topic 5 - stress & strain (1)
TRANSCRIPT
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Chapter 5
STRESS AND STRAIN5.1 INTRODUCTION5.2 STRESS5.3 PLANE STRESS5.4
MORHS CIRCLE FOR PLANESTRESS
5.5 THIN-WALLED PRESSUREVESSELS
5.6 STRAIN5.7 PLANE STRAIN5.8 MOHRS CIRCLE FOR PLANE
STRAIN
5.9 STRAIN ROSETTE
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5.1 Introduction
The notion of stress originates from ourdesire to quantify internal or external forces
distributed, respectively, in a body or along its
boundary. Stresses are those forces distributed
over an infinitesimal unit area cut out of a body in
certain directions or over an infinitesimal unit
area on the bounding surface. The study of
stress-related problems, in general, is referred to
as kinetics.
Stresses may be related to strains in solids or
rates of deformation in fluids through constitutivehypotheses. These hypotheses relate the variation
of stress with respect to the variation of strain,
which we sometimes called stress-strain
relationship. These relationships are important in
the description of mechanical properties ofmaterial, such as hardening, hysteresis etc.
Strain, on the other hand, is a geometrical
concept and is associated with the geometrical
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change of a point in the material. The mechanics
of deformations and strains is referred to as
kinematics.
5.2 Stress
Stress is defined mathematically as
. (5.1)
Thus, the concept of stress is related to the
concept of point in the space. The most general
state of stress at a point may be represented by six
components,
as shown in the figure below:
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Fig. 5.1
5.3 Plane stress
Plane stress is a state of stress in which two faces
of the cubic element are free of stress as shown inFig. 5.2.
Fig. 5.2
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Thus a structural member is defined as in the
state of plane stress when
(5.2)
State of plane stress can occurs
1.In thin plate subjected to forces acting in themidplane of the plate.
2.On thefree-surface of a structural element ormachine component, i.e. at any point of the
surface not subjected to an external force, as
in the cantilever beam shown in Fig. 5.3
Fig. 5.3
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5.3.1 Transformation of plane stress
Consider the conditions for equilibrium of aprismatic element with faces perpendicular to the
x,y,x andy axes, as shown in Fig. 5.4
Equilibrium of force in thex axis yields
(5.3)
Fig. 5.4
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Equilibrium of force in they-axis yields
(5.4)
Solving (5.3) for and (5.4) for , we
have
(5.5)
(5.6)
Recalling now
,
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and
,
we can rewrite (5.5) as
. (5.7)
Similarly, we can rewrite (5.6) as
2cos2sin2'' xy
yx
yx+
=
(5.8)
To obtain , we replace in (5.7) with
that the y-axis forms with the x-axis.
Since and
we have
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(5.9)
Adding (5.7), (5.8) and (5.9) together,
(5.10)
Since , Eq. (5.10) tells us that thesum of the normal stresses exerted on a cubic
element of material is invariant with respect to
the orientation of that material.
5.3.2 Principal stresses
Eqs. (5.7) and (5.8) are in fact parametric
equations of a circle. By eliminating from Eq.
(5.7) and (5.8), we arrive at
.
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Setting
, (5.11)
we now have
(5.12)
This is shown in Fig. 5.5.
Fig. 5.5
This shows us that the Principal stresses occur
on the principal planes of stress with zero
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shearing stresses. The min. and max. stresses are
given respectively as
(5.13)
and the principal plane inclined at an angle
(5.14)
Notice that defines two angles separated by
90o
.
This is shown in Fig. 5.6.
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Fig. 5.6
From (5.12), if , shearing stress is
maximum, i.e.
(5.15)
and this occurs at
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(5.16)
Note that defines two angles separated by 90o
and offset from by 45o
. This is shown in Fig.
5.7.
Fig. 5.7
The corresponding normal stress is given by
(5.17)
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Example: For the state of plane stress shown,
determine (a) the principal planes, (b) the
principal stresses and (c) the maximum shearingstress and the corresponding normal stress.
Soln:
(a)Using Eq. (5.14), .(b)Using Eq. (5.13),
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(c)Using Eq. (5.15), .(d) Using (5.16), .
The corresponding normal stress is
computed using (5.17), which is
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Example: A single horizontal force P of 150lb
is applied to D of lever ABD. Determine (a) the
normal and shearing stresses on an element atpoint H having sides parallel to the x and y axes,
(b) the principal planes and principal stresses at
H.
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Solution
Evaluate the normal and shearing stresses at H.
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5.4 Mohrs circle for plane stress
With the physical significance of Mohrs circlefor plane stress established, it may be applied
with simple geometric considerations. Critical
values can be estimated graphically.
For a known state of plane stress ,
plot the points Xand Yand construct the circlecentered at C, as shown in Fig. 5.8
Fig. 5.8
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The principal stresses are obtained atA andB.
The direction of rotation ofOx to Oa is the same
as CXto CA.
Fig. 5.9
With Mohrs circle uniquely defined, the state of
stress at other orientations may be depicted. For
the state of stress at an angle w.r.t thexy axes,
construct a new diameterXY and an angle
w.r.t.XY.
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Fig. 5.10
Fig. 5.11
Normal and shear stresses are obtained from the
coordinatesXY.
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For centric axial loading on thin plate:
Fig. 5.12
For torsional loading:
Fig. 5.13
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Example: For the state of stress shown, (a)
construct Mohrs circle and determine (b) the
principal planes, (c) the principal stresses and (d)the maximum shearing stress and the
corresponding normal stress.
Soln:
(a)
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The Mohrs circle is thus plotted as shown:
(b) The principal planes are
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(C) The principal stresses are, from Mohrs circle
(d) The maximum shearing stress, from Mohrs
cicle, is . This occurs at
The corresponding normal stress, from Mohrs
circle, is .
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Example: For the state of stress shown,
determine (a) the principal planes and the
principal stresses, (b) the stress componentsexerted on the element obtained by rotating the
given element counterclockwise through 30o
.
Soln:
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(On x-axis, the normal stress is +ve and shear
clockwise. On y-axis, the normal stress is +ve
and shear anticlockwise. Thus XY).
(a)Principal planes and stresses
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(b) Stress components at 30o
counterclockwise
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From Mohrs circle
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5.5 Thin-walled pressure vessels
5.5.1 Cylindrical vessels
The plane stress analysis we studied previously
can be applied to the stress analysis of
thin-walled pressure vessels problems. Why?
Consider a cylindrical vessel of inner radius rand
wall thickness t, containing a fluid under pressure,
Fig. 5.14.
Fig. 5.14
Because the vessel is symmetry, no shearing
stress is exerted on element. The stresses on the
surface of the cylinder are therefore the principal
stresses. The principal stresses on the surface of
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the cylinder is termed
.
We want to determine the stresses exerted on a
small element of wall sides respectively parallel
and perpendicular to the axis of the cylinder, Fig.
5.15.
Fig. 5.15
5.5.1.1 Hoop stress
To obtain the magnitude of hoop stress, taking
force equilibrium in the z-axis, Fig. 5.15, yields
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Solving for gives
(5.18)
5.5.1.2 Longitudinal stress
To obtain the longitudinal stress, we cut a section
perpendicular to the axis of the cylinder.
Fig. 5.16
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Equilibrium of force in x-axis gives
Solving for yields
(5.19)
Remarks:
1.Notice that
2.The limit of validity of Eq. (5.18) and (5.19)must be such that
(5.20)
Notice that the rterm in (5.19) is in fact a meanvalue
(5.21)
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Using (5.21), we see that
.
We can expand the RHS using Binomial theorem
and get
Our argument runs like this:
(a)If , , which cannot be true.(b)If , if we take the first-order
approximation and this cannot be true also.
(c)The only possibility is that . Notethat the argument is insufficient
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since this can also means that .
This is the fundamental idea behind
approximation and analysis.
5.5.1.3 Mohrs circle
Returning now to our problem at hand, the hoop
and longitudinal stresses we found can be plotted
on a Mohrs circle as follows:
Point A corresponds to hoop stress and point B
corresponds to longitudinal stress.
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Fig. 5.17
The maximum in-plane shear stress can be
computed as
(5.22)
The maximum out-of-plane shear stress
corresponds to a 45o
rotation of the plane stress
element around a longitudinal axis, giving
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(5.23)
5.5.2 Spherical vessels
We now consider a spherical vessel of inner
radius rand thickness t, containing a fluid under
gage pressurep, Fig. 5.18.
Fig. 5.18
Because of symmetry, there is no shear stress on
the surface of the sphere. Therefore the stresses
exerted on the sphere must be the principal
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stresses and are of equal magnitude, i.e.
(5.24)
Consider a half-section of the sphere as shown in
Fig. 5.19. We want to find out the stress acting on
the thin wall.
Fig. 5.19
Equilibrium of force in the x-axis
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Therefore,
(5.25)Maximum out-of-plane shearing stress
(5.26)
and this can be shown on a Mohrs circle as
follows:
Fig. 5.20
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With this, we conclude our study on stresses,
plane stress. Next, we look at strain the concepts
associated with it.
5.6 Strain
Strain is a geometrical or kinematical concept
that describes the deformation of a body. Theterm deformation signifies the entire geometric
change by which the points in a body in the initial
state with all loads absent go to another
configuration as a result of the action of loads.
The aforementioned initial state we shall call theundeformed state, and the subsequent state
occurring in the presence of loads we call the
deformedstate. The deformation, so defined, will
be seen to include the following contributions for
each element of a body:
1.Rigid-body translation and rotation2. A dilatation contribution from changes in
geometry associated with the volume change
of the element
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3.A distortion contribution from the remainingchanges in geometry of the element, which
includes e.g. change in angularity betweenline segments. This contribution is
sometimes call deviatoric changes.
Before we define strain, it is important to define
displacement field vector, u. Let us consider the
position vector of a point P in the undeformed
state, . After deformation, point P moves to a
new point P with position vector . The
displacement field is thus defined as
(5.27)
The displacement field is obviously a function of
both and time, t.
Following this, strain is thus defined as
(5.28)
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The concept of strain is seen to be associated with
the concept of point. The most general state of
strain at a point can be represented by sixcomponents,
5.7 Plane strain
Under special conditions, three-dimensional
strain analysis can be analyzed as a
two-dimensional one. This is called plane strain
analysis.
A structural member is in a state of plane strain
when one normal strain and two shear strain
components are zeros, i.e.
(5.29)
In fact, to be strictly plane strain, we also require
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that .
Thus, in laymans term, we say plane strainoccurs when deformations of the material take
place in parallel planes and are the same in each
of those planes.
For example, a plate subjected along its edges to
a uniformly distributed load and restrained fromexpanding or contracting laterally by smooth,
rigid and fixed supports, Fig. 5.21.
Fig. 5.21
Another example is a long bar subjected to
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uniformly distributed transverse loads. State of
plane strain exists in any transverse section not
located too close to the ends of the bar.
Fig. 5.22
A thick-walled circular cylinder constrained at
both ends is also another example of plane strain.
5.7.1 Transformation of plane strain
Similar to the plane stress, it can be demonstrated
that plane strain is invariant under transformation
of coordinate system, i.e.
.
This shows that strain, like stress, is a tensor
quantity that obeys the tensors transformation
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law.
Performing the similar analysis on strain
(Cauchys Tetrahedron Lemma), we can obtain
(5.30)
(5.31)
(5.32)
5.8 Mohrs circle for plane strain
Equations (5.30), (5.31) and (5.32) are
parametric equations of a circle and thus can be
represented by Mohrs method.
Defining now
, (5.33)
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and
, (5.34)
we can plot the strain circle as shown in Fig. 5.23.
Fig. 5.23
The principal axes can be computed as
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, (5.35)
and the maximum and minimum strains can be
evaluated using the formula
(5.36)
The maximum in-plane shearing strain is given
by
(5.37)
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5.9 Strain rosetteThe strain gauge is the most common device for
measuring strain, Fig. 5.24
Fig. 5.24
A single gauge will yield only a normal strain inthe direction of the gauge. Thus in application,
we must use a cluster of gauges or strain rosette
to give pertinent information about the state of
strain at a point, Fig. 5.25.
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Fig. 2.25
If the rosette is aligned at 45o
with respect to each
other, and are obtained directly and is
obtained indirectly as
(5.38)
If the rosette is aligned at some angles
with respect to each other, normal and shearing
strains may be obtained from normal strains in
any three directions via the formula
(5.39)
(5.40)
(5.41)
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Example: A strain rosette aligned 45o
with
respect to each other recorded the following
readings, G1=0.002, G2=0.001 and G3=-0.004.What are the principal strains?
Solution:
We let G1 corresponds tox-axis, G2 to OB, G3 toy-axis
From formula
Notice that (dont get confused!)
To obtain the principal planes, we have
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To obtain the principal strains, we have
Thus .