PTLW Trusses calculations

Hi I’m having troubles understanding how to do Trusses calculations and was wondering if anyone could help or have information that would help me. Thank you. P.S I’m in PTLW POE (Principle of engineering)

It’s been many years since I’ve done it, but if my memory serves me correctly, in a true truss structure, the joints are considered to be pin joints, with no moments. So the equation simply becomes that the sum of the forces at the joints must equal zero, in order for that joint to be in static equilibrium. Also note, the force in a truss member is the same throughout. If you can identify which are in tension and which are in compression, what and where your forces are, and what your angles are, the rest should fall into place rather nicely. If you can provide more details we can provide more help.

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Thank you for posting the picture, that helps a lot.
What specific questions do you have? Can you determine the x,y reaction forces that come from the pin and the roller? Are you able find the interior angle by the 100N by using the leg lengths and a bit of trigonometry? Have you made any free body diagrams yet?

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I can do the leg lengths and find the angle but after that I don’t know what to do or what I should find.

In solving this problem, I see a 100N force being applied at a distance of 8m away from the pin join on the roller support. This is creating an 800 N-m moment about that joint. Since this structure is in static equilibrium, that moment must be balanced by the force in a different member. The vertical member of length 6m does not apply any moment to the joint at the roller support, since it is directly aligned with the joint and thereby has no “moment arm” upon which to act. The only remaining member is the diagonal member. Therefore, the vertical (Y axis) component of the tensile force in the diagonal member is equal to 100N. We also could have arrived at the same conclusion by simply looking at the sum of the forces about the pin joint where the 100N force is applied, and seen that the horizontal member of 8m length does not contribute any upward force to the joint. So therefore, again, the upward component of the tension force in the diagonal member is 100N.

Now, to solve for the rest of it, we need to find the compression force in the horizontal beam, which will be equal to the horizontal component of the tension force in the diagonal beam. In order to do this, we need to sum the moments about the top pin joint. Remember, we have 800N-m of moment being applied to the structure, due to the 100N force being applied at a “moment arm” of 8m. To counter-act this clockwise moment, we need to induce an equal in magnitude and opposite direction (counterclockwise) moment. This is provided by the reactionary support with the rollers. So, we need to apply an 800N-m moment, with a force acting at a “moment arm” distance of 6m, pushing to the right. 800N-m / 6m = 133.3… N. The reaction force at the roller support is 133.3… N to the right, which is equal to the compression in the horizontal beam, which is equal to the horizontal component of the tension in the diagonal beam.

Now from a geometry perspective, I immediately recognize this triangle. First, it’s a right triangle. Second, it’s a 3-4-5 triangle at twice the size, it’s a 6-8-10 triangle. The length of the hypotenuse is 10m. To find the angle between the horizontal member, and the diagonal member, we can use cos(theta) = adjacent/hypotenuse. In other words, cos-1(8/10)=theta. Theta = 36.869 deg. Tension in the diagonal beam = horizontal component / cos(theta). Tension in the diagonal beam = 133.3…N / 0.8 = 166.6… N

Now, for the top support. We previously determined the horizontal component of the tension in the diagonal member to be 133.3…N. This is the horizontal component of the reactionary support at the top joint. The vertical component of the reactionary support at the top joint is simply the vertical component of the tension in the diagonal member, which is 100N. So in other words, the reactionary force at the top joint is simply 166.6…N, directed at an angle of 141.31 degrees from the positive X axis.

The neat part about this problem is that the vertical member carries no tension, because of the fact that this is a right triangle, and the horizontal member is horizontal. If the horizontal member were at any angle other than horizontal, the vertical member would be in either tension or compression.

Someone please check me on this. It’s been probably 15 years since I’ve done one of these, and it’s still early morning here.

This was a fun one. Now to thank me for my efforts, can you give me another problem? I like these.


Can you write this down so I can see it visually like all the points and and math? and thank you I got lots more soo

Im a very visual person so seeing it visually will help.

There are a bunch of really great YouTube videos that might help you along.

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Ok Thank you

let me write it down and ill post what I got to make sure I understand what you said.

This sounds like a good plan. Try to follow through what I wrote as you hand write it out on a piece of blank paper. These are the fundamental equations used in solving something like this.

On top of that, it’s just simple trigonometry, typically sine and cosine.

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It looks like you’ve captured the essence of it in your equations, but your diagram is not correct. Your 166.6N force is drawn in an arbitrary location. There is no such force where you’ve drawn it. Also, your RAy force is pointing the wrong way. The support provides a force directed upward and to the left. Your calculation also does not show how you arrived at the angle.

In solving this problem, it’s probably even easier than I described above. You have an external force downward of 100N. To keep the whole thing from accelerating downward, you need an upward force of 100N. This occurs at the top support. Then, you can calculate the horizontal reaction force using a little simple trig. In this method of solving, we would typically superimpose artificial “x” and “y” axes, one of which is parallel to the diagonal member.

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I would just flip my RAy arrow to go up and to the left and for the angle I did cos^-1(8/10)=36.869

At the load point (where the 100N is hanging), only the load and the diagonal arm can exert vertical forces. They must balance, so (as you have figured the angles), you can calculate the tension in the diagonal. Then, the horizontal element’s force must balance the horizontal component of that diagonal member. As the lower left is free to slide up and down, and the force from the horizontal member is purely horizontal, the tension in the vertical member is zero.


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