stresses in framed structures

[chthulu]

Hi,
In the picture that follows, is fig 9 correct?
Using the scale in pounds shown there, the 70LBS line seems to be correct, where line a seems to be more or less 110 lbs and line b seems to be more or less 130 lbs, but using trigonometry I got something else for a and b.
So, is it possible to get a and b through trigonometry?
If you draw the line for the 70lbs according to the scale in pounds, how do you know where to stop when drawing a? or b? if you start from b.
I mean line a could be any length and fig 8 does not indicate how long line a should be.
What would be the best way to add the sketch of fig 8 together?
I really do not understand even how the 180 and 230 lbs are get.

 

[chthulu]

yes, I now realized that the 70 lbs in fig. 9 is not side c of triangle 1 of fig. 8. I assumed that it was, then I realized that side c is drawn as part of triangle 2 in fig. 9. My mistake.
So what side 70 lbs in fig. 9 represents? Line A to B?

 

[3DDave]

yes, you are right. Ok so you can not copy and paste those triangles of fig. 8

You are to copy and paste junctions (1), (2), (3), (4), (5) as the directions say to, scaled so the matching vectors are the same size. The paste can be any scale you like, but the 70 pound vector is the basis for scaling all the other forces from the remaining vectors.

 

[3DDave]

yes, I now realized that the 70 lbs in fig. 9 is not side c of triangle 1 of fig. 8. I assumed that it was, then I realized that side c is drawn as part of triangle 2 in fig. 9. My mistake.
So what side 70 lbs in fig. 9 represents? Line A to B?

It doesn't represent any side. That is the vector of the force that is applied.

 

[rb1957]

"70 lbs" in fig 9 represents the vertical load at 1. It can be any length. The other two forces at one are in known directions, and so you can draw the triangle, and scale the load from the known vertical load.

I feel that drawing fig 9 to look so much like fig 8 is very misleading. If you wanted to relate fig 9 to fig 8 then draw like node 1.
But fig 9 does have a lot of labelling on it, and I think you can figure it out (but who has time for that !?)

 

[chthulu]

Yes indeed. I did not pay much attention to fig. 9 by realizing that there is already a side called c and therefore the 70 lbs line could not be side c of fig. 8. At the end the 70 lbs line of fig. 9 has nothing to do with any side of fig. 8, therefore you really need to know the angles.

 

[3DDave]

On a drawing one would simply use a parallel transfer to duplicate the slopes, no need to measure the angles, a big advantage of the method. Perhaps they no longer teach how a triangle and a straightedge can be used to do that.

rb1957, it would be difficult for the vector diagram to not resemble the axially loaded member diagram.

 

[GregLocock]

Superficially it looks similar because it is a regular truss with a point load, but in general there is little obvious similitude between a force diagram and the geometry of a structure. Here's an example

Unless you know how to construct a force polygon at a joint I don't see how you can expect to draw a Cremona diagram.

 

[rb1957]

yes, but drawing "70 lbs" in fig 9 to look like AB in fig 8 (even though it isn't) has caused a lot of confusion. And the scale on fig 9 would be clearer if it used multiples of "70 lbs"

and, yes, this is a quite trivial problem to solve, it was just an example solution that the OP had issues with.

 

[3DDave]

The 70 lbs is the basis vector from which all other stress vectors is determined. It has to be where it is and all else exactly attached the way they are.

Were I to try to improve the example I would mark all the features in Figure 9 with a horizontal arrow over each identifier and a note that the arrow indicates the item is a vector. These vectors themselves would have arrows indicating the load into or out of the node. I would have removed the arrows from Figure 8. Finally the joint diagrams would be a separate figure between the first two to indicate the way each works; the nodes should be labeled and not the members.

However, I have no time machine. It's a flaw. My research into graviton diodes has slowed the time travel research and so far the time machine progresses in one direction at 1:1 with local time. Very unsatisfying.