We will deal only with homogeneous and isotropic materials in this
course. An isotropic material is one whose properties are independent of
direction. Thus, if we want to do a test on a specimen of such a material, it
does not matter from which part of the original material or in which direction
we cut our sample out.
For an isotropic material we have to deal with three material properties of
which only two are independent.
These material properties are
The Young's modulus () (also called the modulus of elasticity).
To find the Young's modulus we take a cylindrical specimen and pull on it
uniaxially. The result is a stress-strain plot of the form shown in the
adjacent figure.
The Young's modulus is defined as the ration of stress to strain in the linear
elastic part of the stress-strain curve. Thus
Since the strain is dimensionless, the units of the Young's modulus are the
same as the units of stress, i.e., force/area. In the English system, the
units of the Young's modulus are psi while the SI units are Pa.
Representative values of Young's modulus are psi (10 Msi) or Pa (70 GPa) for
aluminum and psi (29 Msi) or Pa (200 GPa) for steel. The higher the value of
the stiffer a material. Note that stiffness and strength are two different
concepts. Also, the value of Young's modulus does not vary much if the alloying
content of a material is small compared to the amount of pure metal.
Values of Poisson's ratio range between 0.25 to 0.35 for most materials. Also,
note that the shape of the cross section does not matter when we are measuring
the Poisson's ratio as long as the applied stress is axial.
To find the shear modulus directly, we have to do a torsion test on a
cylindrical specimen. A schematic of the stress-strain curve for such a test
is shown in the adjacent figure.
The shear modulus is defined as
where is the shear stress and is the shear strain.
The shear modulus can be expressed in terms of and using the relation
The following table shows the strains that result when a stress is applied in
a particular direction (or on a particular plane).
Resulting strain
Applied stress
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
You can check that these are indeed correct by looking at an infinitesimal element
and seeing what the possible strains are when a particular stress component is
applied.
Notice that normal stresses do not produce shear strains and shear stresses do
not produce axial strains. This is only true for isotropic materials.
One of the advantages of working with linear elastic materials is that we can
apply the principle of superposition. This principle states that the strains
produced by a particular stress state can be added to those produced by a
different stress state to get the strains for a combined stress state
.
Applying the principle of superposition to the results in the table above, we
find that if we have a stress state where all the stress components are active,
then the strains are given by
This relation is called the three-dimensional Hooke's law.
It helps to think of plane stress as the state of stress on a plane with
no stress!
Consider the infinitesimal cube shown in the adjacent figure. Assume that
there is no stress on the plane perpendicular to the -axis. Then moment
equilibrium requires that the out of plane shear stresses are also zero.
In fact we can simplify the diagram and just draw a square instead of a cube
as shown. But keep in mind that the shear stresses still act on a surface
as do the normal stresses.