How to find the forces in a particular framework of light rods

AI Thread Summary
The discussion focuses on calculating forces in a framework of light rods supported at points A and C, specifically seeking to determine the external forces and tensions in each rod. The user successfully resolves vertical forces and moments, arriving at values for Fa and Fc that match book answers, but struggles to find individual tensions T1 and T5 without knowing the angle Ѳ. Participants emphasize the importance of the angle in determining tensions, suggesting that the problem may not be solvable without it. There is a consensus that the book's diagram lacks necessary information, and discrepancies in the provided answers for tensions T6 and T7 are noted, indicating potential errors in the official solutions. The conversation concludes with a call for clarity on the angle and its impact on force calculations.
gnits
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Homework Statement
How to find the forces in a particular framework of light rods
Relevant Equations
Equating of forces
Could I please ask for help in how to do this question.

Is it in fact well formed, can it be solved as it is or do I need more information?

Q. Find the external forces and the force in each rod in the following framework of light rods which is supported and A and C:

forces.png


So, I need to find Fa, Fb and T1, T2, T3, T4, T5, T6 and T7 in terms of W.

I mark all forces as pointing towards the ends of the rods so that in the answers a negative value indicates a force towards the centre of a rod.

I call the length of the sides marked with a single bar "L".

I have labelled the angle Ѳ on the diagram for use in the following.

Taking the framework as a whole and resolving vertically gives:

Fa + Fc = 3W

Taking moments about A gives:

Fc * 3L = W * L + 2W * 2L and so this gives Fc = 5W/3 and so we also now know that Fa = 4W/3. These answers agree with the book answers.

Now, looking at the forces incident at A and resolving vertically we have:

T1 * cos(Ѳ) = 4W/3

and resolving horizontally we have that

T1 * sin(Ѳ) + T5 = 0.

From this, using cos^Ѳ + sin^Ѳ = 1 we can obtain:

16W^2 + 9*T5^2 = 9 * T1^2.

Now this agrees with the book answers for T5 and T1 (which are given individually), but I cannot see how to get them individually.

(In previous questions of this type I had always been provided with the angle of the figure - in this case, if I knew Ѳ then I could obtain T1 and T5 individually).

If I take the book answers for T1 and T5 and use my equations above I get that the principal value of Ѳ = 60 degrees. Does that need to be marked in the diagram for the question to be possible? Or should I be able to see how to derive the angle from the given geometry?

Here are the book answers:

T1 = 8W/3
T2 = 10W/3
T3 = -5*sqrt(3)W / 3
T4 = -sqrt(3)W
T5 = -4*sqrt(3)W/3
T6 = -4*sqrt(3)W/3
T7 = -2*sqrt(3)W/3

Thanks for any help.
 
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Sorry, I don't completely understand your question.
That angle is very important, because the higher the location of node B is, the lower the value of each internal tension should be.
 
Lnewqban said:
Sorry, I don't completely understand your question.
That angle is very important, because the higher the location of node B is, the lower the value of each internal tension should be.
That's very helpful to hear. You see, in the book the diagram is presented without any angle being given. I was thinking that I must be missing something. I don't think that it is solvable without the angle being given. In this case Ѳ = 60⁰.
 
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Is it possible that somehow they give proportions of the formed triangles, or lengths of the bars, or A-B-C-D-C distances?
 
Lnewqban said:
Is it possible that somehow they give proportions of the formed triangles, or lengths of the bars, or A-B-C-D-C distances?
No, nothing else is given. I'll assume that 60⁰ was meant to be given and go from there.

Thanks.
 
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gnits said:
Homework Statement:: How to find the forces in a particular framework of light rods
Relevant Equations:: Equating of forces

T1 = 8W/3
T2 = 10W/3
T3 = -5*sqrt(3)W / 3
T4 = -sqrt(3)W
T5 = -4*sqrt(3)W/3
T6 = -4*sqrt(3)W/3
T7 = -2*sqrt(3)W/3
I am a little bit surprised to see it says
T_1 \neq T_2, T_6 \neq T_7, T_3 \neq T_5
after looking at your sketch which shows axis symmetry wrt vertical line passing B that would suggest
T_1 = T_2, T_6 = T_7, T_3 = T_5
Something seems to be wrong with the answer or the sketch at least.
 
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anuttarasammyak said:
I am a little bit surprised to see it says
T_1 \neq T_2, T_6 \neq T_7, T_3 \neq T_5
after looking at your sketch which shows axis symmetry wrt vertical line passing B that would suggest
T_1 = T_2, T_6 = T_7, T_3 = T_5
Something seems to be wrong with the answer or the sketch at least.
Please, note that loads W and 2W are located in symmetry with respect to the vertical line passing B, but the summation or total force is off-center.

Please, see this, in principle, similar example:
https://mathalino.com/reviewer/engineering-mechanics/problem-407-warren-truss-method-joints
 
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Thanks. I observed Loads W and 2W are put not symmetrical.
 
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anuttarasammyak said:
Thanks. I observed Loads W and 2W are put not symmetrical.
Then your observation was incorrect. Loads W and 2W are positioned symmetrically, but they are obviously not of equal weight and so their weights are not distributed symmetrically.
 
  • #10
If the book answer is right the settings of the problem seem to be
\angle BAE=\frac{\pi}{6}, \angle BED = \frac{\pi}{3}
 
  • #11
anuttarasammyak said:
If the book answer is right the settings of the problem seem to be
\angle BAE=\frac{\pi}{6}, \angle BED = \frac{\pi}{3}
Can I chip in...

Let α = ∠BED = ∠BDE

Resolving vertically at point D gives ##|T_7| sinα = 2W##.
Resolving vertically at point E gives ##|T_6| sinα = W##.
This means ##T_7## should have twice the magnitude of ##T_6##. But the ‘official’ answers (as given in Post #6) say the opposite.

It looks like the official answers for ##T_6## and ##T_7## are the wrong way round. If that’s the case, I think that's consistent with ##α = \frac {\pi}{3}##.
 
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  • #12
Steve4Physics said:
It looks like the official answers for T6 and T7 are the wrong way round.
Thanks for nice observation. For investigation in #10 I made the symmetric case in both load position and load quantity, that is
t_1=t_2=T_1+T_2
t_3=t_5=T_3+T_5
t_6=t_7=T_6+T_7
t_4=2T_4
where new t's are tensions where loads 3W are put on E and D. This process happened to hide that wrong way round.
 
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