Mechanics & Structures University Engineering

Complex Stresses

Chief Noter

July 25, 2021
6 min read
Failure Mohr's Circle Stress Tresca von Mises

Previously, we have only really looked at fairly basic direct stresses (shear stresses only really cropped up in torsion). Obviously, the real world is not so simple.

Stress Transformations
Maximum Normal (Direct) Stresses
Maximum Shear Stresses
Mohr’s Stress Circle
Failure Criteria
Summary

The notation of the shear stresses matters:

$\tau_{xy}$ means the plane faces in the $x$ direction, and the stress in the $y$ direction.

Complimentary shear stresses are when the opposing shear stresses are equal – this is the most common form, else the element would not be in equilibrium and would rotate.

Plane stress state assumes a 2D system of stresses: all direct and shear stresses in the $z$ direction are assumed zero.

Stress Transformations

When an element experiences rotation due to the applied stresses, we measure the angle in the anti-clockwise direction:

$S_n$ is the (normal) direct stress:

$S_n=\frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos{2\theta}+\tau_{xy}\sin{2\theta}$

$S_s$ is the (parallel) shear stress:

$S_s=\frac{\sigma_x-\sigma_y}{2}\sin{2\theta}-\tau_{xy}\cos{2\theta}$

These equations apply only to complimentary shear stresses: $\tau_{xy}=\tau_{yx}$

Maximum Normal (Direct) Stresses

Differentiating the above equation for $S_n$ and setting this to equal zero gives us the angle at which the normal stresses are maximum:

$\tan{2\theta_p}=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}$

$\sin{2\theta_p}=\pm\frac{2\tau_{xy}}{\sqrt{(\sigma_x-\sigma_y)^2+4\tau_{xy}^2}}$

$\cos{2\theta_p}=\pm\frac{\sigma_x-\sigma_y}{\sqrt{(\sigma_x-\sigma_y)^2+4\tau_{xy}^2}}$

$\theta_p$ is the principle angle: the angle at which the maximum direct stresses occur

These maximum direct stresses are known as the principal stresses, $\sigma_1$ and $\sigma_2$ . Substituting the equations for $\sin{2\theta_p}$ and $\cos{2\theta_p}$ into the equation for $S_n$ give us the equation to find these:

$\sigma_{1,2}=S_{n(min,max)}=\frac{\sigma_x+\sigma_2}{2}\pm\frac{1}{2}\sqrt{(\sigma_x-\sigma_y)^2+4\tau_{xy}^2}$

$\sigma_1$ is always the larger of the two
$\sigma_1$ and $\sigma_2$ are always perpendicular to one another

The sum of the orthogonal stresses is constant at any angle

$\sigma_x+\sigma_y=\sigma_1+\sigma_2$

Substituting $\theta_p$ into the above equation for shear stress, $S_s$ , shows us that it must equal zero on this plane.

The shear stresses on a plane of maximum direct stresses is zero.

Maximum Shear Stresses

Differentiating the equation for $S_s$ and setting this to equal zero gives us the angle of the plane of maximum shear stress:

$\tan{2\theta_s}=-\frac{\sigma_x-\sigma_y}{2\tau_{xy}}$

$\sin{2\theta_s}=\pm\frac{\sigma_x-\sigma_y}{\sqrt{(\sigma_x-\sigma_y)^2+4\tau_{xy}^2}}$

$\cos{2\theta_p}=\pm\frac{2\tau_{xy}}{\sqrt{(\sigma_x-\sigma_y)^2+4\tau_{xy}^2}}$

$\theta_s$ is the maximum shear stress angle: the angle of the plane on which shear stresses are a maximum
$\theta_s$ is always 45° from the angle of the maximum normal stresses, $\theta_p$ :

$\tan{2\theta_s}=-\frac{1}{\tan{2\theta_p}}$

Unlike the shear stresses on the plane of maximum direct stresses, the direct stresses on the plane of maximum shear stresses are not zero

Instead, the direct stresses on this plane are given as:

Unlike the shear stresses on the plane of maximum direct stresses, the direct stresses on the plane of maximum shear stresses are not zero.

Instead, the direct stresses on this plane are given as:

$S_{n(min,max)}=\hat\tau=\frac{\sigma_x+\sigma_y}{2}$

Combining Both Planes

The dotted coordinate axes represent the applied direct stresses
The red plane represents the maximum direct stresses (principal stresses)
The grey plane represents the maximum shear stresses

Mohr’s Stress Circle

Mohr’s Stress Circle plots the normal stresses on the x-axis, and the shear stresses on the y-axis. The axis intercepts represent the minimum and maximum values.

Angles on Mohr’s Circle are doubled

$\theta$ is the angle between the applied stress and the coordinate axis
$\theta_p$ is the angle of the plane of principal stresses, relative to the coordinate axis
$\theta'$ is the angle between the applied stress and the plane of principal stresses

Mohr’s Circle in 3D

To construct a Mohr’s stress circle for a 3D system, simply construct it for the two 2D planes and then combine:

Failure Criteria

There are three main failure criteria.

For ductile failure:

Tresca Criterion
Equating the maximum shear stresses
Von Mises Criterion
Equating the maximum strain energy

For brittle failure:

Rankine Criterion
Equating the maximum principle stresses

Tresca Criterion

This assumes that a component will fail when the maximum shear stresses, , reach the yield shear stress, $\tau_Y$ . This is given as half the direct yield stress:

$\tau_Y=\frac{\sigma_Y}{2}$

The maximum shear stress is generally taken as the maximum of the three principal stress differences:

$\hat\tau=\max{[|\frac{\sigma_1-\sigma_2}{2}|,|\frac{\sigma_2-\sigma_3}{2}|,|\frac{\sigma_3-\sigma_1}{2}|]}$

This can be simplified in certain 2D contexts:

Uniaxial Tension	$\hat\tau=\frac{\sigma_1}{2}$
Uniaxial Compression	$\hat\tau=\frac{\sigma_2}{2}$
Biaxial Tension	$\hat\tau=\frac{\sigma_1}{2}$
Biaxial Compression	$\hat\tau=\frac{\sigma_2}{2}$
Both Tension & Compression	$\hat\tau=\frac{\sigma_1-\sigma_2}{2}$

The locus of where failure will occur can be summarised graphically:

In this region, elastic deformation will occur. Outside, plastic deformation and failure occur.

Von Mises Criterion

This uses the root mean square maximum shear stress, $\hat\tau$ , which is representative of strain energy.

$\hat\tau=\sqrt{\frac{1}{3}[(\frac{\sigma_1-\sigma_2}{2})^2+(\frac{\sigma_2-\sigma_3}{2})^2+(\frac{\sigma_3-\sigma_1}{2})^2]}$

For a 2D stress state, we can express the von Mises Equivalent Stress, $\sigma_e$ :

$\sigma_e=\sqrt{\sigma_1^2+\sigma_2^2-\sigma_1\sigma_2}$

Failure occurs when this exceeds the yield stress: $\sigma_e\geq\sigma_Y$

This can also be plotted as a locus, from which it can be seen that the von Mises criterion is more accurate than Tresca, but Tresca is more conservative:

Rankine’s Criterion

This is more crude, and simply assumes that brittle failure will occur whenever the direct stress exceeds the yield stress:

Mohr’s Failure Criterion

This accounts for the fact that many materials are stronger in compression than in tension:

Complementary shear stresses are equal and opposite (equilibrium)
Plane stress states are 2D only
The direct stresses on an element are:
$S_n=\frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos{2\theta}+\tau_{xy}\sin{2\theta}]\[S_s=\frac{\sigma_x-\sigma_y}{2}\sin{2\theta}-\tau_{xy}\cos{2\theta}$
The principle stresses are the maximum direct stresses at a point:
$\sigma_{1,2}=S_{n(min,max)}=\frac{\sigma_x+\sigma_2}{2}\pm\frac{1}{2}\sqrt{(\sigma_x-\sigma_y)^2+4\tau_{xy}^2}$

$\tan{2\theta_p}=\frac{2\tau_{xy}}{\sigma_x-\sigma_y}$
The sum of the orthogonal stresses is constant at an angle.
The shear stresses on an element are:
The Tresca failure criterion equates maximum shear stress to yield shear stress:
$\hat\tau=\max{[|\frac{\sigma_1-\sigma_2}{2}|,|\frac{\sigma_2-\sigma_3}{2}|,|\frac{\sigma_3-\sigma_1}{2}|]}$
The von Mises failure criterion equated strain energy to yield shear stress:
$\hat\tau=\sqrt{\frac{1}{3}[(\frac{\sigma_1-\sigma_2}{2})^2+(\frac{\sigma_2-\sigma_3}{2})^2+(\frac{\sigma_3-\sigma_1}{2})^2]}$

Complex Stresses

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Stress Transformations

Maximum Normal (Direct) Stresses