# Compound Stress and Strain part 1

**Contents**

## 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 is the resultant stress at angle with the normal to the plane on which it acts.##### MISSING IMAGE!

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## 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.

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**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.

- 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 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**.

## Pure Shear

If the Stress on a plane is pure Shear , then there will be an equal complementary Shear Stress on the plane . It is necessary that we should be able to find the Stress components and acting on any plane which is at an angle of to .##### MISSING IMAGE!

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## Pure Normal Stresses On Given Planes.

If the Stresses on are and on , then the forces on the element are proportional to those shown on the diagram.##### MISSING IMAGE!

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## A General Two-dimension Stress System.

If the Stresses acting on the planes and are, and , then the Forces are as shown.##### MISSING IMAGE!

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##### Example - Example 1

**Stress components**and

**resultant Stress**on a plane at 60 degrees to that of the tensile Stress.

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occurs at an angle of

## Principal Planes

It can be seen from equation (6) that there are values of for which is zero, and the Planes on which there are no Shear components are called**Principal**

**Planes**.

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## Principal Stresses.

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

**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.##### MISSING IMAGE!

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## Maximum Shear Stresses.

If and are the principal planes and and are the Principal Stresses, then resolving:##### MISSING IMAGE!

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**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 and are the three Principal Stresses (Compression is shown as being negative). Note that to find the maximum it is necessary to find the greatest difference in the stresses.

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##### Example - Example 2

**magnitude**and

**direction**of the Principal Stresses and calculate the

**maximum Shear Stress**.

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- The
**maximum Shear Stress**is

## 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 . There will also be a Strain in all directions perpendicular to and the final shape of the bar will be as shown by the dotted lines on the diagram.##### MISSING IMAGE!

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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).

**Poisson**'

**s Ratio**and the symbol used is or e.g. Lateral Strain = or

##### Example - Example 3

**changes in dimension**. and Poisson's ratio = .

**Decrease in dimension**is**Decrease in dimension**is