Find deflection and slope of a cantilever beam with a point load (double integration method)
ฝัง
- เผยแพร่เมื่อ 7 ก.พ. 2025
- This mechanics of materials tutorial goes over an example using the double integration method to find the deflection and slope of a cantilever beam with a single point load.
If you found this video helpful, please consider supporting my work on Patreon:
/ engineer4free
Looking for more mechanics of material tutorials? Check out all of the videos that I made:
TH-cam: • Mechanics of Materials
Website: www.engineer4f...
Looking for software? I highly recommend checking out SkyCiv. They make a full suite of online structural analysis software and tools that are useful for both students and professionals, including calculators for beams, trusses, frames, moment of inertia and more. It’s great for checking your work:
bit.ly/skyciv-e4f
Lastly, don’t forget to subscribe on TH-cam and join the Engineer4Free mail list:
Subscribe: / @engineer4free
Mail List: www.engineer4f...
Thanks for watching, I hope it helps!
This man consistently saves my life every semester, when I'm not a broke student, 100% supporting this mans on Pateron.
Appreciate it Vanessa!!! For now you can just tell some friends about my vids 🙂🙂, that would be a big help to me too!
It's 9:34pm right now and I'm having a final tomorrow at 8:30am, your video just explained to me almost an entire chapter in 13 minutes.
mine is in 2 hours
@@tturi2 Oofff best of luck mate. I just picked up my diploma yesterday, that program year was fucked but it went better afterwards, you got this!
@@nazariimoroz6216 thanks man, I've completed my diploma and doing my honours
This was really clearly explained, thank you so much!!!!
thank you for all the videos...it has been very helpful
Glad to hear it, thanks for commenting!
On the cut beam with the positive moment and shear directions (1:30), why is P(x) negative? Wouldn't P force a positive moment on the beam since it would rotate counterclockwise?
Who on earth figured this stuff out, 10 years after graduating I'm still at awe..
Yeah it's impressive to think about!
Well you can always learn something new.
Thank you very much. It is very useful
Thanks Dom! Glad you liked it =)
Careful, dy/dx is only theta for small dy/dx, it is tan(theta) in general
thanks
Hello there sir. At 1:21 you wrote the moment M to be positive counter-clockwise. So shouldn't M be equal to Px and not -Px? As P will be exerting force counter-clockwise as well?
Hey Naved. M here is short for M(x), which is the internal moment equation in terms of x. On that little diagram at 1:21, P is oriented down. v (the internal shear) is pointing down, but that's just its sign convention, it needs to be equal and opposite to P, so if we had billed it in, it would be pointing up, with a magnitude of P'. Now we have a force couple, P and P', seperated by the distance x. The force couple causes a counter clockwise moment. The internal moment, M, is drawn in in the positive convention, being CCW. But in reality, because there is a CCW moment caused by the force couple, which is equal to Px, then for moment equilibrium, the internal bending moment at x must be equal and opposite, so it is - Px. A different way to think of it is that because the moment from the force couple is CCW, then simply the sense of the internal moment must be opposite, so it's CW. And for equilibrium, they just have the same magnitude of +/- Px. Hope that helps to clarify
do you have an example for cantilever-beam (fixed at left) loaded with 3 concentrated loads
Tq very helpful in online exam 😂
Why do we show the internal bending moment and the internal shear force acting in the positive direction on the free-body diagram, although it should be directed oppositely?
If my BMD looks inverted to you (like you are expecting the positive values to be displayed below the axis) then that’s fine. Depending on which country you’re from, the convention of which side to label as positive will differ. If mine are upside down to you, then just carry on with the way you are learning, and check against the inverted form of mine 👍
@@Engineer4Free I would like to share my rapture towards what you are doing! I'm currently studying a structural mechanics course of the next year cause I won't be able to do so at the due time. Frankly speaking, but for you, I would have already given up this idea twice for sure. So, I just want to say that you are a unique and amazing person! As soon as I earn some money, I will definitely join the Patreon system. Thank you!!!
@@alexeychechin1803 🙂 thanks for the kind words and glad I'm able to help you!!
What is the technical name for the differential equation if this beam is statically determinant?!
Mmmm this is a linear first order ordinary differential equation. Heres a short video that I made that talks a little bit about the classification of differential equations: www.engineer4free.com/4/how-to-classify-differential-equations and you can also check out www.engineer4free.com/differential-equations for more methods solving specific types of ODEs
@@Engineer4Free Thank you! One more question. How can I derive that equation if I wanted to find a deflection at any point. I know you have a table on your website for any loaction 0
Hi
You are taking the clockwise moment to be positive. Can I take the anti clockwise moment to be positive ? Is it going to give me the same correct answer ?
I have drawn the internal bending moment as counterclockwise, not clockwise. I do this at 1:21. This follows the positive sign convention for beam bending (see this video: www.engineer4free.com/4/internal-force-sign-convention) and should be followed when using the double integration method
Good
👍
Is there anywhere where I can check what the boundary conditions are as I’m still confused how you know what they are
is it the deflection point that will always be negative?
It depends which way the actual beam is bending I suppose. If we had a point load pushing up on the end of this cantilever, it would deflect upwards.
How do we prove for cantilever deflection equation for any point
What if you change the x direction from the wall to P?
Thanks for sharing a great video. I am still learning and don't know much about integration. Will also check your website if you have some materials to study. Could you please explain more detailed why the terms become zero at 11:30?
Hey thanks for watching! We are technically solving a second order differential equation in this video, so a good understanding of integration definitely helps. tl;dr the solution to a differential equation is the function y in terms of x, so in this case y=(1/EI)((-P(x^3)/6)+(P(L^2)x/2)-(P(L^3)/3)). That function y(x) in these problems describes the position at every point on the deflected structure. So basically if you make a graph where the origin is at the free end of the beam when it is undeflected, the function y(x) = (1/EI)((-P(x^3)/6)+(P(L^2)x/2)-(P(L^3)/3)) will return the y value of the deflected structure for and x value on the span (ie, it's just basically plotting a side view of the deflected structure on the graph). Once you have determined the function y(x) all you need to do is input the value of x that you are curious about into all of the x's in the function, and then the function will spit out a value for y at that value of x. So when we want to find the displacement at the free end of the beam (at x = 0) then we just plug in zeros for all of the x's in y=(1/EI)((-P(x^3)/6)+(P(L^2)x/2)-(P(L^3)/3)). That makes the terms with x's become zeros because of simple multiplication and we find that y(0)=-P(L^3)/3EI. If you plug in x = L and then simplify, you'll notice that the entire function will equal zero, which matches the boundary condition observed from the diagram at 2:00, same with the slope function, if you plug theta = L, that equation for theta will equal zero, which also confirms the second boundary condition. Does that clarify it? I wouldn't worry too much about how differential equations and integration works, if you made it to this step then its just plugging in values of x into y(x) to get y at that x.
Of course, it's very obvious now when you told it. I was thinking way too complicated. Thank you very much for your answer, didn't expect such a good answer from someone who has 21k subscribers, hehe.
No worries, glad it helps!!
How would I derive for deflection if I have the same beam but the point load is not at the tip, its somewhere at distance 'x'?
Just pretend the beam is x metres long then and from 0 to x the process is identical. From x to the end, the beam will actually have no curve, it will be straight, and the slope anywhere in this region will equal the slope at x. So you can find deflection from 0 to x with the method in this video, and then because you know the deflection at x, then just some simple trigonometry will allow you to find the deflection from x to the end.
@@Engineer4Free Thank you very much! Liking and Subscribing is the least I could do to express my appreciation, I thank you again.
Very informative. I have a small question about a cantilever beam which is half composite (x=L/2) and the other half polymer ((Rem)x=L/2). How can I calculate its deflection when a point load is applied?
You would have to break it up into two problems basically. The curvature and deflection in the first half will depend on the flexural rigidity in that region, and then use the slope and deflection at midpoint as the boundary condition to base the zecond half off with the new flexural rigidity. Doesn't sound like any fun!
which software interphase u are using..? can anyone tell?
Hey Rajdeep, you can find a full list of all the hardware and software that I use at engineer4free.com/tools
How do you also consider the horizontal displacement?
We consider this type of problem to have zero horizontal displacement because there are no horizontally applied loads. We assume that the the deflection at all points is entirely in the y direction.
Engineer4Free yes, but for large deflection, horizontal displacement becomes important doesn’t it? What would be your approach in this case?
At the undergraduate level, if the problem is statically determinate, then just ignore it. If an undergraduate level problem is structurally indeterminate, then you do need to consider the axial forces that develop from what you have described. However, the axial forces in that sense are just used in the calculation of horizontal reaction forces at connections, not the displacements at free ends (if there are no horizontally applied loads). I have a whole course on various methods available for analyzing structurally indeterminate beams, you can check it out here: engineer4free.com/structural-analysis You are ahead of most people by realizing that in real life there should be some horizontal displacement, but just enjoy the simplifications and don't worry about it unless you pursue graduate studies in structural engineering :)
What if we have 2 forces?
Depends on the exact loading and structure. I’d recommend getting familiar with all the methods on engineer4free.com/structural-analysis
Excuse me Sir. Do you have any video on deriving the geometric stiffness matrix derivation Galerkin (36 3L 36 3L)
Hey, sorry I do not
Where is the example sir ?
why is x equal to L. shouldnt X equal L/X
Can you point to the time that you are confused about. x is how far we are along the beam, and can range from 0 to L.
@@Engineer4Free it's fine now I figured it out afterwards lol
What if you swapped the positions so that the wall is now at the origin, and the beam extends to the right?
If the beam is oriented as you described and you need to derive the expression using the double integration method then honestly the easiest way to approach that is set x = 0 at the free end of the beam which would now be on the right side. Define leftward as the positive direction for x and take the exact same approach mathematically just with the image and positive direction switched. If you don't need to derive the expression, then just go to a table like this one: www.engineer4free.com/extras/beam-slope-and-deflection-table or the one in your textbook. The deflection at the free end of a cantilever beam with a single point load acting down perpendicular at the free end will always be y = -PL^3/3EI regardless of which way the beam appears to be oriented in the drawing. On that note, the way the diagram is drawn is really just a result of where the observer is viewing the beam from. You can view a diving board from its "left" side or its "right" side and if you drew a 2D diagram of your view of the diving board than one would look like the set up in my video and one would look like the set up in your problem. Either way it's still the same diving board, so doing something like setting the positive x direction to be leftward on a 2D drawing might feel a little awkward, but it's the same as setting it rightward when viewed from the other side. Hope that makes sense!
Pool
-pL^3/6+pL^3/2+c2
You say c2=-pL^3/3 how sir prove ti
Yes I was understanded sir tnks 100%
Why can’t you just use numbers
See videos 35 - 39 here engineer4free.com/mechanics-of-materials for some more examples with numbers
why would you call a slope theta and put it like an angle threw me off bro
Hmmm it might be dialect. Slope can refer to something like sloped hill. like, "hey, the ground over there is on a slope, it's a hill, the hill has a 30 degree slope." So when I say slope in this case, I'm referring to the angle of the deformed structure after it's loaded. The angle theta can actually be used in the general case to refer to the angle of the deformed structure at any point x. Does that make sense?