NowLast time, we discussed torsional stress in round bars.
This time, we will explain thermal stress.
In addition to thermal stress, residual stress, which is somewhat related, will also be explained.
This lesson will mostly be an application of what we've already covered, and there won't be much new information to learn or understand. However, it's still quite important, so please take your time and work through it carefully.
Thermal stress
Before explaining thermal stress, I think you can see from experience that when an object is heated, it expands.
To put it in extreme terms, if you heat water in its ice state, it becomes a liquid, and if you heat a liquid, it becomes a gas, and its volume increases exponentially.
The objects and materials dealt with in mechanics of materials and machinery are fundamentally the same as water; they expand when heated.
Let's first consider this expansion.
Let's consider the setup shown in the following diagram.
A round bar of length L0 is fixed to a wall at both ends, and a sensor (like a weighing scale) is installed at the fixed point to measure the load, applying heat from the temperature t0 of the round bar.
Normally, a round rod would expand when heated, but since the rod is sandwiched between walls, it's physically impossible for it to expand.
But thisThe tendency to stretch acts as a force on the wall, and since the wall does not move, a reaction force acts on the round bar. The stress generated by this reaction force is thermal stress.It is.
It's difficult to understand from just the text, so please look at the following diagram.
Let's take a moment to summarize the situation.
• Original length of the round bar L0
The actual length of the round bar that was supposed to stretch: L1
- The temperature of the round bar is initially t0, then heated to t1
The load measured by the load measuring device installed on the wall is P
The reaction force acting on the round bar is R = P
The amount of elongation that the round bar should have stretched is λ = L1 - L0
And
Actually, the number we can find out from this isOriginal length L0withThe temperature difference is t1-t0withMeasuring instrument load PFrom this, we can seeReaction force RThese are the three.
Therefore, we will use these three factors to find the reaction force R.
hereThe new concept of the coefficient of linear expansion α (alpha)Use the number.
This is easyThe coefficient of linear expansion α indicates the rate at which an object expands (stretches) when its temperature rises.That's it. Since it represents a proportion, it doesn't have units like distortion.
Let's get started by setting up the equation. Using the coefficient of thermal expansion α, the elongation λ of the round rod can be expressed as follows.
The elongation of the rod is calculated as follows: λ = L1 - L0 = α (coefficient of linear expansion) × L0 (original length of the rod) × (t1 - t0) (temperature difference of the rod).
nextThe amount λ that was supposed to expand is prevented from expanding by the wall, so the round rod is compressed.This will be the case.
This meansMechanics of Materials for Beginners 1I want you to remember that.
If you forget about the wall for a moment and consider λ as the amount of compression when an external force R is applied to a round bar, you should be able to understand it.
Here, if we know the strain and elastic modulus from the relationship of elasticity, we can determine the thermal stress, and if we know the area of the round bar, we can determine the measured load P=R from the thermal stress.
Let's look at it step by step. First, we find the strain ε of the round bar.Note that since we are considering compressive strain here, the denominator in the strain calculation will be the elongated length L1.
The strain of the round bar is ε = \frac{L0-L1}{L1}=\frac{-λ(elongation of the round bar)}{L0+λ(elongation of the round bar)} =\frac{-1}{1+\frac{λ}{L0}}×\frac{λ}{L0}
Here's a bit of a cheat: Since the elongation λ is very small relative to the original length L0, we can consider $ \frac{λ}{L0}$ to be 0.
Such interpretations and omissions occur frequently in engineering.
Why thenUnlike academic research where rigorous solutions are sought, in many cases, we want solutions that can be used in real-world situations.This is because... I will explain the nuances of this later in the Numerical Calculation Course (tentative title).
Now let's go back and look at the formula. In the end, the strain ε of the round bar is given by the following formula.
The strain of the round bar is given by ε = -\frac{λ(elongation of the round bar)}{L0(original length of the round bar)}
The thermal stress σ of a round bar can be expressed using the coefficient of linear expansion α, with E being the elastic modulus of the round bar, as follows:
$ Stress σ of a round bar = E (elastic modulus) × ε (strain of the round bar) = E × -\frac{λ}{L0}$
$ = -E × α (coefficient of linear expansion) (t1 - t0) (temperature difference of the round rod) $
Becomes
Since it's negative, compressIt can be seen that this is the case.
If we let A be the cross-sectional area of the round bar, then the reaction force R of the round bar, measured by the measuring instrument and given by the external force P, is equal to a compressive force and can be expressed by the following equation.
Reaction force R = -σ × A (cross-sectional area) = AEα(t1 - t0)
It is represented as follows.
Naturally, we have discussed the case when heated, but when cooled, the signs reverse and tensile stress is generated.
However, in my experience, the amount of shrinkage when most mechanical materials are cooled is considerably smaller than the coefficient of linear expansion when heated, so I haven't really worried about it.
Perhaps it's because the author was a designer of internal combustion engines.
However, even though the engine is designed to withstand environments ranging from -15°C to 40°C, we have never worried about shrinkage due to cooling.
However, the one that came up hereThe coefficient of thermal expansion α is an extremely important concept.
The coefficient of linear expansion α is a material-specific physical property, and in addition to determining thermal stress, it is essential to always consider how much the material expands and how its dimensions change when using machinery at high temperatures.
Incidentally, the coefficient of thermal expansion α is usually listed in the material properties, but if you don't know it, you can ask your material supplier or other relevant parties.
Any reputable company should have the data.
Residual stress
Were you surprised to suddenly encounter such a difficult term as residual stress?
This concept is simple, but it's a very important characteristic, so I really want you to understand it.
We have explained thermal stress up to this point.I think you now understand that when heat is applied, an object expands, and in some cases, stress is generated.
Furthermore, if an object is left alone and returns to its original temperature, it should return to its original size and the stress should disappear.
However,In reality, some stress continues to be generated even after the temperature returns to its original level. This is called residual stress.
A typical example of residual stress is compressive residual stress, as explained in the section on thermal stress.
The residual stress due to thermal stress, as explained above, can be represented in the following diagram.
In the case shown in the diagram above, compressive residual stress remains, and the residual stress is added to the original material's strength when applied to tension.
Residual stress can be generated not only by heat but also by striking an object. The diagram shows an example.
This is one of the reasons why parts made using the forging method, which is generally known as the manufacturing process, are very strong.
When processing with heat in this way, residual stress frequently occurs during manufacturing processes such as quenching, annealing (though not solely due to residual stress), and work hardening, so keep that in mind.
An interesting process is shot peening (WPC treatment, a registered trademark of Fuji Kihan Co., Ltd., Fuji Seisakusho Co., Ltd., and Fuji WPC Co., Ltd., and practically an industry standard). Roughly speaking, it's a process that increases the strength of a molded part by applying thousands of tiny steel balls to the surface, generating compressive residual stress.
In my experience, depending on the conditions, fatigue intensity can increase by about 2%.
Well, I'll explain the heat treatment, surface treatment, metal materials, and manufacturing methods in detail later in the mechanical design course.
Incidentally, I've seen some theoretical formulas for residual stress, but I didn't really understand them. I've used some of them in my work, so I'll introduce them later as techniques (like Minor's rule).
Summary
Summary of this time
Objects dealt with in mechanical design and materials mechanics expand and contract with temperature changes.
The coefficient of linear expansion α is defined as the rate at which an object expands when heat is applied, and it is a value unique to that object.
- Some of the stress generated in an object due to heat or other factors remains, and this stress is called residual stress.
When designing machinery, consider changes in the ambient temperature of the machine and understand the dimensional changes due to linear expansion.
become.
Next time, we'll cover things commonly used in mechanical design.Analysis of Circular Stress and TrussesI would like to explain this.
Once that's done, we can finally explain the stress in beams, and for the time being, we can conclude our study of mechanics of 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.
Comment: