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Just like stresses, we can have direct and shear strains.

  • Direct stresses, e or \varepsilon
    These are the relative change in length
    Positive if the length increases
  • Shear stresses, \gamma
    These are the change in angle
    Positive if angle decreases

Reference Strains

In 3D cartesian, there are 6 reference strains: e_x,e_y,e_x,\gamma_{xy},\gamma_{yz},\gamma{zx}

In 2D cartesian, there are three. These are derived from two displacement directions, u and v:

    \[e_x=\frac{\partial u}{\partial x} \qquad e_y=\frac{\partial v}{\partial x} \qquad \gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}\]

    \[\frac{\partial^2\gamma_{xy}}{\partial x \partial y}=\frac{\partial^2 e_x}{\partial y^2}+\frac{\partial^2e_y}{\partial x^2}\]

In 3D polar coordinates, there are also six:

    \[e_r=\frac{\partial u_r}{\partial r} \qquad e_\theta=\frac{u_r}{r} \qquad e_x=\frac{\partial u_z}{\partial z}\]

    \[\gamma_{zr}=\frac{\partial u_z}{\partial r}+\frac{\partial u_r}{\partial z}\]

  • For axisymmetric cases with no variation in hoop stress:

        \[\gamma_{r\theta}=\gamma_{\theta z}=0\]

  • For axisymmetric cases with a single displacement, u:

        \[e_r=\frac{\partial u_r}{\partial r}=\frac{\partial}{\partial r}(re_\theta)\]

Plane Strain

In very long components (like pipes), the deformation in the long direction (the axial direction, z, in pipes) is restricted. This leads to plane strain:

    \[e_x, e_y, \gamma_{\theta r} \neq 0 \qquad e_z,\gamma_{\theta z},\gamma_{rz}=0\]

Strain Transformations

The direct strains e_x and e_y and the shear strain \gamma_{xy} are experienced by the element above, at \theta to the x-axis. This causes it to deform in such a way that it experiences an overall normal (direct) strain, e_n, and an overall shear strain, e_s:

    \[e_n=\frac{e_x+e_y}{2}+\frac{e_x-e_y}{2}\cos{2\theta}+\frac{\gamma_{xy}}{2}\sin{2\theta}\]

    \[\frac{e_s}{2}=\frac{e_x-e_y}{2}\sin{2\theta}-\frac{\gamma_{xy}}{2}\cos{2\theta}\]

Exactly like direct stresses, the sum of the direct strains is constant, regardless of angle:

    \[e_n+e_{n+90}=e_x+e_y\]

Principal (Maximum Direct) Strains

Just as with stresses, the maximum direct strains are known as the principal strains:

    \[e_{1,2}=\frac{e_x+e_y}{2}\pm\frac{1}{2}\sqrt{(e_x-e_y)^2+\gamma_{xy}^2}\]

The direction in which these act respective to the x-axis is \theta_p:

    \[\tan{2\theta_p}=\frac{\gamma_{xy}}{e_x-e_y}\]

The shear strain on the plane of principal strains is zero.

Maximum Shear Strains

The minimum and maximum values of shear strain in the x-y plane are given by:

    \[\frac{e_{s(min,max)}}{2}=\pm\frac{1}{2}\sqrt{(e_x-e_y)^2+\gamma_{xy}^2}\]

    \[e_{s(min,max)}=\pm(e_1-e_2)\]

This happens at the angle \theta_s:

    \[\tan{2\theta_s}=-\frac{e_x-e_y}{\gamma_{xy}}\]

These minimum and maximum shear strain values are at 45° to the plane of maximum direct strain

Young’s Modulus & Shear Modulus

Just like we can use Young’s modulus, E, to convert between direct stresses and strains, the shear modulus, G, can be used for shear stresses and strains:

    \[\sigma=Ee \qquad \tau = G\gamma\]

Mohr’s Strain Circle

This is much the same as Mohr’s stress circle, but with strains (wow):

Strain Gauges & Rosettes

Strain gauges measure strain by measuring the resistance in a wire as it is stretched:

    \[R=\frac{\rho L}{A}\]

A gauge factor, G, links the change in resistance to a strain:

    \[\Delta e=\frac{\Delta R}{GR}\]

To increase the accuracy, Wheatstone bridges are used.

Strain Gauge Rosette


A strain gauge rosette measures the strain in three directions. From this, the strains in the orthogonal directions can be calculated using Mohr’s Strain Circle:

  1. Draw the three measured strain values relative to an arbitrary horizontal axis
    Note that they must be drawn numerically accurately along the axis, so will not necessarily be in order A-B-C from left to right but should be in increasing order of strain.
  1. Draw lines at the corresponding angles from an arbitrary point on the central strain line
    Ensure the angles a relative to the vertical line, from below
  1. Construct perpendicular bisectors for the chords BA and BC
    The intersection of these is the midpoint of the circle, so use this to draw a circle that passes through all three points (A, B & C)
  1. Move the horizontal axis up to go through the circle midpoint
    The original middle point, point B, must be moved to the top of the circle
  1. Draw the radii from the circle centre to each of the three points
    The angles between the radii should be twice the angles between the strain readings on the rosette
  1. The principal strains and maximum shear strains can now be read off the graph
    As can the strains in the x- and y-directions, at an angle of  from radius A (  being the angle of the strain rosette)

Useful Strain Relations

From Hooke’s Law, we can relate strains and stresses. This is extremely useful when dealing with real life applications, where generally only the applied direct stresses are known.

    \[e_x=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_x)]\]

    \[e_y=\frac{1}{E}[\sigma_y-\nu(\sigma_x+\sigma_z)]\]

    \[e_z=\frac{1}{E}[\sigma_z-\nu(\sigma_x+\sigma_y)]\]

We can use these equations to get stresses from strains – helpful when we only have strain gauge data:

    \[\sigma_x=\frac{E(e_x+\nu e_y)}{1-\nu^2}\]

    \[\sigma_y=\frac{E(e_y+\nu e_x)}{1-\nu^2}\]

The shear modulus is used to convert between shear stresses and strains:

    \[\gamma_{xy}=\frac{\tau_{xy}}{G}\]

This is related to the Young’s modulus:

    \[E=2G(1+\nu)\]

  • The strains on a deformed element are

        \[e_n=\frac{e_x+e_y}{2}+\frac{e_x-e_y}{2}\cos{2\theta}+\frac{\gamma_{xy}}{2}\sin{2\theta}\]


        \[\frac{e_s}{2}=\frac{e_x-e_y}{2}\sin{2\theta}-\frac{\gamma_{xy}}{2}\cos{2\theta}\]

  • The principle strains are the maximum direct strains at a point:

        \[e_{1,2}=\frac{e_x+e_y}{2}\pm\frac{1}{2}\sqrt{(e_x-e_y)^2+\gamma_{xy}^2}\]


        \[\tan{2\theta_p}=\frac{\gamma_{xy}}{e_x-e_y}\]

  • The maximum shear strains are:

        \[\frac{e_{s(min,max)}}{2}=\pm\frac{1}{2}\sqrt{(e_x-e_y)^2+\gamma_{xy}^2} \qquad e_{s(min,max)}=\pm(e_1-e_2)\]


        \[\tan{2\theta_s}=-\frac{e_x-e_y}{\gamma_{xy}}\]

  • The key relation between stresses and strains used for strain gauges:

        \[\sigma_x=\frac{E(e_x+\nu e_y)}{1-\nu^2}\]


        \[\sigma_y=\frac{E(e_y+\nu e_x)}{1-\nu^2}\]

Failure Materials Mohr's Circle Strain

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