Thick Cylinders: Stress Analysis Guide

Hey guys, let's dive into the fascinating world of thick cylinders and how we analyze the stresses within them. This is a super important topic in engineering, especially when you're dealing with high pressures or heavy-duty applications. Think of things like pressure vessels, pipelines, and even components in aircraft engines. Understanding the stress distribution in these thick-walled structures is crucial for ensuring safety, preventing failures, and optimizing material usage. We're going to break down a common scenario involving a steel cylinder shrunk onto a bronze one, and figure out the maximum hoop stresses. It's not as complicated as it sounds, and by the end of this, you'll have a solid grasp of the principles involved.

Understanding Hoop Stress in Thick Cylinders

So, what exactly is hoop stress? Imagine a cylinder under internal pressure. The pressure pushes outwards, trying to burst the cylinder. The material of the cylinder wall resists this outward push. Hoop stress, also known as circumferential stress, is the stress that acts tangentially to the cylinder's circumference. It's essentially the stress that holds the cylinder together against the internal pressure. In thin-walled cylinders, we often assume this stress is uniform across the wall thickness. However, when the ratio of the inside radius to the outside radius is significant (generally greater than 0.5), we enter the realm of thick cylinders, and things get a bit more interesting. In thick cylinders, the hoop stress is not uniform. It's highest at the inner surface, where the pressure is greatest, and decreases gradually towards the outer surface. This non-uniform distribution is a key concept we need to work with. Analyzing this requires us to consider the radial stress as well, which acts perpendicular to the wall, from the inside to the outside. Both radial and hoop stresses contribute to the overall stress state within the cylinder wall. The formulas derived from Lame's equations are the go-to tools for calculating these stresses. They allow us to determine the precise stress values at any point within the cylinder wall, which is vital for designing components that can withstand the expected operating conditions without yielding or fracturing. We'll be using these principles to solve our specific problem, so keep this distinction between thin and thick cylinders in mind!

The Shrunk-On Assembly Scenario

Now, let's talk about the specific problem at hand: a steel cylinder shrunk onto a bronze cylinder. This is a classic engineering challenge that introduces an initial state of stress even before any internal pressure is applied. Here’s the setup: we have a steel cylinder with an inside diameter of 150 mm, and it's shrunk onto a bronze cylinder. This bronze cylinder has an inside diameter of 100 mm and an outside diameter of 150 mm. The key here is the 'shrunk-on' part. This usually means the steel cylinder was heated, expanded, placed over the cooler bronze cylinder, and then as the steel cooled, it contracted, creating a tight interference fit. This interference fit generates compressive stresses in the bronze cylinder (at its outer surface) and tensile stresses in the steel cylinder (at its inner surface). These stresses are present before we even consider any internal pressure. So, when we later apply internal pressure, we need to account for this initial stress state. The question asks about the maximum hoop stresses in both the steel and bronze cylinders, implying that these stresses should not exceed certain limits. This usually means we need to find the stresses due to the interference fit and the stresses due to any applied internal pressure and then sum them up to find the total maximum hoop stress in each material. The limits are often given as material properties, like the yield strength or ultimate tensile strength, but here they are given implicitly as values that may not be exceeded. This implies we need to calculate the stresses and compare them to some allowable values, or perhaps the problem statement is incomplete and these limits are to be provided later. For now, let's focus on the stress calculation itself.

Applying Lame's Equations for Stress Calculation

To tackle this problem rigorously, we turn to Lame's equations. These equations are the cornerstone for analyzing stresses in thick-walled cylinders subjected to internal and external pressures. Lame's equations provide expressions for radial stress ($\\sigma_r$) and hoop stress ($\\sigma_h$) at any radial distance '$r$' within the cylinder wall. For a thick cylinder with internal radius '$a$' and external radius '$b$', subjected to internal pressure '$p_i$' and external pressure '$p_o$', the general forms are:

$\\sigma_r =\\frac{A}{r^2} - B$ $\\sigma_h =\\frac{A}{r^2} + B$

Here, '$A$' and '$B$' are constants that are determined by the boundary conditions – specifically, the pressures applied at the inner and outer surfaces, and the dimensions of the cylinder. In our specific problem, we have two cylinders, steel and bronze, and the interference fit introduces an initial pressure between them. Let's denote the interface pressure between the steel and bronze cylinders as '$p_{int}$'.

For the bronze cylinder:

• Inner radius ($a_{bronze}$) = 50 mm (since ID = 100 mm)
• Outer radius ($b_{bronze}$) = 75 mm (since OD = 150 mm)
• Boundary conditions: At $r = a_{bronze}$ (inner surface), $\\sigma_r = 0$ (assuming no internal pressure in bronze initially). At $r = b_{bronze}$ (outer surface), $\\sigma_r = -p_{int}$ (compressive pressure from the steel shrink fit).

For the steel cylinder:

• Inner radius ($a_{steel}$) = 75 mm (since ID = 150 mm, this is the interface with bronze)
• Outer radius ($b_{steel}$) = ? (This is not given, but for calculating hoop stress within the steel cylinder, we need its outer radius. Let's assume for now we are primarily concerned with the stresses at the interface or within the specified dimensions. If the problem implies the steel cylinder's outer diameter is also 150mm, it would be a very thin steel shell shrunk onto a thick bronze one, which is unlikely. Let's assume the problem intends for us to focus on the stresses due to the interface pressure within the steel cylinder where its inner radius is 75mm. If an outer diameter for the steel cylinder was provided, say $b_{steel}$, then the boundary conditions for the steel cylinder would be: At $r = a_{steel}$ (interface), $\\sigma_r = p_{int}$. At $r = b_{steel}$ (outer surface), $\\sigma_r = 0$ (assuming no external pressure on the steel). However, the problem statement only gives the inside diameter of the steel cylinder as 150mm, which is confusing. Let's re-read: "A steel cylinder with an inside diameter of 150 mm is shrunk onto a bronze cylinder with an inside diameter of 100 mm and an outside diameter of 150 mm." This implies the steel cylinder's inner diameter is 150mm. This means the bronze cylinder's outer diameter is 150mm, and the steel cylinder fits over it. So the interface radius is 75mm. The steel cylinder has an ID of 150mm (radius 75mm) and fits onto the bronze cylinder's OD of 150mm (radius 75mm). This means the interference fit is happening exactly at the 150mm diameter. The steel cylinder must have an outer diameter larger than 150mm to be considered 'thick'. Let's assume the steel cylinder has an outer radius $b_{steel}$.

Let's clarify the radii:

• Bronze ID = 100 mm => $a_{bronze}$ = 50 mm
• Bronze OD = 150 mm => $b_{bronze}$ = 75 mm
• Steel ID = 150 mm => $a_{steel}$ = 75 mm

This means the bronze cylinder is inside the steel cylinder, and the interface is at r = 75 mm. The steel cylinder's OD is not given, which is a problem for determining the constants A and B for the steel cylinder. However, the question asks about the hoop stresses in both cylinders. The hoop stress in the bronze cylinder at its inner surface is $\\sigma_{h,bronze}(a_{bronze})$ and at its outer surface is $\\sigma_{h,bronze}(b_{bronze})$. Similarly for the steel cylinder.

Let's assume the problem is primarily interested in the stresses at the interface and possibly the stresses on the inner surface of the steel and outer surface of the bronze due to the interference fit. The 'shrunk-on' process creates an interference pressure, $p_{int}$, at the interface (r = 75 mm).

• For the bronze cylinder, at $r = b_{bronze} = 75$ mm, $\\sigma_r = -p_{int}$. At $r = a_{bronze} = 50$ mm, $\\sigma_r = 0$ (assuming no internal pressure on the bronze itself).
• For the steel cylinder, at $r = a_{steel} = 75$ mm, $\\sigma_r = p_{int}$. At its outer radius $b_{steel}$, $\\sigma_r = 0$ (assuming no external pressure on the steel).

We would use these boundary conditions to solve for the constants A and B for each cylinder individually. Then, we'd calculate the hoop stresses at the relevant locations. The 'maximum hoop stresses' usually refer to the highest tensile or lowest compressive hoop stress experienced in each material.

Calculating Stresses Due to Interference Fit

The core of this problem lies in determining the interface pressure ($p_{int}$) caused by the shrink-fitting process. This pressure is what creates the initial stresses in both cylinders. We need to relate this interface pressure to the interference between the two cylinders. The interference is the difference in diameters (or radii) that existed before assembly, which is now taken up by the elastic deformation of the materials.

Let's assume the diameters were measured at room temperature. The bronze cylinder has an OD of 150 mm, and the steel cylinder has an ID of 150 mm. For a shrink fit, the steel cylinder's ID must be smaller than the bronze cylinder's OD when both are at the same temperature. The difference is the interference. However, the problem states the steel ID is 150 mm and the bronze OD is 150 mm. This phrasing is unusual for a shrink fit unless it implies the assembly results in a 150mm interface diameter, and there was an interference that caused the steel ID to expand and the bronze OD to compress to meet at 150mm. Let's assume this is the case and there is an interface pressure $p_{int}$ acting.

We need to know the modulus of elasticity (E) and Poisson's ratio (v) for both steel and bronze to calculate the deformation. Let's use typical values:

• Steel: $E_{steel} \\approx 200$ GPa, $v_{steel} \\approx 0.3$
• Bronze: $E_{bronze} \\approx 100$ GPa, $v_{bronze} \\approx 0.3$

For a thick cylinder under pressure, the radial expansion/contraction at the outer surface due to an internal pressure $p_i$ and external pressure $p_o$ is given by: $\\delta_{outer} = \\frac{b}{E} \\left( \\frac{1+v}{2} \\right) \\left( \\frac{A}{b^2} + B \\right)$ which simplifies using $A$ and $B$ based on pressures to $\\delta_{outer} = \\frac{b}{E} \\left( \\frac{(1+v)}{2} \\frac{p_i b^2 - p_o a^2}{b^2 - a^2} + \\frac{(1-v)}{2} \\frac{p_i - p_o}{1} \\right)$.

A more direct way to relate interference to interface pressure uses the radial displacement at the interface. The interference ($\\delta$) is the difference in radii that must be closed by the pressure.

$\\delta = \\text{interference at interface} = (r_{steel, inner} - r_{bronze, outer})$ where $r_{steel, inner}$ and $r_{bronze, outer}$ are the radii before assembly. After assembly, they meet at $r_{interface} = 75$ mm.

The radial expansion of the bronze's outer surface due to $p_{int}$ is $\\Delta r_{bronze} = \\frac{b_{bronze}}{E_{bronze}} \\left( \\frac{1+v_{bronze}}{1} \\right) \\left( \\frac{A_{bronze}}{b_{bronze}^2} + B_{bronze} \\right)$. Using boundary conditions for bronze ($\\sigma_r(a_{bronze})=0, \\sigma_r(b_{bronze})=-p_{int}$): $A_{bronze} = \\frac{p_{int} a_{bronze}^2 b_{bronze}^2}{b_{bronze}^2 - a_{bronze}^2}$, $B_{bronze} = \\frac{p_{int} b_{bronze}^2}{b_{bronze}^2 - a_{bronze}^2}$ (Note: here $p_o = p_{int}$, $p_i = 0$ is incorrect. It's $\\sigma_r(a_{bronze})=0, \\sigma_r(b_{bronze})=-p_{int}$)

Let's use the radial displacement formula directly: The increase in outer radius of bronze due to internal pressure $p_{int}$ is $\\Delta b_{bronze} = \\frac{b_{bronze}}{E_{bronze}} \\left( \\frac{p_{int} b_{bronze}^2}{b_{bronze}^2 - a_{bronze}^2} \\right) \\left(1 + v_{bronze} \\right)$.

The decrease in inner radius of steel due to external pressure $p_{int}$ is $\\Delta a_{steel} = \\frac{a_{steel}}{E_{steel}} \\left( \\frac{p_{int} a_{steel}^2}{b_{steel}^2 - a_{steel}^2} \\right) \\left(1 + v_{steel} \\right)$.

These two deformations must sum up to the initial interference. If we assume the initial interference caused the bronze OD to be slightly larger than 150mm and the steel ID to be slightly smaller than 150mm, and they match at 150mm (75mm radius) after assembly: $\\text{Interference} = \\Delta b_{bronze} + \\Delta a_{steel}$

$\\text{Interference} = \\frac{b_{bronze} p_{int}}{E_{bronze}} \\left( \\frac{b_{bronze}^2}{b_{bronze}^2 - a_{bronze}^2} \\right) (1+v_{bronze}) + \\frac{a_{steel} p_{int}}{E_{steel}} \\left( \\frac{a_{steel}^2}{b_{steel}^2 - a_{steel}^2} \\right) (1+v_{steel})$

Here, $a_{bronze} = 50$ mm, $b_{bronze} = 75$ mm, $a_{steel} = 75$ mm. We are missing $b_{steel}$ (outer radius of steel). This is a critical piece of information. If $b_{steel}$ is not provided, we might have to assume the question is focused only on the stresses at the interface or that the steel cylinder is also