In the previous article, we actually used the multiple integral method to calculate the deflection of a beam.
While the deflection of most beams can generally be calculated using the multiple integral method, the calculation is cumbersome, so I will explain a more intuitive and simpler method.
As I've mentioned many times before, simulations have become easy and accessible these days, so you'll probably never actually need to use the method I'm going to introduce.
However, simply remembering the concept can often be helpful in predicting structural deformation when creating layouts and sketches, so I would like you to understand it.
Solution using the area moment method
The new method I'm going to introduce is called the area moment method, but the name alone doesn't really explain it, so I'll explain it step by step.
Let's set up the usual example problem. A load P is applied to the tip of a cantilever beam of length l. The coordinate system is such that the leftmost point is the origin.
We will now determine the deflection of this.
The multiple integral method involves determining the bending moment at an arbitrary point from the balance of external forces and moments, but this time it's a little different.
Proof of the area moment method
First, select an arbitrary interval AB. Here, set the deflection angle at point A as θA and the deflection as yA, and set point B similarly.
First, let's recall the deflection angle and the deflection formula that we introduced previously.
$ \frac{dy}{dx}=θ=-\int\frac{M}{EI}dx+C1 (where C1 is the integration constant)$
$ y=\intθdx=-\int\int\frac{M}{EI}dxdx+C1x+C2 $
Let's use the two equations above to find the deflection angle θB and deflection yB in the example problem.
The boundary condition for the deflection angle is θA when x is xA, so the boundary condition C1 is θA.
$ θB=θA-\int_{xA}^{xB}\frac{M}{EI}dx $
$ \int_{xA}^{xB}\frac{M}{EI}dx $ represents the area of the bending moment diagram (BMD).
The boundary condition for deflection is that when x is xA, the amount of deflection is yA, so the boundary condition C2 is yA (C1 is θA, x is xB-xA).
$ yB=-\int\int\frac{M}{EI}dxdx +θA(xB-xA)+yA $
The problem here is $ \int\int\frac{M}{EI}dxdx $.
The rules for double integrals dictate that the same variable (dx in this equation) cannot be repeated, so we split the variable into two: dξ and dx (ξ has no special meaning). It's tedious, but we also need to split the integration range into ξ and x, so ξ goes from xA to x and x goes from xA to xB (double integrals require stepping).
The formula then becomes as follows:
$ yB=yA+θA(xB-xA)-\int_{xA}^{xB}\int_{xA}^{x}\frac{M}{EI}dξdx $
First, let's solve the double integral $ \int_{xA}^{xB}\int_{xA}^{x}\frac{M}{EI}dξdx $ (this part is arithmetic).
Calculation of the double integral $ \int_{xA}^{xB}\int_{xA}^{x}\frac{M}{EI}dξdx$
The middle part of the double integral, $ \int_{xA}^{x}\frac{M}{EI}dξ$, becomes a function of x when integrated, so we tentatively denote it as f(x).
$\int_{xA}^{xB}\int_{xA}^{x}\frac{M}{EI}dξdx=\int_{xA}^{xB}f(x)dx $
Now let's find the function that is not visible. The invisible function is the 1 in $ \int_{xA}^{xB}f(x)×1dx $. In the integral expression, 1 can be considered as the derivative of x, so let x be g(x) and 1 be g'(x).
$ \int_{xA}^{xB}\int_{xA}^{x}\frac{M}{EI}dξdx=\int_{xA}^{xB}f(x)dx=\int_{xA}^{xB}f(x)g'(x)dx(' represents differentiation) $
Here, we apply integration by parts from high school mathematics, $ \int f(x)g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$, to perform the calculation.
$ \int_{xA}^{xB} f(x)g'(x)dx=[f(x)g(x)]_{xA}^{xB}-\int_{xB}^{xA} f'(x)g(x)dx $
Substituting f(x) and g(x) back in, we get the following equation. Since f'(x) is the derivative of f(x), it becomes $ \frac{M}{EI}$, which is the derivative of $ \int_{xA}^{x}\frac{M}{EI}dξ$.
$ [f(x)g(x)]_{xA}^{xB}-\int_{xB}^{xA} f'(x)g(x)dx=[\int_{xA}^{x}\frac{M}{EI}dξ・x]_{xA}^{xB}-\int_{xA}^{xB}\frac{M}{EI}xdx $
Now let's calculate $ [\int_{xA}^{x}\frac{M}{EI}dξ・x]_{xA}^{xB} $
$ [\int_{xA}^{x}\frac{M}{EI}dξ・x]_{xA}^{xB}=\int_{xA}^{xB}\frac{M}{EI}dξ・xBー\int_{xA}^{xA}\frac{M}{EI}dξ・xA $
$ \int_{xA}^{xA}\frac{M}{EI}dξ・xA $ is zero because the integration range is from xA to xA.
$ [\int_{xA}^{x}\frac{M}{EI}dξ・x]_{xA}^{xB}=\int_{xB}^{xA}\frac{M}{EI}dξ・xB $
Therefore
$ [\int_{xA}^{x}\frac{M}{EI}dξ・x]_{xA}^{xB}-\int_{xA}^{xB}\frac{M}{EI}xdx=\int_{xB}^{xA}\frac{M}{EI}dξ・xB -\int_{xA}^{xB}\frac{M}{EI}xdx $
Here, $ \int_{xB}^{xA}\frac{M}{EI}dξ $ and $ \int_{xB}^{xA}\frac{M}{EI}dx $ have the same meaning but different variable names, so they can be rewritten as follows.
$ \int_{xB}^{xA}\frac{M}{EI}dξ・xB-\int_{xB}^{xA}\frac{M}{EI}xdx =\int_{xA}^{xB}\frac{M}{EI}(xB-x)dx $
That finally finished the calculation of the double integral.
Meaning of $ \int_{xA}^{xB}\frac{M}{EI}(xB-x)dx $
Let's consider the meaning of $ \int_{xA}^{xB}\frac{M}{EI}(xB-x)dx $.
Here, let xB-x in $ \int_{xA}^{xB}\frac{M}{EI}(xB-x)dx $ be Y.
$ \int_{xA}^{xB}\frac{M}{EI}(xB-x)dx=\int_{xA}^{xB}\frac{M}{EI}Y・dx\frac{dY}{dx}=-\int_{xA}^{xB}\frac{M}{EI}YdY=\int_{xB}^{xA}\frac{M}{EI}YdY $
Does this $ \int_{xB}^{xA}\frac{M}{EI}YdY $ look familiar?
As explained in the diagram, since Y is set as xB-x, it becomes the first moment of area about the line segment B1-B2 of BMD ($ First moment of area \int_{A}ydA $).
It's important to remember that the centroid can be found by dividing the first moment of area by the area.
If xG is the centroid in the x-axis direction from the line segment B1-B2 of BMD, then the following equation holds.
$ \int_{xB}^{xA}\frac{M}{EI}YdY=AxG(A is the area of BMD)=\int_{xA}^{xB}\frac{M}{EI}(xB-x)dx $
In other words, the double integral $ \int_{xA}^{xB}\int_{xA}^{x}\frac{M}{EI}dξdx $ becomes a simple integral $ \int_{xA}^{xB}\frac{M}{EI}(xB-x)dx $ and finally becomes just AxG (note that xG is the distance from the line segment B1-B2).
This AxG is called the area moment, hence the method is called the area moment method.
The relationship between the first moment of area and the centroid is explained in the article below.
Tawami yB
Now we return to the initial deflection yB.
$ yB=yA+θA(xB-xA)-\int_{xA}^{xB}\int_{xA}^{x}\frac{M}{EI}dξdx, θB=θA-\int_{xA}^{xB}\frac{M}{EI}dx $
Expressing the above equation in terms of AxG results in the equation below.
$ yB=yA+θA(xB-xA)-AxG, θB=θA-A $
Ultimately, deflection can be calculated using a simple formula if you know the boundary conditions, the area of the bending moment diagram, and the centroid.
Therefore, you don't need to go to the trouble of solving the second differential equations that I've introduced so far.
Let's actually try calculating the deflection using the area moment method.
Let's actually try using the area moment method.
Find the deflection of a cantilever beam of length l with a load P applied to its tip.
The bending moment is simply $ \frac{P}{EI}(lx)$.
If θA and θB are the angles of deflection at both ends of the beam, then θA is fixed so it is 0.
$ θB=-A=-\frac{Pl^2}{2EI} $
Similarly, yB
$ yB=-AxG=\frac{Pl^3}{3EI} $
It can be easily calculated as follows.
The section on determining the typical deflection of a beam calculates the deflection of a beam under the same conditions, so please compare them.
When the relationship between bending moments can be easily understood in this way, the deflection can be easily determined.
The relationship between typical curves, area, and centroid.
I've included diagrams of typical bending moment shapes, their areas, and their relationship to the centroid, so please feel free to use them.
When you look at the external forces acting on a structure and can easily imagine the bending moment, recalling the area moment method and visualizing it will make it easy to imagine the amount of deformation.
Method for solving deflection problems using singular functions
Up until now, we have been calculating the deflection when a single type of load is applied to a beam.
However, in reality, beams often have multiple types of loads applied to them. Even in such cases, it is possible to solve the problem using the multiple integral method, but it becomes quite complicated.
Therefore, using something called a singular function makes it a bit easier to find, so I will explain that.
What are singular functions?
This isn't that difficult; it's just a rule, so please just keep it in mind.
The properties of singular functions are shown in the following figure.
It's very simple: the function is valid only when the value inside the angle brackets is positive; otherwise, it's treated as 0.
When you make it into a diagram, $ ^0 $ ($ x^0 = 1 $)
Well, these singular functions are more commonly used in control engineering and logical formulas. I might explain control engineering in more detail later, so I'll cover that in more detail then.
Now, let's look at how to express the bending moment M using singular functions. Consider the case where the bending moment M0, load P, and uniformly distributed load q are acting.
Substituting this formula for each force acting on the beam into the equation will make it a little easier to find.
Let me briefly explain an example of a uniformly distributed load.
An example of a uniformly distributed load is the distance x over which no load is applied.
The treatment of bending moments under uniformly distributed loads is described on the page below, so please take a look if you are unsure.
As an example of a solution using this singular function, let's consider a double-supported beam as shown in the figure below.
Using the singular functions in the table above, you should be able to quickly determine the bending moment.
Bending moment M = RA + P ^1-M0 ^0 - \frac{q}{2} ^2 $
From here, you can either solve the second differential equation or use the method of area moment.
Summary
In summary, this article introduced the surface moment method and a method using singular functions.
There is only one pointIn the area moment method, the area of the bending moment curve and the position of the centroid are related to the deflection angle and deflection amount.That's all.
All you need to do is remember that there was such a solution, and you'll be fine.
Especially these days, since these things can be easily obtained through simulations, you probably won't have many opportunities to use what I've introduced here, but it's good to have a wide range of knowledge, so it's fine to understand it.
However, this is different for students; if they can't do this, I think it will be difficult for them to solve the beam deflection problem within the time limit in the unit acquisition exam.
I could just use the multiple integral method to solve it, but there are probably other exam questions, so I think that would be a bad idea.
Next time, I'd like to introduce some tips for solving statically indeterminate beam problems, and if possible, even cover combined beams (made of different materials).
To those who found this article helpful in understanding design:
While there is basically no textbook covering this content, and it is my own original work, I will introduce the textbook that I have been using since I was a student.
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