Engineering Structure Mechanics-Three Pinned/Hinged Arches

[pj33]

Homework Statement

Calculate the reaction Forces at A and B

Relevant Equations

Section Method
Equilibirum of moments

The Example 2 at this site ' http://ocw.ump.edu.my/pluginfile.php/1435/mod_resource/content/2/Chapter%205%20%E2%80%93%20Three%20Pinned%20Arch.pdf ' (page 14):
It is parabolic hence the equation is y = (x^2)/10.
I used the section method between A-D and applied equilibrium of moment to find a relation between the vertical and horizontal reaction at A and then I used again the section method between A-C and again equilibrium of moments.

When i solved the simultenoeous equations I didnt get the right answer, is there a reason why the section method didnt work on a 3hinged arch?

 

[Mark44]

Homework Statement:: Calculate the reaction Forces at A and B
Homework Equations:: Section Method
Equilibirum of moments

The Example 2 at this site ' http://ocw.ump.edu.my/pluginfile.php/1435/mod_resource/content/2/Chapter%205%20%E2%80%93%20Three%20Pinned%20Arch.pdf ' (page 14):
It is parabolic hence the equation is y = (x^2)/10.
I used the section method between A-D and applied equilibrium of moment to find a relation between the vertical and horizontal reaction at A and then I used again the section method between A-C and again equilibrium of moments.

When i solved the simultenoeous equations I didnt get the right answer, is there a reason why the section method didnt work on a 3hinged arch?

I looked through the pdf but I don't see what you are talking about. I did not find the equation $y = \frac {x^2}{10}$ anywhere in this document.

 

[pj33]

I looked through the pdf but I don't see what you are talking about. I did not find the equation $y = \frac {x^2}{10}$ anywhere in this document.

Sorry i didnt explain my thought,
It says it is parabolic, so I set the origin at C with y-positive downwards and x-positive towards the right.
I used y = ax^2, where i set x=5 and y = 5/2

 

[Mark44]

Please post an image of just the page you're talking about, not the whole pdf. It's difficult to find a particular page, since the pages aren't numbered in the pdf.

 

[pj33]

It is a bit hard to do this because I am on a phone now, I am talkign about Example 2

 

[pj33]

Here is a pdf with the example and the answers

 

[Mark44]

To be honest, I'm not following all of the calculations in the pdf. One thing that the author is doing that's different from what you're doing is to put the origin in a different place. Based on example 1 (I realize you're asking about example 2), the author puts the origin at the left end of the bridge, not at the apex of the bridge. Also, the y-axis is positive upward, not downward as you have described. This makes the equation of the bridge $y = 0.1x(x - 10)$.

One thing that isn't clear to me in the image is where it says 16kN/m.

 

[pj33]

To be honest, I'm not following all of the calculations in the pdf. One thing that the author is doing that's different from what you're doing is to put the origin in a different place. Based on example 1 (I realize you're asking about example 2), the author puts the origin at the left end of the bridge, not at the apex of the bridge. Also, the y-axis is positive upward, not downward as you have described. This makes the equation of the bridge $y = 0.1x(x - 10)$.

One thing that isn't clear to me in the image is where it says 16kN/m.

I tried my way to solve it, I my equation describes the same curve as yours, I just need it to calculate the distances.
On the pdf it uses equilibrium of moments over the whole curve, I try to use the method of sections and moment equilibirum by I didnt get the same answer, is there a reason for that? I was sure that would work but it didnt and I don't know why.
For the last bit of 16kN/m I thing it just means that there is an extra vertical force exerted there, hence that force is equal to 16*5kN and it is exerted at 2.5m right to A.

Are there any restrictions on the Method Of Sections? Can I use the method of section on joints?

 

[jrmichler]

I worked it slightly differently from the example, and got the same answers as the handout. Here's how I did it:

1) Considering the entire arch, a sum of moments about A results in a vertical force component at B of 20 kN.

2) The vertical force component at A can be found by either sum of moments about B or by sum of vertical forces. Either way results in the same answer: 60 kN.

3) Make an FBD of segment BC, and calculate sum of moments about C to get the horizontal force component at C of 40 kN.

4) The horizontal force component at B is equal and opposite to the horizontal force component at C.

5) The horizontal force component at A can be found the same way, or by realizing that it is equal and opposite to the horizontal force component at B.
All of this is exactly the same as in your Doc2.pdf, just expanded.

6) Use the equation of a parabola that goes through the points A, B, and C to get the XY coordinates of Point D, which is 2.5 to the right of A (as is given), and 1.875 m up from A (as noted in the handout). I worked directly from the equation of a parabola starting a C with the positive direction downward to make the calculation easier (Y = 0.1 * X^2).

7) Make an FBD of segment AD. Here I did it differently from the example. I used only the horizontal and vertical force components to get the horizontal and vertical force components at D, then sum of moments about A to get the moment at D. I got the same answer as in the handout.

8) The force component at D parallel to the arch can be found by calculating the slope of the arch at D, then summing the force components at D in that direction. I got the same answer as in the handout.

I did not look at the equations in the handout for calculating the coordinates and slope at D, but worked with the simple equation (Y = 0.1 * X^2), and it's derivative. When solving this type of problem, keep certain basic principles in mind:

1) Sum of forces always equals zero, regardless of the coordinate system. So use the coordinate system that makes the problem easiest to solve.

2) When the desired result is force components in a particular direction, it may be easier to calculate all forces in some other direction, then find the desired force components as the last step.

3) FBD's always work, and you can cut a part anywhere you want to make an FBD.

 

[vela]

On the pdf it uses equilibrium of moments over the whole curve, I try to use the method of sections and moment equilibirum by I didnt get the same answer, is there a reason for that?

The glib answer is "because you did it wrong." Unfortunately, it's impossible to say where you made mistakes because you didn't show your work.

 

[pj33]

I worked it slightly differently from the example, and got the same answers as the handout. Here's how I did it:

1) Considering the entire arch, a sum of moments about A results in a vertical force component at B of 20 kN.

2) The vertical force component at A can be found by either sum of moments about B or by sum of vertical forces. Either way results in the same answer: 60 kN.

3) Make an FBD of segment BC, and calculate sum of moments about C to get the horizontal force component at C of 40 kN.

4) The horizontal force component at B is equal and opposite to the horizontal force component at C.

5) The horizontal force component at A can be found the same way, or by realizing that it is equal and opposite to the horizontal force component at B.
All of this is exactly the same as in your Doc2.pdf, just expanded.

6) Use the equation of a parabola that goes through the points A, B, and C to get the XY coordinates of Point D, which is 2.5 to the right of A (as is given), and 1.875 m up from A (as noted in the handout). I worked directly from the equation of a parabola starting a C with the positive direction downward to make the calculation easier (Y = 0.1 * X^2).

7) Make an FBD of segment AD. Here I did it differently from the example. I used only the horizontal and vertical force components to get the horizontal and vertical force components at D, then sum of moments about A to get the moment at D. I got the same answer as in the handout.

8) The force component at D parallel to the arch can be found by calculating the slope of the arch at D, then summing the force components at D in that direction. I got the same answer as in the handout.

I did not look at the equations in the handout for calculating the coordinates and slope at D, but worked with the simple equation (Y = 0.1 * X^2), and it's derivative. When solving this type of problem, keep certain basic principles in mind:

1) Sum of forces always equals zero, regardless of the coordinate system. So use the coordinate system that makes the problem easiest to solve.

2) When the desired result is force components in a particular direction, it may be easier to calculate all forces in some other direction, then find the desired force components as the last step.

3) FBD's always work, and you can cut a part anywhere you want to make an FBD.

Thank you very much, I think my mistake before when I was using a certain segment, it was that i ignores the bending moment.
Where you say ''I used only the horizontal and vertical force components to get the horizontal and vertical force components at D, then sum of moments about A to get the moment at D.'', you mean that you calculated the bending moment right?

 

[pj33]

The glib answer is "because you did it wrong." Unfortunately, it's impossible to say where you made mistakes because you didn't show your work.

The calculations were easy to do, my concept was wrong but I couldn't say which part was wrong, as it seems I ignored the bending moment.