# Truss Structure dilemma

1. Jan 22, 2008

### crustation

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

I spent a few hours trying out this question, so I probably need help. This is a seven pin-jointed structure mounted on the wall, with the following information given:
1) S1 and S2 are wall supports, but S2 is assumed to have no friction with the wall i.e. no vertical support provided to P4 from the wall.
2) The weights W are of the same mass, and in this question is of not much relevance to the answer.
3) The whole system is in static equilibrium, and idealization is assumed.

The question requires us to determine if each rod is in compression or tension.

2. Relevant equations
Nil.

3. The attempt at a solution
After much resolving and equating (which I shall not post here due to lack of space, I have determined the following:
1) Rod P1P3 is in tension.
2) Rod P1P2 is in compression.
3) Rod P2P3 is in compression.
4) Rod P3P5 is in tension.
5) Rod P2P4 is in compression.
6) Rod P2P5 is in tension.
7) There are no forces along P4P5.

This is after checking that each point P1 to P4 is in static equilibrium, and that the tension and compression on each rod is equal and in opposite directions.

I determined that there is no force along P4P5 as there is no vertical component from the wall, and since the rods are assumed to be massless, there should only be a compressional force acting on P4 to counter the reaction force from the wall.

The problem comes when I try to resolve P5. As the tension along rod P3P5 and the reaction force from the wall act in the same direction, I am missing a horizontal component acting towards the wall on P5.

But if I made a mistake along P2P5 try to balance out the horizontal component on P5 with compressional force from P2P5, there would be no upward-acting force on P2 to balance the downward-acting forces on P2.

Moreover, friction (or some upward acting force) has to be present at P5 in order to balance the downward acting component along P2P5.

Last edited: Jan 22, 2008
2. Jan 22, 2008

### PhanthomJay

This may be your error. The tension force from P3P5 acting on joint P5 acts in the opposite direction of the reaction force S1 from the wall on P5 (they each pull away from the joint).

Yes, you can find this upward force (from S1) from the equilibrium condition (sum of all vertical external forces acting on the system= 0).

Your tenson /compression /zero force members analysis otherwise appears correct.

3. Jan 22, 2008

### crustation

Ok.. but what do you call the upward force acting at P5 if it does exist? Is it friction or just a vertical component of another force acting in the "2 o'clock" direction? That is my dilemma because I don't know what to name and how to define it.

Secondly, the tension force from P3P5 does act away from P5, as does the Reaction force from the wall on P5 i.e. from "right" to "left". So what is the force going from "left" to "right"?

so in other words, what I mean is
R(rxn force from wall) + Tension(P3P5) + Tension(P2P5)cos45 all act in the same "sense" (right to left) and direction (parallel to same unit vector), but what is the force that counters the sum of all these and acts from "left to right"?

4. Jan 23, 2008

### PhanthomJay

The first step in solving truss problems is to find the reactions by applying the equilibrium equations to the system (sum of Fy = 0 and sum of moments about a point = 0). If you do that, you will find that the wall at support S1 exerts not only a horizontal force on P5, but also a vertical upward force on P5 equal to 2W. This upward force is likely transmitted through a bolted connection rather than friction, but in any case, it is called the upward component of the reaction force at S1. It is an external force. In turn, this force (as well as the horizontal component of S1) introduces internal forces into the connecting members (solved by isolating that joint) . Then the NET vertical force at P5 (and net horizontal force at P5) must be 0.

You still appear to be mixed up in your directions. By summing moments about S2, you should note that the horizontal reaction force S1 from the wall on P5 acts from left to right, not right to left.

So this should be corrected in light of the above to read:
R_x (which is the horizontal component of the S1 force from wall) - Tension(P3P5) - Tension(P2P5)cos45 = 0 (summing forces in the horizontal direction), AND now write an equation which sums forces in the vertical direction.

Last edited: Jan 23, 2008
5. Jan 23, 2008

### crustation

yeah that's what I thought, it's the only possible force that will balance out the horizontal components. So do you mean that since the overall structure is acting on the wall at about 67degrees from the vertical downwards, the reaction force from the wall acting on the point itself is the "balancing force"?

6. Jan 23, 2008

### PhanthomJay

Im not sure where your 67 degree figure is coming from unless your scaling from the drawing, but it doesn't matter much. The balancing forces at the supports....both supports.....come from the equilibrium equations for the system, sum of forces = 0 and sum of moments = 0. You have in this problem a single horizontal reaction force at the lower support, and both a horizontal and vertical reaction force at the upper support. You might want to assume for practice that W=1 newton, and that the length of the horizontal members between joints is 1m, and that the length of the vertical members between joints is 1m, and tell me what you get for the reaction forces at S1 and S2(magnitude and direction).

7. Jan 24, 2008

### crustation

oh ok i just realised i made a fundamental conceptual error. i always thought newton's third law meant that reaction forces have to act perpendicularly from the wall through the point, but it just means reaction force is always equal and opposite to the force acting on it.

in this case, the downwards force around the "7 - 9 o'clock" direction is being exerted on the wall, hence the wall "pulls back" and exerts and equal force on P5 with equal magnitude and opposite direction. is that right?

8. Jan 24, 2008

### PhanthomJay

Yes, that is more or less correct. I say more or less, because you shouldn't confuse Newton 3 (If A exerts a force on B, then B exerts an equal and opposite force on A, that is, these 'paired' forces act on different objects but don't imply equilibrium) with Newton 1 (sum of forces on any one object or joint = 0, for equilibrium), but that's perhaps a topic for another day.