work?
Well I've attached some sketches. Ignoring the weight of the plank.
Has the door or plank laid so it is spanning across two tables. In the centre is a 160 lb load.
In order to maintain vertical equilibrium the support reactions at A and B must together equal this load.
By taking moments we can show that they must be equal and half the 160. If the 160 load where placed elsewhere we could use moments to calculate the reactions.
There are no horizontal forces acting.
This condition is know as 'simply supported'
This shows what really happens, but exaggerated. The plank deflects as shown. The top surface becomes shorter and is in compression. The bottom surface is stretched in tension.
The reactions are still the same as the loads have not changed.
Suppose we cut out a section of plank, immediately to the right of the load. This forms what is known as a free body diagram of the left hand part of the plank.
Now if we look at the vertical equilibrium of this isolated piece of plank we see that the missing piece must have been exerting an upward force on the left hand chunk of plank as the only other two vertical forces acting are the load and the reaction at A.
This force must act upwards and I have shown it with a half head arrow.
Similarly there must be a downward force acting on the remaining length of plank at the right hand end.
This time I am showing the missing bit of plank as a free body diagram.
Since it was exerting an upward force on the left hand piece in Fig3, the left hand piece must be exerting a downward force on it (action and reaction). And the same on the other side.
I have shown these on the sketch.
Since these vertical forces are equal an opposite they form an anti clockwise couple.
Now comes the clever bit.
We know that this chunk of plank is in equilibrium.
Therefore there must be another couple acting on it, opposite to the vertical forces in Fig 4.
This is provided by a pair of equal and opposite horizontal forces shown in the sketch.
These correspond to the compression and tension noted in Fig 2 and are the method by which the structure (plank) distributes the loads imposed upon it.
I will not prove, just state that most of the tension is transmitted near the bottom surface and most of the compression near the top.
This is why the cellular door type construction would work, so long as you could laminate sufficient layers of cardboard to be equivalent to the veneer skin of the door. This would also provide sufficient surface strength for the teacher to walk on.
No doubt you will have questions.
|Feb28-10, 06:56 PM||#11|
Okay, took a while to get that all down, amazingly I understood most of it.
1.) How would I calculate, or how would I figure out an optimal distance of how much of the bridge should be touching the table?
2.) Seeing how I am just building a large corrugated beam, could I add my "useless" arch design?
3.) To build the cell walls, I would assume I would still have to roll the cardboard up to get an adequate compression member.
Thanks for all you help.
|Mar1-10, 06:25 PM||#12|
I'm glad page1 made some sense, we can move on to page2. Sorry I don't have time for pretty.
If you go on to study any subject involving mechanics of materials you will become very familiar with the type of analysis I have been showing.
I have done it in a simplified form to avoid discussing internal stresses, bending moments and other more advanced stuff.
If you can, get hold of a copy the small book
Introducing Structures by A J Francis is a wonderfully comprehensible introduction. He is one of the few Architects who really understand structures and can put it across in an easy way.
Anyway to continue. You haven't indicated what sort of cardboard is available - you will perhaps need to discuss with your teacher. I am assuming you are going with the cellular 'door' approach. (Corrugated will not do see Fig 8.)
Shows a section of the construction with laminated skins several sheets thick and cardboard tube cells, probably simlar to the centres of toilet rolls.
Shows a cutaway drawing of the arrangement. The stack of tubes is much stronger for this application than the typical corrugated cardboard construction which looks like Fig 8 (b)
Shows alternative spacing arrangements can you see why (a) representing the cells is stronger than (b) ? A good point to discuss with your teacher or in your report.
This answers your question about abutments / bearing areas. Since someone will walk across the walkway and could stand anywhere, the space he occupies has to be able to support him, on one foot as he walks. This space is just under 1ft of beam and is labelled A.
Since one end reaction is equal to the full 160 lbs when the man is standing at that end the reaction will need a similar area to bear upwards on. So the bearing length will be about 1 ft.
Here we go back to the analysis in Figs 3 nd 4 with some numbers.
The couple (clockwise this time) is formed by the 80 lb reaction and the (160 - 80) forces acting wherever the man is standing. The magnitude of this couple is therefore 80d where d is the distance from the support.
This reaches a maximum when the man is halfway.
This couple or moment is countered by the horizontal forces in the skins. Now this couple has a substantially shorter lever arm I have labelled e. Since the Tension = Compression the couple = Ce or Te.
Thus Ce = 80d.
C = 80d/e
Thus the magnification factor if d is 42 inches and e is 2 inches is 21 times or 1680 lbs.
Now you can see why engineers like beams to be as deep as practicable.
You can get an idea of the tensile strength of your cardboard laminations by making a strip and measuring its breaking strength.
The cellular construction should be enough to stabilise the compression skin against buckling which is the reason why thin cardboard has difficulty with compression.
You should discuss this with your teacher. Remember the 1680 lbs (or whatever) is the total force. The stress will depend upon the width of the walkway as stress = Force/area.
|Mar1-10, 07:33 PM||#13|
Please correct me if I am wrong, but for this cellular door application here is that I would mean I would create many "toilet paper rolls" and stack them vertically?
Or b.) I am just stacking non corrugated cardboard on top of another to create one large beam?
|Mar2-10, 02:51 AM||#14|
A standard door is around inch and a quarter to inch and a half thick.
Since the veneer skins will be stronger than your cardboard I suggested looking at a figure of around 2 inches for your beam.
To be quite clear.
The main forces are in the skins. A single sheet of cardboard will be inadequate so I suggest gluing several together like plywood. Simply stacking them would not be enough they would not act as one.
I suggested some experiments to determine how many layers, sheets or laminations or plies your would need.
The purpose of the arrary of tubes is to keep the skins apart. The further you can separate them the less the skins are stressed. You will need to stand the tubes up like a single layer of tins of beans and glue them to the skins and to each other.
They obviously have to transmit the vertical force from the top to the bottom.
Did you look at the question I posed for Fig 8? The cellular construction is also important as it prevents the whole structure collapsing due to both the horizontal and vertical forces acting.
Take your time and work it through. Also talk it through with your teacher. You can learn a lot from this exercise.