SA12: Beam Deflection: The Double Integration Method
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- เผยแพร่เมื่อ 16 ต.ค. 2024
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good day maam/sir. i am a second year student studying civil engineering and currently taking strength of materials as part of our curriculum. your lectures are a big help.. i am now studying the deflection of beam, using three methods(double integration, moment area, and superposition). i would like to request that you please upload lectures on ths two remaning method(moment area, and superposition). thank you, i hope that my request will considered.. god bless. by the way, i am a Filipino engineering student
Very good presentation and explanation.
Are you able to findout work efficiency of aneroid barometer as its size increased
One kilometers cubic cylinder cell of aneroid barometer feel how much pressure and doing how much work ???
Please do some math over it
CAN WE HAVE PDF OF OTHER LECTURES
Thank you Soo much...
Please do the same for singularity method...🙏🙏🙏
What will be the deflection formula for steel cylindrical pipe, hand rail with two posts on two side?With a point load? Is it the same way?
Yes, the same principle applies. However, to determine the moment equation for the pipe (beam), the system needs to be analyzed as a frame. That is, the pipe/beam cannot be assumed to be simply supported.
@@DrStructure thank you very much for the clarification i was confused whether it will be a beam with fixed end or frames with fixed end.
Moment of inertia of hollow pipe will be pi*(D^4-d^4)/64
I am confused about the formula regarding the deflection(WL^3/48EI) according to me,but as you mentioned it will be considered as a frame.
I will look at it.
Hi at 11:20 you are saying d theta is the difference between the slopes but the figure on the right shows d theta to be theta 1 + theta 2. Am I missing any sign convention rule. Please help.
Note that in the diagram to the right the two slopes on the beam have different directions. The left one is obtained by trurning the beam clockwise, the right one is the result of a counterclockwise rotation in the beam. So, one is positive and the other is negative. When subtracting one from the other, we end up adding their magnitudes.
They don’t overlap. You can view it as you explained, one is placed to the left of G and one is placed to the right of G.
Thanks a lot....
Where can i find the last part with varying I(x)
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Hello, the theta at 2:25 is the theta 1 at 5:16 right? When substituting back, why not theta/2 ? Please help.
Yes, to the first part of your question. For your second question, you need to say a bit more. I don't see any back substitution taking place.
Dr. Structure Sorry for my poor expression. I think I understand it now. Thank you very much for your video and help!
You are welcome!
Please make a video on castigliano theorem.
Can anyone provide me a link where I can learn things like this what happens at 9:06? I mean I don't understand at all how to simplify the equations and it's really bothering since I understand the idea quite well otherwise. Thanks.
To learn how to simplify equations or how to solve them you need to look into algebra. A introductory college level math course on algebra should give you adequate knowledge and practice for the type of problems you encounter here. For getting comfortable with differentiation and integration problems, you need to look into calculus.
at 8:03, why can't you take the p out of the integral? I thought p is a constant
In both cases (videos), P is constant. Whether we leave such a constant to the right of the integration symbol or move it to the left of the symbol is not significant. That is, Integral (P) = P Integral (1). Or, Integral (P x) = P Integral (x).
The only difference that we would experience is in the integration constant. For example, here, since we left P inside the integral, the expression for v became:
V = (1/EI)( px^3/6 - pLx^2/2 + C1 x + C2)
If we take p out as a constant before integrating, the expression for deflection becomes:
V = (p/EI)(x^3/6 - Lx&2/2 + D1 x + D2).
The only difference between the two equations would be in the values for the integration constants. But that does not matter at all, since if we multiply p by what is inside the parenthesis in the second equation, we get the first equation. That is, C1 = p D1 and C2 = p D2.
it'd be best to use different symbols between tetha in radian and degree.
Why didn't you take M(X) as negative? It's direction is Anti-clockwise
The sign convention is rather arbitrary. We can assume either clockwise or anti-clcokwise to be positive, as long as we use our chosen convention consistently throughout the calculations.
why the moment value is not px? i couldnt understand how the moment value became px-pl. pls explain
@07:46, the beam is fixed at the left end with a counterclockwise moment of PL in place. Further, there is an upward vertical reaction force of P acting at the support. Both the reaction bending moment at the reaction vertical force effect internal bending moment in the beam.
If we cut the beam at some arbitrary point, say x, from the left end of the beam, then bending moment at that point is a function of the reaction forces I mentioned above.
Since PL is already a bending moment, we don't multiply it by x, we simply carry it over to the cut point. The vertical force, however, needs to be multiplied by x in order to determine the moment the force causes about the point.
Hence, the total bending moment at the cut point is: PL - Px. The negative sign implies the two terms are acting in opposite directions, one is acting in the counterclockwise direction and the other in the clockwise direction.
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at 9:42, why you did not consider a moment from the pin joint in FBD?
But there is no moment at that pin. A pin or roller support at an end of a beam has zero moment.
@@DrStructure thankyou for your response ... !
I am really grateful that I have come across your lectures it was really very helpful.
Thankyou very much
You're welcome. Thanks for the feedback.
Why d^2 x not dx^2 or (dx)^2 in the double integral.....?
please help
+VIVEK SINGH That is just a matter of notation. It simply means dx dx, for double integral with respect to x.
Thanks Dr. Structure ... Got it.
You all are doing a great job. Keep uploading.
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hello, why is the theta=Pi/18 at 1:38? Thanks.
theta = 10 degrees
1 degree = pi/180 radians
theta (in radian) = 10 (pi/180) = pi/18
Thank you very much!
You are welcome!
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iI really appreciate what your are doing in those videos. however, at the end of this video, should the moment equation be 500x-100x^2 or not? i don't get from where come 50X^2 you wrote. thank you
The distributed load is 100 N/m, the length of the segment is x. This makes the equivalent concentrated load 100x. This load is placed at the center of the segment, at x/2 from either end. So, the moment arm would be x/2. Therefore, the moment of 100x load is: (100x)(x/2) = 50 x^2.
Why is the moment - M(x) not equal to PL-PL?
Please be more specific. Where?
At 7:40 why is the moment equation: M(x)=Px-PL, I thought because the length was L and not x the equations would be: M(x)= PL-PL=0? Thanks
The moment value at x = L would be PL - PL. That is not what we are looking for here. We are not looking for the value of moment at a specific point, we want to write an equation for the moment for the entire beam, an equation in terms of x where x could take any value between 0 and L. That equation is M(x) = PL - Px. This can be used to determine moment at any point along the length of the beam, at x = 0, we get PL; at x = L/2 (mid-point of the beam) , we get PL - P (L/2); at x = L, we get PL - PL; ...
The double integration method requires the use of the moment equation, not the value of moment at a specific point.
Then why did you wrote M(x) = Px - PL in the video.That equation is wrong.
Why is the moment equation for the beam is wrong? Please explain.
Thanks a lot
Why we need to draw a tangent to the elastic curve 0:53
Because the beam’s deflection can be determined using the slope (tangent) of the curve.
thanks for your hardwork
what is meant by beam deflection?
+Lyrex Ices It means deformation due to bending. In the case of a horizontal beam subjected to a vertical load, it means vertical displacement of the beam at a given point. If the beam moves (displaces) downward at a point by, say 2 mm, then it has a deformation/deflection of 2 mm at that point.
+Dr. Structure thank you
10:55 - where does the equation -12500/3 come from?
@10:55 we have:
v(x) = (250 x^3/3 - 25 x^4/6 + c1 x + c2)/EI
The boundary condition is:
v(10) = 0
Substituting 10 for x in the above equation and setting it to zero, we get:
(250,000/3 - 250000/6 + 10 c1 + c2)/EI = 0
Since c2 = 0, the above equation simplifies to:
250,000/3 - 250000/6 + 10 c1 = 0
Or,
c1 = -12500/3
Thank you very much!
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V(10)=0 I got C1=-25000/3
v = (1/EI)(250 x^3/3 - 25 x^4/6 + c1 x + c2)
set c2 = 0 and x = 10,
v = (1/EI)(250000/3 - 250000/6 + 10 c1)
or,
v = (1/EI)(250000/3 - 125000/3 + 10 c1)
or,
v = (1/EI)(125000/3 + 10 c1).
Solve for c1 by setting v = 0.
(1/EI) (125000/3 + 10 c1) = 0
or, c1 = -12500/3
@@DrStructure
Why not solve for C1 using equation of first integral of Moment equation.? Set V=0 x=10....is it posible to do that way?
@@Mr.P539 The first integral of M/EI gives us the equation for the slope of the beam. If the slope of the beam is known at a specific point, then yes, we can use that equation to determine the integration constant C1. In the case of a simply supported beam, the only known boundary conditions are the end deflections; neither of the end slopes is known.