How far will the middle point of the beam sag under load?

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  • #1
HAF
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Homework Statement:

There is a wooden beam with given size. We put 1kg in the middle point of beam. Question is how much will the middle point lower itself? Young modulus is given.

Relevant Equations:

F/S = E.ε
I tried to use F/S = E.ε to find the new length after deformation and than use pytagoras theorem in right angled triangle to find how much did the middle point lower.
But it was said to me that my method is wrong and that there exists some formula to solve it. Can you please tell me where could I possibly find the formula?
 

Answers and Replies

  • #2
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I tried to use F/S = E.ε to find the new length after deformation and than use pytagoras theorem in right angled triangle to find how much did the middle point lower.
But it was said to me that my method is wrong and that there exists some formula to solve it. Can you please tell me where could I possibly find the formula?
This seems to me like a beam bending question, for which you will need to use:
[tex] EI\frac{d^2 v}{dx^2} = M(x) [/tex]

Where:
- E = Young's Modulus
- I = Moment of Area of the cross-section of the beam
- v = deflection
- x = Distance along the beam from reference
- M = Bending Moment

Have you heard of the bending moment before? If not, I would suggest you look into that a bit more as it would take a while for me to type it all out. There should be some good content on youtube.

If you have heard of it, can you think of an expression for the bending moment in terms of x and then solve the differential equation?

[EDIT]: When solving the differential equation, think about the supports you have to help you determine boundary conditions.

Hope that is of some help
 
  • #3
HAF
57
6
This seems to me like a beam bending question, for which you will need to use:
[tex] EI\frac{d^2 v}{dx^2} = M(x) [/tex]

Where:
- E = Young's Modulus
- I = Moment of Area of the cross-section of the beam
- v = deflection
- x = Distance along the beam from reference
- M = Bending Moment

Have you heard of the bending moment before? If not, I would suggest you look into that a bit more as it would take a while for me to type it all out. There should be some good content on youtube.

If you have heard of it, can you think of an expression for the bending moment in terms of x and then solve the differential equation?

Hope that is of some help
Thank you very much! No I haven't heard about it so I should check it. So is it solvable from the low amount of information that is given?
 
  • #4
368
73
Thank you very much! No I haven't heard about it so I should check it. So is it solvable from the low amount of information that is given?
No problem, if you want an exact numerical answer, then you will need the dimensions of the beam to work out the second moment of area. Otherwise, you can form an algebraic expression.

Do you know what types of supports you have? Fixed support, 'propped up support', etc?

For the bending moment, I have heard of something called McCauley's method that might help (note: I haven't used it, but it might help you get an expression for M(x)).
 
  • #5
HAF
57
6
No problem, if you want an exact numerical answer, then you will need the dimensions of the beam to work out the second moment of area. Otherwise, you can form an algebraic expression.

Do you know what types of supports you have? Fixed support, 'propped up support', etc?

For the bending moment, I have heard of something called McCauley's method that might help (note: I haven't used it, but it might help you get an expression for M(x)).
The beam is supported on the ends of it.
Length of it is 100cm. The other two dimensions are 5cm and 0,5cm. The cross section is rectangular.

I'm a bit confused. Hopefully I wil somehow solve it.
 
  • #6
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73
The beam is supported on the ends of it.
Length of it is 100cm. The other two dimensions are 5cm and 0,5cm. The cross section is rectangular.

I'm a bit confused. Hopefully I wil somehow solve it.
If you give it a go and post your working/ attempts with any questions, we can help you.
 
  • #7
haruspex
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something called McCauley's method
The basic method is to work from one end of the beam, calling that x=0. At position x, consider the sum of moments there due to torques and forces at points <x. Ignore anything at >x. This gives you M(x), and you can apply the differential equation in post #2.

There is a useful notation: for a vertical downward force F applied at x', write F[x-x'] for the moment at x, where it is understood that the square brackets mean to use a value of zero if the term inside is negative. In this way, you can write a single equation that is valid for all x.

Next step is to integrate to find the slope of the beam as a function of x. Don't forget the integration constant. You can use any known boundary conditions, like slopes at the beam ends, and continuity of slope across load points.
Integrating the [] terms presents no complications.

Integrate a second time to obtain displacement, again applying boundary and continuity conditions to resolve the unknowns.

Note that the differential equation is only valid for small deflections. It ignores lengthwise contractions that result.
 
  • #8
PhanthomJay
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I tried to use F/S = E.ε to find the new length after deformation and than use pytagoras theorem in right angled triangle to find how much did the middle point lower.
But it was said to me that my method is wrong and that there exists some formula to solve it. Can you please tell me where could I possibly find the formula?
I don’t see how or why you are asked to find deflections without first studying moments. Are you just supposed to look up the formula thru a web search for deflection of a simply supported beam under mid point loading in order to compare measured versus theoretical results? What sort of a course is this?
 
  • #9
HAF
57
6
I don’t see how or why you are asked to find deflections without first studying moments. Are you just supposed to look up the formula thru a web search for deflection of a simply supported beam under mid point loading in order to compare measured versus theoretical results? What sort of a course is this?
I'm a high school student and we made this experiment at school and I just wanted to know if I could calculate it with some formula. But I think that I don't have enough knowledge to solve it by myself. Is there anybody who would like to solve it for me please?
 
  • #10
HAF
57
6
If you give it a go and post your working/ attempts with any questions, we can help you.
I decided to solve it on my own. But I have one question. How could I calculate the Moment of Area of the cross-section of the beam? That's the only thing I have no idea what it is and how to calculate it.
 
  • #11
HAF
57
6
I found this. How can I make from Ix and Iy one I to put it to the equation? Just by pytagoras theorem?
Screenshot_20190423-145733_DuckDuckGo.jpg
 
  • #12
HAF
57
6
So for the deflection I found a formula P.L^3/(48.E.I)

The right answer should be 3,33cm
I calculated Ix and put it there to the formula and what I got is 3,39cm.

Is my work at least a bit correct?
 
  • #13
368
73
So for the deflection I found a formula P.L^3/(48.E.I)
That is correct

I calculated Ix and put it there to the formula and what I got is 3,39cm.
So for the second moment of area, we take this about the axis passing through the centroid of the cross-section as that is the axis which the bending moment is being calculated about. You could get the correct second moment of area by using the parallel axis theorem in reverse or just looking it up. The latter yields [itex] I_{centroid} = \frac{A h^2}{12} [/itex] where h is the height of the cross-section and A is the cross-sectional area.

Try adjusting your moment of area and re-doing the calculation. However, dividing by a smaller [itex] I [/itex] will increase your deflection, so I would double check your young's modulus (E) number to ensure it is correct.

Hope that is of some use.

[Edit] I have just re-read an earlier post and see the dimensions of the beam are not the same, thus requiring you to know the orientation of the beam. Therefore, the second moment of area will depend on the orientation of the beam so you know which dimension is the height and which one is the base.
 
  • #14
HAF
57
6
That is correct



So for the second moment of area, we take this about the axis passing through the centroid of the cross-section as that is the axis which the bending moment is being calculated about. You could get the correct second moment of area by using the parallel axis theorem in reverse or just looking it up. The latter yields [itex] I_{centroid} = \frac{A h^2}{12} [/itex] where h is the height of the cross-section and A is the cross-sectional area.

Try adjusting your moment of area and re-doing the calculation. However, dividing by a smaller [itex] I [/itex] will increase your deflection, so I would double check your young's modulus (E) number to ensure it is correct.

Hope that is of some use.

[Edit] I have just re-read an earlier post and see the dimensions of the beam are not the same, thus requiring you to know the orientation of the beam. Therefore, the second moment of area will depend on the orientation of the beam so you know which dimension is the height and which one is the base.
Yeah no worries I know which one is the base and which one is height. Thank You for your help guys. I solved it with your help finally :)
 
  • #15
368
73
Yeah no worries I know which one is the base and which one is height. Thank You for your help guys. I solved it with your help finally :)
Happy to be of help. Out of interest, how was your teacher expecting you to solve this?
 
  • #16
HAF
57
6
Happy to be of help. Out of interest, how was your teacher expecting you to solve this?
He didn't give me this problem as a homework. I wanted to solve it without his help. He just told me that my working is wrong and to try it again. :)
 

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