Compound Stress and Strain part 1

introduction

Drawing reference from the section on http://codecogs.izyba.com/reference/engineering/materials/direct_stress.php?search=Direct%20Stress "Direct Stress" , we know that a direct load applied along the axis of a structural member produces a stress that is normal to the axis. However, stresses are not always induced along a direction normal to the axis, but also along inclined planes, in the form of shear stress. It is very common therefore, to find that both Direct Stress and Shear Stress occur at the same time, and as a result, the Stress across any section will be neither normal nor tangential to the plane of that section.

Oblique Stress.

If $\displaystyle f_r$ is the resultant stress at angle $\displaystyle \phi$ with the normal to the plane on which it acts.

To find the normal and tangential components $f$ and $s$ then by inspection:

$\phi = \tan^{-1}\displaystyle\frac{s}{f}$ And $\;\;\;\;\;\;\;f_r = \sqrt{f^2 + s^2}$

Several important particular cases can now be considered.

Simple Tension

The bar in the diagram is under the action of pure Tensile Stress $\displaystyle f$ along its length. Any transverse section such as $AB$ will have pure normal Stress acting on it. The problem is to find the Stress acting on any plane $AC$ at an angle $\displaystyle \theta$ to $AB$. This Stress will not be normal to the plane and can be resolved into two components $\displaystyle f_\theta$ and $s_\theta$.

The diagrams show the Stresses acting along the three planes of the triangular prism $ABC$. There can be no stress on the plane $BC$ which is a longitudinal plane of the bar. The Stress component $\displaystyle s_\theta$ must act up the plane for equilibrium. The thickness of the prism is $t$.

It can be seen from the above equations that the maximum normal Stress occurs at $\displaystyle \theta = 0$ and is of course equal to the applied stress $f$. The maximum Shear Stress occurs at $\displaystyle \theta = 45^{0}$ and has a magnitude of $\displaystyle \frac{1}{2}f$ . On these planes there is also a normal component equal to $\displaystyle \frac{1}{2}f$ . The variation of Stress components with $\displaystyle \theta$ is given by equations (1),(2) and (3).

Note $\displaystyle f_\theta$ is zero when $\theta = 90^{0}$ and $s_\theta = 0$ when $\theta = 0$ and $90^{0}.$ The resultant Stress is at a maximum when $\displaystyle \theta = 0$

Notes on the diagrams.

• In most problems the Stress is varying from point to point in the member and it is necessary to consider the equilibrium of an element which is sufficiently small to give a value at a particular point.

• It can be seen that the results are independent of the thickness $t$ of the element being considered and for convenience it may be assumed to be unity.

• As all the figures will always be right angle triangles, there will be no loss in generality by assuming that the Hypotenuse is of unit length.

By making use of the above it can be seen that the areas on which the Stresses act are proportional to 1 $AC$, $sin\theta$ and $cos\theta$, and future figures will show the forces acting on the element.

Pure Shear

If the Stress on a plane $AB$ is pure Shear $s$, then there will be an equal complementary Shear Stress on the plane $BC$. It is necessary that we should be able to find the Stress components $\displaystyle f_\theta$ and $s_\theta$ acting on any plane $AC$ which is at an angle of $\displaystyle \theta$ to $AB$.

Following normal convention the Applied Shear Stresses will be shown acting towards $B$ and $\displaystyle s_\theta$ acting up the plane $AC$. In accordance with the note at the end of the last paragraph, the area of the plane $AC$ is unity. The Forces acting on the plane are shown on the diagram.

In the above system the Normal component $\displaystyle f_\theta$ has maximum and minimum values of $+ s$ (tension) and $- s$ (compression) on planes which are at $\displaystyle \pm 45^{0}$ to the applied Shear, and on these planes the tangential component $\displaystyle s_\theta$ is zero. This shows that at a point where there is pure Shear Stress on any two given planes at right angles, the action across the planes of an element taken at $\displaystyle 45^{0}$ to the given planes is one of equal tension and compression.

Pure Normal Stresses on Given Planes.

If the Stresses on $BC$ are $\displaystyle f_x$ and on $AB$ $\displaystyle f_y$, then the forces on the element are proportional to those shown on the diagram.

It can be shown that $\displaystyle f_\theta$ varies between the limits of $\displaystyle f_x$ and$\displaystyle f_y$ which are its maximum and minimum values. However, $\displaystyle s_\theta$ has a maximum value equal numerically to half the difference between the given normal Stresses and occurring on a plane at $\displaystyle 45^{0}$ to the given planes. This is of some significance when calculating the maximum Shear Stress in any complex system and it will be found that $\displaystyle f_x$ and $f_y$ correspond to the Principal Stresses.

A General Two-dimension Stress system.

If the Stresses acting on the planes $AB$ and $BC$ are$\displaystyle f_y$, $f_x$ and $s$, then the Forces are as shown.

Principal Planes

It can be seen from equation (5) that there are values of $\theta$ for which $\displaystyle s_\theta$ is zero, and the Planes on which there are no Shear components are called Principal Planes.

Principal Stresses.

The Stresses on the Principal Planes will be pure Normal (Tension or Compression) and their values are called the Principal Stresses.

A Shorter Method for Principal Stresses.

On the assumption that Principal Planes (i.e. planes on which the shear stress is zero) exist, it is possible to use a shorter analysis to determine their position and the value of the associated Principal Stresses.

Let $AC$ be a principle plane, and let $\displaystyle f$ be the Principal Stress acting upon it. Then $\displaystyle f_x$, $f_y$ and $\;\;s$ are the Stresses on the planes $BC$ and $AB$ (See diagram)

Maximum Shear Stresses.

If $AB$ and $BC$ are the principal planes and $\displaystyle f_1$ and $f_2$ are the Principal Stresses, then resolving:

$s_\theta = (f_2\times \cos\theta )\;\sin\theta - (f_1\times\sin\theta )\;\cos\theta = \frac{1}{2}(f_2 - f_1)\;\sin\;2\theta$

Note: This can be compared to the paragraph on Pure Normal Stress.

The maximum value of $q$ is given by:

$q = \frac{1}{2}(f_2 - f_1) = \frac{1}{2}\sqrt{(f_x - f_y)^2 + 4s^2}$

Thus the maximum Shear Stress is half the algebraic difference between the Principal Stresses.

As all solids have in fact three dimensions, there must be three Principal Stresses although in many cases the third stress is zero.

In calculating the maximum Shear Stress by taking one-half the algebraic difference between the Principal Stresses, the zero Stress will be of importance if the other two are of the same type (Tensile or Compressive). The following figure illustrates this where $\displaystyle f_1\;\;,\;\;f_2$ and $f_3$ are the three Principal Stresses (Compression is shown as being negative). Note that to find the maximum $q$ it is necessary to find the greatest difference in the stresses.

Poisson's Ratio

If a bar is subjected to a longitudinal Stress, there will be a Strain in the direction of the Stress. The value of this Strain is $E$. There will also be a Strain in all directions perpendicular to $f$ and the final shape of the bar will be as shown by the dotted lines on the diagram.

For an elastic material it has been found that the Lateral Strain is proportional to the Longitudinal Strain and is of the opposite type.

If the Stress is beyond the Elastic Limit and the Total Longitudinal Strain is $e$ then the "elastic" portion is approximately $\displaystyle\frac{f}{E}$ and the "plastic" portion is $e - \displaystyle\frac{f}{E}$. For plastic deformation Poisson's Ratio may be taken as $0.5$, i.e. there is no change in either density or volume, and hence:

Total Lateral Strain = $- \sigma \times\displaystyle\frac{f}{E} - 0.5\left(e - \displaystyle\frac{f}{E} \right)$