Efficient way to solve truss by hand

[arestes]

Homework Statement

Hello. Given the statically determinate truss shown in the image, solve for all forces in the members, in terms of the load P.

Homework Equations

Equations of equilibrium.
Node analysis.
Section analysis.

The Attempt at a Solution

I was able to solve it by solving first for the vertical components of the external reactions.
Taking moments at the left support:

I get 7/3 P for the vertical component of the reaction at the support on the right. The vertical reaction at the support on the left is therefore 5/3 P.

There's a relationship between the horizontal components of the supports: they're opposite for equilibrium.
Of course, there are no more equations to solve for the horizontal components of the reactions at both supports, only this relationship. That's why going for a node analysis, there's an unknown I have to drag all the way to the end. This led me to a solution after painstakingly finishing all nodes. I checked that this truss could be solved using an online truss solver. (it's not statically indeterminate. There are 8 members (the ones in the middle don't cross) + 4 unknowns because of the supports=12 unknowns. There are 6 nodes X 2= 12 equations.)

Is there a clever cut to make this problems easier?

 

[tech99]

Homework Statement

Hello. Given the statically determinate truss shown in the image, solve for all forces in the members, in terms of the load P.

Homework Equations

Equations of equilibrium.
Node analysis.
Section analysis.

The Attempt at a Solution

I was able to solve it by solving first for the vertical components of the external reactions.
Taking moments at the left support:

I get 7/3 P for the vertical component of the reaction at the support on the right. The vertical reaction at the support on the left is therefore 5/3 P.

There's a relationship between the horizontal components of the supports: they're opposite for equilibrium.
Of course, there are no more equations to solve for the horizontal components of the reactions at both supports, only this relationship. That's why going for a node analysis, there's an unknown I have to drag all the way to the end.This led me to a solution after painstakingly finishing all nodes. I checked that this truss could be solved using an online truss solver. (it's not statically indeterminate. There are 8 members (the ones in the middle don't cross) + 4 unknowns because of the supports=12 unknowns. There are 6 nodes X 2= 12 equations.)

Is there a clever cut to make this problems easier?

What is the rectangle at the bottom - it seems to be obscuring text. Is it just a text box?

 

[arestes]

What is the rectangle at the bottom - it seems to be obscuring text. Is it just a text box?

oh, yes. It's just a text box that i deleted so I could insert a translation into English.

 

[PhanthomJay]

Perhaps the problem is in error and assumes a roller joint at one support, and thus, is statically determinate. If both supports are pinned, you need to know the sizes (areas) of each member , which is not given, unless you assume they are all the same size. I am not aware of a shortcut to determine the pin forces if the structure is externally statically indeterminate as shown, with the pins at each support. You have to calculate displacements using virtual work method, by first removing one of the pins and replacing it with a roller, then solving for the horizontal force at that support required to make the displacement of that joint equal to zero.

 

[tech99]

I tried to draw a Bow diagram for it and cannot make it work. I think the structure has redundant members which prevent the diagram being drawn. Also, I cannot see how the nodes with 4 forces acting can be solved this way.

 

[PhanthomJay]

I tried to draw a Bow diagram for it and cannot make it work. I think the structure has redundant members which prevent the diagram being drawn. Also, I cannot see how the nodes with 4 forces acting can be solved this way.

There are no redundant members, and, excluding supports, not more than 3 unknowns at a node, and the truss is internally statically determinate, but it is indeterminate externally due to the 4 unknown reactions at the supports. You must first solve for the reactions using indeterminate analysis, then once you find them, the internal member forces can be found. Since you can't do an indeterminate analysis without knowing the member sizes, I believe the problem may have intended for one of those supports to be a roller, which makes the problem statically determinate internally and externally, and solvable with the equilibrium equations.

 

[arestes]

Hello;
Thanks for the ideas. I should look into the virtual work method and Bow diagrams (any ideas of where to learn best by examples these methods quickly?)

Also, I see that member sizes are not really specified. I asked a colleague who proposed this problem some time ago and he said he remembers that he solved assuming symmetric shape, which is also what I assumed. (see attached pic) but he could not remember what method he used. He also pointed out that going node by node (dragging one of the unknowns till the end) would solve the problem... too long a method.

Also, I think none of the supports was intended to be a roller since that would mean there are 8+3 unknowns =11 unknowns but there are six nodes, which make up to 12 equilibrium equations and one member should be redundant. Is that right?
So... so far the most efficient method would be using the indeterminate analysis for solving for the reactions? Also, does this method need any knowledge of elastic properties of the members?
thanks

 

[PhanthomJay]

Hello;
Thanks for the ideas. I should look into the virtual work method and Bow diagrams (any ideas of where to learn best by examples these methods quickly?)

here's a site for virtual work :
http://www.public.iastate.edu/~fanous/ce332/force/truss.html
Rather tedious.

Also, I see that member sizes are not really specified. I asked a colleague who proposed this problem some time ago and he said he remembers that he solved assuming symmetric shape, which is also what I assumed. (see attached pic) but he could not remember what method he used. He also pointed out that going node by node (dragging one of the unknowns till the end) would solve the problem... too long a method.

I don't see how you can solve it this way...the truss is externally statically indeterminate

Also, I think none of the supports was intended to be a roller since that would mean there are 8+3 unknowns =11 unknowns but there are six nodes, which make up to 12 equilibrium equations and one member should be redundant. Is that right?

I don't trust any of these determinancy equations...they don't work all the time and sometimes refer to internal indeterminancy, not external...and there are several different ones...confusing at best.

So... so far the most efficient method would be using the indeterminate analysis for solving for the reactions? Also, does this method need any knowledge of elastic properties of the members?
thanks

That is the only way I know how to do it, and since deflection is a function of load, length, A and E, you need to know the elastic modulus of the members (E) and the cross section area of the members (A), unless they are all of the same material and same area, in which case the A and E terms cancel out when solving for the horizontal end reactions. The horizontal end reactions will be equal and opposite, the left horiz reaction points right and the right horiz reaction points left. Ihaven't worked out the answer because the method is time consuming...I have a suspicion that the end horizontal reactions are about 2P...assuming same material and sizes for all members...I wonder what the computer says, not that I trust it either.