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8/10/2019 MOM2E chap2A
1/16
2005 Pearson Education South Asia Pte Ltd
2. Strain
1
CHAPTER OBJECTIVES
Define concept of normal
strain
Define concept of shear
strain
Determine normal andshear strain in
engineering applications
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2005 Pearson Education South Asia Pte Ltd
2. Strain
2
CHAPTER OUTLINE
1. Deformation2. Strain
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2. Strain
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Deformation
Occurs when a force is applied to a body
Can be highly visible or practically unnoticeable
Can also occur when temperature of a body is
changed
Is not uniform throughout a bodys volume, thus
change in geometry of any line segment within
body may vary along its length
2.1 DEFORMATION
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To simplify study of deformation
Assume lines to be very short and located in
neighborhood of a point, and
Take into account the orientation of the line
segment at the point
2.1 DEFORMATION
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Normal strain
Defined as the elongation or contraction of a linesegment per unit of length
Consider lineABin figure below
After deformation, schanges to s
2.2 STRAIN
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Normal strain
Defining average normal strainusing avg(epsilon)
As s 0, s 0
2.2 STRAIN
avg=s s
s
=
s s
s
lim
BA along n
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Normal strain
If normal strain is known, use the equation to
obtain approx. final length of a shortline segment
in direction of nafter deformation.
Hence, when is positive, initial line will elongate,if is negative, the line contracts
2.2 STRAIN
s (1 + )s
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2.2 STRAIN
Units
Normal strain is a dimensionless quantity, asits a ratio of two lengths
But common practice to state it in terms ofmeters/meter (m/m)
is small for most engineering applications, sois normally expressed as micrometers permeter (m/m)where1m = 106
Also expressed as a percentage,e.g.,0.001 m/m = 0.1 %
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2.2 STRAIN
Shear strain
Defined as the change in anglethat occurs
between two line segments that were originally
perpendicularto one another
This angle is denoted by (gamma) andmeasured in radians (rad).
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2.2 STRAIN
Shear strain
Consider line segmentsABandACoriginating
from same pointAin a body, and directed along
the perpendicular nand taxes
After deformation, lines become curves, such thatangle between them atAis
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2.2 STRAIN
Shear strain
Hence, shear strain at pointAassociated with n
and taxes is
If is smaller than /2, shear strain is positive,otherwise, shear strain is negative
nt=
2
lim
BA along n
C A along t
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Cartesian strain components
Using above definitions of normal and shear strain,
we show how they describe the deformation of the
body
2.2 STRAIN
Divide body into smallelements with
undeformed dimensions
ofx,yandz
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Cartesian strain components
Since element is very small, deformed shape of
element is a parallelepiped
Approx. lengths of sides of parallelepiped are
(1 + x) x (1 + y)y (1 + z)z
2.2 STRAIN
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Cartesian strain components
Approx. angles between the sides are
2.2 STRAIN
2 xy
2 yz
2 xz
Normal strains cause a change in its volume
Shear strains cause a change in its shape
To summarize, state of strain at a point requires
specifying 3 normal strains; x, y, zand 3 shear
strains of xy,yz,xz
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Small strain analysis
Most engineering design involves applications
for which only small deformationsare allowed
Well assume that deformations that take place
within a body are almost infinitesimal, so normalstrainsoccurring within material are very small
compared to 1, i.e.,
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Small strain analysis
This assumption is widely applied in practical
engineering problems, and is referred to as
small strain analysis
E.g., it can be used to approximate sin = , cos= and tan = , provided is small
2.2 STRAIN