• Assuming that the load P is
directed along the x axis we
have
where A is the c.s.a of the
bar,
From Hooke’s law,
• Where E is the modulus of
elasticity of the material.
All materials considered will be assumed to be both homogeneous
and isotropic, i.e., their mechanical properties will be assumed
independent of both position and direction
Poisson’s ratio
The condition of strain under an
axial load applied in a direction
parallel to the x axis:
Example
A 500-mm-long, 16-mm-diameter rod made of a homogenous,
isotropic material is observed to increase in length by 300 mm, and to
decrease in diameter by 2.4 mm when subjected to an axial 12-kN
load. Determine the modulus of elasticity and Poisson’s ratio of the
material.
The cross-sectional area of the rod is
2.12 MULTIAXIAL LOADING; GENERALIZED
HOOKE’S LAW
Choosing this axis as the x axis, and denoting by P the internal force at a given
location, the corresponding stress components were found to be
structural elements subjected to loads acting in the directions of the
three coordinate axes and producing normal stresses σx , σy , and σz
which are all different from zero. ( called a multiaxial loading )
we are concerned here only
with the actual deformation of the
element, and not with any possible
superimposed rigid-body displacement.
We can assume the side of the cube to be equal to unity
Under the given multiaxial
loading, the element will
deform into a rectangular
parallelepiped of sides
equal, respectively
Deformation of cube under multiaxial
principle of superposition.
This principle states that the effect of a given combined loading on a
structure can be obtained by determining separately the effects of the
various loads and combining the results obtained, provided that;
1. Each effect is linearly related to the load that produces it.
2. The deformation resulting from any given load is small and
does not affect the conditions of application of the other loads.
the components of
strain corresponding to
the given multiaxial
loading are:
generalized Hooke’s
law for the multiaxial loading of a
homogeneous isotropic material
Example
The steel block shown is subjected to a uniform pressure on all its faces.
Knowing that the change in length of edge AB is -1.2 x 103 in., determine
(a) the change in length of the other two edges,
(b) the pressure p applied to the faces of the block. Assume E = 29x106
psi and υ = 0.29.
2.14 SHEARING STRAIN
Shr. str τxy , τyz, and τzx present (as
well, of course, τ yx, τ zy, and τ xz ).
▪ These stresses have no direct
effect on the normal strains and,
as long as all the deformations
involved remain small, they will
not affect the derivation nor the
validity of the previous relations.
▪ The shearing stresses, however,
will tend to deform a cubic
element of material into an
oblique parallelepiped.
General state of stress.
Consider Cubic element subjected to
only
shearing stresses.
Two of the angles formed by the four faces
under stress are reduced from
Deformation of cubic
element due to shearing stresses.
➢ Plotting successive values of τxy against the corresponding values
of γxy, we obtain the shearing stress-strain diagram for the
material under consideration.
➢ For values of the shearing stress that do not exceed the
proportional
The following group of equations representing the generalized
Hooke’s law for a homogeneous isotropic material under the most
general stress condition.
Example: A rectangular block of a material with a modulus of rigidity G
= 90 ksi Eis bonded to two rigid horizontal plates. The lower plate is fixed,
while the upper plate is subjected to a horizontal force P (Fig.below).
Knowing that the upper plate moves through 0.04 in. under the action of the
force, determine (a) the average shearing strain in the material, (b) the force
P exerted on the upper plate.
Remember for a slender bar subjected to an axial tensile
load P directed along the x axis will elongate in the x direction
and contract in both of the transverse y and z directions. If
εx denotes the axial strain, the lateral strain is expressed as
εy = εz = -υ εx , where υ is Poisson’s ratio
axial load P causes normal and shearing stresses of equal magnitude on four
of the faces of an element oriented at 45º to the axis of the member.
▪ τmax on a plane forming an angle of 45º with the axis of the load.
▪ Hooke’s law for shearing stress and strain that the shearing strain γ´
associated with the element of is also maximum: γ´ = γm
▪ consider for this purpose the prismatic element obtained by
intersecting the cubic element by a diagonal plane
Solving for γm, we write:
▪ Since εx << 1, the denominator in the expression obtained can be
assumed equal to one; we have
Recall that σx = P/A and τm = P/2A, where A is the c.s. a of the
member. It thus follows that σx / τm = 2 , Substituting this value:
OR