Material Mechanics for Beginners 25: Deformation of a Rod Due to Its Own Weight and Deformation Due to Centrifugal Force (Rod of Equal Strength, Stress in a Propeller, Stress in a Rotating Body)

In the previous lessons, we explained the principles of material mechanics up to fracture, giving us a comprehensive overview of the subject.

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A Beginner's Guide to Destruction: Destruction Theory, One-Shot Destruction, Fatigue, Creep, Brittleness We've explained fracture multiple times so far. This covers all the basics of material fracture. For beginners who want to learn material mechanics step-by-step, please refer to this index…

I've tried to explain things with a particular emphasis on the underlying concepts.

From this point on, I will explain how to solve stress problems that require a slightly unconventional approach.

In particular, this book deals with problems where stress cannot be determined without solving differential and integral calculus or differential equations.

However, it's not particularly difficult; the important thing is the way of thinking, and the calculation of the formulas is just a bonus.

These are the usual lines, but instead of memorizing them,I think it's fine if people just recognize that there are problems that require a special method of solving, and then remember and refer back to that method when they need to use it.

Let me explain.

Deformation due to the weight of the rod

In this discussion, we will consider the stress generated by the weight of a fairly long rod (with a constant circular cross-section) suspended from the ceiling.

Up until now, we have ignored the weight of the components, considering it to be quite small in relation to external forces. However, here we will consider the effect of their own weight.

As always, when considering an example problem, let's consider a rod with length L, cross-sectional area A, density ρ [kg/m^3], acceleration due to gravity g [m/s^2], and a load P acting on its tip.

First, if we simply consider the balance of external forces, it will be as follows.

This is fundamental to high school physics.

Now, let's consider a slightly unusual approach. First, let's assign a coordinate system to the rod, with its tip at 0 and the ceiling at L. Then, let's consider the balance of forces on the rod in a small interval dx at a given coordinate x.

Then the equation for the balance of forces becomes as follows:

$ Rx = P + ρgAx (density × volume) $

In other words, The external force acting on a certain coordinate point x is the load P plus the weight of the rod extending to point x.

Therefore, stress is expressed by the following formula.

$ σx=\frac{P+ρgAx}{A} =\frac{P}{A}+ρgAx $

From this formula, the generation of the rodThe maximum stress occurs when x is L, that is, at the point where it is suspended from the ceiling, and the magnitude is as follows:

$ σmax=\frac{P}{A}+ρgAL $

If we consider the required cross-sectional area for the rod, with σmax being the yield stress σa of the material, we can transform the equation into the following.

$ A=\frac{P}{σa-ρgx} $

Therefore, the closer you get to the base from which the rod is suspended, the thicker the required cross-sectional area becomes.

ObviouslyIf the yield stress σa is the same size as ρgL at L where ρgx is maximized, the required cross-sectional area becomes infinite, which is practically impossible.

Next, let's consider the extension of the rod.

We consider the extension as λ, but here we take a slightly special approach and consider the extension dλ of a small interval dx at coordinate point x.

Let the strain be ε and the elastic modulus of the material be E.

$ εdx=dλ, ε=\frac{dλ}{dx}=\frac{σx}{E} $

Therefore, the extension dλ of the small interval can be summarized by the following equation.

$ dλ=\frac{σx}{E}dx=\frac{1}{E}(\frac{P}{A}+ρgx)dx $

Therefore, integrating this from coordinate 0 to L will give the total extension λ.

$ λ=\int{dλ} =\frac{1}{E}\int_0^L{(\frac{P}{A}+ρgx)}dx=\frac{PL}{AE}+\frac{ρgL^2}{2E} $

If we plot the extension λ on the vertical axis and the length of the bar on the horizontal axis, we get the following graph.

Becomes

Equal Strength Stick

From the explanation so far, you should understand that the stress generated in the rod affected by the attendant differs depending on the coordinates.

The stress generated becomes considerably larger, especially as you get closer to the root.

In other words, for a rod with the same cross-sectional area, the longer it gets, the larger the required cross-sectional area at the base becomes, resulting in a considerably thicker overall design that is quite wasteful in terms of cost and weight.

thereYou'll notice that if you vary the cross-sectional area according to the coordinates and adjust it to the minimum cross-sectional area required at each coordinate, the rod becomes considerably lighter.

in this wayA rod in which the stress generated throughout the rod is equal is called a rod of uniform strength and is very commonly used in industry.

Now, let's find out what kind of cross-sectional area changes result in rods of equal strength.

First, let's consider an example where a load P is acting on the tip of a rod of length L whose cross-sectional area changes.

Let the density be ρ and the acceleration due to gravity be g, and consider the coordinate system as above.

Now, this involves another unusual approach: let A and A+dA be the cross-sectional areas at coordinates x and x+dx, respectively.

At this timeTo create rods of equal strength, we must make the stress generated at coordinate x equal to the stress generated at coordinate x+dx, which gives the following equation:(Let the generated stress be denoted as σ).

$ (A+dA)σ=Aσ＋ρgAdx(self-weight acting between dx)$

$ \frac{dA}{A}=\frac{ρg}{σ}dx $

This is an equation called a differential equation, and if we solve it for now, we get the following (C', C are integration constants).

$ \log A=\frac{ρg}{σ}x+C`. A=Ce^{(\frac{ρg}{σ}x)} $

The boundary conditions for determining the integration constant are as follows: If the cross-sectional area of ​​the lowest end of the rod is A0, then at x=0 C=A0=$\frac{P}{σ}$. The formula for the cross-sectional area of ​​a rod of equal strength is as follows:

$ A=A0e^{(\frac{ρg}{σ})x}=\frac{P}{σ}e^{(\frac{ρg}{σ})x} $

Becomes

In other words, The overall shape of the rod is determined once the external force P and the cross-sectional area A0 at the bottom of the rod are determined.

Furthermore, the elongation is simple, and the total elongation λ is equal to the stress throughout.

$ λ=\frac{σ}{E}L $

Becomes

Therefore, the shape of the rods of equal strength is as follows.

In practice, strictly ensuring the cross-sectional area is a power of the natural logarithm e would make drawing instructions, manufacturing, and inspection considerably more difficult, so the shape is made to be closer to that of a rod with uniform strength.

Up to this point, we have mainly explained the tensile load caused by the rod's support, but the same idea holds true for compression when the rod is upright, and the resulting equation simply has a reversed sign.

Based on this idea, massive structures like Tokyo Tower and Tokyo Skytree, which are heavily influenced by their own weight, often have a shape resembling a rod of equal strength when viewed closely.

If you're interested, please do a search and take a look at its exterior shape.

In actual machinery, especially in the case of automobile engines, which are my area of ​​expertise, the external forces are often overwhelmingly greater than the weight of the structural components, so it's not something that's used very often. However, I'd like you to keep in mind that "something like this exists" (I think it's often used in architecture, large stationary machinery, ships, etc.).

This article omits mathematical explanations, but we plan to offer mathematics courses in the future, so please look forward to explanations of differential and integral calculus, differential equations, logarithms, and other related topics.

Stress on an object subjected to centrifugal force

Next, let's consider the generation of stress due to centrifugal force.

This meansRotating mechanical elements are extremely important in almost any machine, and they are used in a considerable number of applications, so it's definitely worth understanding them.

Let's start with a simple example to get a better understanding.

The example assumes that an object of mass m is rotating at an angular velocity ω [rad/s] at a distance L from a certain axis.

If we ignore the mass of the rod connecting the axis to the mass m and let the cross-sectional area be A, the stress σ generated in the rod is as follows:

$Centrifugal force P due to mass m = mLω^2$

$ Generated stress σ=\frac{P}{A}=\frac{WL}{A}ω^2 $

Becomes

like thisExternal forces acting inside an object in proportion to its mass, such as centrifugal force, inertial force, and the self-weight explained above, are called physical forces.

このKeep in mind that when dealing with stresses caused by physical forces, the method often involves taking a coordinate system, finding the balance of forces over a small interval, and then integrating or solving differential equations, similar to how you would solve the effect of self-weight.

Next, let's consider the case where centrifugal force due to the object's own weight acts on it.

First, let's consider an example where a rod with a uniform cross-section (cross-sectional area A) and a total length of 2L rotates at an angular velocity ω around an axis perpendicular to the rod with a center O as the axis of rotation.

Here, let λ be the extension of half the rod and ρ be its density.

Since this stress is caused by an object force, as usual, we set the rotation axis O as the coordinate origin and the x-coordinate to be positive towards the tip.

Next, the centrifugal force dP of a mass ρAdξ (density × cross-sectional area × length) in a small interval dξ located at a distance ξ (qi) from the axis of rotation is expressed by the following equation.

$ dP = ρAdξ・ξω^2 (radius of rotation × angular velocity squared) $

The tensile force Px generated at a certain distance x from the axis of rotation is the sum of the centrifugal forces generated by each infinitesimal interval from x to L, so we perform the following integral.

$ Px=\int {dP}=\int_x^L{ρAω^2ξ}dξ=\frac{ρAω^2}{2}(L^2-x^2) $

Therefore, the stress is given by the following equation, since the cross-sectional area is A.

$ σx=\frac{Px}{A}=\frac{ρω^2}{2}(L^2-x^2) $

From thisThe maximum stress generated by the centrifugal force of a rotating rod occurs at x=0 on the axis of rotation O.

$ σx=\frac{ρω^2L^2}{2}$

Becomes

Furthermore, the elongation λ of the rod can be found by substituting $dλ=\frac{σx}{E}dx$ into the above equation and integrating from x = 0 to L, since the elongation εdx=dλ over a small length dx is ε = Eσx.

$ λ=\int {dλ}=\int_0^L{\frac{σx}{E}}dx=\int_0^L{\frac{ρω^2}{2E}(L^2-x^2)}dx=\frac{ρω^2}{2E}\displaystyle\left[L^2x-\frac{x^3}{3}\right]_0^L=\frac{ρω^2L^3}{3E} $

Becomes

Up to this point, we have only calculated the extension λ from the center (x=0) to one end (x=L), so the extension of the entire rod (total length of the rod is 2L) is double that, $ \frac{2ρω^2L^3}{3E}$.

Here's an interesting characteristic:The stress and elongation generated by a rotating object are determined entirely by density, length, and rotational speed (the elastic modulus also plays a role in elongation), and have absolutely no relation to the cross-sectional area.

In other words, the stress on a rotating object does not depend on its shape, whether it is a rod, a circle, or a gear.It is determined by the material of the object (elastic modulus, density), its size (radius), and its rotational speed.

Furthermore, the equation for the generated stress, $σx=\frac{ρω^2}{2}(L^2-x^2)$, can be graphed as follows.

Obviously, stress decreases with the square of the distance from the axis of rotation, so when designing a rotating object,If you don't add plenty of material around the rotating axis, it will break.

A typical example of this type of machine is the propeller of an airplane or helicopter; if you look closely, you'll see that it has plenty of material at the base and becomes thinner towards the tip.

Therefore, when reducing the weight of components such as the flywheel (inertia maintenance device), gears, and balancers in automobiles, it's perfectly fine to leave some material around the rotating shaft and remove a significant amount of material from the rest of the assembly.

However, devices that maintain inertial mass, such as flywheels,Since the mass is concentrated at the tip of the rotating object, make sure to leave sufficient wall thickness around the axis of rotation.

Furthermore, as can be seen from the formula, centrifugal force increases in proportion to the square of the rotational speed, so even when the rotational speed is high, such as with speed-increasing gears (for example, the secondary balancer of an engine), the wall thickness around the rotating shaft must be sufficiently ensured to prevent damage.

AnywaysIt's helpful to keep in mind that the stress on a rotating object decreases with the square of the distance and increases with the square of the rotation speed.

This concludes our explanation.

Next time, we will explain the stresses that occur in simple spherical and cylindrical pressure vessels.

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Strength of Materials for Beginners 26: Stress in Thin-Walled Pressure Vessels (Cylindrical Pressure Vessels, Spherical Pressure Vessels, and Pressure Vessels in General) In the previous article, we introduced a method for calculating stress using body forces as a special example of stress resolution in mechanics of materials. This time, we will introduce another special example of stress calculation: the stress calculation of a container subjected to internal pressure. ...

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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