Compound Stress and Strain part 1

The results of both tensile and shear stress acting together.

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Contents

Introduction
Oblique Stress.
Simple Tension
Pure Shear
Pure Normal Stresses On Given Planes.
A General Two-dimension Stress System.
Principal Planes
Principal Stresses.
A Shorter Method For Principal Stresses.
Maximum Shear Stresses.
Poisson's Ratio
Page Comments

Introduction

Drawing reference from the section on 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 $\inline \displaystyle f_r$ is the resultant stress at angle $\inline \displaystyle \phi$ with the normal to the plane on which it acts.

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

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

Several important particular cases can now be considered.

Simple Tension

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Tension is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object. It is the opposite of compression. As tension is the magnitude of a force, it is measured in newtons (or sometimes pounds-force) and is always measured parallel to the string on which it applies.

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

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

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The equations of equilibrium can be used to solve for $\inline \displaystyle f_\theta$ and $\inline s_\theta$ . Resolving in the Direction of $\inline \displaystyle f_\theta$ :

$f_\theta \timesAC\times t = f\times AB\times t\times \cos\theta$

Re-arranging,

$\;\;\;\;\;\;\;f_\theta = f\times\left(\frac{AB}{AC} \right)\times \cos\theta = f\times \cos^2\theta$

(2)

Resolving in the Direction $\inline \displaystyle s_\theta$ , $\inline s_\theta\times AC \times t = f\times AB\times t\times \sin \theta$

Re-arranging, $\inline \;\;\;\;\;\;\;s_\theta = f\times \left(\displaystyle\frac{AB}{AC} \right)\times \sin\theta$

$\therefore\;\;\;\;\;\;\;s_\theta = f\times \cos\theta \;\sin\theta = \frac{1}{2}f\times \sin2\theta$

(3)

The Resultant Stress,

$f_r + \sqrt{f_\theta ^2 + s_\theta ^2} = f\;\sqrt{\cos^4\theta + \cos^2\theta \times \sin^2\theta } = f\;\cos\theta$

(4)

It can be seen from the above equations that the maximum normal Stress occurs at $\inline \displaystyle \theta = 0$ and is of course equal to the applied stress $\inline f$ . The maximum Shear Stress occurs at $\inline \displaystyle \theta = 45^{0}$ and has a magnitude of $\inline \displaystyle \frac{1}{2}f$ . On these planes there is also a normal component equal to $\inline \displaystyle \frac{1}{2}f$ . The variation of Stress components with $\inline \displaystyle \theta$ is given by equations (2),(3) and (4).

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

A shear stress is defined as the component of stress coplanar with a material cross section. Shear stress arises from a force vector perpendicular to the surface normal vector of the cross section.

The important Result is that: In simple Tension (or Compression) the maximum Shear Stress is equal to half of the applied Stress.

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 $\inline t$ 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 are 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 $\inline AC$ , $\inline sin\theta$ and $\inline cos\theta$ , and future figures will show the forces acting on the element.

Pure Shear

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

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

Resolving in the direction of $\inline \displaystyle f_\theta$ ,

$f_\theta = (s\;\cos\theta )\times \sin\theta + (s\;\sin\theta )\times \cos\theta = s\times \sin2\theta$

Resolving in the direction of $\inline \displaystyle s_\theta$ ,

$s_\theta = (s\;\sin\theta )\times \sin\theta - (s\;\cos\theta )\times \cos\theta = s\times \cos2\theta$

Note $\inline \displaystyle s_\theta$ acts down the plane for $\inline \theta\; <\;45^{0}$

Thus, $\inline \;\;\;\;\;\;\;f_r = \sqrt{f_\theta ^2 + s_\theta ^2}$ at an angle $\inline 2\theta$ to $\inline s_\theta$

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In the above system the Normal component $\inline \displaystyle f_\theta$ has maximum and minimum values of $\inline + s$ (tension) and $\inline - s$ (compression) on planes which are at $\inline \displaystyle \pm 45^{0}$ to the applied Shear, and on these planes the tangential component $\inline \displaystyle s_\theta$ 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 $\inline \displaystyle 45^{0}$ to the given planes is one of equal tension and compression.

Pure Normal Stresses On Given Planes.

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

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Resolving in the direction of $\inline \displaystyle f_\theta$ ,

$f_\theta = (f_y\;\cos\theta)\;\cos\theta + f_x\;\sin\theta\times\sin\theta= f_y\;\cos^2\theta + f_x\;sin^2\theta$

Resolving in the direction of $\inline \displaystyle S_\theta$

$S_\theta = f_y\;cos\theta \times \sin\theta - f_x\;\sin\theta \times \cos\theta= \frac{1}{2}(f_y - f_x)\;\sin2\theta$

It can be shown that $\inline \displaystyle f_\theta$ varies between the limits of $\inline \displaystyle f_x$ and $\inline \displaystyle f_y$ which are its maximum and minimum values. However, $\inline \displaystyle s_\theta$ has a maximum value equal numerically to half the difference between the given normal Stresses and occurring on a plane at $\inline \displaystyle 46^{0}$ 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 $\inline \displaystyle f_x$ and $\inline f_y$ correspond to the Principle Stresses.

A General Two-dimension Stress System.

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

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Resolving in the direction of $\inline \displaystyle f_\theta$ ,

$f_\theta = (f_y\;\cos\theta )\times\cos\theta + (f_x\;\sin\theta )\times\sin\theta + (s\;\cos\theta )\times\sin\theta \;+(s\;\sin\theta )\times\cos\theta$

$\therefore\;\;\;\;\;\;f_\theta = f_y\left(\frac{1 + \cos2\theta }{2} \right) + f_x\left(\frac{1 - \cos2\theta }{2} \right) + s\;\sin2\theta$

$\therefore\;\;\;\;\;\;f_\theta = \frac{1}{2}(f_y + f_x) + \frac{1}{2}(f_y - f_x)\;\cos2\theta + s\;\sin2\theta$

(5)

Resolving in the direction of $\inline \displaystyle s_\theta$ ,

$s_\theta = (f_y\;\cos\theta )\sin\theta - (f_x\;\sin\theta )\;\cos\theta + (s\;\sin\theta )\sin\theta \;-(s\;\cos\theta )\cos\theta$

$\therefore\;\;\;\;\;s_\theta = \frac{1}{2}(f_y - f_x)\sin2\theta - s\;\cos2\theta$

(6)

Example:

[imperial]

Example - Example 1

Problem

If the Stresses on two perpendicular planes which pass through a point $\inline P$ are $\inline 6\;tons/sq.in.$ tension; $\inline 4\;tons/sq.in.$ compression and $\inline 3\;tons/sq.in.$ in Shear, find the Stress components and resultant Stress on a plane at 60 degrees to that of the tensile Stress.

Workings

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The above diagram shows the forces on the element.

Resolving,

$f_\theta = (6\times \cos 60)\times \cos60 - (4\times\sin60)\times \sin60 +$

$+(3\cos60)\times \sin60\; + (3\sin60)\times\cos60$

From which $\inline f_\theta = 1.1\;tons\;in.^{-2}$

And

$s_\theta = (6\cos60)\times\sin60 + (4\sin60)\times \cos60 - (3\cos60)\times\ cos60 +$

$+(3\sin60)\times\sin60$

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From which $\inline s_\theta = 5.83\;tons\;in.^{-2}$

$\therefore\;\;\;\;\;\;\;f_r = \sqrt{1.1^2 + 5.83^2} = 5.93\;tons\;in^{-2}$

$\inline \displaystyle f_r$ occurs at an angle of $\inline \phi = \tan^{-1}\displaystyle\frac{5.83}{1.1} = 80^{0}\;15'$

Principal Planes

It can be seen from equation (6) that there are values of $\inline \theta$ for which $\inline \displaystyle s_\theta$ is zero, and the Planes on which there are no Shear components are called Principal Planes.

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From equation (6)

$\inline \tan2\theta = \pm \displaystyle\frac{2s}{f_y - f_x}$ when $\inline s_\theta = 0$

This will give two values of $\inline \displaystyle 2\theta$ which differ by 180 degrees, and hence there are two values of $\inline \displaystyle \theta$ which differ by 90 degrees, i.e., the Principal Planes are at Right Angles.

From the diagram it can be seen that:

$\sin2\theta = \pm \frac{2s}{\sqrt{(f_y - f_x)^2 + 4s^2}}$

(7)

And,

$\cos2\theta = \pm \frac{(f_y - f_x)}{\sqrt{(f_y - f_x)^2 + 4s^2}}$

(8)

where the signs are to be taken as both positive or both negative (giving the Values for $\inline \displaystyle 2\theta$ )

Principal Stresses.

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

From Equation (5) and using equations (7) and (8),

Principal Stresses = $\inline \displaystyle\frac{1}{2}\;(f_y + f_x)\;\pm \displaystyle\frac{\displaystyle\frac{1}{2}(f_y - f_x)^2}{\sqrt{(f_f - f_x)^2 + 4s^2}}\;\pm \displaystyle\frac{s\times 2s}{\sqrt{(f_y - f_x)^2 + 4s^2}}$

$= \displaystyle\frac{1}{2}(f_y + f_x)\;\pm \frac{\displaystyle\frac{1}{2}[(f_y - f_x)^2 + 4s^2]}{\sqrt{(f_y - f_x)^2 + 4s^2}}$

$= \frac{1}{2}(f_y + f_x)\;\pm \frac{1}{2}\sqrt{(f_y - f_x)^2} + 4s^2$

The Importance of Principle Stresses is that they are the Maximum and Minimum values of Normal Stress in the two dimensions under consideration. When they are of opposite type they give the numerical values of the maximum tensile and compressive 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 Principle Stresses.

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

Resolving in the direction of $\inline \displaystyle f_x$ ,

$f\times \sin\theta = f_x\times \sin\theta + s\times \cos\theta$

$\therefore\;\;\;\;\;\;\;f - f_x = s\times \cot\theta$

(9)

Similarly resolving in the direction of $\inline \displaystyle f_y$ ,

$f\times\cos\theta = f_y\times \cos\theta + s\times \sin\theta$

$\therefore\;\;\;\;\;\;\;f - f_y = s\times \tan\theta$

(10)

It is now possible to eliminate $\inline \displaystyle \theta$ from equations (9) and (10) to give:

$(f - f_x)(f - f_y) = s^2$

In any numerical problem it is advisable to substitute the values at this stage and then solve the quadratic for the two values of the Principal Stresses. However mulitplying out the equation becomes:

$f^2\;-(f_x\;+f_y)f\;+f_xf_y - s^2 = 0$

Solving the quadratic,

$f = \frac{1}{2}(f_x + f_y)\;\pm\; \frac{1}{2}\sqrt{(f_x + f_y)^2 - 4\;f_x\;f_y + 4s^2}$

Or,

$\;\;\;\;\;f = \frac{1}{2}(f_x + f_y)\;\pm \;\frac{1}{2}\sqrt{(f_x - f_y)^2 + 4s^2}$

The values of $\inline \displaystyle \theta$ for the Principal Planes are of course found by substituting in equations (9) and (10).

Maximum Shear Stresses.

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

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

Hence, the maximum Shear Stress $\inline q$ occurs when $\inline 2\theta = 90^{0}$ , i.e. on planes at $\inline \displaystyle\45^{0}$ to the Principal Planes.

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

The maximum value of $\inline q$ is given by:

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

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 $\inline \displaystyle f_1\;\;,\;\;f_2$ and $\inline f_3$ are the three Principal Stresses (Compression is shown as being negative). Note that to find the maximum $\inline q$ it is necessary to find the greatest difference in the stresses.

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Example:

[imperial]

Example - Example 2

Problem

At a Section in a beam the tensile stress due to bending is $\inline 500\;lb./sq.in.$ and there is a shear stress of $\inline 2000lb./sq.in.$

Determine from first principles the magnitude and direction of the Principal Stresses and calculate the maximum Shear Stress.

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Workings

Let $\inline AC$ be a principal Plane and $\inline BC$ the plane on which the Bending Stress acts. There is no Normal Stress on $\inline AB$ since it is a longitudinal plane of the Beam. The forces are shown on the diagram.

Resolving in the direction of $\inline AB$ ,

$f\;\sin\theta = 5000\sin\theta + 2000\cos\theta$

$\therefore\;\;\;\;\;\;\;f - 5000 = 2000\cot\theta$

Resolving in the direction of $\inline BC$ ,

$f\cos\theta = 2000\sin\theta$

$\therefore\;\;\;\;\;\;f = 2000\tan\theta$

Eliminating $\inline \displaystyle \theta$ ,

$f(f - 5000) = 2000^2$

$\therefore\;\;\;\;\;\;f^2 - 5000f\;-4,000,000 = 0$

Solving the quadratic,

$\inline f = 5700$ or $\inline - 700$

i.e. The Principal Stresses are $\inline 5700\;lb./sq.in.$ in tension and $\inline 700\;lb./sq.in.$ in compression. The third stress is zero.

Using these values the directions of the Principal Planes are given by:

$\inline \theta = 70^{0}40'$ and $\inline 160^{0}40'$ (Differing by $\inline 90^{0}$ )

The Maximum Shear Stress $\inline \displaystyle \frac{1}{2}\;5700 - (- 700) = 3200\;lb.in^{-2}$

and the Planes of Maximum Shear are at $\inline \displaystyle 45^{0}$ to the Principal Planes i.e. $\inline \displaystyle 25^{0}40'$ and $\inline 115^{0}40'$ .

Solution

The maximum Shear Stress is $\inline 3200\;lb.in^{-2}$

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 $\inline E$ . There will also be a Strain in all directions perpendicular to $\inline f$ and the final shape of the bar will be as shown by the dotted lines on the diagram.

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Poisson's ratio is the ratio, when a sample object is stretched, of the contraction or transverse strain (perpendicular to the applied load), to the extension or axial strain (in the direction of the applied load).

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

The Ratio = Lateral Strain / Longitudinal Strain

when it is produced by a single Stress is called Poisson's Ratio and the symbol used is $\inline \displaystyle\frac{1}{m}$ or $\inline \sigma$

e.g. Lateral Strain = $\inline - \displaystyle\frac{1}{m}\times \frac{f}{E}$ or $\inline - \sigma\times \displaystyle\frac{f}{E}$

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

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

Example:

[imperial]

Example - Example 3

Problem

A steel bar $\inline 10\;ins.$ long and with a rectangular cross section of $\inline 1\;in.$ by $\inline 2\;ins.$ is subjected to a uniform Tensile Stress of $\inline 12\;tons/sq.in.$ along its length.

Find the changes in dimension.

$\inline E = 13,200\;tons/sq.in.$ and Poisson's ratio = $\inline 0.3$ .

Workings

Longitudinal Strain = $\inline \displaystyle\frac{f}{E} = \displaystyle\frac{12}{13,200}$

Therefore, the increase in length = $\inline \displaystyle\frac{12}{13,200}\times 10 = 0.0091\; inches.$

Lateral Strain = $\inline - \left(\displaystyle\frac{1}{m} \right)\times \displaystyle\frac{f}{E}\;= - 0.3\times\displaystyle\frac{12}{13,200}$

Decrease in $\inline 1\;in.$ dimension = $\inline \left(\displaystyle\frac{0.3\times 12}{13,200} \right)\times 1 = 0.00027\;in.$

Decrease in $\inline 2\;in.$ dimension = $\inline \left(\displaystyle\frac{0.3\times 12}{13,200} \right)\times 2 = 0.00054\;in.$

Solution

Decrease in $\inline 1\;in.$ dimension is $\inline 0.00027\;in.$
Decrease in $\inline 2\;in.$ dimension is $\inline 0.00054\;in.$