Statics Project Help: Simplifying Distributed Loads & Finding External Forces

In summary, the project involves analyzing a statics problem with a distributed load on top of a bridge. The weight of the members at the nodes should be distributed as 2 equal loads, one at the top and one at the bottom. To find the external forces, one can calculate reactions by summing moments at any support.
  • #1
ftballguy8282
2
0
I am working on a statics project and need a little help on what to do next. A link to the project diagram and description is below:

http://i227.photobucket.com/albums/dd178/ftballguy8282/Statics2.jpg

I've simplified the distributed load on top of the bridge to: 2400 lbs above point A, 4000 lbs above point B, 3200 lbs above points C, D, and E, 4000 lbs above point F, and 2400 lbs above point G. The tip says to distribute the weight of the members at the nodes. I'm not sure what exactly that means. Also, I'm unsure of what I should use to start out with for the weight. I know obviously that the stronger the member the greater the weight, but I'm not sure how to start. Should I just use the 1" diameter pipe for each member and use that as a baraomter to go higher or lower? Once I figure out how to use the draw the weight, what would be the easiest way to find the external forces with so many known forces? Maybe moments about point G? Thanks for any help!
 
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  • #2
Sorry I forgot to add the new one with all the measurements on it. From A to B is 15', B to C, C to D, D to E, and E to F are all 10', and F to G is 15'.
 
  • #3
ftballguy8282 said:
I am working on a statics project and need a little help on what to do next. A link to the project diagram and description is below:

http://i227.photobucket.com/albums/dd178/ftballguy8282/Statics2.jpg

I've simplified the distributed load on top of the bridge to: 2400 lbs above point A, 4000 lbs above point B, 3200 lbs above points C, D, and E, 4000 lbs above point F, and 2400 lbs above point G. The tip says to distribute the weight of the members at the nodes. I'm not sure what exactly that means. Also, I'm unsure of what I should use to start out with for the weight. I know obviously that the stronger the member the greater the weight, but I'm not sure how to start. Should I just use the 1" diameter pipe for each member and use that as a baraomter to go higher or lower? Once I figure out how to use the draw the weight, what would be the easiest way to find the external forces with so many known forces? Maybe moments about point G? Thanks for any help!
Your distribution of the roof loads is very good. The members weight can be distributed as 2 equal loads, one at the top and one at the bottom of each member, but they are so small compared to the roof load that I wouldn't worry about them initially. You can calculate reactions by summing moments at about anyone support, but since the truss is symmetrical in geometry and load, the vertical reactions at each support are equal, and I don't see any primary horizonal reactions at either support.
 

FAQ: Statics Project Help: Simplifying Distributed Loads & Finding External Forces

How do you simplify distributed loads in a statics project?

To simplify distributed loads in a statics project, you can divide the distributed load into smaller, concentrated loads. Then, you can use the principle of superposition to determine the total effect of the distributed load on the structure.

What is an external force in a statics project?

An external force in a statics project is a force that acts on a structure from outside of the system. These forces can include forces from objects that the structure is supporting or forces from the environment, such as wind or water.

How do you find external forces in a statics project?

To find external forces in a statics project, you can use the equations of equilibrium, which state that the sum of all forces and moments acting on a structure must equal zero. By setting up and solving these equations, you can determine the magnitude and direction of the external forces.

What is the difference between a distributed load and a concentrated load?

A distributed load is a force that is spread out over a certain area or length of a structure, while a concentrated load is a single force applied at a specific point. Distributed loads are typically more complex and require additional calculations to determine their effects on a structure, while concentrated loads are simpler to analyze.

Why is it important to simplify distributed loads in a statics project?

Simplifying distributed loads in a statics project is important because it allows for easier analysis of the structure. By breaking down a distributed load into smaller, concentrated loads, the equations of equilibrium can be applied more easily and accurately. Simplifying distributed loads also helps to determine the most critical points of a structure, where the maximum forces and stresses may occur.

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