Circular stress and trusses explained for beginners (press-fit, interlocking, statically indeterminate problems)

This time, as mentioned in the final explanation of the previous lesson on thermal stress, we will explain annular stress and simple trusses.

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Thermal stress and residual stress explained for beginners Last time, we discussed torsional stress in round bars. This time, we'll explain thermal stress. We'll also discuss residual stress, which is somewhat related to thermal stress. This time, we'll build upon many of our previous discussions…

The first half, which deals with annular stress, forms the basis of press-fitting and fitting, representative methods used in joining parts, which are essential in mechanical design, so I would like you to understand it.

Regarding the trusses in the latter half of the explanation, it would ideally be better to explain them after covering bending stress and beam deflection, but I will explain them now because there was a request and it's a pattern that frequently appears on university exams.

Okay, let's begin.

Problem setting for annular stress, and the meaning of press-fitting.

This is very easyPress-fitting is a process where a shaft and a disk with a hole are prepared, and the shaft is passed through the hole in the disk. In this process, the diameter of the hole in the disk is narrower than the diameter of the shaft.

In other words, you force the shaft into the hole.

This means the axis will be fixed in place and will not move even with some force.

thisPress-fitThat.

At this timeThe overlap between the hole diameter and the shaft diameter is called the overlap, and its setting is defined by what is called the fit tolerance in JIS (Japanese Industrial Standards).

Well, let's put the formalities aside and look at a diagram.

It's something like this.

To be serious, the typical press-fitting method involves creating a jig, fixing the cylindrical part in place, and then using a press machine to push the shaft in.

Now, let's set the dimensions of the ring for the purpose of this explanation.

Let r be the radius of the midpoint between the hole and the outer circumference of the ring, let t be the thickness of the ring, and let l be its length. In this case, the thickness t is considerably smaller than the radius r.

Now let's actually calculate the stress in the ring.

Circular stress

Let's first consider the deformation of the ring when the shaft is press-fitted.

In mechanical design, it's crucial to roughly visualize the movement and deformation of objects before diving into detailed calculations, so make it a habit to think this way.

The ring bulges because a thick shaft is forced into a narrow hole. Note that it bulges in both the radial and circumferential directions, but does not deform in the longitudinal direction.

In reality, the shaft is also crushed and compressed, but for the purpose of explaining cyclic stress, we will focus only on the annular portion. Later, we will create a separate section on press-fit calculations to explain this in more detail.

Now that we have a visual representation of the deformation, let's consider the load acting on the ring.

Since the shaft presses uniformly and equally into the hole in the ring, let's consider that the force applied per unit area of ​​the hole in the ring (for example, every 1 mm²) is q (N).

This type of load is called a uniformly distributed load. In this case, it is a uniformly distributed load q(N/$ mm^2 $).

The next logical step would be to consider the balance of forces, but in a situation like this where the load is widely distributed, it's difficult to consider the whole picture at once.

Therefore, just like with torsional stress, we isolate a small section, consider it, and then add it all together later.

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Twisting techniques explained for beginners (twisting of round rods and hollow round rods) As explained at the end of the last post about what the elastic region is, this time we will explain torsional stress. Shafts are objects used very frequently in mechanical design; both springs and screws experience large torsional forces…

Now let's cut out a part of the ring.

First, cut out a Baumkuchen-like piece from the ring, with a radius r and a small angle dθ.

Next, since the Baumkuchen is 1 unit long, cut off a piece that is 1 unit long.

This results in an extremely small Baumkuchen with radius r, angle dθ, and length 1.

We will consider the balance of forces in this cropped form.

First, a uniformly distributed load q generated by the press-fitting process applies a force of q × r (radius) × dθ (angle) × 1 (length) in the radial direction.

Next, since it is also pulled in the circumferential direction, if we consider the force in the circumferential direction as T, the equilibrium of forces will be as shown in the following diagram.

Since forces are vectors, they cannot be added together directly because their directions are different. Therefore, if we separate the circumferential force T into circumferential and radial forces, the following equation holds true.

$ qrdθ-2×Tsin(\frac{dθ}{2}) =0 $

Here's a bit of a cheat: since it's a very small interval, dθ is very small, so we can assume that $ sin(\frac{dθ}{2})≈\frac{dθ}{2} $ (sin can be considered as y=x when it's close to 0).

$ T=qr $

Becomes

Now that we understand the balance of forces, let's consider the internal force, stress.

When we look at the balance of forces, we see that it is being pulled in the circumferential direction, so tensile stress is generated in the circumferential direction, and the following equation holds true.

The cross-sectional area A of a small segment can be calculated by multiplying the thickness t of the ring by its unit length of 1.

$ Stress σ = \frac{T}{A(cross-sectional area)}=\frac{qr}{t} $

This stress is called stress, but the name doesn't really matter.The force required to fix the ring in place at the desired press-fit setting will not be exerted unless the stress is below the yield point of the ring-shaped material.

If this stress exceeds the tensile strength, the ring will break while the shaft is being inserted.

This is a very important stress, so make sure you understand it well.

Normally, press-fit calculations involve determining the uniformly distributed load q using Poisson's ratio and the elastic modulus, based on the deformation of the shaft and the hole in the annular ring. However, since this explanation focuses on annular stress, the method for determining the uniformly distributed load q has been omitted. However, press-fit calculations are a very important topic in machinery, so please wait for a separate, dedicated page to cover them.

Putting that aside, now that we've determined the stress, we can also determine the strain.

Since the stress is in the circumferential direction, the strain is also expressed as circumferential strain ε, and if the elastic modulus of the material of the ring is E, then the following equation holds.

$Circumferential distortion ε =\frac{σ}{E}=\frac{qr}{Et}$

Furthermore, if we let r' be the average radius of the deformed ring, the following equation holds:

Circumferential distortion ε = \frac{2πr'-2πr}{2πr}=\frac{r'-r}{r}

this means,By measuring the dimensions of the ring before and after press-fitting, the strain can be determined and the resulting stress can be calculated.

Given this characteristic, when using press-fitting, it is a good idea—or rather, absolutely essential—to measure the diameter before and after press-fitting.

This concludes the explanation of the stress and press-fitting principles for rings.

Next, I will explain trusses.

What is a truss?

A truss is a type of structural framework, but I think you'll understand it better if you look at the following diagram.

Which would cause less deformation: pushing on a box-like frame, or inserting a rod into the middle of the box?

These rod-like structures are called trusses, or in mechanical design, reinforcing ribs. In my area of ​​expertise, automobiles and motorcycles, a typical example is the truss frame, which was commonly used in older racing machines (the truss frame is still in use in NASCAR).

Now, if we were to represent the box-like structure shown in the diagram using material mechanics, it would look like this.

Each joint between rods is called a node, and there are specific symbols for them. Even the same node can be rotated or fixed, and there are various ways of representing them, so the way they are written is standardized. However, in mechanical design drawings and layouts, simplified diagrams are rarely used (everything is represented realistically), so I don't think it's that important (I don't know about architecture or civil engineering).

The loads, stresses, and strains on each rod can be determined using the knowledge we've discussed so far, but let me give you just one example.

Examples of solutions for truss problems (statically indeterminate problems)

The title might seem unfamiliar, and it mentions statically indeterminate problems, but don't worry, I'll explain it by solving a simple example.

First, let's set up an example problem.

In the following truss structure, a load P acts at node O. The cross-sectional area of ​​members AO and CO is A1, and the elastic modulus is E1, while members BO are similarly A2 and E2, respectively.

We will determine the stress and deformation of each component.

Did you get a good idea of ​​the example? Let's get started solving it.

I always try to say this, but firstIt is important to consider the image of the deformation.

The formula is secondary.

Let's think about it.

They simply grow in their own way.

Here, let P1 be the tensile force (load, not stress) generated in members AO and CO, and P2 be the tensile force generated in member BO. Furthermore, let λ1 be the elongation of members AO and CO, and similarly let λ2 be the elongation of BO.

Once you have an image in mind, start as usual.We seek the balance of external forces.

The following equation for vertical equilibrium can be derived.

$ Load P =P2+2P1cos(θ) $

We want to find the tensile forces P1 and P2 from here, but you can see that we can't solve it because there is only one equation for tensile forces with two variables. Therefore, we need to set up another equation related to P1 and P2.

The key point to focus on is the amount of elongation of each component!

The key point is to formulate a relational equation by considering the amount of elongation of each component.

$ λ1 = λ2 × cos(θ) $

Strictly speaking, the angle θ before the material stretches and the angle after stretching are different, but we assume the stretch is negligible and treat them as the same.

As described aboveOther than the balance of forcesToProblems that cannot be solved without finding the relationship between the deformation of an object are called statically indeterminate problems.That.

Basically, by setting up the equations for the equilibrium of external forces, the internal forces, stress and strain, can be determined. This is called a statically determinate problem.(The name doesn't matter.

If you realize that the problem cannot be solved simply by considering the relationships between forces, you can usually solve it by setting up equations related to the deformation of the object, especially its extension.

With practice, you'll quickly be able to tell whether a base is static or statically indeterminate, but it's understandable that you won't know at first. Everyone starts somewhere.

Let's go back to the example. From here, we just need to find the amount of deformation and the loads P1 and P2 from the strain.

First, the deformation amount can be solved using the relationship of elasticity (assuming elongation ≈ strain).

$ λ1=\frac{P1}{A1E1}×\frac{l}{cos(θ)} $

$ λ2=\frac{P2}{A2E2} $

Substituting this into the equation relating the elongation of each component,

$ \frac{P1l}{A1E1cos(θ)}=\frac{P2l}{A2E2}×cos(θ)、　P1=\frac{A1E1cos^2(θ)}{A2E2}×P2 $

Substituting these values ​​into the force equilibrium equation, we can find P1 and P2.

$ P1=\frac{Pcos^2(θ)}{(\frac{A2E2}{A1E1})+2cos^3(θ)}、P2=\frac{p}{1+2(\frac{A1E1}{A2E2})cos^3(θ)} $

At this point, the only thing left to do is substitution, so we'll stop here.

While there are issues of static and non-static bases when determining trusses and stresses, they are not particularly important, and I recommend keeping the following in mind.

I think this will solve most of the problems.

For those who have forgotten about elasticity, click here.

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Elasticity and stress-strain diagrams (plasticity, material properties) that even beginners can understand Continuing from last time, we'll explain the very important concepts of elasticity and the basic properties of materials. This explanation is also extremely important in mechanical design, so please try your best to follow along. Even beginners can understand...

Despite all this explanation, the ability to find exact solutions is not particularly required in mechanical design. Moreover, this is especially true now that computer simulation (CAE) has become advanced and readily available at low cost.

The important thing is to have a correct understanding of deformation and external forces, as well as a grasp of material mechanics.

Therefore, please don't get discouraged if you can't do the calculations. Understanding the meaning is what's important.

indeterminate problem

I've already tackled a statically indeterminate problem in the truss assignment, but let me explain statically indeterminate problems in a little more detail here.

First, let me explain the terms statically definite and statically indefinite problems.

A statically determinate problem is one in which the stress can be determined solely from the equation for the balance between the external force acting on the object and the stress within the object.

A statically indeterminate problem refers to a problem where the stress cannot be determined solely from the equation for the balance between the external force acting on the object and the internal stress.

This might sound like a Zen riddle and be difficult to understand, so let's illustrate it with diagrams.

The diagram below shows examples of statically definite and statically indeterminate problems side by side.

I think you'll see the difference clearly (in the statically indeterminate problem, the two cylinders are firmly fixed to each other).

The statically determinate problem is simple: a cylinder with cross-sectional area A1 and elastic modulus E1 is pulled by a load P, so the stress is defined as follows.

$ Stress σ = \frac{P}{A1} $

Next, let's solve the statically indeterminate problem.Since two cylinders are pulled by a load P, each cylinder experiences a shared load..

Since the loads borne by each cylinder are not specifically known, if we let P1 be the load on the blue cylinder and P2 be the load on the green cylinder, the following equation holds true.

$ P = P1 + P2 $

ObviouslyThe equation above (equilibrium of forces) alone does not tell us the stress in each cylinder. This is a statically indeterminate problem.

To solve this, since we can't formulate any more force equations, we'll use the extension of the cylinder (geometry) to create the equations.

Since the two cylinders are firmly fixed to each other, the elongation when pulled by a load P is the same (let the elongation be λ). This can be expressed as follows:

The elongation (elongation ≈ distortion) of each cylinder is determined by Hooke's Law.

$ The elongation (distortion) of the blue cylinder is \frac{P1}{A1E1}, and the elongation (distortion) of the green cylinder is \frac{P2}{A2E2} $

The extensions of the two cylinders are the same.

$ \frac{P1}{A1E1}=\frac{P2}{A2E2} $

Becomes

Now that we have two equations for the two variables P1 and P2, we can solve them (simply by substituting the values).

$ 荷重P1＝\frac{A2E2P}{A1E1+A2E2}、荷重P2=\frac{A1E1P}{A1E1+A2E2} $

Problems that require equations other than force-related equations, following this process, are called statically indeterminate problems.

From an arithmetic perspective, if a system of equations requires the same number of equations as the number of variables we want to solve, then there will be no solution. Therefore, we need as many equations as there are loads and stresses we want to find. When the necessary equations cannot be satisfied by force equilibrium alone, we supplement the equations with deformation amounts (geometry) such as elongation.

For example, if you want to know the loads P1, P2, and P3, you'll need three equations: one for force equilibrium, two for moment equilibrium, and then you'll need a third equation based on the deformation to make it work.

Now that we have a better understanding of statically determinate and statically indeterminate solutions, let's consider which type of problem is more common in reality. In my experience, statically indeterminate problems are more frequent.

From the author's perspective, the visual difference is that statically indeterminate problems often involve the addition of reinforcing structures or trusses compared to statically determinate problems.

Since this may be difficult to understand from words alone, I will show some examples in the diagram below.

Have you grasped the concept?

Summary

This has gotten long, but to summarize...

This time, we'll only touch on trusses, so we'll omit the summary.

Next time, we'll finally get to explaining beams.

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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…

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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