# Why the forces in the arch are carried to the ground

1. Apr 20, 2014

### mcastillo356

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

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
First law of static equilibrium. I must prove that the vertical forces on every part of the system is zero: ∑FAy=0, ∑FBy=0, ∑Cy=0. And then add them to explain how the system works

3. The attempt at a solution
∑FAy = -mAg + FBAysenθ + FCAysenθ = 0
∑FBy = -mBg - FABysenθ + FBN = 0
∑FCy = -mCg - FACysenθ + FCN = 0

Where FBAysenθ is the vertical component of the force B exerts on A; -mAg is the weight of A; FBN is the normal force grounds exerts on B.

∑FAy + ∑FBy + ∑FCy = 0 = -mAg -mBg - mCg + FBN + FCN.

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

¿Is it right?.

2. Apr 20, 2014

### 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.

3. Apr 20, 2014

### 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 $\sum{F_{Ay}}$, wich represents my notation for the vertical forces on $A$.

#### Attached Files:

• ###### Arch.pdf
File size:
20.4 KB
Views:
62
Last edited: Apr 20, 2014
4. Apr 20, 2014

### 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.

Last edited: Apr 21, 2014
5. Apr 21, 2014

### 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?.

#### Attached Files:

• ###### Arch 2.pdf
File size:
14.5 KB
Views:
71
6. Apr 21, 2014

### 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.

Last edited: Apr 21, 2014
7. Apr 21, 2014

### 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!

8. Apr 21, 2014

### mcastillo356

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

9. Apr 21, 2014

### 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".

10. Apr 23, 2014

### 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.

#### Attached Files:

• ###### arch3.pdf
File size:
18.9 KB
Views:
65
11. Apr 23, 2014

### paisiello2

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.

12. Apr 23, 2014

### 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!

#### Attached Files:

• ###### arch3.pdf
File size:
25.8 KB
Views:
35
13. Apr 24, 2014

### 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.

14. Apr 24, 2014

### 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!

15. Apr 27, 2014

### 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!

16. Apr 27, 2014

### 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 text book to see in detail how to take moments for a body in equilibrium.

17. May 26, 2014

### 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!

#### Attached Files:

• ###### arch3.pdf
File size:
29.8 KB
Views:
39
18. May 27, 2014

### 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.

Last edited: May 27, 2014
19. Jun 10, 2014

### 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.

#### Attached Files:

• ###### Arco momentos 3.pdf
File size:
26 KB
Views:
45
20. Jun 13, 2014

### paisiello2

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.

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

Yes!

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

Last edited: Jun 13, 2014