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

[gnits]

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:

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.

 

[Lnewqban]

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.

 

[gnits]

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.

 

[Lnewqban]

Is it possible that somehow they give proportions of the formed triangles, or lengths of the bars, or A-B-C-D-C distances?

 

[gnits]

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.

 

[anuttarasammyak]

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.

 

[Lnewqban]

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

 

[anuttarasammyak]

Thanks. I observed Loads W and 2W are put not symmetrical.

 

[pbuk]

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.

 

[anuttarasammyak]

If the book answer is right the settings of the problem seem to be
$$\angle BAE=\frac{\pi}{6}, \angle BED = \frac{\pi}{3}$$

 

[Steve4Physics]

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 $α = \frac {\pi}{3}$.

 

[anuttarasammyak]

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.