Last time, I explained the tensile aspect of single-shot destruction, which is the fundamental principle of destruction.
This time, we'll look at the deformation of members when subjected to shear force and compression.
In the destruction list, it falls in the middle section from the top of the "one-shot destruction" category.
Since the main deformation caused by shear force is torsion, we will focus our explanation on a round bar shaft.
Before that, let me explain what happens when a compressive load is applied to a component.
Deformation of a member due to compressive stress
We have explained that the deformation of a member due to compressive stress is basically buckling.
The condition for buckling to occur should be that the cross-sectional area of the member is sufficiently small relative to its length.
Now let's consider what happens when compressive stress is applied to a member that is short and thick enough not to buckle.
Let's try applying compressive stress to a short, thick round bar.
Basically, dislocations do not occur during compression, so it does not cause destruction. This is because only the arrangement of atoms is crushed, preventing the atoms from moving.
The deformation simply involves the member shortening and shrinking, while the cross-sectional area increases. It will not break even under extremely large compressive stresses.
However, in the case of hard and brittle materials (such as concrete and cast iron), fracture occurs in the same way as tensile fracture, with sliding lines (Lüders lines) forming within the material. This is because atoms slip diagonally and horizontally within the compressed and expanded material.
However, the fracture load on the sliding surface due to compression is far greater than that due to tensile failure, so it is not something that needs to be given much consideration in mechanical design.
However, that doesn't mean that any member that doesn't buckle when subjected to compressive load or stress can be used in any way.
It doesn't break, but it yields completely. In other words, applying stress above the yield point results in plastic deformation.I end up doing that.
For example, suppose a short round bar is pressed against a board to generate compressive stress. If the bar is removed after applying a large compressive stress, the part of the board that was pressed against the bar will be indented.
このIndentations caused by plate yielding are a major problem in mechanical design.This will be the case.
A prime example is the seating surface of a bolt.If a bolt is overtightened into a base material that is weak, such as aluminum (bolts are basically made of iron), the seating surface of the bolt will yield and the seating surface will dent.
At this point, the bolt loses its function and may loosen or fail to fasten properly, so caution is necessary (bolts should generally be used within the elasticity of the material; plastic tightening and angle tightening are exceptions).
Then.What is the yield point when subjected to compressive stress? In industrial materials, it is surprisingly almost the same as the yield point in a tensile test.That's how it will be.
Therefore, to subject the material to compressive stress without buckling, it is basically fine as long as care is taken not to exceed the material's yield point.
Most materials have their tensile yield strength and 0.2% proof stress listed in the specifications, so as long as you adhere to those, there shouldn't be any problems.
Next, I will explain the main topic: failure due to shear force.
Breakdown due to shear force
The primary deformation caused by shear force is torsion.
Therefore, we will focus on twisting.
Here, a torque-torque curve is used to represent the fracture characteristics of a round bar. The vertical axis represents torque, and the horizontal axis represents the torsion angle.
This characteristic remarkably resembles the stress-strain diagram of a tensile test, clearly showing a distinct elastic region, yield point, and plastic region before fracture.
This characteristic is typical of viscous materials (such as tempered S30C and S35C).
However, this graph shows the relationship between torque and torsional angle, so we will consider shear force in more detail.
Determining the shear yield point from the yield torque
Consider an extremely thin, hollow round bar. Let the average diameter of the bar be d, and the wall thickness be h, with a torque Ts applied. Let τs be the shear force across the cross-section of the thin-walled round bar at that time.
Then, the following equation holds true from the balance between shear force and torque.
$ Ts=\frac{d}{2}πdyτs (area of thin-walled section × shear force). τs=\frac{2}{πhd^2}Ts $
If the torque Ts is the yield torque of the axis, then the shear force τs becomes the shear yield point.
Generally, measuring the yield point of shear stress is difficult by simply pulling on a plate or a rod, and is often done by twisting a round bar.
However, actually creating extremely thin-walled round bars is difficult, making measurement challenging, so solid round bars are often used for measurement.
If you've forgotten, please refer to this.
Determining the yield torsional moment and shear yield point for a solid round bar.
Let's consider what happens to the shear force when a yield torque Ts is applied to a round bar.
The maximum shear force on a round bar occurs at the outermost part where the strain is greatest, and the maximum shear force τ0 is given by $τ0=\frac{16T}{πd^3}$.
このIf the maximum shear force τ0 reaches the shear yield point τs that we calculated earlier, does that mean dislocations will occur and the material will enter the plastic region? Not necessarily.
Upon closer examination, in the case of a round bar, even if the shear force on the outer surface reaches the yield point, the internal strain is smaller than that of the outer surface, so the shear force will naturally be smaller there as well.
In other words, Only the outermost layer of material yields; the internal structure still generates shear forces within its elastic range.
So how does the entire cross-section of a round bar yield? If you apply an even larger torque, the twist angle increases while remaining constant at a certain torque.
What's happening here is that dislocations are occurring within the cross-section of the round bar. However, as explained earlier, these dislocations start from the outer circumference of the round bar and gradually move towards the center.
What's interesting here is that while dislocation is progressing within the cross-section, the torque does not increase. In other words, the shear force within the cross-section remains the same (just like in tension).
The torque at this point is called the yield torsional moment and is denoted by Ts.
The diagram below illustrates this.
For example, if a yield torsional moment Ts is applied to a round bar of diameter d, the shear force within the cross-section becomes uniformly τs, and the following equation holds.
$ Ts=\int_{0}^{\frac{d}{2}}{(τs2πrdr)r}=2πτs\int_{0}^{\frac{d}{2}}{r^2dr}=\frac{πd^3}{12}τs $
$ τs=\frac{12}{πd^3}Ts $
For some reason, experiments show that the shear yield point determined for an extremely thin-walled round bar is the same as the shear yield point determined for a solid round bar.
What's interesting here is that at the moment the round bar begins to yield, when the torsional torque reaches Ts, the shear force at the outermost circumference is $τ0=\frac{16Ts}{πd^3}$. However, while dislocations are progressing at the torsional torque Ts, the shear force is uniformly $τs=\frac{12}{πd^3}Ts$.
When we rearrange this relationship, the following equation holds true.
$ τ0 = \frac{4}{3}τs $
In other words, dislocation occurs only after the maximum shear force on the round shaft reaches 1.33 times the shear yield point.
Due to this characteristic, the fracture surface of the round bar has a very interesting shape.
When a shaft is made from a decent material, such as drawn material, the atoms are often neatly aligned vertically (there are few crystal defects, which is where the skill of the material manufacturer comes into play).
for that reasonDislocations tend to occur radially. Moreover, the shear force on the outer circumference when a dislocation begins is 1.33 times the shear yield point, so the dislocation progresses rapidly up to a certain point, causing a frayed, split-like line. This line is also called a Lüders line.
This is visible to the naked eye, but it becomes even more visible when immersed in a special solution. More details will be provided in the test section. To the naked eye, it appears as if the shaft is wrinkled.
Even if the shaft isn't completely destroyed, if you see this Lüders wire, it means it has completely surrendered, so that shaft is basically unusable and lacks strength.It is determined that...
Let's try applying even more torque to generate a larger shear force.
When a torque TB is applied to a round bar, in a ductile material, fracture progresses from the outer circumference to the interior, similar to yielding, and during this process, the torque TB remains constant, generating a uniform shear force τB at the point of breakage.
Therefore, the following equation holds true, similar to the relationship in the case of surrender.
$ τB=\frac{12}{πd^3}TB $
This concludes the explanation of shaft failure due to shear force.
Fracture due to shear force in brittle materials (cast iron and other cast materials)
Brittle materials (often cast iron) and easily damaged materials are not typically used in applications like shafts due to their inherent properties.
Now let's consider how to determine the shear yield point and shear strength.
I explained this in most of the previous lesson. We'll be using the Lüders wire.
To measure this, a round bar is pulled. This creates a 45-degree slip surface. If the stress (yield point) at the time the slip surface is created is σs and the shear force is τs, then the following equation holds true.
$ σs ≈ 2τs $
Similarly, when fracture occurs, if the fracture stress is σT and the shear force is τT,
$ σT ≈ 2τT $
The result can be found.
For detailed calculations, please refer to the previous Lüders line example.
Summary
This section summarizes failure due to compression and shear forces.
- When a compressive force is applied to a member that does not buckle, it deforms but does not break.
- When compressive force is applied to a component that does not buckle, if it is made of a brittle material (often a cast iron material), it will break at the sliding surface.
- The material will not break under compression, but it will yield. The yield point is approximately the same as in the tensile test.
The shear yield point is determined by the torsion of the shaft.
- Axial displacement begins when the shear force is approximately 1.33 times greater than the yield point, causing Lüders lines to form on the axis.
The yield point τS under shear can be determined from the yield torsional moment Ts of the axis and the diameter d of the axis as $τs = \frac{12}{πd^3}Ts$.
Similarly, the shear fracture force τB can be calculated from the fracture moment TB and the diameter of the axis d as $τB=\frac{12}{πd^3}TB$.
The shear yield point of brittle materials (common in cast materials) is given by $σs≈2τs$, where σS is the tensile yield stress from the slip plane.
Similarly, for brittle materials, the fracture stress at the time of fracture can be calculated as $σT ≈ 2τT$, where σT is the fracture stress and τT is the shear force.
In particular, when designing a machine for this topic, one thing to be careful about is surface yielding due to compressive force.
This meansIf the screw calculations are wrong, or if the expected load is applied when objects come into contact, it will easily yield and fail.
I've seen this kind of mistake quite a few times. Everyone is so confident that the compression will be fine that they forget to surrender.
The axis is,While strength calculations are standard practice during the design phase, it's crucial to carefully observe the tested materials. If Lüders lines are visible, even without deformation or failure, yielding occurs, indicating insufficient strength.
It's easy to overlook because it's not visible unless you look closely, so be careful. Also, normally the testers do the crack check (applying powder to the part and letting it soak with an etching solution to make microscopic cracks visible), but the designers should also carefully check it themselves.
It's incredibly important to not only destroy the parts but also to check on the parts after testing is complete, so make sure you go and check them no matter how busy you are.
Next time, we'll introduce the final stage of single-shot failure: failure due to bending stress.
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: