Analysis of systems in equilibrium: articulated structures

[Guillem_dlc]

Homework Statement

Solve the following articulated structure.

Relevant Equations

$\sum F=0, \sum M=0$

Figure:

My attempt at a Solution:

We calculate REACTIONS:
$$\sum F_x=0\rightarrow \boxed{Bx=0}\qquad \sum F_y=0\rightarrow By+A=50$$
$$\sum M_B=0\rightarrow \boxed{A=25\, \textrm{kN}}\quad \boxed{B=25\, \textrm{kN}}$$
KNOT B:

$$\alpha \rightarrow \alpha =90-40=50\, \textrm{º}$$
$$\beta \rightarrow \beta =50+\arctan \left( \dfrac36 \right)-90=13,43\, \textrm{º}$$
$$\sum Fx=0\rightarrow \cancel{Bx}+TBD\cos \beta +TBC\sin \alpha=0$$
$$\sum Fy=0\rightarrow TBD\sin \beta +TBC\cos \alpha=15$$
$$\boxed{TBD=-25,69\, \textrm{kN},\quad TBC=32,62\, \textrm{kN}}$$
Could you look at this one? The solution they give is:

 

[erobz]

Are you sure the reaction at $A$ is only vertical? What is the connection type at point $A$?

 

[erobz]

Are you sure the reaction at $A$ is only vertical? What is the connection type at point $A$?

Nevermind. I can make out the little rollers. It is a vertical reaction force at $A$.

Can you please provide the math that concluded:

$$\sum M_B=0\rightarrow \boxed{A=25\, \textrm{kN}}\quad \boxed{B=25\, \textrm{kN}}$$

I get that $A \neq B_y$ when I take the moment $\circlearrowright^+ \sum M_{B} = 0$.

 

[haruspex]

I also disagree with your moment equation.
The easy way is, first, to note that the weights at A and B are irrelevant, then to resolve each of the others into a force parallel to AB and one parallel to CE.

 

[erobz]

I also disagree with your moment equation.
The easy way is, first, to note that the weights at A and B are irrelevant, then to resolve each of the others into a force parallel to AB and one parallel to CE.

(Alternatively) I took all the distances from $B$ perpendicular to the applied forces.

 

[Guillem_dlc]

(Alternatively) I took all the distances from $B$ perpendicular to the applied forces.

But this is already what I have done, isn't it?

We calculate REACTIONS:
$$\sum Fx=0\rightarrow \boxed{Bx=0}\qquad \sum Fy=0\rightarrow By+A=50\quad \boxed{B_y=25\, \textrm{kN}}$$
$$\sum M_B=0=12\cos \alpha A-10(3\cos \alpha +6\cos \alpha +9\cos \alpha +12\cos \alpha)\rightarrow \boxed{A=25\, \textrm{kN}}$$
KNOT B:
$$\alpha \rightarrow \alpha =90-40=50\, \textrm{º}$$
$$\beta \rightarrow \beta =50+\arctan \left( \dfrac36 \right)-90=13,43\, \textrm{º}$$
$$\sum Fx=0\rightarrow \cancel{Bx}+TBD\cos \beta +TBC\sin \alpha=0$$
$$\sum F_y=0\rightarrow TBD\sin \beta +TBC\cos \alpha =15$$
$$\boxed{TBD=-25,69\, \textrm{kN},\,\, TBC=32,62\, \textrm{kN}}$$

 

[haruspex]

$$By+A=50\quad \boxed{B_y=25\, \textrm{kN}}$$

To see that By and A cannot be equal, consider moments about C.

 

[erobz]

But this is already what I have done, isn't it?

We calculate REACTIONS:
$$\sum Fx=0\rightarrow \boxed{Bx=0}\qquad \sum Fy=0\rightarrow By+A=50\quad \boxed{B_y=25\, \textrm{kN}}$$
$$\sum M_B=0=12\cos \alpha A-10(3\cos \alpha +6\cos \alpha +9\cos \alpha +12\cos \alpha)\rightarrow \boxed{A=25\, \textrm{kN}}$$
KNOT B:
$$\alpha \rightarrow \alpha =90-40=50\, \textrm{º}$$
$$\beta \rightarrow \beta =50+\arctan \left( \dfrac36 \right)-90=13,43\, \textrm{º}$$
$$\sum Fx=0\rightarrow \cancel{Bx}+TBD\cos \beta +TBC\sin \alpha=0$$
$$\sum F_y=0\rightarrow TBD\sin \beta +TBC\cos \alpha =15$$
$$\boxed{TBD=-25,69\, \textrm{kN},\,\, TBC=32,62\, \textrm{kN}}$$

I don’t need to see all the other math( that math checks out), just the moment about B is the issue.

You didn’t take moments correctly.

To get you on the right track: what is the perpendicular distance from B to the point of application of the first 10kN load? It’s not $3 \cos \alpha$ like you suggest.

 

[Guillem_dlc]

I don’t need to see all the other math( that math checks out), just the moment about B is the issue.

You didn’t take moments correctly.

To get you on the right track: what is the perpendicular distance from B to the point of application of the first 10kN load? It’s not $3 \cos \alpha$ like you suggest.

$3\cos 40$ meant sorry

 

[erobz]

$3\cos 40$ meant sorry

It's not that either.

We are looking for the distance indicated below:

 

[Lnewqban]

Homework Statement:: Solve the following articulated structure.
Relevant Equations:: $\sum F=0, \sum M=0$

Figure:
View attachment 316522
My attempt at a Solution:

We calculate REACTIONS:
$$\sum F_x=0\rightarrow \boxed{Bx=0}\qquad \sum F_y=0\rightarrow By+A=50$$
$$\sum M_B=0\rightarrow \boxed{A=25\, \textrm{kN}}\quad \boxed{B=25\, \textrm{kN}}$$

If you could turn the truss so its bottom chord were horizontal, the locations of the loads would be symetrical respect to the supports, and Ay could be equal to By.
Nevertheless, your truss is not horizontal.

Using a building stairs, two men are carrying a heavy sofa upstairs.
Which person will feel more weight, the one holding a sofa at the higher end or at the lower end?

For this type of problem, carefully calculating the support’s reactions is very important to accurately calculate the internal compression and tension loads of the structure members.
Otherwise you waste your work while propagating any error.

 

[Guillem_dlc]

It's not that either.

We are looking for the distance indicated below:

View attachment 316572

Would this be the case?

So the moment with respect to B, it won't go well, will it? Because it will give me $25$.

 

[erobz]

Would this be the case?
View attachment 316574

No. Its going to be larger than $3 \rm{m}$. What you have shown is smaller.

Draw a perpendicular bisector from point $D$ to line $BA$, yielding point $D'$ on line $BA$. Use the resulting triangle $BDD'$ ( side lengths ) and the angle which you originally call $\beta$ in the OP to find the distance from $B$ perpendicular to the first vertical $10 \, \rm{kN}$ force.

 

[Guillem_dlc]

$3/\cos \beta$

 

[erobz]

$3/\cos \beta$

Sorry, but no. Please see the figure I updated. We are looking for the length of the red double arrow. Use the triangle $BDD'$ to determine the length of $BD$. Then use $BD$ and the angle $\beta$ to determine the length of the red double arrow.

 

[Guillem_dlc]

View attachment 316574

I see it as: $\dfrac{3}{\cos \beta}\cdot \cos \mu$. I don't know it.

 

[erobz]

I see it as: $\dfrac{3}{\cos \beta}\cdot \cos \mu$. I don't know it.

Where is angle $\mu$? If you are saying it is the angle between $BD$ and $BA$ then you have some stuff flipped around.

 

[erobz]

Anyhow, you don't need to introduce $\mu$ (wherever you have defined it). You calculate length $BD$ directly from Pythagorean Theorem.

 

[Guillem_dlc]

Where is angle $\mu$?

It's $\varphi$, sorry:
$$\dfrac{3}{\cos \beta}\cdot \cos \varphi$$

 

[erobz]

It's $\varphi$, sorry:
$$\dfrac{3}{\cos \beta}\cdot \cos \varphi$$

Ok, for future reference you are changing the angle names you established in the OP. That is unnecessarily confusing. Try to avoid that in the future.

That is the proper length, but it's preferable to just calculate $BD$ from Pythagorean Theorem here.

$$ L = BD \cos \varphi = \sqrt{ 3^2 + 1.5^2} \cos \varphi $$

I made edits to this. I was mixing up $BD$ and the desired length. Hope I didn't cause confusion. $L$ is the perpendicular distance (moment arm for the applied load) we are after.

 

[Guillem_dlc]

Ok, for future reference you are changing the angle names you established in the OP. That is unnecessarily confusing. Try to avoid that in the future.

That is the propper length, but it's preferable to just calculate $BD$ from Pythagorean Theorem here.

$$BD = \sqrt{ 3^2 + 1.5^2} \cos \varphi $$

OK, perfect. Thank you

 

[erobz]

OK, perfect. Thank you

Now do the rest of them in a similar fashion.

 

[Lnewqban]

Could you just project the truss and the forces onto a horizontal line (perpendicular to the direction of gravity and forces) and treat it as a loaded beam supported at both ends?

 

[erobz]

Could you just project the truss and the forces onto a horizontal line (perpendicular to the direction of gravity and forces) and treat it as a loaded beam supported at both ends?

That’s effectively what I have done to solve it I believe. Or did you have a more in depth transformation/analysis in mind for the total result?

 

[Guillem_dlc]

Could you just project the truss and the forces onto a horizontal line (perpendicular to the direction of gravity and forces) and treat it as a loaded beam supported at both ends?

I don't understand the question.

 

[Lnewqban]

I don't understand the question.

A horizontal projection of the real thing will be proportional regarding distribution of forces, I believe.
Just like @erobz has shown us above.

 

[Lnewqban]

@Guillem_dlc , did you fully solve this problem?

 

[haruspex]

A horizontal projection of the real thing will be proportional regarding distribution of forces

Regarding the external forces, right?

 

[Lnewqban]

Regarding the external forces, right?

Yes, indeed.
Internal forces require a different approach.
It seems to me that those internal forces were never calculated for this problem, but I may be missing it.

 

[haruspex]

those internal forces were never calculated for this problem, but I may be missing it.

The question asks for the whole structure to be solved, and post #1 makes a start on the internals, though predicated on a false calculation of the external forces.