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 a stress solution method in mechanics of materials.

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Material Mechanics for Beginners 25: Deformation of a Rod Due to Its Own Weight and Deformation Due to Centrifugal Force (Rods of Equal Strength, Stress on a Propeller, Rotation...) In the previous lessons, we covered the entire field of materials mechanics, including fracture. I tried to emphasize the underlying concepts. From this lesson onward, we'll introduce a slightly more specialized method…

This time, as a special example of stress calculation, we will introduce the stress calculation of a container subjected to internal pressure.

A typical example of a specific tool, though rarely seen these days, is the spherical gas tank.

A slightly more niche example is a tool used to contain expanding gases in sealed containers, such as the pressurized cabin of an airplane or the gasoline tank of a car.

A more familiar example, though perhaps less common these days, is a pressure cooker, which traps high-pressure steam inside, making it a tool to which this stress calculation method applies.

In this blog, airsoft gun magazines and gas cylinders are also considered pressure vessels.

This article will introduce methods for calculating stress in cylindrical and spherical pressure vessels, which are typical types of pressure vessels.

Thin-walled cylindrical pressure vessel

First, the following diagram shows what a thin-walled cylindrical pressure vessel actually looks like.

You can think of it as having high-pressure gas inside the tea caddy.

Let's consider what kind of stresses are acting on this pressure vessel.

This time, as with last time, we will use a special approach.

First, since it's difficult to consider the stress all at once, we cut out a small portion from the outer circumference of the cylindrical pressure vessel.

As can be seen in the diagram, in the small section on the outer circumference of the cylindrical pressure vessel,The pressure of the internal gas generates stress that resists the force that causes it to expand in the circumferential direction.

Furthermore, while the gas pressure is not a direct force acting on each minute section, connecting these minute sections vertically will eventually lead to the top and bottom surfaces.

The top and bottom surfacesBecause the gas pressure causes it to expand vertically, stress is generated to withstand this movement.(The vertical direction is called the axis direction.)

このStress in the circumferential direction is called circumferential stress and is represented by σt.

On the other handStress in the axial direction is called axial stress and is represented by σz.

This covers the basics of the stresses that occur in cylindrical pressure vessels.

Now let's consider the force balance in a cylindrical pressure vessel in more detail.

Consider circumferential stress

Let's start by considering the circumferential stress σt.

Here, we'll use a special approach, but first, let's consider this cylindrical pressure vessel as a ring with a small section b as shown in the diagram.

Here, let r be the inner diameter of the cylinder, t be the wall thickness of the container, and b be the cut length (width), and assume that the pressure P due to the gas inside is uniform.

I'm sorry, but we'll have to further divide the ring of the infinitesimal interval b in half and consider that.

Since this halved small segment b of the ring tends to expand due to internal pressure, the following forces act on the cross-section cut in half, where σt is the circumferential stress.

The thickness of the cut surface is t and the length is b (width), so the area is tb and the load is area × stress.A load of σtbt occurs at two locations.

Next, consider a small sector-shaped section from the container that has been cut in half, as shown in the following diagram.

Next, to consider the force acting on the small segment of the cut-out sector, we multiply the pressure P by the inner circumference length rdθ of the sector's area and the length b (width).

$ dP=P(brdθ) $

become.

Let's take another look at the cylindrical container cut from above, keeping in mind the pressure dp acting on the sector.

In this case, the load acting on the cut surface is 2σtbt, and when the pressure dP acting in the small section is decomposed in the direction of the cut surface, it becomes dPsinθ, which is in the same direction as the load generated on the cut surface, so an equation for force equilibrium can be set up.

Considering the balance of forces from here, the total pressure dPsinθ acting over the small section balances the cut section load 2σtbt, so the following equation holds.

$ 2btσt=\int{dp}sinθ=2\int_0^{\frac{π}{2}}{Pbrsinθ}dθ=2Pbr $

Becomes

This equation can be rewritten as an equation for circumferential stress σt, and it becomes as follows:

$ σt=\frac{Pr}{t} $

Thus, the circumferential stress of the cylindrical pressure vessel can be determined.

Let's consider axial stress.

Next, let's consider the remaining axial stress σz.

This is quite simple; just like with circumferential stress, let's consider it by cutting a cylindrical container into a small section b.

Here, if we let σz be the axial stress, t be the wall thickness of the cylindrical container (similar to the circumferential stress), r be the radius of the inner diameter, and P be the pressure due to the gas, then the forces acting are as shown in the following figure.

A helpful way to understand this is to connect small intervals along the axis.

From the diagram, there are two types of forces acting on the system.

The force acting on an infinitesimal interval is 2πrtσz (circumference is radius × 2π and area is thickness t).

$ The sum of the pressures on the top and bottom lids Pπr^2 $

These two forces are in balance, so the following equation holds true.

$ 2πrtσz=Pπr^2 $

If we rearrange this into an equation for axial stress σz,

$ σz=\frac{Pr}{2t} $

Thus, the axial stress σz can be determined.

Now that we have determined the circumferential stress σt and the axial stress σz, we can express their relationship in an equation as follows.

Circumferential stress σt = 2-axial stress σz

NextCircumferential stress is twice as strong as axial stress.

In other words, when designing a thin-walled cylindrical container, you only need to be concerned with the magnitude of the circumferential stress σt. If σt is within the yield stress of the material, then the axial stress σz will naturally also fall within the yield stress.

Furthermore, you can see that when a cylindrical container is destroyed due to an increase in internal gas pressure caused by environmental changes such as temperature, the destruction is almost always circumferential.

Conversely, if the thickness of the entire container is uniform, it is highly unlikely that the contents will be blown off towards the top and bottom lids due to internal pressure.

If it ruptures in the circumferential direction, fragments will scatter all around, which is very dangerous, so the material and thickness must be carefully determined and designed.specification.

As an example of this cylindrical container design, I performed stress calculations on the cylindrical container, or magazine, of a CO2 gas gun. If you're interested, please take a look.

Let's try calculating the strength of a CO2 magazine.

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Let's try calculating the strength of a CO2 magazine. As of this writing (early 2021), the safe handling of CO2 magazines for gas guns is becoming a bit of a topic of discussion. Considering the current focus on environmental issues, the number of CO2 gas gun users is likely to increase...

Stress calculation for spherical pressure vessels (gas tanks, etc.)

Next, let's consider the stresses in spherical pressure vessels, which are the next most common shape after cylindrical pressure vessels.

The basic concept is the same for a spherical shape as for a cylindrical shape.

Similar to the cylindrical type, let's consider the circumferential and axial stresses by cutting out a small section from a spherical pressure vessel.

Actually, when it becomes a spherical shape, the stress generated does not change depending on the circumferential direction and the axial direction as it does in a cylindrical shape.Basically, pressure is applied equally in all directions inside the container, so the stress is the same in both the axial and circumferential directions.

Let's consider the equilibrium of forces, with σt being the stress generated here.

We will also cut out a small section, but in the case of this spherical shape, we will consider a small section that will be a cross-section as shown in the following figure.

The following diagram illustrates the forces acting on the small section that has been cut out.

Here, let r be the inner radius of the sphere, t be the wall thickness of the container, and P be the internal pressure.

First, in the cut surface of the spherical container that has been cut in half, just like with a cylindrical container, a force of 2πrtσt (area = circumference × wall thickness t) acts on the small section cut out in the cross-section due to pressure, which is exerted by the gas pressure P that causes the container to expand. This force is dP.

As things stand, the directions of dP and 2πtrσt are different, so we cannot set up an equation for force equilibrium. Therefore, we perform the following force direction transformation on dP.

This allows us to formulate the equation for the balance of forces.

The equation for force equilibrium is such that the sum of the forces obtained by cutting out small sections of a cross-section from angle θ 0 to $\frac{π}{2}$ is equal to 2πrtσt.

I'm sorry, but it involves integration.

Now let's consider the dP part of dPsinθ in more detail.

Since dP is the force generated by the gas pressure in a small section of a cross-section, it is necessary to determine the inner area of ​​the cross-section.

The area of ​​the inside of this cross-section is as shown in the following diagram.

The area of ​​the inside of the cross-section is 2πrcosθ × rdθ.

Applying pressure to an area results in a force, and the sum of the forces across a cross-section becomes the following integral.

The sum of the forces acting on the cross-section is = \int{dPsinθ}=\int_0^{\frac{π}{2}}{(Psinθ)(2πrcosθ)(rdθ)}=πPr^2

Furthermore, the following equation holds true based on the balance of forces.

$ 2πrtσt=πPr^2 $

Therefore, the stress σt is as follows:

$ σt=\frac{Pr}{2t} $

In other words, considering the stress generated in response to the pressure P caused by the gas, a container shape that is less advantageous than a cylindrical shape would result in lower stress.

However,ballThe drawback of using molded containers is that they are difficult to manufacture and take up a lot of space, making them difficult to adopt.

Concepts of stress generated in pressure vessels

Up to this point, we have considered the stresses that occur in typical pressure vessels, namely cylindrical and spherical types.

From here, I will explain a concept that can be applied to any type of pressure vessel.

Consider a pressure vessel of any shape, and then cut that pressure vessel in half.

Furthermore, we consider the force acting on the minute section ds of the cut container and the force acting on the cross-section of the halved container.

First, as before, when you cut a cross-section, the force obtained by multiplying the stress σt that withstands the force of the container trying to expand by the area of ​​the cross-section is the force (σt × cross-sectional area).

On the other hand, the force acting on a small interval ds is also Pds, since it is the pressure P multiplied by the cross-sectional area.

As it is, the directions are different and we cannot set up an equation for force equilibrium, so we decompose Pds and make it in the same direction as the force acting on the cut surface to get Pds sinθ.

The sum of the forces acting on the infinitesimal interval ds is obtained as follows integral.

The sum of forces acting over a small interval = \int{Psinθds}

Since it's difficult to calculate when the variable being integrated is s, we'll make it clearer using the following formula.

$ ds sinθ=dx $

Substituting this into the sum of forces,

The sum of forces = \int_A^B{Pdx} (line integral) = P × length between points A and B.

Generalizing this further, the above equation becomes:Since we considered it in two dimensions, we multiplied the pressure by the length of section AB, but in three dimensions, the section becomes an area rather than a line.

In other words, The sum of the forces acting on a small section is the internal pressure multiplied by the area of ​​the cut cross-section.

In other words, the relationship shown in the following diagram holds true for any container of any shape.

In other words, regardless of the shape of the container, if you know the area of ​​the cut cross-section, the area crossing the cut surface, and the internal pressure P, you can calculate the stress σt generated in the container.

this is,This is a very useful concept and is often used in stress calculations for containers with limited space and complex shapes, such as car gasoline tanks, so keep it in mind.

I'd like you to try this yourself as practice, but the stress on a spherical shape, as explained above, can be calculated very easily using this concept.

If you're interested, I encourage you to try it out yourself using a notepad or something similar; you'll understand why.

This concludes the explanation of the concept of stress in pressure vessels.

For now, let's consider this the end of the fundamentals of mechanics of materials.

If there are any specific requests for solutions to particular problems, I plan to write articles about them.

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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