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Thin Walled Cylinders and Spheres

An analysis of thin walled Cylinders and Spheres, including wire wound tubes.

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Contents

Thin Walled Cylinders Under Pressure.
Thin Spherical Shells Under Internal Pressure.
Cylindrical Shells With Hemispherical Ends.
Volumetric Strain On The Capacity Of A Cylinder
A Tube Under A Combined Loading.
Wire Winding Of Thin Walled Cylinders.
Rotational Stresses In Thin Cylinders.
Page Comments

Thin Walled Cylinders Under Pressure.

The three principal Stresses in the Shell are the Circumferential or Hoop Stress, the Longitudinal Stress, and the Radial Stress.

If the Cylinder walls are thin and the ratio of the thickness to the Internal diameter is less than about $\inline \displaystyle\frac{1}{20}$ , then it can be assumed that the hoop and longitudinal stresses are constant across the thickness. It may also be assumed that the radial stress is small and can be neglected. In point of fact it must have a value equal to the pressure on the inside surface and zero at the outside surface. These assumptions are within the bounds of reasonable accuracy.

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A cylinder is the central working part of a reciprocating engine or pump, the space in which a piston travels.

The above cylinder has an internal diameter $\inline d$ and a wall thickness of $\inline t$ . If the applied internal pressure is $\inline p$ , then the Hoop stress is $\inline \ f_1$ and the Longitudinal stress is $\inline f_2$ .

Section the above cylinder through a diametral plane and consider the equilibrium of the resulting half cylinder, where $\inline f_1$ acts upon an area of $\inline 2\;t\;l$ . The resultant vertical pressure force is found from the projected horizontal area $\inline d\;l$ .

Clearly, the Pressure force upwards is equal to the hoop Stress times the area over which it acts.

Hence, $\inline p\times d\times l = f_1\times 2\times t \timesl$

$\therefore\;\;\;\;\;\;\;f_1 = \frac{p\times d}{2\;t}$

Now consider the equilibrium of a section cut by a transverse plane. The longitudinal Stress $\inline \displaystyle f_2$ acts on an area of approximately $\inline \displaystyle \pi \times d\times t$ (Note the diameter taken should really be the mean diameter), and the pressure $\inline p$ acts upon a projected area of $\inline \displaystyle \pi \frac{d^2}{4}$ (Note this is true no matter the actual shape of the end of the cylinder).

Equating longitudinal Forces:

$f_2\times \pi \times d\times t = p\times \pi \times \frac{d^2}{4}$

$\therefore\;\;\;\;\;\;\;f_2 = \frac{p\times d}{4\;t}$

(2)

Where long cylinders or tubes are braced or carried in brackets, the longitudinal Stress may be much less than given by equation (2) and is sometimes neglected.

Thin Spherical Shells Under Internal Pressure.

As in the previous section the radial Stress will be neglected and the circumferential or hoop Stress is assumed to be constant.

A spherical shell is a generalization of an annulus to three dimensions. A spherical shell is the region between two concentric spheres of differing radii.

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By symmetry the two principal Stresses are equal and the stress in any tangential direction is equal to $\inline f$ . From the diagram it can be seen that:

$f\times \pi \times d\times t = p\times \pi \times \frac{d^2}{4}$

where $\inline d$ is the internal diameter. The equation can be re-written as:

$f = \frac{p\times d}{4\;t}$

Cylindrical Shells With Hemispherical Ends.

Pressure is an effect which occurs when a force is applied on a surface. Pressure is the amount of force acting on a unit area. The symbol of pressure is $\inline P$ .

Let the thickness of the cylinder walls be $\inline \displaystyle t_1$ and the thickness of the hemisphere $\inline \displaystyle t_2$ . It is assumed that the internal diameter of both are the same.

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If the internal pressure of the shell is $\inline \displaystyle p_1$ then:

The Stresses in the cylinder walls are:

Hoop Stress $\inline f_1 = \displaystyle\frac{p\times d}{2\;t_1}$

Longitudinal Stress $\inline f_2 = \displaystyle\frac{p\times d}{4\;t_1}$

Then, Hoop Strain

$e_1 = \left(\frac{1}{E} \right)\left(f_1 - \frac{f_2}{m} \right)= \left(\frac{p\times d}{4\;t_1\;E} \right)\left(2\;-\frac{1}{m} \right)$

The Stresses in the hemispherical ends are:

Hoop Stress $\inline f = \displaystyle\frac{p\times d}{4\;t_2}$

Hoop Strain $\inline e = \left(\displaystyle\frac{1}{E} \right)\left(f - \displaystyle\frac{f}{m} \right)= \left(\displaystyle\frac{p\times d}{4\;t_2\;E} \right)\left(1\;-\displaystyle\frac{1}{m} \right)$

If there is to be distortion under pressure, at the junction between the cylinder and the ends, $\inline e_1 = e$

i.e. $\inline \;\;\;\;\;\;\;\displaystyle\frac{2 - \displaystyle\frac{1}{m}}{t_1} = \displaystyle\frac{1 - \displaystyle\frac{1}{m}}{t_2}$ Or $\inline \;\;\;\;\;\;\;\;\;\;\displaystyle\frac{t_2}{t_1} = \displaystyle\frac{1 - \displaystyle\frac{1}{m}}{2 - \displaystyle\frac{1}{m}}$

Taking a value for Poisson's ratio of $\inline 0.3$ , $\inline \;\;\;\;\;\;\;\;\;\;\displaystyle\frac{t_2}{t_1} = \displaystyle\frac{7}{17}$

Note that the maximum Stress will then occur in the hemispherical ends, i.e.

$f = \frac{p\times d}{4\;t_2} = \left(\frac{17}{7} \right)\left(\frac{p\times d}{4\;t_1} \right)$

Which is greater than the hoop stress $\inline \displaystyle f_1$ in the cylinder. For equal maximum stress $\inline \displaystyle \frac{t_2}{t_1}$ should equal $\inline 0.5$ .

Volumetric Strain On The Capacity Of A Cylinder

The capacity of a cylinder is $\inline \displaystyle \frac{\pi \;d^2\;l}{4}$ , so if the dimensions increase by $\inline \displaystyle \delta\;d$ and $\inline \delta \;l$ there will be an increase in volume and:

Volume is the quantity of three-dimensional space enclosed by some closed boundary, for example, the space that a substance (solid, liquid, gas, or plasma) or shape occupies or contains.

Volumetric Strain = $\inline \displaystyle\frac{[(d + \delta \;d)^2(l + \delta \;l) - d^2l}{d^2l}$

$= \frac{[d^2l + d^2\,\delta l + 2\delta d\,\delta l + 2\,\delta d\,d\,\delta l + (\delta d)^2\,l + (\delta d)^2\;\delta \,l - d^2\;l]}{d^2\,l}$

Neglecting the products of small quantities:

Volumetric Strain = $\inline 2\times \displaystyle\frac{\delta d}{d} + \displaystyle\frac{\delta l}{l}$

= 2 $\inline X$ Diametral Strain + Longitudinal Strain

Since the length of the Circumference = The diameter $\inline X$ a constant

Therefore, Volumetric Strain = 2 $\inline X$ Hoop Strain + Longitudinal Strain

It should be noticed that this is the sum of the linear Strains in three Principal directions. It is worth comparing this with "Engineering Materials: Compound Stress and strain."

Similar reasoning for a Spherical Shell shows that:

Volumetric Strain = 3 $\inline X$ Hoop Strain

i.e., to increase the capacity it is only necessary to multiply the Volumetric Strain by the original Volume.

Example:

[imperial]

Example - Example 1

Problem

A Boiler drum consists of a cylindrical portion $8\;ft.$ long, $4\;ft.$ in diameter and $1\;in.$ thick.It is closed at both ends by hemispherical shells.

In a hydraulic test to $1000\;lb./sq.in$ , how much additional water will be pumped in after an initial filling at atmospheric pressure? Assume that the circumferential Strain at the junction of cylinder and hemisphere is the same for both.

For the drum material $\displaystyle E = 30\times 10^6\;lb.in.^{-2}\;\;:\;\;\frac{1}{m} = 0.3$

For water $\;\;\;\;\; K = 0.32\times10^6\;lb.in.^{-2}$

Workings

For the Cylinder, Hoop Stress,

$f_1 = \frac{p\;d}{2t} = \frac{1000\times 4\times 12}{2\times 1}= 24,000\;lb.in.^{-2}$

Longitudinal Stress, $f_2 = \displaystyle\frac{p\;d}{4t} = 12,000lb.in.^{-2}$

Hoop Strain, $e_1 = \left(\displaystyle\frac{1}{E} \right)\left(f_1 - \displaystyle\frac{f_2}{m} \right) = 12,000\times\displaystyle\frac{1.7}{E}$

Longitudinal Strain, $e_2 = \left(\displaystyle\frac{1}{E} \right)\left(f_2 - \displaystyle\frac{f_1}{m} \right) = 12,000\times \displaystyle\frac{0.4}{E}$

Therefore, Increase in Capacity

$= (2e_1 + e_2)\times V= \frac{12,000\times3.8}{30\times 10^6}\times\frac{\pi \times48^2}{4}\times 96 = 265\;in.^3$

(1)

where $V$ represents the Volume

For the hemispherical ends:

The question states that the hoop Strain in the ends is the same as in the cylinder thus:

Hoop Strain $\;e = e_1$

Increase in capacity = $3e\times Volume$

$= \frac{12,000\times 5.1}{30 \times10^6}\times\frac{\pi \times 48^3}{6} = 125\;in.^3$

(2)

As the pressure on the water originally inside the cylinder increases, there is a loss in volume which is made up with additional water.

Decrease in the Volume of the original water is given by:

$\left(\frac{p}{K} \right)\times V = \frac{1000}{0.32\times 10^6}\left(\frac{\pi \times 48^2\times 96}{4} + \frac{\pi \times 48^3}{6} \right)= 723\;in.^3$

(3)

where $V$ represents the Volume

The Additional Volume of Water required is given by the increase in volume of the cylinder; the end shields and the effects of the compression of the water. i.e. Equations (1)+(2)+(3)

$265 + 125 + 723 = 1113\;in^3\;at\;1000\;lb.in.^{-2}$

or $\;1117\;in^3$ at atmospheric pressure

Solution

The additional water must be $1113\;in^3\;at\;1000\;lb.in.^{-2}$ or $\;1117\;in^3$ at atmospheric pressure.

A Tube Under A Combined Loading.

Example:

[imperial]

Example - Example 4

Problem

A thin cylindrical tube of $3\;in.$ internal diameter, $0.2\;in.$ thick, is closed at the ends and subjected to an internal pressure of $800\;lb./sq.in.$ A Torque of $\displaystyle 4500\pi\;lb.in.$ is also applied to the tube.

Determine the maximum and minimum principle Stresses and the maximum shearing Stress.

Workings

Hoop Stress = $\displaystyle\frac{800\times 3}{2\times 0.2} = 6000\;lb.in.^{-2}$

Longitudinal Stress = $\displaystyle\frac{800\times 3}{4\times 0.2} = 3000\;lb.in.^{-2}$

Shear Stress on transverse Planes = Torque / (Mean radius $X$ Area) $= \displaystyle\frac{4500\pi }{1.6\,(\pi \times 3.2\times 0.2)} = 4400\;lb.in.^{-2}$

Note. This assumes that the Stress is uniform.

The Maximum and Minimum Principal Stresses. (See the pages on "Compound Stress and Strain")

Maximum and Minimum Principal Stress = $\displaystyle\frac{1}{2}(f_x + f_y)\;\pm \displaystyle\frac{1}{2}\;\sqrt{(f_x - f_y)^2 + 4s^2}$

$= \frac{1}{2}\times 9000\;\pm\;\sqrt{3000^2 + 4\times 4400^2}= 4500\;\pm \;4600$

$= 9100$ and $- 100\;lb.in.^{-2}$

The Maximum Shear Stress is given by:

$\frac{1}{2}\;\sqrt{(f_x - f_y)^2 + f_s^2} = 4600\;lb.in.^{-2}$

Solution

The maximum and minimum principle Stresses are $= 9100$ and $- 100\;lb.in.^{-2}$
The maximum shearing Stress is $4600\;lb.in.^{-2}$

Wire Winding Of Thin Walled Cylinders.

One way to strengthen a thin walled tube against an Internal Pressure is to wind the outside with wire under tension. This puts the tube into compression and consequently reduces the Hoop Stress. In many applications the maximum Stress will be in the wire, which must be made of high-tensile material.

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The method of analysis can be broken down into a number of stages. It is assumed that the diameter of the wire is $\inline d$ and that it is closely wound onto the cylinder with a tension $\inline T$ .

Consider the wire to be replaced by an equivalent cylindrical shell of thickness $\inline \displaystyle t_w$ . Note that this gives the longitudinal cross sectional area, i.e.

$t_w\times d = \frac{\pi \;d^2}{4}\;\;\;\;\;\;\text{or}\;\;\;\;\;t_w = \frac{\pi \;d}{4}$

(3)

Let the Initial Tensile Stress in the wire $\inline \displaystyle f_w = \frac{4\;T}{\pi d^2}$
Let the initial Compressive Stress in the cylinder be $\inline \displaystyle f_1$ . Then for equilibrium:

$f_1\times t = f_w\times t_w$

(4)

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When an internal pressure $\inline p$ is applied, let the Stresses be $\inline \displaystyle f_w'$ tension in the wire, and $\inline \displaystyle f_1'$ tensile Hoop Stress in the cylinder.

For equilibrium:

$f_1'\times 2t + f_w'\times 2t_w = p\;D$

(5)

fig. 229

The final longitudinal Stress in the Cylinder is $\inline f_2 = \displaystyle\frac{p\;D}{4t}$

or a smaller value for long tubes which are externally supported (See the first paragraph).

As the wire and cylinder remain in contact, the change in Hoop Strain due to the application of the internal pressure must be the same for both. i.e.,

$\left(\frac{1}{E} \right)\left[\left(f_1' - \frac{f_2}{m} \right) + f_1 \right] = \left(\frac{1}{E_w} \right)\left(f_w' - f_w \right)$

Note that $\inline \displaystyle f_1$ is compressive and that the wire is under Stress in one direction only.

Example:

[imperial]

Example - Example 5

Problem

A copper tube of $1.5\;in.$ external diameter and $1.4\;in.$ internal diameter, is closely wound with steel wire of $0.03\;in.$ diameter.

Stating clearly the assumptions made, estimate the tension at which the wire must have been been wound if an internal pressure of $300\;p.s.i.$ produces a tensile circumferential stress of $\displaystyle 1000\;lb.in.^{-2}$ in the tube.

$E_s = 1.6\times E_c$

Workings

The equivalent Wire thickness. (see equation (2)

$t_w = \frac{\pi d}{4} = 0.0236\;in.$

If $\displaystyle f_w\;$ is the winding stress in the wire, the initial Hoop Stress in the tube ( see equation (3)

$f_1 = \left(\displaystyle\frac{t_w}{t} \right)\times f_w = 0.472\;f_w$ compression.

If the final stresses are $\displaystyle f_w'$ and $f_1'$ the equilibrium equation (4) gives:

$f_1'\times 0.1 + f_w'\times0.0472 = 300\times 1.4$

But $\;\;\;\;\;\;f_1' = 1000\;lb.in.^{-2}$

$\therefore\;\;\;\;\;\;f_w' = \frac{420 - 100}{0.0472} = 6780\;lb.in.^{-2}$

Equating the change in the Hoop Strain for the wire and tube and neglecting longitudinal stress in the tube.

$\frac{6780 - f_w}{E_s} = \frac{1000 + f_1}{E_c}$

Substituting for $\displaystyle f_1$

$6780 - f_w = 1600 + 1.6\times0.472\;f_w$

$\therefore\;\;\;\;\;f_w = \frac{5180}{1.755} = 2960\;lb.in.^{-2}$

The winding Tension $= 2960\times \displaystyle\frac{\pi d^2}{4} = 2.08\;lb.in.$

Solution

The tension is $2.08\;lb.in.$

Rotational Stresses In Thin Cylinders.

When a Cylinder rotates about its axis a centrifugal force will occur in its walls, which will produce a Hoop Stress $\inline \displaystyle f$ . This stress may be assumed to remain constant at any given angular velocity.

Let the cylinder have a mean radius $\inline r$ , a wall thickness of $\inline t$ and made of material with a density of $\inline \displaystyle \rho$ . Assume that the angular velocity of revolution is $\inline \displaystyle \omega$ .

The centrifugal force on an element of unit length which subtends an angle $\inline \displaystyle \delta \theta$ at the centre of the cylinder ( see diagram) equals:

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$\left(\frac{\rho \,r\,\delta \theta \times t}{g} \right)r\,\omega ^2$

Resolving Radially:

$2\,f\,t\times \sin\frac{1}{2}\delta \theta = \rho \,r^2\times \omega ^2\times t\,\frac{\delta \theta }{g}$

Since $\inline \;\;\;\;\;\sin\displaystyle\frac{1}{2}\delta \theta \rightarrow \displaystyle\frac{1}{2}\delta \theta$

$f = \rho \,r^2\;\frac{\omega ^2}{g}$

The above analysis can be applied , approximately, to rim-type flywheels.

Example:

[imperial]

Example - Example 6

Problem

A flywheel is required with a moment of inertia of $\displaystyle 6000 \;lb.ft^2$ . It is to run at $250\;r.p.m.$ and the maximum stress is not to exceed $\displaystyle 600 \;lb.in.^{-2}$ . Neglecting the inertia of the spokes and assuming a width of $5\;in.$

Find the thickness of the rim. The Density is $\displaystyle 0.26 \;lb.in.^{-3}$

Workings

The Maximum radius is determined by the Stress.

Hence,

$600 = \frac{\rho \,r^2\;\omega ^2}{g}= \frac{0.26\times r^2}{g}\left(\frac{250\times 2\pi }{60} \right)^2$

Re-arranging and substituting in values (Note the "12" converts $g$ into inches/seconds squared)

$r = \left( \frac{600\times 32.2\times 12}{0.26} \right)\times \left(\frac{60}{250\times 2\pi } \right)^2 = 1301$

$\therefore\;\;\;\;\;\;r = 36.07\;inches$

For the first approximation assume that the mean radius is $34.5\;inches$ . This assumes that the thickness of the rim is approximately $3.14\;inches$ .

Moment of Inertia = $6000\;lb.ft^2 = 6000\times 12^2\;lb.in.^2$

Also the Moment of inertia = the mass $X$ the radius of gyration $k$ squared. Having set the mean radius at $34.5\;in.$ the inner radius will be $36.07 - 3.14 = 32.93 in.$

From which, $\;\;\;\;\;\;k^2 = \displaystyle\frac{36.07^2 + 32.93^2}{2} = 1192.72\;in^2$

$\therefore\;\;\;\;\;\;\;6000\times 144 = 2\pi \times 34.5\times 5\times t\times 0.26\times 1192.72$

From which:

$t = 5.62\;inches$ and The inner radius = $30.45\;in.$

Thus the corrected value of the Mean Radius is $\displaystyle 36.07 - \frac{5.62}{2} = 33.26\;in.$

$k^2 = \frac{R^2 + r^2}{2} = \frac{36.07^2 + 30.45^2}{2} = 1114.12\; in.^2$

Substituting this back in the original equation for the Moment of Inertia

$6000\times 144 = (2\pi \times 5\times 33.26\times t\times 0.26)\times 1114.12$

$\therefore\;\;\;\;\;\;t = \frac{6000\times 114}{1114.12\times 2\pi \times 5\times 33.26\times 0.26} = 2.57\;in.$

This value can now be used to correct the original guess.

i.e. Mean radius = $36.07 - \displaystyle\frac{2.57}{2} = 34.79\; in.$

And Inner radius = $36.07 - 2.57 = 33.5\; in.$

Thus the corrected value of $k$ can be used and is given by:

$k^2 = \frac{36.07^2 + 33.5^2}{2} = 1211.65\;in.^2$

Thus, $\;\;\;\;\;\;6000\times 144 = 2\pi \times 34.79\times 5\times t \times 0.26 \times 1211.65$

$\therefore\;\;\;\;\;\;\;\;t = \frac{6000\times 144}{2\pi \times 34.79\times 5\times 0.26\times 1211.65} = 2,5\;in.$

Which approximately satisfies the assumption of $34.79\;in.$ as the mean radius.

Solution

The thickness is $t=2,5\;in.$