# Truss Calculations

by Paco_Mexicano
Tags: calculations, truss
 P: 3 I mainly am having trouble accurately calculating the force on members in a truss. The reason being is sort of common, my teacher skimmed through the lecture on these calculations, and left the entire class in a fog. I've made my best attempts at researching on the Internet, but so far no real luck. Sure, there a methods on solving the equations, but I would like to know HOW you do it. http://cabot.k12.ar.us/Schools/chs/F...lculations.htm The link leads to our exact worksheet containing the problems we need to solve. Now, in regards to the very first truss equation for the top left pin-joint, my teach wrote down: (point A= top left pin-joint / point B=top right pin-joint / point C= bottom left pin-joint) Sigma Force X= 0 -Force AB= 0 Sigma Force Y= 0 -60+ Force AC= 0 Force AC= 60 lbs. I hope you guys understood what I'm trying to say, as it's not easy to write most of this. But what I need to know is, HOW those equations work and where my teach is getting them from. I know how the free body diagrams work, but how do they tie in with the equations? I found out Trigonometry is also used to determine lengths of the members along with angles, but how do they tie in as well? As you can see, to most of you this is a snap, but to me, it is extremely difficult being which I have very few resources at my disposal. I can see that the first problem is explained can be fairly easy, but I do not have any idea how to do the rest of the problems, like #2,3, and 4. Now, I'm not asking you do these problems for me, but please explain to me how I do it. You guys seem like very smart people, and I'd be obliged to any assistance at all.
 P: 33 The equations of statics tie into your free body diagram by mathmatically describing the the situation dipicted in the diagram. Beause you are dealing with elements that are in a "static" state, this means that they are in equlibrium. This means that all the forces in the x-direction should have a total sum to zero. Same goes for the y-direction. This also applies to moments occuring on the body about a certain point. A more mathematical representation would be: $$\Sigma F_x = 0$$ $$\Sigma F_y = 0$$ $$\Sigma M = 0$$ You would use these in conjuction with methods known as "Method of Sections" and "Method of Points". What your instructor did in his analysis was that he took point "A" and drew it's free body diagram. In the diagram you would have the applied force of 60 lbs in the negative y-direction, the force in the member between points "A" and "B" acting in the positive x-direction, and the force in the member between point "A" and "C" acting in the positive y-direction. I am assuming a standard coordinate system with y is positive in the up direction and x is positive going to the right. With this free body diagram he applied the two force equlibrium conditions to find the two unknown forces in the diagram. Because there are only 2 unknowns you only need 2 equations to find your unknowns.
 P: 3 For as little as I know in this problem, I do know what you are saying. I understand that the forces must zero out, otherwise you would have a net force, in which the object, or point, would move. And we know, that shouldn't happen in a bridge. I think I may understand your answer, as it seems obvious now that if point A has 60 lbs. pressing down on it, then member "AC" should have 60 lbs. of compression, because it is directly below the input force, or load force I shoud say. But let's take a look at problem #2, which points labeled A,B, and C along the top three points, and D, and E along the bottom, all in a left-to-right fashion. A B C D E Like that. How would I get the calculations to find out what the force of member "AD" is, with point "A" having 500 lbs. of load force pressing down on it? First, I can see having to solve for angles, and with the triangle being 2x2, its obvious that the angles will be 45. Now, is there a certain formula I have to follow now in order to get the answer to the member force? You know, like the angle formula would've been: Tan -1 (Angle) = 2/2 Is there a special formula I need to figure out the member force? If so, then what is the relation to the formula (like a certain pin joint force would been needed along with certain angles or whatever)? I greatly appreciate your help, as this problem is starting to actually clear up and make sense, but not fully yet.
HW Helper
PF Gold
P: 5,920

## Truss Calculations

 Quote by Paco_Mexicano For as little as I know in this problem, I do know what you are saying. I understand that the forces must zero out, otherwise you would have a net force, in which the object, or point, would move. And we know, that shouldn't happen in a bridge. I think I may understand your answer, as it seems obvious now that if point A has 60 lbs. pressing down on it, then member "AC" should have 60 lbs. of compression, because it is directly below the input force, or load force I shoud say. But let's take a look at problem #2, which points labeled A,B, and C along the top three points, and D, and E along the bottom, all in a left-to-right fashion. A B C D E Like that. How would I get the calculations to find out what the force of member "AD" is, with point "A" having 500 lbs. of load force pressing down on it? First, I can see having to solve for angles, and with the triangle being 2x2, its obvious that the angles will be 45. Now, is there a certain formula I have to follow now in order to get the answer to the member force? You know, like the angle formula would've been: Tan -1 (Angle) = 2/2 Is there a special formula I need to figure out the member force? If so, then what is the relation to the formula (like a certain pin joint force would been needed along with certain angles or whatever)? I greatly appreciate your help, as this problem is starting to actually clear up and make sense, but not fully yet.
It is always a good idea to first calculate the reaction forces at the roller or pin support points before proceeding with the joint analyses. And always remember that the resultant forces in a truss member are axial along the member axis.
For problem 2, at joint A, isolate the joint and examine the forces. You have 500 pounds applied down, an unknown force in member AB applied horizontally (x direction only), and an unknown force in member AD along its axis (the force in AD has x and y components, Fsintheta and Fcostheta, which in this case are equal due to the 45 degree angle). Now do your sum of forces in y direction: The 500 pound force applied force is resisted by the y-component of the AD force, which must be 500 pounds up. That means the x component of the AD force is 500 pounds horizontal left. This 500 pounds left must be balanced by a 500 pound right force acting in member AB. So AB has a force of 500 pounds tension, and AD has a compression force of 500(root 2) = 707 pounds. Proceed to the other joints using the simplest ones first.
 P: 3 Well, thank you for your help. I can understand that this isn't the easiest thing to learn, but hopefully I can fully understand it one day. Thanks again for your help, and I'll turn here for help if I ever get stuck again. Adios

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