Beam deflection calculations for beginners: cantilever beams, double-sided cantilever beams

In the previous lesson, we explained the basics of how to determine the deflection of a beam. This time, we will actually calculate the deflection of some typical beams.

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Beam deflection explained for beginners (deflection curve, amount of deflection, and deflection angle) I've explained beams over the past three installments. This time, as a culmination of that, I'd like to calculate the deflection of a beam. Determining this beam deflection is very important; without it, the machine...

There are two basic types of beams: cantilever beams and simply supported beams (double-supported beams).

As I mentioned last time, even complex structures, when broken down, are usually composed of these two things.

Therefore, by memorizing the deflection angles and deflection amounts of the two beams, it becomes easy to predict the deformation of most structures during design studies.

Especially during the layout and sketching stages, precise values ​​aren't necessary; what's crucial is understanding the direction and degree of deformation.

It's not that difficult, so I really hope you'll understand it. Also, since it's an important topic, I'll explain it as carefully as possible.

Deflection of a cantilevered beam

First, let's explain cantilever beams. We will explain two scenarios: one where a uniformly distributed load is applied to the beam, and another where the load is applied to the tip of the beam.

When a uniformly distributed load is applied

Let's set up an example problem as usual. A uniformly distributed load q [for example, in N/mm] (total load is ql) is applied to the entire length l of a cantilever beam. The coordinate system is set with the tip of the beam as the origin.

First, we find the bending moment M at an arbitrary point C. As a recap from last time, the deformation due to shear force is minimal, so we ignore it and do not consider it here.

As usual, we consider the balance of forces in the cross-section cut at an arbitrary point C, and then consider the balance of moments.

Since we will ignore shear force in this case, we will only calculate the moment.

We need to find the moment at an arbitrary point C, but since a uniformly distributed load q is applied to the entire beam, it cannot be calculated simply.

Therefore, we can find the answer by considering the moment acting on a small interval dξ at a distance ξ from point C and summing (integrating) it up to point C.

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Shear force and bending moment (SFD, BMD) of beams explained for beginners As explained in the previous lesson on annular stress and trusses, we will now begin our discussion of beams. Beam analysis is one of the most important aspects of mechanical design and is frequently used. Furthermore, it is covered in the first half of material mechanics…

The moment acting on point C at distance ξ is calculated as distance × force, so $qdξ × ξ$ and added up to the distance x from point C to the origin, so the bending moment M is as follows.

Furthermore, as can be seen in the diagram, ξ starts from the x-point and goes to the origin, so the integration range is x~0 (the orientation of the coordinate system of ξ is opposite to that of x).

In $M=\int_x^0qdξ×ξ=\frac{q}{2}x^2$, the sign is negative because the moment is counterclockwise (the direction that deforms the beam upwards).

Now that we know the bending moment, we can substitute it into the differential equation for deflection that we derived last time.

$ \frac{d^2y}{dx^2}=-\frac{M}{EI}=-\frac{q}{2EI}x^2 $

If we integrate this then

$ θ=\frac{dy}{dx}=\frac{q}{2EI}(\frac{x^3}{3}+C1) $

$ y=\frac{q}{2EI}(\frac{x^4}{12}+C1x+C2) $

Becomes

Now let's consider the important boundary conditions that we discussed last time.

In this cantilever beam, the fixed end (right end) is forcibly a wall and the displacement is 0, so at X=l, $ \frac{dy}{dx}=0 $ and y=0 $, and substituting these into the equation...

$ C1=-\frac{l^3}{3},C2=\frac{l^4}{4} $

Becomes

From these, the deflection and angle of deflection of the beam can be calculated using the following formulas.

$ θ=\frac{dy}{dx}=\frac{q}{6EI}(x^3-l^3) $

$ y=\frac{q}{24EI}(x^3-4l^3x+3l^4) $

Therefore, the maximum deflection angle θmax and the maximum deflection amount ymax are thought to occur at the tip (x=0).

$ θmax=-\frac{ql^3}{6EI} $

$ ymax=\frac{ql^4}{8EI} $

Becomes

Thus, determining the boundary conditions, maximum deflection angle, and deflection amount requires visualizing the movement of the beam.

Therefore, instead of simply memorizing the explanations given so far, let's deepen our understanding of physical phenomena and be able to visualize them.

If you forget, just look it up. I want you to be able to understand the meaning immediately when you look it up.

Next, let's find the deflection when a load P is applied to the tip of a cantilever beam.

When a load is applied to the tip of the beam

First, as usual, let's set up an example problem. A cantilever beam of length l has a load P applied to its tip. The coordinate system is set with the tip A as the origin.

As usual, we find the balance of forces and moments at an arbitrary point C. Shear forces can be ignored in deflection.

Consider the bending moment around a point C at a distance x from an arbitrary origin.

From the figure, the bending moment is $M=-Px$, so substituting this into the differential equation for deflection,

$ \frac{d^2y}{dx^2}==-\frac{M}{EI}=\frac{P}{EI}x $

From here, we can find the angle of deflection θ and the amount of deflection y.

$ θ=\frac{P}{EI}(\frac{x^2}{2}+C1) $

$ y=\frac{P}{EI}(\frac{x^3}{6}+C1x+C2) $

The boundary conditions are that at the fixed ends, x=l, θ and y are 0.

$ C1=-\frac{l^2}{2}. C2=\frac{l^3}{3} $

Therefore, the deflection angle θ and deflection y of the beam are,

$ θ=\frac{dy}{dx}=\frac{P}{2EI}(x^2-l^2) $

$ y=\frac{P}{6EI}(x^3-3l^2x^2+2l^3) $

Therefore, the maximum deflection angle θmax and the maximum deflection amount ymax occur at tip A.

$ θmax=-\frac{pl^2}{2EI} $

$ymax=\frac{Pl^3}{3EI}$

This is something useful to remember.

When you break down structures, you'll frequently encounter cantilever beams, so try to visualize them quickly. While cantilever structures aren't ideal, they're sometimes functionally necessary. In those cases, design carefully.

Next is the simply supported beam.

Deflection of a simply supported beam (double-supported beam)

Here, we will introduce two patterns of simply supported beams. One is the case where a load is applied at any point on the beam, and the other is the case where only bending moments are applied to both ends of the beam.

When a load is applied to any point on the beam

Let's set up an example problem as usual. We want to find the deflection of a beam of length l supported at both ends A and B, with a load P applied at a distance a from point a (the distance from point B is la, but we'll use b for simplicity). Point A is the origin of the coordinate system.

First, consider the balance of external forces.

$ P = RA + RB $

Next, from the balance of moments around point A and point B...

$ Pa=RBl, Pb=RAI $

From here, RA and RB are,

$ RA=\frac{Pb}{l}, RB=\frac{Pa}{l} $

Becomes

Next, let's consider the equilibrium of internal forces. If we take point A as the origin of the coordinate system, there are two possible scenarios: when an arbitrary point x is before the position of load P, and when it is beyond the position of load P.

As shown in the diagram, the bending moment M differs depending on the position x.

$ M1=RAx=\frac{Pb}{l}x (0≦x≦a) $

$ M2=RAx-P(xa)=\frac{Pa}{l}(lx) (a≦x≦l) $
*Let's try using $P=\frac{Pb+Pa}{l} and l=a+b$

Substituting this into the differential equation for deflection,

$ \frac{d^2y}{dx^2}=-\frac{M1}{EI}=-\frac{Pb}{EIl}x (0≦x≦a) $

$ \frac{d^2y}{dx^2}=-\frac{M2}{EI}=-\frac{Pa}{EIl}(lx) (a≦x≦l) $

Solving this using the multiple integral method gives

$ θ1=\frac{dy}{dx}=-\frac{Pb}{EIl}(\frac{x^2}{2}+C1) (0≦x≦a) $

$ θ2=\frac{dy}{dx}=-\frac{Pa}{EIl}(-\frac{(lx)^2}{2}+C2) (0≦x≦a) $

$ y1=-\frac{Pb}{EIl}(\frac{x^3}{6}+C1x+C3) (0≦x≦a) $

$ y2=-\frac{Pa}{EIl}(\frac{(lx)^3}{6}-C2(lx)+C4) (a≦x≦l) $

Here the boundary condition is C1,I ended up with four C2, C3, and C4.

If there are four unknowns, there are four equations needed to solve it, so there are four boundary conditions.

As a fun fact, to create an equation with n variables, you need n equations (if there is one variable x, you only need one equation; if there are two variables x and y, you need two equations).

This is obvious if you think about it: since the beams are connected, even if the equation is different, the angle of deflection and the amount of deflection should be the same when x=a.

From this

$ C3=C4=0 $

$ θ1=\frac{a^2b}{2}+C1b=-\frac{ab^2}{2}+C2a=θ2 $

$ y1=\frac{a^2}{6}+C1=\frac{b^3}{6}-C2=y2 $

From this

$ C1=-\frac{a(a+2b}{6}, C2=\frac{b(2a+b)}{6} $

The angle of deflection and the amount of deflection are expressed by the following formulas.

$ θ1=\frac{dy}{dx}=\frac{Pb}{6EIl}(-3x^2+a(a+2b)) (0≦x≦a) $

$ θ2=\frac{dy}{dx}=\frac{Pa}{6EIl}(3(lx)^2-b(2a+b)) (a≦x≦b) $

$ y1=\frac{Pb}{6EIl}(-x^3+a(a+2b)x) (0≦x≦a) $

$ y2=\frac{Pa}{6EIl}(-(lx)^3+b(2a+b)(lx)) (a≦x≦b) $

If the load P is at the center of the beam, then $ a = b = \frac{l}{2} $

The maximum amount of deflection occurs at the center of the beam.

$ ymax=\frac{Pl^3}{48EI} $

Becomes

This is a fairly important result, so keep it in mind, just like with cantilevered beams.

When bending moments are applied to both ends of a beam

Let's start by setting up an example problem.

A cantilevered beam of length l is subjected to bending moments MA and MB (MB > MA) at both ends. The coordinate system is set with the left end of the beam as the origin.

As usual, when we consider the balance of external forces

$ RA + RB = 0 $

Considering the balance of moments around point B,

$ RAl+MA-MB=0 $

From these

$ RA=-RB=\frac{MB-MA}{l} $

Next, we can determine the bending moment M at an arbitrary position (ignoring shear force).

$ M=RAx+MA=\frac{MB-MA}{l}x+MA $

Substituting this into the differential equation for deflection and using the multiple integral method,

$ \frac{d^2y}{dx^2}=-\frac{1}{EI}(\frac{MB-MA}{l}x+MA) $

$ \frac{dy}{dx}=-\frac{1}{EI}(\frac{MB-MA}{2l}x^2+MAx+C1) $

$ y=-\frac{1}{EI}(\frac{MB-MA}{6l}x^3+\frac{MA}{2}x^2+C1x+C2) $

Next, considering the important boundary conditions, when x=0 and l, the deflection is y=0.

$ C1=-\frac{(2MA+MB)l}{6},C2=0 $

Therefore, if we substitute this into the angle of deflection θ and the amount of deflection y,

$ θ=\frac{dy}{dx}=-\frac{1}{6EI}(-\frac{3(MB-MA)}{l^2}x^2-\frac{6MA}{l}x+(2(MA+MB)) $

$ y=\frac{1}{6EI}(-\frac{MB-MA}{l^2}x^3-\frac{3MA}{l}x^2+(2MA+MB)x) $

The angles of deflection at both ends are

$ θA=\frac{(2MA+MB)l}{6EI} $

$ θB=\frac{(MA+2MB)l}{6EI} $

with accommodation.If the bending moment MA = MB = M

$ θmax=θA=θB=\frac{Ml}{2EI} $

The maximum deflection occurs when $ x = \frac{l}{2} $

$ ymax=\frac{Ml^3}{8EI} $

Becomes

Summary

In this explanation, we calculated the deflection of a typical beam using the multiple integral method.

Basically, as long as the boundary conditions are known, the amount of deformation can be calculated for any structure, no matter how complex (regardless of the difficulty of the mathematical solution).

Especially now that simulations have advanced, the barrier to mathematical difficulty and complexity has lowered, so it's all the more important to pay attention to boundary conditions.

Next time, I'll introduce a method for calculating deflection that differs from the multiple integral method. While it's not used very often, it's good to understand the concept, so please check it out.

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Beam deflection, area moment method, and singular functions explained for beginners In the previous article, we actually used the multiple integral method to calculate the deflection of a beam. While the multiple integral method can basically be used to calculate the deflection of most beams, the calculation is cumbersome, so we'll explore a more intuitive and simpler method...

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.

Basic Mechanics of Materials (Revised Edition)

created by Rinker

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