Why the forces in the arch are carried to the ground

[mcastillo356]

¿Why the forces in the arch are carried to the ground?

Homework Statement

Hello, I want to know if my argue is right. The question is to justificate why the forces in the arch are carried to the ground.
I suppose a three voussoir's rounded arch. Just the arch, no structure supported. I call A to the keystone, B the springer down on the left, and C to the springer down on the right.
I place the origin of coordinates down in the middle.
This time I only care about vertical forces.

Homework Equations

First law of static equilibrium. I must prove that the vertical forces on every part of the system is zero: ∑F_Ay=0, ∑F_By=0, ∑_Cy=0. And then add them to explain how the system works

The Attempt at a Solution

∑F_Ay = -m_Ag + F_BAysenθ + F_CAysenθ = 0
∑F_By = -m_Bg - F_ABysenθ + F_BN = 0
∑F_Cy = -m_Cg - F_ACysenθ + F_CN = 0

Where F_BAysenθ is the vertical component of the force B exerts on A; -m_Ag is the weight of A; F_BN is the normal force grounds exerts on B.

∑F_Ay + ∑F_By + ∑F_Cy = 0 = -m_Ag -m_Bg - m_Cg + F_BN + F_CN.

In conclusion, normal forces over B and C support all the weight.

¿Is it right?.

 

[paisiello2]

Difficult to follow your argument without a diagram and without some definitions:

1) what is a three voussoir's rounded arch?
2) what is θ?

Of course normal force supports the weight but there is something more important that allows the weight to be carried over the span of the arch.

 

[mcastillo356]

Sorry, here is the drawing, hope it makes sense. It's an arch with a circular form, made with three stones, as I describe on the image. $\theta$ is the angle I was thinking about.

In the picture I've drawn the forces exerted on $A$, and also written $\sum{F_{Ay}}$, which represents my notation for the vertical forces on $A$.

 

[paisiello2]

Ok, if we assume there is negligible friction between the stone blocks then your first 3 questions look right. However, it's not the whole story. You also have to sum the forces in the "x" direction and sum the moments for each block

I don't know what you're trying to do with your 4th equation.

 

[mcastillo356]

Sorry, I didn't mention I considered no friction. And I didn't consider horizontal forces. I was only concerned about vertical forces.

The 4th equation intends to be the sum of all vertical forces exerted on the three stones, one by one. My intention is to figure out what happens on the system, on the whole arch, in relation to the vertical forces.

In this sum, some forces cancel each other:

$+F_{CAy}\sin\theta-F_{ACy}\sin\theta=0$ (Diagram below)

$+F_{BAy}\sin\theta-F_{ABy}\sin\theta=0$

So the fourth equation means (from my point of view) that the weight of the arch should wind up at the bases of the arch.

As you can see, my approach is not complete, but I want to know if it justifies that the weight ends up at the bottom of both bases.

You suggest to calculate the sum of the moments for each block. Is there any torque?.

 

[paisiello2]

Your result that the weight of the arch is supported by the base is rather trivial. This is true of all structures regardless of whether it is an arch or not.

Even if you are not concerned with horizontal forces or moments, the structure is. Go ahead and sum the horizontal forces and the moments. It will be instructional.

 

[mcastillo356]

Let's go!

For the arch to stand up there must be a horizontal equilibrium:

$\sum{F_{Ax}}=+F_{BA}\cos\theta-F_{CA}\cos\theta=0$

$\sum{F_{Bx}}=-F_{AB}\cos\theta+f_{eB}=0$

$\sum{F_{Cx}}=+F_{AC}\cos\theta-f_{eC}=0$

$f_{eB}$ and $f_{eC}$ means static friction.

And also a moment equilibrium I am not sure of. I've been searching on the internet, whith no results. I was going to try, but I prefer to ask before: ¿could you say which is the moment equilibrium for the blocks?.

Thanks!

 

[mcastillo356]

Wait, I've found information. Let's see if I can manage.

 

[paisiello2]

OK, for the arch to be in equilibrium means that there must be a horizontal force: in this case you called it friction but any kind of horizontal reaction force to resist the thrust of the arch. This is what makes an arch an arch as opposed to just a big beam spanning over some distance.

As for the moment, draw a free body diagram of block C and take moments about some convenient point. You already found out the components of the reaction force at the base. Make an assumption about the location "x" of this reaction force say from point +3 and solve for "x".

 

[mcastillo356]

First I think that $A$ can be seen as a particle, so there is not any torque on it.

$C$ remains a mistery to me, as like as $B$. I've tried to describe, and, intuition says me that normal force on $C$ restrains the twist horizontal forces create. Here is the diagram. I don't dare to suppose which is the resultant moment about any point.

 

[paisiello2]

Go ahead and dare.

You drew the normal force FCN in the middle of the block. In reality we don't know the exact location unless we equilibrate moments about some point.

 

[mcastillo356]

Fine!

I've drawn the resultant moment about the center of gravity. Here it is, in the diagram. How does it look like?

Thanks!

 

[paisiello2]

I don't what you did here exactly. Looks like you made a number of assumptions about the locations of the reactions forces which are not necessarily true.

You might want to get hold of an engineering mechanics book to find out how to properly resolve forces into moments.

 

[mcastillo356]

Ok. I will try first to ask my physics teacher. The doubt doesn't belong to the subjects discussed, but I am really interested on it.

Thanks!

 

[mcastillo356]

Hi again, paisiello2 and everybody. I've tried hard, but no way. My teacher says me it doesn't belong to him to answer this question. Could anybody tell me how to resolve forces into equilibrated moments for C, taken as a free body?

Thanks!

 

[paisiello2]

I gave you a suggestion in post #4 how to do it. However, it sounds like you might also need to review your teacher's notes and/or a textbook to see in detail how to take moments for a body in equilibrium.

 

[mcastillo356]

Hello. I think I am closer to the solution. In the diagram, total torque is calculated respect the origin of coordenates, down in the middle of the span. How does it look like?

Thanks!

 

[paisiello2]

It is not right.

The force F from block A should act at the top not at the bottom the way you are showing it.

Also see post #4 for location of normal force. The friction force should act at the same location as the normal force.

 

[mcastillo356]

Helo paisiello2:

Can I take as valid this analysis of the forces exerteted on voussoir $C$ just before calculate total torque?

1- The center of gravity is the center of simmetry;
2- The normal force exerted by the ground is opposite to the sum of $\vec{F_y}_{AC}$ and $m_{C}\vec{g}$, considerated as a system of parallel forces;
3- The normal force exerted by the $A$ voussoir on $C$ is as shown in diagram, as well as friction force.

 

[paisiello2]

Can I take as valid this analysis of the forces exerteted on voussoir $C$ just before calculate total torque?

The way you are showing the forces is unorthodox. However, I think it looks acceptable for the purposes of determining the location of the ground reaction force.

1- The center of gravity is the center of simmetry;

I am Not sure what you mean by this exactly. But it looks like you have drawn it correctly.

2- The normal force exerted by the ground is opposite to the sum of $\vec{F_y}_{AC}$ and $m_{C}\vec{g}$, considerated as a system of parallel forces;

Yes!

3- The normal force exerted by the $A$ voussoir on $C$ is as shown in diagram, as well as friction force.

it looks correct except for the unknown location of the ground reaction force and friction force.

 

[mcastillo356]

Hello. paisiello2, I thought you were on holidays. The analisys is not good. The center of gravity is not at the center of simmetry, its slightly outwards. I must integrate, and its a lot of work (I am on holidays). Any clue?

 

[paisiello2]

Center of symmetry of what exactly? Of the C block?

 

[mcastillo356]

Center of symmetry of what exactly? Of the C block?

Yes, the C block. First of all I must beg you a pardon for my poor english, and because i write like everybody knew what am i thinking about.

The center of symmetry of the block C of the arch i am thinking about (three equal blocks' arch) is not the center of gravity(in this case when talk about center of gravity i am talking about center of mass, because i suppose uniform density).

I did not draw it in the diagram, sorry.

 

[mcastillo356]

paisiello2, I was wrong: the center of gravity is at the center of simmetry. I am still looking after the total torque. Your words made me think. Hope to have your help. It will be neccesary. Forgive my english once again. Please answer.

 

[paisiello2]

Not sure that the center of gravity is at the center of symmetry (whatever that definition might be) but let's assume that we can approximate it to be so.

Please go back and re-read my post #20 regarding the corrections I noted. I think I explained the procedure clearly but let me know if you have any specific questions and I will try to help you if I can.

 

[mcastillo356]

Hello paisiello2. Is the ground force well located?. I've considered it as opposite to $\vec{F}_{yAC}$ and $m_{C}\vec{g}$, as shown in diagram at message #19. And I've considered $\vec{F}_{yAC}$ and $m_{C}\vec{g}$ as parallel forces. So the resultant module is the sum of modules, and the point of application is given by this ecuation: $F_{yAC}\;d_1=m_{C}g\;d_2$, where $d_1$ is the distance from $\vec{F}_{yAC}$ to the point of application of the sum of the two forces, and $d_2$ is also the distance from $m_{C}\vec{g}$ to the sum of the two forces. $\vec{F}_{NC}$ is opposite and equal in module. I've moved $\vec{F}_{yAC}$ and $m_{C}\vec{g}$ along their line of action to the ground, where the ground acts.

The center of mass is at the center of simmetry. It can be calculated by polar integration: consider a circular sector with angle $\pi/3$ which axis of symmetry is $X$ axis. The coordenate $Y$ is zero, and $X$ coordenate is:

$\displaystyle\frac{1}{\frac{\pi}{6}(r_1^2-r_0^2)}\displaystyle\int_{r_0}^{r_1}\int_{-\pi/6}^{\pi/6}\rho^2\cos\theta\;d\theta\;d\rho$

$r_1=2$, and $r_0=1$, so the result is approximately 1.49.

Regarding the total torque, the piece named $C$ might rotate, given the enough weight to the piece $A$, around the point $O$, but it is not the case: The arch with which we are working is a three pieces arch, same in shape and weight. When I say, for example, that piece $A$ weights 0.5, it could mean Newtons or pounds; it could be expressed to scale, or not. When I draw the arch from 1 to 2, it could be any measure: feet, meters...the arch works. I will do the calculations:


$\vec{M}_O(\vec{F}_{AC})=\vec{M}_O(\overrightarrow{F_x}_{AC})+\vec{M}_O(\overrightarrow{F_y}_{AC})$;

$\vec{F}_{AC}=(1'18-0'75,1'05-1'3)=(0'43,-0'25)$;

$\overrightarrow{OM}=(0'75-2,1'3-0)=(-1'25,1'3)$;

$\vec{M}_{O} (\vec{F}_{AC})=\overrightarrow{OM}\land{\vec{F}_{AC}}=\begin{vmatrix} \vec{i}&\vec{j}&\vec{k}\\-1'25&1'3&0\\0'43&-0'25&0\end{vmatrix}=-0'25\vec{k}$.

$\overrightarrow{ON}=(1'3-2,0'75-0)=(-0'7,0'75)$;

$m_{C}\vec{g}=(1'3-1'3,0'25-0'75)=(0,-0'5)$;

$\vec{M}_{O}(m_{C}\vec{g})=\overrightarrow{ON}\land{m_{C}\vec{g}}=\begin{vmatrix}\vec i&\vec{j}&\vec{k}\\-0'7&0'75&0\\0&-0'5&0\end{vmatrix}=0'35\vec{k}$.

The ground force restrains the counterclockwise rotation, so the total torque is $\vec{0}$

 

[mcastillo356]

Hello, if I place $\vec{F}_{NC}$ as shown in diagram, everything seems right:

1- Sum of forces:

$\sum{\vec{F}_{yC}}=m_{C}\vec{g}+\vec{F}_{yAC}+\vec{F}_{NC}=\vec{0}$

$\sum{\vec{F}_{xC}}=\vec{F}_{xAC}+\vec{f}_{eC}=\vec{0}$

2-Total torque about O and Q (because it might be clockwise or counterclockwise rotation).$\vec{f}_{eC}$ does not generate torque:

About O:

$\vec{M}_{O}(\vec{F}_{AC})=\vec{M}_{O}(\overrightarrow{F_x}_{AC})+\vec{M}_{O}(\overrightarrow{F_y}_{AC})$;


$\vec{F}_{AC}=(1'18-0'75,1'05-1'3)=(0'43,-0'25)$;


$\overrightarrow{OM}=(0'75-2,1'3-0)=(-1'25,1'3)$;


$\vec{M}_{O}(\vec{F}_{AC})=\overrightarrow{OM}\land{\vec{F}_{AC}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-1'25&1'3&0\\0'43&-0'25&0\end{vmatrix}=-0'25\vec{k}$.


$\overrightarrow{ON}=(1'3-2,0'75-0)=(-0'7,0'75)$;


$m_{C}\vec{g}=(1'3-1'3,0'25-0'75)=(0,-0'5)$;


$\vec{M}_{O}(m_{C}\vec{g})=\overrightarrow{ON}\land{m_{C}\vec{g}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-0'7&0'75&0\\0&-0'5&0\end{vmatrix}=0'35\vec{k}$.

$\overrightarrow{OP}=(-0'13,0)$;

$\vec{F}_{NC}=(0,0'75)$;

$\vec{M}_{O}(\vec{F}_{NC})=\overrightarrow{OP}\land{\vec{F}_{NC}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-0'13&0&0\\0&0'75&0\end{vmatrix}=-0'10\vec{k}$.

Torque about Q:

$\overrightarrow{QM}=(-0'25,1'30)$;

$\vec{F}_{AC}=(0'43,-0'25)$;

$\vec{M}_{Q}(\vec{F}_{AC})=\overrightarrow{QM}\land{\vec{F}_{AC}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-0'25&1'30&0\\0'43&-0'25&0\end{vmatrix}=-0'50\vec{k}$.

$\overrightarrow{QN}=(0'3,0'75)$;

$m_{C}\vec{g}=(0,-0'5)$;

$\vec{M}_{Q}(m_{C}\vec{g})=\overrightarrow{QN}\land{m_{C}\vec{g}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0'3&0'75&0\\0&-0'5&0\end{vmatrix}=-0'15\vec{k}$.

$\overrightarrow{QP}=(0'87,0)$;

$\vec{F}_{NC}=(0,0'75)$

$\vec{M}_{Q}(\vec{F}_{NC})=\overrightarrow{QP}\land{\vec{F}_{NC}}=\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\0'87&0&0\\0&0'75&0\end{vmatrix}=0'65\vec{k}$.


I don't really know if it is right. The only thing I am sure of is the place of the center of mass of the pieces of my arch, because I've got the proof. Just wanted to publish my last effort. I've already got back to the subject I left almost three months ago.

Thanks a lot!