How do I find deflection in this beam?

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The discussion focuses on calculating the deflection at point C of a beam with a roller connection at B and an external pin at A. Key parameters include the modulus of elasticity (E = 2.1 × 105 N/mm2) and the area moment of inertia (I = 57.9 × 106 mm4). The distances AB (3 meters) and BC (2 meters) are critical for the calculations. The final calculated deflection at point C is approximately 5.92 mm, confirmed by multiple methods and calculations.

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How do I find deflection (displacement) in point C at the end of the beam? Answer is to be given in mm. There is a roller connection (only Fy) in B and an external pin in A (Fy and Fx). The module of elasticity and area moment of inertia is given.

2vwzre8.jpg


Relevant equations:
1. http://en.wikipedia.org/wiki/Deflection_(engineering)
2. http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf


Here is a solution to a similar problem: https://dl.dropboxusercontent.com/u/11241083/beer_mecanica_4e_solucionario_c09_a.pdf
 
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First, you are going to need to know E, I, and LAB and LBC.
 
I tried solving this using the unit load method.
I = 57.9 * 10^6 (mm^4)
E = 2.1 * 10^5 (N/mm^2)
AB distance = 3 meters
BC distance = 2 meters

https://dl.dropboxusercontent.com/u/11241083/Solution_Beam_Problem.pdf
 
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raymanmusic: I currently got an answer different from your post 3 answer for deflection of point C, using a different method.
 
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raymanmusic said:
How do I find deflection (displacement) in point C at the end of the beam? Answer is to be given in mm. There is a roller connection (only Fy) in B and an external pin in A (Fy and Fx). The module of elasticity and area moment of inertia is given.

2vwzre8.jpg


Relevant equations:
1. http://en.wikipedia.org/wiki/Deflection_(engineering)
2. http://www.advancepipeliner.com/Resources/Others/Beams/Beam_Deflection_Formulae.pdf


Here is a solution to a similar problem: https://dl.dropboxusercontent.com/u/11241083/beer_mecanica_4e_solucionario_c09_a.pdf
You failed to use the template which was provided. Do not do this again, we have rules for a reason and they must be followed.

Answers discussed in OP's first thread.

https://www.physicsforums.com/showthread.php?t=711176
 

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raymanmusic: I now also used a second method, and I got exactly the same answer I got using a completely different method in post 4. Therefore, I am about 99 % certain I have the correct answer.

You did excellent work in post 3, but it appears you made one main mistake, as follows. Delete your unit load at x2 = 666,67 mm, and then try your solution again. You do not want to find deflection at x2 = 666,67 mm, and you therefore should not place a unit load at x2 = 666,67 mm. Try again.

By the way, always maintain four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits. E.g., -7,556 kN, not -7,56 kN.
 
Thanks for helping me out nvn. Do I place the unit load where I want to find deflection? What about the triangle load? Should I ignore it when I calculate mx2? What did you get as your final answer in mm?
 
raymanmusic: Yes, place the unit load where you want to find deflection. Yes, ignore the triangular load when you compute mx1 and mx2. I will let you know if you obtain the correct final answer, if this is not a test question.
 
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I got approximately 5.922 mm as my final answer. Is this close to the correct answer?
 
  • #10
Excellent work, raymanmusic. Your answer is correct (5.921 54, which rounds to 5.92 mm).

By the way, N/mm^2 is called MPa. If a derived unit has a standardized special name, then only the standardized[/color] special name should be used. E.g., 210 000 MPa, not 210 000 N/mm^2.
 
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  • #11
Thanks for the help nvn. I'll ask you if I have more problems I'm stuck with.
 
  • #12
deflection101 came with a formula for this case,Where is it?
 
  • #13
Yc=[wb^3(5L-2b)+30PLb^2] / 90
 
  • #14
given: w=10kn/m
p=8kn/m
a= 3
b=2
EI=1 (to simplify)

Ans.:yc= - 72 (-160/3 - 56/3)
 
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  • #15
it doesn't get any better than this! yc= - 72
 
  • #16
EIy(x)=wx/360b(3x^4-15bx^3+30b^2X^2-30b^3X-20ab^3)+Px/6(x^2-3bx-2ab)
EIy(2)=-72
y=- 72/EI EI=read above
 

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