Last time, we explained a typical second moment of area.
This time, we will introduce some typical examples of needle deflection.
It's not an exaggeration to say that mechanical structures, in their simplified form, are composed of multiple beams. Therefore, when performing strength calculations, the issue of beams cannot be avoided.
Furthermore, strength calculations are unavoidable when designing machinery. In other words, it's an important concept that is always used.
However, you don't need to solve everything on your own. Just understand the meaning and look it up if you forget. Please refer to the previous pages for the meanings.
From here, we will introduce typical solutions after understanding the meaning of beam deflection.
Typical beam deflection
In the following explanation, we will assume that all beams have a uniform cross-section (with a second moment of area of I) and an elastic modulus of E. The coordinate system will be unified with the left end of the beam as the origin, with downwards being +y and rightwards being +x.
When calculating composite materials, refer to this and use the second moment of area of the equivalent cross-section; all the results I will introduce from now on will be applicable.
A cantilevered beam that receives a load at its tip.
Deflection y and deflection angle θ of a cantilevered beam receiving a load P at its tip.
$ Deflection y=\frac{Pl^3}{6EI}(3-\frac{x}{l})\frac{x^2}{l^2} $
$ Maximum deflection ymax = \frac{Pl^3}{3EI} (x=l) $
$ Deflection angle θ=\frac{Pl^2}{2EI}(2-\frac{x}{l})\frac{x}{l} $
$ Maximum deflection angle θmax = \frac{Pl^2}{2EI} (x=l) $
This is a basic form of beam deflection and is used quite often.
In reality, a gear is attached to the end of the shaft, and when that gear rotates with a certain torque, a load is applied to the shaft as a reaction force, making it behave like a cantilever beam.
While deflection is important, calculating bearing lifespan requires the load at the base of the beam.
A cantilever beam subjected to a uniformly distributed load q.
Deflection y and deflection angle θ of a cantilever beam uniformly subjected to a uniformly distributed load q (N/mm, etc.)
$ Deflection y = \frac{ql^2}{24EI}x^2(6-4\frac{x}{l}+\frac{x^2}{l^2})$
$ Maximum deflection ymax = \frac{ql^4}{8EI} (x=l) $
$ Deflection angle θ=\frac{ql^2}{6EI}x(3-3\frac{x}{l}+\frac{x^2}{l^2}) $
$ Maximum deflection angle θmax = \frac{ql^3}{6EI} (x=l) $
This is a beam that can be applied to components that receive fluid (and thus are subjected to pressure).
These are mechanical components called reed valves or relief valves, consisting of a hole and a leaf spring that seals the hole. When the leaf spring is subjected to pressure and bends, a gap is created in the hole, allowing fluid to flow.
In other words, it's like a movable lid that operates under pressure.
This is a feature that is always present in the engines I specialize in. I also think it's commonly used in kitchens, bathrooms, and washrooms—things that people often don't even notice in their daily lives.
A cantilevered beam with load P applied to its center.
Deflection y and deflection angle θ of a cantilevered beam with a load P applied at its center $ \frac{l}{2}$.
$ Deflection y = -\frac{Pl^3}{96EI}(\frac{x^3}{l^2}-3x), (x≦ \frac{l}{2})$
$ Deflection amount y=-\frac{Pl^3}{12EI}(1-\frac{l}{x})(\frac{x^2}{l^2}-2\frac{x}{l}+\frac{1}{4}), (\frac{l}{2}≦x) $
$ Maximum deflection ymax = \frac{Pl^3}{48EI} (x = \frac{l}{2}) $
The general formula for the angle of deflection is omitted.
$ Maximum deflection angle θmax = \frac{Pl^2}{16EI} (x=0,l) $
This involves things like a gear in the middle of an axle or placing an object on a stand, which are common in everyday life, so I won't give specific examples, but it's important.
A cantilevered beam when a moment M is applied to both ends of the beam.
Deflection y and deflection angle θ of a cantilevered beam in which moments M are applied to both ends in opposite directions.
$ Deflection y = \frac{M}{2EI}x(lx) $
$ Maximum deflection ymax = \frac{Ml^2}{8EI}, (x = \frac{l}{2}) $
$ Deflection angle θ = \frac{M}{2EI}x(l-2x) $
$ Maximum deflection angle θmax = \frac{Ml}{2EI} (x=0, l) $
This is another common type of needle deflection.
For example, two separate walls are reinforced by connecting them with ribs. The walls are subjected to some force and begin to collapse. The collapse of the walls forces the ribs to deform, creating a moment.
If you think about it, there are probably a lot of them around us. For example, I think a lot of the concrete blocks sold at home improvement stores are this shape.
A cantilevered beam that receives a uniformly distributed load q.
Deflection y and deflection angle θ of a cantilevered beam uniformly subjected to a uniformly distributed load q (N/mm).
$ Deflection amount y=\frac{ql^4}{24EI}\frac{x}{l}(1-2\frac{x^2}{l^2}+\frac{x^3}{l^3}) $
$ Maximum deflection ymax = \frac{5ql^4}{384EI}, (x = \frac{l}{2}) $
$ Deflection angle θ=\frac{ql^3}{24EI}(1-6\frac{x^2}{l^2}+4\frac{x^3}{l^3}) $
$ Maximum deflection angle θmax = \frac{ql^3}{24EI}, (x=0, l) $
This type of structure is used for lids that receive fluids. While it can't be treated exactly as the same beam, this concept applies to the bottom of a container (though in reality it's a surface, making calculations complicated).
These types of containers are often treated as pressure vessels or thin-walled containers. I intend to explain this in detail in a separate section.
Summary
Actually, the solution to the problem I mentioned is explained in the article below.
However, if you were to perform all the calculations in an actual design setting, you wouldn't have enough time, so please use this as a reference.
In reality, they are often subjected to multiple complex loads, but when you break them down one by one, they are surprisingly just combinations of simple loads like the ones I mentioned earlier.
In that case, you can easily solve it by using singular functions or the superposition method that I introduced previously.
Having explained all that, in real-world design work, strength calculations are often performed using simulations, so I think it's quite rare to find the exact solution using a scientific calculator or manual calculations.
However, running simulations unnecessarily multiple times during the layout and sketching stages is a waste of time and will likely anger the person in charge of simulations. Ideally, we want to decide on the layout in just one calculation.
Visualization is crucial for this, so I encourage you to take on the challenge of developing your visualization skills through this course.
Next time, we'll finally discuss buckling, a prime example of destruction.
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.
Comment: