In the previous lessons, we explained failure due to shear force.
This time, we'll explain the final stage of destruction caused by bending, which is a one-shot destruction.
Looking at the table for one-hit destruction, it falls at the very bottom.
Let's get started with the explanation.
Review of stress caused by bending
Let's briefly review the stresses generated by bending.
Typical examples of deformation due to bending include the deflection of beams and the buckling of long columns.
Let's review what happens when a bending moment is applied to a component.
A bending moment M is applied to both ends of a rectangular timber with a square cross-section, height h, and width b. This causes the member to deflect downwards in the diagram.
At that time, we consider what kind of stress is occurring in the cross-section of the member.
For example, if we consider cutting the member in the diagram in the middle, the lower part of the member will be pulled, while the upper part will be compressed.
In other words, tensile stress is generated at the lower end and compressive stress is generated at the upper end.
Although the diagram jumped ahead, tensile and compressive stresses can be determined by the section modulus Z.
The section modulus Z is calculated as $Z=\frac{I}{h}$ (second moment of area I, section height h from the neutral plane), and if the tensile stress is σp and the compressive stress is σc, then
$ σp=\frac{M}{Z} $
$ σc=-\frac{M}{Z} $
Becomes
This deformation decreases as you move towards the center of the cross-section, so each stress also decreases and becomes zero in the middle. (Neutral plane)
Furthermore, since these stresses occur in opposition to the bending moment (not force), the sum of all the stresses equals zero.
However, when the moments of each stress are added together, they balance the bending moment acting on the member.
These are the basic characteristics of stress due to bending moment.
Fracture due to bending moment
Let's consider fracture caused by bending moment.
If the member breaks when a bending moment MB is applied, then, as we reviewed earlier, the maximum tensile stress acting on the member can be determined by the section modulus Z, and therefore the fracture stress can be determined.
$ σB=\frac{MB}{Z} $
It can be easily calculated as follows.
Many people might think that this formula allows us to calculate the bending moment at which a member breaks from its tensile strength σs, but that is not the case.
Like a twistEven when the maximum tensile stress reaches the tensile strength or yield point, the internal stress of the member is less than or equal to these levels, so it does not break or yield.
To see how bending causes failure, we need to look at something called a bending-deflection diagram, similar to stress-strain diagrams and stress-torsion diagrams.
When a bending moment is applied to a ductile material until it breaks, the following graph is obtained.
Similar to torsion, when the bending moment Ms is greater than the bending moment calculated from the tensile yield point of the member, the moment does not increase, and only the deflection increases.
Yes, thisDislocations are progressing within the material while the bending moment Ms remains constant.
Once the dislocations within the member are complete, the bending moment increases again in accordance with the deflection, leading to plastic deformation and then fracture.
Let's take a closer look at the internal stress at this time.
As with torsion, we assume that the stress generated within the member during the dislocation caused by the bending moment Ms is constant.
Consider a member of a certain length with a rectangular cross-section, width b, and height h, under a bending moment Ms, and where dislocation is progressing.
If we let the stress be σs, then, as we have reviewed, the sum of the stress moments generated inside the cross-section is equal to the bending moment Ms, so the following equation holds.
Moment Ms = Force over infinitesimal interval dz (σs × bdz (area of infinitesimal interval)) × integral from the neutral plane to the edge over distance z
$ Ms=2\int_{0}^{\frac{h}{2}}(σsbdz)z=2bσs\int_{0}^{\frac{h}{2}}zdz=\frac{bh^3}{4}σs $
$ σs=\frac{4}{bh^2}Ms $
become.
Also, if we consider the stress at the end of the member just before yielding to be σ0, then the following equation holds using the bending moment Ms and the section modulus Z.
$ Z=\frac{I}{\frac{h}{2}}=\frac{bh^2}{6}$
$ I=\frac{bh^3}{12} $
$ σ0=\frac{Ms}{Z}=\frac{6Ms}{bh^2} $
Compared to the σs mentioned earlier,
$ σ0 = \frac{3}{2}σs $
NextThe material will not yield until the surface stress reaches 1.5 times the yield point of the member.
However, this is only true for extremely viscous materials; in reality, the actual value is a bit lower.
When a bending moment-deflection test is actually performed, the graph will look like this.
in this wayIt's difficult to judge because there's no clear yield point, and it deforms gradually.
Therefore, in reality, it depends on the componentIf we set the yield point as σs and the generated stress as σ0, then it is used when σs < σ0 < 1.5σs.
If you're a thoughtful supplier of materialsThe bending strength is listed in the specifications sheet, so you can check it there. If the data isn't available, they should be able to provide it if you ask.
If no data is available anywhere, the only option is to perform a bending-deflection test (although bending strength data is available for most materials).
At that time, the test will be conducted in a manner similar to the following diagram.
The amount of deflection can be determined by attaching strain gauges to the measurement points of the member, and the load can be determined by attaching load cells to the bases at both ends.
It's not that difficult and doesn't take much time, so if you don't have the data, I think it's worth trying.
If requested, I will explain the detailed testing method.
Summary of one-shot fracture due to bending stress
Let's summarize the case of instant failure due to bending stress.
- Single-shot failure due to bending moment occurs because of the tensile stress generated within the material by the moment.
- In a single fracture caused by bending moment, the tensile yield point of the material and the bending yield point of the member do not coincide.
- When dislocations occur within a member due to deformation caused by bending moment, the stress inside the member remains constant.
- An ideally ductile material will not yield up to a stress 1.5 times its yield point.
For practical materials, a clear yield point cannot be observed through bending-deflection tests, so the yield point should be considered to be less than 1.5 times the material's yield point.
The bending strength of the material is generally listed in the specifications.
If you don't know the bending strength of the material, try performing a bending-deflection test.
Becomes
Even though we call it bending, if you think about it carefully, it ultimately boils down to failure due to tensile stress.
weak crown,It's just tensile stress, occurring in a unique way, so let's not overthink it.
In practice, structural design is rarely based solely on bending strength.
However, I will explain this now.One of the important values used in fatigue failureSo, make sure you understand this.
Summary of one-shot destruction
I've now explained all the basic one-shot destruction techniques.
The types are as follows:
I'll leave the individual reviews to those sections, but to summarize the whole thing...
• Materials should generally be handled below their yield point under any stress.
- For materials where the yield point is not clearly defined, handle them at a yield strength of 0.2% or less.
Regardless of the type of stress, stress-strain diagrams, torque-angle diagrams, and moment-deflection diagrams are important, so make sure you can read them.
After testing, always inspect the parts to ensure they haven't been destroyed or surrendered.
If a part breaks during testing, carefully examine the fracture surface to determine what kind of stress caused the failure.
With a standard design, it's highly unlikely that a single-shot failure will occur (the material is designed to be below its tensile yield point).
So, when does a component break down instantly? It breaks instantly when a part is damaged due to fatigue or corrosion, resulting in a load greater than expected being placed on the component, or when it receives an unintended impact load.
If anythingSecondary formsIt is.
HoweverIf we can't identify the cause of the destruction, we won't be able to take any action or implement countermeasures, and nothing will progress.
If you come across a scene of destruction (it's good to pay attention to products other than your own), actively go and investigate to develop the skill of identifying the cause of the destruction.
Next time, I will explain fatigue failure, which is extremely important.
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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