Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

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STRAIN TRANSFORMATIONS
Hence, the largest possible value for Poisson’s ratio is ν = 0.5. A Poisson’s ratio of ν = 0.5
is found in materials such as rubber and soft biological tissues. For common engineering
materials, Poisson’s ratio is generally less than 0.5, since most materials demonstrate some
change in volume when subjected to stress. A negative value of Poisson’s ratio is theoretically
possible, but materials with such a value are atypical. Several foams and certain crystals
have negative Poisson’s ratios.
Since ν < 0.5 for most engineering materials, stretching a material in one direction
will increase its volume. For example, consider an axially loaded rod in tension where the
axial normal stress is σ x > 0 and the transverse normal stresses are σ y = σ z = 0. From Equation
(13.22), as long as ν < 0.5, e must be greater than zero, and thus, the volume of the
stretched rod will be greater than the volume of the unstretched rod.
Special case of Hydrostatic pressure. When a volume element of a material
is subjected to the uniform pressure of a fluid, the pressure on the body is the same in all
directions. Further, the fluid pressure always act perpendicular to any surface on which it
acts. Shear stresses are not present, since the shear resistance of a fluid is zero.
When an elastic body is subjected to principal stresses such that σ x = σ y = σ z = −p, this
state of stress is referred to as the hydrostatic stress state. In that case, Equation (13.22) for
the volumetric strain reduces to
3(1−
2 ν)
e =− p
E
The term modulus defines a ratio between a stress and a strain. Here, our stress is the
hydrostatic compressive stress −p and the strain is the volumetric strain. We can now define
a new modulus, the bulk modulus, as
Notice that, for many metals,
ν ≈ 1 3 . Thus, the bulk
modulus K for many metals is
approximately equal to the
elastic modulus E.
p E
K = − =
e 3(1−
2 ν)
(13.23)
Note that a material under hydrostatic pressure can only decrease in volume; accordingly,
the dilatation e must be negative and the bulk modulus is a positive constant.
ExAmpLE 13.5
y
p
An aluminum alloy [E = 73 GPa; ν = 0.33] block is subjected
to a uniform pressure of p = 35 MPa as shown. Determine
(a) The change in length of sides AB, BC, and BD.
(b) The change in volume of the block.
z
p
p
A
160 mm
C
B
p
p
120 mm
D
x
200 mm
Plan the Solution
This is a case of hydrostatic stress, so the stresses σ x , σ y , and
σ z are each equal to −p. Equations (13.16) will be used to
calculate the strains e x , e y , and e z for the three directions, and
from these strains and the initial dimensions of the block, the
changes in length will be found. The change in volume will be
found from the initial volume and the dilatation e, as defined
in Equation (13.21).