Mechanics & Structures University Engineering

Analysing Stress & Strain

Chief Noter

July 30, 2021
6 min read
Hooke's Law Strain Stress Stress Analysis Young's Modulus

This notes sheet looks at the most common applications of stress & strain: pressure vessels, Poisson’s ratio, hydrostatic stress and volumetric strain.

Poisson’s Ratio
Rectangular Cross-Sections
Cylindrical Thin-Walled Pressure Vessels
Spherical Thin-Walled Pressure Vessels
Summary

As we know, engineering stress, $\sigma$ , and engineering strain, $\varepsilon$ are given by:

$\sigma=\frac{F}{A} \quad \quad \varepsilon=\frac{x}{L}$

$F$ is the force applied
$A$ is the cross-sectional area
$x$ is the extension
$L$ is the initial gauge length

Stress can be normal to the surface or parallel to it: the latter is called a shear stress.

Young’s Modulus, $E$ , is a material property relating the two, and is an indicator of stiffness:

$E=\frac{\sigma}{\varepsilon}$

The units of both stress and Young’s modulus are Pascals, Pa.

Poisson’s Ratio

When a bar is loaded in one direction (uniaxial tension), not only does it extend in that direction, it
contracts in cross-section.

For a rectangular section with a tensile load applied in the x-direction, the strain in the
perpendicular (y- and z-directions) is directly proportional to the negative of the strain in the xdirection.

This is known as Poisson’s Ratio, and the constant of proportionality is a material property, $\nu$ :

$\varepsilon_y=\varepsilon_z=-\nu\varepsilon_x$

The same applies for a circular section.

Different materials will have a different Poisson’s Ratio:

$\nu < 0.5$ The volume increases as the specimen is stretched in one direction
$\nu = 0.5$ There is no change in volume as the specimen is stretched in one direction
$\nu > 0.5$ The material is auxetic: the cross-section increases under uniaxial loading

Rectangular Cross-Sections

The strain in each direction of a rectangular cross-section can be found if the Poisson’s ratio,
Young’s modulus, and stresses in each direction are known:

$\varepsilon_x = \frac{1}{E} [\sigma_x - \nu(\sigma_y + \sigma_z)] + \alpha \Delta T$

$\varepsilon_y = \frac{1}{E} [\sigma_y - \nu(\sigma_x + \sigma_z)] + \alpha \Delta T$

$\varepsilon_z = \frac{1}{E} [\sigma_z - \nu(\sigma_x + \sigma_y)] + \alpha \Delta T$

The term on the end, $\alpha\Delta T$ represents thermal expansion. $\alpha$ is a material constant: the coefficient of linear thermal expansion, and $\Delta T$ is the change in temperature.

Hydrostatic Stress & Volumetric Strain

When the normal stress applied in all directions is the same, we call it hydrostatic stress:

$\sigma_H=\sigma_x=\sigma_y=\sigma_z$

This means the strain in each direction is the same too, resulting in a change of volume. The ratio
of change in volume to initial volume is called the volumetric strain:

$\varepsilon_V=\frac{\Delta V}{V} = \varepsilon_x + \varepsilon_y+\varepsilon_z$

This gives rise to yet another material property, bulk modulus, $K$ :

$K=\frac{\sigma_H}{\varepsilon_V}$

From the equation for volumetric strain, the change in volume can be found as:

$\Delta V = \varepsilon_V\times V$

This applies only to isotropic materials in the elastic region.

Isotropic means the material deforms uniformly. Wood and bone are examples of anisotropic
materials.

Cylindrical Thin-Walled Pressure Vessels

Pipes and tubes can be represented as cylindrical thin-walled pressure vessels, and we often need
to know the strength of these vessels so that we know what the maximum permissible pressure
inside it is.

cylindrical pressure vessels, stress & strain in pipes, hoop stress, axial stress

The key assumption for all thin-walled pressure vessels is that the thickness of the wall is
significantly smaller than the internal diameter, and as such is negligible.

Cylindrical Stresses

A cylindrical pressure vessel has three stresses: axial, along its centreline, hoop stress, around the
circumference, and radial stress pointing outwards. To simplify things, this is taken as zero, as it
is so much smaller than the hoop stress.

Hoop stress is given as:

$\sigma_\theta=\frac{PR_i}{t}$

And axial stress is given as half of the hoop stress:

$\sigma_z=\frac{\sigma_\theta}{2}=\frac{PR_i}{2t}$

Open-Ended Thin-Walled Pressure Vessels

In an open-ended cylinder, the axial stress is zero. Take a pipe with a piston on one or both ends
as an example: the pressure force pushes the piston in/out instead of transferring it to the pressure
vessel itself

Cylindrical Strain

Cylindrical pressure vessels also experience three strains.

Axial strain is given as the change in length over initial length:

$\varepsilon_z=\frac{\Delta L}{L}$

Hoop strain is given as the change in radius over the initial radius:

$\varepsilon_\theta = \frac{\Delta r}{r}$

Radial strain is the increase in wall thickness over the initial thickness, though we assume this to
be zero as the change in thickness is generally miniscule:

$\varepsilon_r = \frac{\Delta t}{t} \approx 0$

Using Hooke’s Law, we can also define these strains as:

$\varepsilon_z = \frac{1}{E} [\sigma_z - \nu(\sigma_\theta + \sigma_r)] + \alpha \Delta T$

$\varepsilon_\theta = \frac{1}{E} [\sigma_\theta - \nu(\sigma_z + \sigma_r)] + \alpha \Delta T$

$\varepsilon_r = \frac{1}{E} [\sigma_r - \nu(\sigma_\theta + \sigma_z)] + \alpha \Delta T$

Remember that we are taking radial stress & strain as zero

Volumetric Strain in Cylinders

The volumetric strain in a thin-walled cylinder is the sum of twice the hoop strain and the axial
strain:

$\varepsilon_V = \frac{\Delta V}{V} = 2\varepsilon_\theta + \varepsilon_z$

From which the change in volume can be found.

Spherical Thin-Walled Pressure Vessels

The hoop stress in a spherical thin-walled pressure vessel is half that in a cylindrical pressure
vessel:

$\sigma_\theta = \frac{PR_i}{2t}$

This makes them twice as strong as their cylindrical counterparts, and as such should be used
whenever possible (as less thickness is required, so less material).

Spheres also experience radial stresses, $r$ , and perpendicular hoop stresses, $\phi$ .

Assume the two hoop stresses (θ & φ) equal one another
Assume radial stress is zero

Spherical Strains

The three strains experienced by spherical pressure vessels are in the two hoop directions, as well
as in the radial direction.

Again, the two hoop strains are given as the change in radius over the initial radius:

$\varepsilon_\theta = \varepsilon_\phi = \frac{\Delta r}{r}$

Radial strain is the same as for cylinders, and we assume it is zero.

Using Hooke’s Law, we can also define these strains as:

$\varepsilon_\theta = \frac{1}{E} [\sigma_\theta - \nu(\sigma_r + \sigma_\phi)] + \alpha \Delta T$

$\varepsilon_\phi= \frac{1}{E} [\sigma_\phi- \nu(\sigma_\theta + \sigma_r)] + \alpha \Delta T$

$\varepsilon_r = \frac{1}{E} [\sigma_r - \nu(\sigma_\theta + \sigma_\phi)] + \alpha \Delta T$

Remember that we are taking radial stress & strain as zero

Volumetric Strain in Spheres

The volumetric strain in a thin-walled sphere is three times the hoop strain:

$\varepslion_V = \frac{\Delta V}{V} = 3\varepsilon_\theta$

From which the change in volume can be found.

Stress is given as $\sigma=\frac{F}{A}$
Strain is given as $\varepsilon = \frac{x}{L}$
Young’s modulus is given as $E=\frac{\sigma}{\varepsilon}$
Poisson’s ratio is given as $\varepsilon_y = \varepsilon_z = -\nu\varepsilon_x$
When the stress in all three directions is the same, the stress is hydrostatic, $\sigma_H$
Volumetric strain, $\varepsilon_V$ is the sum of all strains
Bulk Modulus, $K$ , is given as $\frac{\sigma_H}{\varepsilon_V}$
Hoop stress is given as $\sigma_\theta = \frac{PR_i}{t}$
Axial stress is half the hoop stress: $\sigma_z = \frac{\sigma_\theta}{2}$

Analysing Stress & Strain

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Poisson’s Ratio