Strain Transformation
Lecture 4-b
Mechanics of Materials
(Chapter 10)
By
R. C. Hibbeler
Course: Strength of Materials-II
Instructor: Dr. Shahzad Saleem
Department of Civil Engineering
University of Engineering and Technology, Taxila
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10.3
Mohr’s Circle—Plane Strain
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10.3 Mohr’s Circle—Plane Strain
Since the equations of plane-strain transformation are mathematically
similar to the equations of plane-stress transformation, we can also solve
problems involving the transformation of strain using Mohr’s circle.
Like the case for stress, the parameter θ in Eqs. 10–5 and 10–6 can be
eliminated and the result rewritten in the form
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10.3 Mohr’s Circle—Plane Strain
Equation 10–13 represents the equation of Mohr’s circle for strain. It has
a center on the ϵ axis at point C(ϵavg, 0) and a radius R. Notice how these
equations compare with Eqs. 9–11 and 9–12.
Here ϵx', ϵavg, and γxy/2 replace σx', σavg, and τxy.
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10.3 Mohr’s Circle—Plane Strain
Procedure for Analysis
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10.3 Mohr’s Circle—Plane Strain
Principal Strains
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10.3 Mohr’s Circle—Plane Strain
Maximum In-Plane Shear Strain
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10.3 Mohr’s Circle—Plane Strain
Strains on Arbitrary Plane
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Example 10.4
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Example 10.4
The state of plane strain at a point is
represented by the components ϵx =
250(10-6), ϵy = -150(10-6), and γxy =
120(10-6). Determine the principal
strains and the orientation of the
element.
SOLUTION:
Construction of the Circle
The center of the circle C is located on
the ϵ axis at
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Example 10.4
SOLUTION:
Construction of the Circle
Since γxy/2 = 60(10-6), the reference
point A (θ = 0°) has coordinates
A(250(10-6), 60(10-6)). From the
shaded triangle in Fig. 10–10a, the
radius of the circle is CA ; that is,
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Example 10.4
SOLUTION:
Principal Strains
The ϵ coordinates of points B and D
represent the principal strains. They
are
The direction of the positive principal
strain ϵ1 is defined by the
counterclockwise angle 2θp1, measured
from the radial reference line CA (θ =
0°) to the line CB. We have
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Example 10.4
SOLUTION:
Principal Strains
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Example 10.5
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Example 10.5
The state of plane strain at a point is
represented by the components ϵx =
250(10-6), ϵy = -150(10-6), and γxy =
120(10-6). Determine the maximum inplane shear strains and the orientation
of the element.
SOLUTION:
Construction of the Circle
The circle has been established in the
previous example.
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Example 10.4
SOLUTION:
Maximum In-Plane Shear Strain
Half the maximum in-plane shear
strain and average normal strain are
represented by the coordinates of point
E or F on the circle. From the
coordinates of point E,
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Example 10.4
Maximum In-Plane Shear Strain
To orient the element, we can
determine the clockwise angle 2θs1
measured from CA (θ = 0°) to CE.
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Example 10.6
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Example 10.6
The state of plane strain at a point is
represented by the components ϵx = 300(10-6), ϵy = -100(10-6), and γxy =
100(10-6). Determine the state of strain
on an element oriented 20° clockwise
from this reported position.
SOLUTION:
Construction of the Circle
The ϵ and γ/2 axes are established in
Fig. 10–12a. The center of the circle is
on the ϵ axis at
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Example 10.6
Construction of the Circle
The reference point A has coordinates
A(-300(10-6), 50(10-6)). The radius CA
determined from the shaded triangle is
therefore
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Example 10.6
Strains on Inclined Element
Since the element is to be oriented 20°
clockwise, we must establish a radial
line CP, 2(20°) = 40° clockwise,
measured from CA (θ = 0°), Fig. 10–
12a. The coordinates of point P (ϵx',
γx'y'/2) are obtained from the geometry
of the circle.
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Example 10.6
Strains on Inclined Element
Since the element is to be oriented 20°
clockwise, we must establish a radial
line CP, 2(20°) = 40° clockwise,
measured from CA (θ = 0°), Fig. 10–
12a. The coordinates of point P (ϵx',
γx'y'/2) are obtained from the geometry
of the circle.
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Example 10.6
Strains on Inclined Element
The normal strain ϵy' can be
determined from the ϵ coordinate of
point Q on the circle, Fig. 10–12a.
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10.4
Absolute Maximum Shear Strain
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10.4 Absolute Maximum Shear Strain
In Sec. 9–5 it was pointed out that in the
case of plane stress, the absolute
maximum shear stress in an element of
material will occur out of the plane when
the principal stresses have the same sign,
i.e., both are tensile or both are
compressive.
A similar result occurs for plane strain.
For example, if the principal in-plane
strains cause elongations, Fig. 10–13a,
then the three Mohr’s circles describing
the normal and shear strain components
for elements oriented about the x', y', and
z' axes are shown in Fig. 10–13b.
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10.4 Absolute Maximum Shear Strain
By inspection, the largest circle has a radius R = (γx'z') max /2. Hence,
This value gives the absolute maximum shear strain for the material. Note
that it is larger than the maximum in-plane shear strain, which is (γx'y')max
= ϵ1 - ϵ2.
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10.4 Absolute Maximum Shear Strain
Now consider the case where one of
the in-plane principal strains is of
opposite sign to the other in-plane
principal strain, so that ϵ1 causes
elongation
and
ϵ2
causes
contraction, Fig. 10–14a. Mohr’s
circles, which describe the strains
on each element’s orientation about
the x', y', z' axes, are shown in Fig.
10–14b. Here
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10.4 Absolute Maximum Shear Strain
We may therefore summarize the
above two cases as follows. If the
in-plane principal strains both have
the same sign, the absolute
maximum shear strain will occur
out of plane and has a value of γmaxabs = ϵmax.
However, if the in-plane principal
strains are of opposite signs, then
the absolute maximum shear strain
equals the maximum in-plane shear
strain.
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10.4 Absolute Maximum Shear Strain
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Example 10.7
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Example 10.7
The state of plane strain at a point is
represented by the strain components
ϵx = -400(10-6), ϵy = 200(10-6), and γxy
= 150(10-6). Determine the maximum
in-plane shear strain and the absolute
maximum shear strain.
SOLUTION:
Maximum In-Plane Shear Strain
From the strain components, the center
of the circle is on the ϵ axis at
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Example 10.7
SOLUTION:
Maximum In-Plane Shear Strain
Since γxy/2 = 75(10-6), the reference
point A has coordinates (-400(10-6),
75(10-6)). As shown in Fig. 10–15, the
radius of the circle is therefore
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Example 10.7
SOLUTION:
Absolute Maximum Shear Strain
From the above results, we have ϵ1 =
209(10-6), ϵ2 = -409(10-6). The three
Mohr’s circles, plotted for element
orientations about each of the x, y, z
axes, are also shown in Fig. 10–15.
It is seen that since the principal inplane strains have opposite signs, the
maximum in-plane shear strain is also
the absolute maximum shear strain;
i.e.,
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