# Statics: bad solution to simple truss problem regarding determinancy

1. Nov 4, 2009

### pnd910

1. The problem statement, all variables and given/known data

The attached solution was given as an answer to a simple truss problem in my sophomore level Statics class. We were to solve for the forces in each member of the truss, using the method of sections for FE and BC, and by joints for the rest. The Hibbeler 12th edition we use indicates that external reactions should be solved for from the free body diagram of the whole truss before using the method of sections. The problem is not from the text. The truss is supported by a pin at point A, with two reactions (Ax,Ay), and has a roller at point C with one reaction (Cy) and a roller at D with one reaction (Dy). There is a 60 Kip force (in x) at point F, and no other external forces. Our (first year) professor solved for Ax=0 using the whole truss free body diagram, then solved Ay=0 using the section free body diagram, since the whole truss free body diagram was indeterminate (3 unknowns, 2 equations left). We ended up with a roller whose reaction force is 80 Kip down at point C, which is not physically possible, since it is not attached to the roller surface (wouldn't be a roller if it was, right). I am sure this solution is wrong, and that his method of solution "created" the 80 Kip force downward at point C, but he says it is valid.
2. Relevant equations
Sum of forces =0, sum of moments =0

3. The attempt at a solution
I solved for the forces the way we were shown in class, got the "correct" answers, and noted on my homework that the "correct answers" were not correct, that the truss was statically indeterminate and could not be solved for (assuming no deformation in the truss), and that I do not believe that the reaction at C is down, or that FA is in compression, or that FE pushes back harder than it is pushed on. My concerns were not addressed except to say that it probably means the roller weighs 80 Kip, or that we need 80 Kip at C for equilibrium. The truss was in equilibrium to begin with. It is not urgent, but if I am missing something here, I'd like to know what it is. Could someone either confirm or deny his solution, and why? Thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Nov 5, 2009

### PhanthomJay

My first question is : Why is member FC or EB missing? You don't have a truss without them. There's no sense in continuing. Also, is that a support at C, or an applied load at C??

3. Nov 5, 2009

### pnd910

You would have to ask my professor why members FC or EB are missing, it is his homemade problem. As per my post, it is a roller for support at point C. I also agree there is no point in continuing, since the reaction forces can't be solved for with the three equilibrium equations. Bear in mind I am hoping to show this problem statically indeterminate and unsolvable with just sum of forces (x,y) and sum of moments. My fifteen classmates seemed to think it was just fine, but I think it's a load of.... Thanks for taking a look!

4. Nov 5, 2009

### pnd910

Sorry, the diagram has his solved support reactions given in it, and is likely confusing. The original problem had the 60 K force in x at point F, reactions Ax (to the left) and Ay (upward) at A, reaction Cy (upward at C) and reaction Dy (upward at D). Point A is a pin connection, C and D are rollers.

5. Nov 7, 2009

### pongo38

It's not a practical truss because of the missing diagonal in bay EFBC, although it appears in the calcs. Also I make the practice of distinguishing actions from reactions by using a different style of arrow - say, solid for a reaction, open for an action. That way, the origins of the forces at C and D would be clear. If you have reactions at A, at C and at D, the problem is statically indeterminate. Probably D is a load, and C a reaction. If there is a dispute, always check equilibrium up down and moments. If it is not in external equilibrium then there is no point in looking at it internally. Anyway, on your first page, the equation taking moments about D is missing the vertical reaction at A, and that has not been determined to be zero. Hence the 80 at C must be in error.

6. Nov 8, 2009

### PhanthomJay

Note that if diagonal EB is added, you get a stable truss, and the solution, although statically indeterminate, is solvable, by noting that the the reaction support at roller C is 0. This is done by removing that support, and noting that the vertical member at that the unknown force vertical reaction force at C back in at that joint, it's deflection too must be zero, thus R_C = 0.