The results of both tensile and shear stress acting together.
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!
23287/Compound-SnS-p1-0010.png cannot be found in /users/23287/Compound-SnS-p1-0010.png. Please contact the submission author.
To find the normal and tangential components

and

then by inspection:

And
Several important particular cases can now be considered.
Simple Tension
MISSING IMAGE!
23287/Compound-SnS-p1-0001.png cannot be found in /users/23287/Compound-SnS-p1-0001.png. Please contact the submission author.
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

along its length. Any transverse section such as

will have pure normal Stress acting on it. The problem is to find the Stress acting on any plane

at an angle

to

. This Stress will not be normal to the plane and can be resolved into two components

and

.
The diagrams show the Stresses acting along the three planes of the triangular prism

. There can be no stress on the plane

which is a longitudinal plane of the bar. The Stress component

must act up the plane for equilibrium. The thickness of the prism is

.
MISSING IMAGE!
23287/Compound-SnS-p1-0002.png cannot be found in /users/23287/Compound-SnS-p1-0002.png. Please contact the submission author.
The equations of equilibrium can be used to solve for

and

. Resolving in the Direction of

:
Re-arranging,
Resolving in the Direction

,
Re-arranging,
The Resultant Stress,
It can be seen from the above equations that the maximum normal Stress occurs at

and is of course equal to the applied stress

. The maximum Shear Stress occurs at

and has a magnitude of

. On these planes there is also a normal component equal to

. The variation of Stress components with

is given by equations (
2),(
3) and (
4).
Note

is zero when

and

when

and

The resultant Stress is at a maximum when
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
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

,

and

, and future figures will show the forces acting on the element.
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!
23287/Compound-SnS-p1-0003.png cannot be found in /users/23287/Compound-SnS-p1-0003.png. Please contact the submission author.
Following normal convention the Applied Shear Stresses will be shown acting towards

and

acting up the plane

. In accordance with the note at the end of the last paragraph, the area of the plane

is unity. The Forces acting on the plane are shown on the diagram.
Resolving in the direction of

,
Resolving in the direction of

,
Note

acts down the plane for
Thus,

at an angle

to

MISSING IMAGE!
23287/Compound-SnS-p1-0004.png cannot be found in /users/23287/Compound-SnS-p1-0004.png. Please contact the submission author.
In the above system the Normal component

has maximum and minimum values of

(tension) and

(compression) on planes which are at

to the applied Shear, and on these planes the tangential component

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

to the given planes is one of equal tension and compression.
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!
23287/Compound-SnS-p1-0005.png cannot be found in /users/23287/Compound-SnS-p1-0005.png. Please contact the submission author.
Resolving in the direction of

,
Resolving in the direction of
It can be shown that

varies between the limits of

and

which are its maximum and minimum values. However,

has a maximum value equal numerically to half the difference between the given normal Stresses and occurring on a plane at

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

and

correspond to the Principle Stresses.
A General Two-dimension Stress System.
If the Stresses acting on the planes

and

are

,

and

, then the Forces are as shown.
MISSING IMAGE!
23287/Compound-SnS-p1-0006.png cannot be found in /users/23287/Compound-SnS-p1-0006.png. Please contact the submission author.
Resolving in the direction of

,
Resolving in the direction of

,
[imperial]
Example - Example 1
Problem
If the Stresses on two perpendicular planes which pass through a point

are

tension;

compression and

in Shear, find the
Stress components and
resultant Stress on a plane at 60 degrees to that of the tensile Stress.
Workings
MISSING IMAGE!
23287/Compound-SnS-p1-0007.png cannot be found in /users/23287/Compound-SnS-p1-0007.png. Please contact the submission author.
The above diagram shows the forces on the element.
Resolving,
From which
And
MISSING IMAGE!
23287/Compound-SnS-p1-0008.png cannot be found in /users/23287/Compound-SnS-p1-0008.png. Please contact the submission author.
From which
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.
MISSING IMAGE!
23287/Compound-SnS-p1-0009.png cannot be found in /users/23287/Compound-SnS-p1-0009.png. Please contact the submission author.
From equation (
6)

when
This will give two values of

which differ by 180 degrees, and hence there are two values of

which differ by 90 degrees, i.e., the Principal Planes are at Right Angles.
From the diagram it can be seen that:
And,
where the signs are to be taken as both positive or both negative (giving the Values for

)
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 =
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.
MISSING IMAGE!
23287/Compound-SnS-p1-0011.png cannot be found in /users/23287/Compound-SnS-p1-0011.png. Please contact the submission author.
Let

be a principle plane, and let

be the Principal Stress acting upon it. Then

,

and

are the Stresses on the planes

and

(See diagram)
Resolving in the direction of

,
Similarly resolving in the direction of

,
It is now possible to eliminate

from equations (
9) and (
10) to give:
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:
Solving the quadratic,
Or,
The values of

for the Principal Planes are of course found by substituting in equations (
9) and (
10).
Maximum Shear Stresses.
If

and

are the principal planes and

and

are the Principal Stresses, then resolving:
MISSING IMAGE!
23287/Compound-SnS-p1-0012.png cannot be found in /users/23287/Compound-SnS-p1-0012.png. Please contact the submission author.
Hence, the maximum Shear Stress

occurs when

, i.e. on planes at

to the Principal Planes.
Note: This can be compared to the paragraph on Pure Normal Stress.
The maximum value of

is given by:
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

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.
MISSING IMAGE!
22109/untitled_4.jpg cannot be found in /users/22109/untitled_4.jpg. Please contact the submission author.
[imperial]
Example - Example 2
Problem
At a Section in a beam the tensile stress due to bending is

and there is a shear stress of
Determine from first principles the
magnitude and
direction of the Principal Stresses and calculate the
maximum Shear Stress.
MISSING IMAGE!
23287/Compound-SnS-p1-0013.png cannot be found in /users/23287/Compound-SnS-p1-0013.png. Please contact the submission author.
Workings
Let

be a principal Plane and

the plane on which the Bending Stress acts. There is no Normal Stress on

since it is a longitudinal plane of the Beam. The forces are shown on the diagram.
Resolving in the direction of

,
Resolving in the direction of

,
Eliminating

,
Solving the quadratic,

or
i.e. The Principal Stresses are

in tension and

in compression. The third stress is zero.
Using these values the directions of the Principal Planes are given by:

and

(Differing by

)
The Maximum Shear Stress
and the Planes of Maximum Shear are at

to the Principal Planes i.e.

and

.
Solution
- 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!
23287/Compound-SnS-p1-0016.png cannot be found in /users/23287/Compound-SnS-p1-0016.png. Please contact the submission author.
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

or
e.g. Lateral Strain =

or
If the Stress is beyond the Elastic Limit and the Total Longitudinal Strain is

then the "elastic" portion is approximately

and the "plastic" portion is

. For plastic deformation Poisson's Ratio may be taken as

, i.e. there is no change in either density or volume, and hence:
Total Lateral Strain =
[imperial]
Example - Example 3
Problem
A steel bar

long and with a rectangular cross section of

by

is subjected to a uniform Tensile Stress of

along its length.
Find the
changes in dimension.

and Poisson's ratio =

.
Workings
Longitudinal Strain =
Therefore, the increase in length =
Lateral Strain =
Decrease in

dimension =
Decrease in

dimension =
\times&space;2&space;=&space;0.00054\;in.)
Solution
- Decrease in
dimension is 
- Decrease in
dimension is