How does physics apply to wooden truss bridges?
106060
Many physics priciples apply.\nYou got your tensil strength, this is your pulling streingth <- ->\nthen, your shear strenght _> <-(left one is lower)like holding a tooth pick between your thumb and index finger and pushing on the middle with your other hand\nand compretion strength -> <- \nthese factors determin wood's ability to withstand load. normaly, bridge trusses use triangles because it focuses on tensile and compretion strenght(whigh wood does well with)\nenginering sites are the best place to look up on these things
862327
it depends...most arch bridges are made of steel and have rigid connections.  Plates are bolted outside of the steel strcuture or the steel members are simply welded together.  Now with better fabrication and transporation technology arches can be one piece.
37625
I used superglue to build a minature bridge out of toothpick for a class.  It weighed the least amout of all the bridges in the class and it held the brick (which was the assignment).  If you don't care about the weight other glues will work, but the will absorb into the wood more and make your bridge heavier.  Also, work on an old surface that it is okay to get glue on.  I had to use a razor blade to get it off my desk.  Using my desk helped make it straighter and stronger.  Good luck.
856591
You left out a LOT of info, like\n1) Loaded as column or beam?\n2) If beam, concentrated or distributed load?\n3) What kind of wood?\n4) Answer based on engineering equations or building code?
1097253
i'm not an engineer but i would say that there are more than one methods. i think you have to make each carrier autonomic,which means it can react as an indepenent part of the bridge in forces like wind or earthquake. i don't know if i helped, i hope i'm not beyond the subject of your question? if i'm sorry
237010
Try buying a copy, try a local structural design firm, try a local engineering school, try a local library.\n\nIf you know enough to use the load factors, you should know enough to find them.
254925
I'm assuming that the pivot point will land between the 15 and 30 Kg loads because it seems more logical. If you try putting the X distance between the 20 and 15 loads, you'd probably get a negative distance. \n\nActually, the weights have to be multiplied by the pull of gravity (9.81m/ss) to make them actual forces but then again, they would all just cancel out.\n\ni hope you understand the crude diagram\n\n|-------10------|       - distance where weight of beam is applied\n|---------x-------|    - pivot point dist from left\n ____________________  - beam (20 m)\n|_________|__________|\n20_______15_________30 - applied forces (weights)\n\n20kg(X) + 15kg(X-10m)  = 30kg(20m-X)\n20kgX + 15kgX - 150kgm = 600kgm - 30kgX\n20kgX + 15kgX + 30kgX  = 600kgm + 150kgm\n                65kgX  = 750kgm\n                    X  = 750kgm/65kg\n                    X  = 11.53846 m from left end
944537
If you idealize the motion without any resistance and an idealized spring representing the bungee cord, then the potential energy lost in the drop of the student from the bridge to the river is\nm*g*y(0)\n\nThe energy then stored in the bungee at the moment the velocity of the student is zero, is:\n.5*k*(y(0)-y(e))^2\n\nthis includes the earth, the student, and the bungee cord, and the relative vertical distance from the bridge to the river.\n\nThe other factors such as the moon and the sun will have negligible effect on the student's motion.\n\nAir resistance could be a factor depending on the distance from the bridge to the river.  Since air resistance creates a drag force, there will be energy loss due to this force which will reduce the extension of the bungee cord resulting in a hand in the water versus a face plunk.  F=d*v^2\nThe terminal velocity of a skydiver presenting maximal area to the fall is about 40 m/s, which requires more than 800 feet of free fall.  Since the maximum velocity occurs in the early stages of deccelartion due to the force of the bungee, air resistance is unlikely to be a major factor.  The highest bridge in the world is Royal Gorge, Colorado U.S.A. , which is 950 feet from road surface to the gorge floor.  If the bored college student forgets to attach the cord, they will plunge into the ground at about 95 miles per hour.\n\nHeating the bungee is unlikely to be a factor since the cord is relatively light and suseptible to melting.  The amount of heat that would be measurable as a loss in potential energy would likely cause spontaneous combustion in the cord.\n\nj
337551
It depends.  Wood floats on water, as does ice.  Rocks sink.
896679
A mature tree trunk may support a greater force than a fine steel wire but intuitively we feel that steel is stronger than wood. To predict the forces which a structure of any size, made of any material, can sustain without failing, we need the concept of mechanical stress.\n\nThe definition of stress originates in two observations of the behaviour of a one-dimensional body subject to uniaxial loading, for example, a steel wire under uniaxial tension:\n\n   1. When the wire is pulled, it stretches. Up to a certain load, the stretch ratio (current length / initial length) is proportional to the load divided by the cross-sectional area of the wire. We therefore define the stress as σ = F/A.\n   2. Failure occurs when the load exceeds a critical value for the material, the tensile strength multiplied by the cross-sectional area of the wire, Fc = σt A.\n\nThese observations suggest that the quantity that affects the deformation and failure of materials is stress.\n\nIn the more general setting of continuum mechanics, stress is a measure of the internal distribution of force per unit area that balances and reacts to the external loads or boundary conditions applied to a body. Stress is an example of a type of quantity called a second-order tensor. In three dimensions, a second-order tensor can be represented by a 3×3 square matrix (containing nine components). However, in the absence of body moments, the stress tensor is symmetric and can be fully specified by six components. These six components are sometimes written in Voigt notation as a 6×1 column matrix which is often, misleadingly, called the stress vector.
16047
In Christmas with the Krank's, there is a scene where a guy is setting up a snowman on the roof. The snowman tips over, colliding with the man inelastically (momentum), sliding down the roof (inclined plane/force), and together they fall hard on the ground (energy). I think that should work; I did this project last year as a junior. Good luck with Hinsdale Central's infamous cinema physics project!
905049
First assume that we neglecting any losses from air resistance or internal friction.\n\nWith that simplification, imagine two states:\n1) you are on the bridge 100m above the water and at rest\n2) you are suspended by the spring at its point of maximum extension, and therefore with zero velocity\n\nSince we assumed that there were no losses, the energy of both of these states is equal. Where is that energy located?\n -->1: your gravitational potential energy (no kinetic energy, no stored elastic energy)\n--> 2: your remaining gravitational potential energy, and the stored elastic energy in the spring (by definition, no kinetic energy)\n\nSo what has happened is that the spring has stored elastic potential energy that is equal to your loss of gravitiational potential energy. That's the only way the energies of the two states are the same.\n\nNow you can set up some equations. Let's say that you start at height H0 and end up at height H1.\n\nchange in PEg= m * g * (H1-H0)\nchange in PEk = 1/2 * k * [(H1-H0)-30m]^2\nwhere we have accounted for the 30m long unstretched length of the spring.\n\nNow by observing that -PEg = PEk, and plugging in the given values for m and k and the known value for g, you can readily solve for the change in height (H1-H0) (it should be < 0) It's a quadratic, so you expect to find one answer that is not physically realizable - reject that one, and select the other one.\n\nDoes that help?
237576
Structural engineering is the science and art of designing and  making, with economy and elegance, buildings, bridges, frameworks and other similar structures so that they can safely resist the forces to which they may be subjected.
547508
No. \nThe system is static and the sum of the forces equal zero, but there definately are forces flowing through the bar.  Replace the bar with a piece of plastic and then what happens.  The system is no longer in equilbruim, becomes dynamic and the plastic gets torn to shreads.  The same would be true if say you used a piece of wire.  Just because something is not moving does not mean there are no forces flowing through it.
110767
Yes, a pretty minimal effort at writing a question, but for what it's worth:\nUse cross bracing (making triangles) to make structure stronger.\nUse an adequate glue that really wets the wood, and lap or splint joints not simple butt joints.\nGet more strength from each stick by having as many as possible edge-down not face-down.
848001
Look up Vector Mechanics for Engineers and Scientists.  This deals with basic Newtonian physics as applied to static structures.\n\nBone up on chemistry.  It is easy to forget your chemistry, and as far as I know you will be required to take a few semesters of basic Chem, even if you are not planning on being a Chem Eng.\n\nDo not lose your Maths either!  Grab a pre-calc or a calculus book and keep practicing.
472579
Depends on the structural design.  Shear will take place near to supports.  Load pressures will take place all the way down to the foundations (getting larger all the way.\n\nSome structures will tend to push out (like a pitched roof) and some will push down (like a flat roof). \n\nIf you need anything more specific drop me a note.