A
fied
using the
will define these
STRAIN
various types of
Normal Strain. If we define the normal strain as the change in
length of a line per unit length, then we will not have to specify the actual
length of any particular line segment. Consider, for example, the line AB,
which is contained
within
undeformed
66
C H Athe
PTER 2
S T R A I N body shown in Fig. 2–1a. This
line
lies
along
the
n
axis
and
has
an original length of ¢s. After
formed body
deformation,
points
A
and
B
are
displaced
to A¿ and B¿, and the line
Strain
(a)
becomes a curve having a length of ¢s¿, Fig. 2–1b. The change in length
BİRİM DEFORMASYON
Strain
of to
thedescribe
line is therefore
¢s¿ - ¢s.
we define
the average
normal
In order
the deformation
of If
a body
by changes
in length
of strain
using
the
symbol
P
(epsilon),
then
avg
Deformasyon:
ısı etkisiinvs.the
altında
şekilwe
vedescribe
boyutlarında
meydana of
gelen
line
segments and Yük,
the changes
angleselemanların
betweenInthem,
will
order
to
the deformation
a body by changes in len
ange the body’s develop
the concept
of strain.
measured
by
değişikliklere
deformasyon
denir. Strain is actually line
segments and the changes in the angles between them, w
n and they experiments, and once the strain is obtained, it will be shown in the next
ation,
develop the concept of strain. Strain is actually measur
For example, a chapter how 2it can be related to the stress acting
within the
body.
n
experiments,
and once the strain is obtained, it will be shown in th
œ
¢s - ¢s
B¿
when stretched,
Pavg =
chapter how it can
be related to the stress acting within the body.
(2–1)
s occur
B
¢s
!s¿ when a
ation
of a body Normal Strain. If we define the normal strain as the change in
¿
2.2 STRAIN
67
67
nged. A typical length of a line per unit length, then we!swill not have to specify the actual 2.2 STRAIN
Normal Strain. If we define the normal strain as the cha
A
of caused by the length of any particular line segment. Consider, for example, the line AB,
lengthe.g.,
of a line per unit length, then we will not have to specify the
for experimental
work,
strain
is expressed
astoshown
a percent,
As point
Bwithin
is chosen
and closer
point
A,
the length
of the line
is
contained
the closer
undeformed
body
inlength
Fig.
2–1a.
This
for experimental
work,
660.001which
C
H A P T E Rstrain
2
S
T RisA I Nexpressed as a percent, e.g.,
of
any
particular line segment. Consider, for example, the li
-6
Strain
m>m
=
0.1%.
480110
2
As
an
example,
a
normal
strain
of
can
be
-6
not0.001
be uniform
will
become
shorter
and
shorter,
such
that
¢s
:
0.
Also,
this
causes
to
line lies
along
the na axis
andstrain
has of
an480110
original2 length
of ¢s.
After B¿within
m>m = 0.1%.
As an
example,
normal
can be
body
which
is
contained
the undeformed body shown in Fig. 2–1
-6
rmed
body
2
480110
in.>in.
480
mm>m,
as
,
or
0.0480%.
Also,
one
can
state
-6
tryreported
of any reported
line
approach
A¿,
such
that
¢s¿
:
0.
Consequently,
in
the
limit
the
normal
deformation,
points
A andorB0.0480%.
are displaced
to A¿
B¿,
and
the
line the n axis and has an original length of ¢s.
2 in.>in.
480110
mm>m,
as
, 480
Also, of
one
canand
state
line
lies
along
to
describe
the
deformation
of
a
body
by
changes
in
length
(b)
Undeformed
480
mA(480
this answer
as
simply
“micros”).
s length.
Hence,
strain
at
point
and
in
theof
direction
n is The change in length
becomes
am
curve
having
a length
¢s¿,body
Fig.of2–1b.
#
480
this and
answer
simply
(480
“micros”).
deformation, points A and B are displaced to A¿ and B¿, and t
ments
theaschanges
in the
angles
between them,
will
(a) we Strain
manner,
g. 2–1 we will of the line is therefore ¢s¿ - ¢s. If
we define the average
normalastrain
becomes
curve having a length of ¢s¿, Fig. 2–1b. The change in
the
concept
of
strain.
Strain
is
actually
measured
by
Note the
before
and
after
positionsthen
of three
Birim
ocated in the using
2
theDeformasyon:
symbol
Pavg
(epsilon),
2 of
of
the
line
is
therefore
¢s¿
¢s. If we
define
the average norma
ents,
and
once
the
strain
is
obtained,
it
will
be
shown
in
the
next
different
line
segments
on
this
rubber
In only
order cause
to describe
the deformation
of a body
by -changes
in length
Shear
Strain.
Deformations
not
line
segments
to
hanges
will
also
Shear
Strain.
not
only
line segments to
2.2 STRAIN
67
membrane
which isacting
subjected
to tension.
The
the
Pavg (epsilon),
then
how
it can
beelongate
related
toDeformations
the
stress
within
thecause
body.
line
segments
the using
changes
in the angles
between
them, we will
or
contract,
but
they
also
cause
them
todirection.
change
If
wesymbol
nt.elongate
For
example,
¢s¿and
- direction.
¢s
or contract,
also
cause
to
change
If
we
verticalbut
linethey
is lengthened,
the them
horizontal
line
P
=
lim
(2–2)
develop
the
concept
of
strain.
Strain
is
actually
measured by
select
two
line
segments
that
are
originally
perpendicular
to
one
another,
tion,
whereas
it
B : A along
2 two line segments
is shortened,
inclined perpendicular
line changes
select
thatand
arethe
originally
ton one¢s
another,
n
experiments,
and
once
the
strain
is
obtained,
it
will
be
shown
in the next
œ
xperimental
work,
strain
is
expressed
as
a
percent,
e.g.,
then
the
change
in
angle
that
occurs
between
these
two
line
segments
is
its
length
and
rotates.
¢s two
--6 ¢s
then the change
in
anglethe
thatnormal
occurs strain
between
these
lineinsegments is
lB¿ Strain.
IfAs
wean
define
as
the
change
Pangle
(2–1)
chapter
how
itbe
can
be related
to the
stress acting within the body.
m>m
= 0.1%.
480110
2 can
example,
a normal
strain
of =
g
referred
to
as
shear
strain.
This
is
denoted
by
(gamma)
and
is
avg
B angle is denoted by¢s
gB¿actual
referred
to-6as
shearthen
strain. This
(gamma) and is
¢sœ - ¢s
a line
per unit
will
not
have
towhich
specify
thedimensionless.
2length,
in.>in.
mm>m,
ed
as 480110
, 480 we
or 0.0480%.
Also,are
one
can state
always
measured
in(rad),
radians
(rad),
For example,
Pavg =
always
measured
in
radians
which
are
dimensionless.
For
example,
any particular
line
Consider,
for
line AB,from the same point
¢s
!s¿
480segment.
mthe
nswer
as simply
(480
“micros”).
consider
line
segments
ABexample,
and) is
AC the
originating
!s andP AC
Hence,
when
(or
Poriginating
the
initial
will elongate, whereas if
considerwithin
the line
segments
AB
from
the
sameline
point
avg in positive
contained
the
undeformed
body
shown
Fig.
2–1a.
This
Strain.
If we define the normal strain as the change in
A inand
a body,
directed
along
perpendicular
n and
A¿Normal
Pand
isAnegative
linethe
contracts.
A in athe
body,
directed
thethe
perpendicular
n¢s.
andAfter
t axes,
Fig.t axes,
2–2a.Fig. 2–2a.
along
n axisdeformation,
and hasalong
anthe
original
length
of
length
of
a
line
per
unit
length,
then
we
will not
After
ends
of
both
lines
are
displaced,
and
the
lines
2 the actual
normal
strain
isdisplaced,
a dimensionless
since
it isline
a ratio
of have to specify
point
B isthat
chosen
closer
and
closer
to line
point
the
length
of
the
After
deformation,
theNote
ends
of
both
lines
are
and A,
thequantity,
lines
ion,
points
A Deformations
and As
B are
displaced
to A¿
and
B¿,
and
the
length
of
any
particular
line
segment.
Consider,
for
example,
the line AB,
r themselves
Strain.
not
only
cause
line
segments
to
u¿,
themselves
become
curves,
such
that
the
angle
between
them
at
A
is
two lengths.
Although
thissuch
is thethat
case,
isat
sometimes
stated
inB¿
of a closer and closer
become
shorter
andthe
shorter,
¢sit:
0.
Also,
thisAs
causes
tochosen
become
curves,
such
that
angle
between
them
A is u¿,
point
Bterms
is
to point
a curve
having
awill
length
of ¢s¿,
Fig.
2–1b.
The
change
inis
length
which
contained
within
the
undeformed
body
shown
in
Fig.
2–1a.
ThisA, the length of t
ate
or contract,
but
they
also
cause
them
to
change
direction.
If
we
Fig.
2–2b.
Hence
the
shear
strain
at
point
A
associated
with
the
n
and
t
dy
ratio
ofstrain
length
units.
IfA
the
systemwith
is used,
for length
approach
A¿,
such
that
¢s¿
:
0.SI
Consequently,
limit
the unit
normal
Fig.
2–2b. Hence
the
shear
at
point
associated
theinthen
nthe
andthe
t basic
"
"
will
become
shorter
and
shorter,
such
that
¢s
: 0. Also, this cause
etwo
is
therefore
¢s¿
¢s.
If
we
define
the
average
normal
strain
line
lies
along the n axis
and has Pan
of ¢s. After
line segments
that are
originally
perpendicular
tofor
one
another,
axes becomes
is point
the
meter
(m).
Ordinarily,
most
applications
willoriginal
be ¢s¿length
Deformed
body
strainUndeformed
at
Abody
and
in the
direction
of
n is engineeringapproach
axes becomes
A¿,
such
that
:
0.
Consequently,
symbol
P
(epsilon),
then
points
Amicrometers
and B gelen
are displaced
to A¿ andBirimsizdir
B¿, and the line in the limit the
he changeavg
in angle
thatvery
occurs
between
these twodeformation,
lineBirim
segments
is in
(a)
Normal
birim
deformasyon:
boyda
meydana
boy
değişimidir.
small,
so
measurements
of
strain
are
per
meter
(b) 65
strain
at
point
A
and
in
the
direction
of n
aand
curve
having a length of ¢s¿,
Fig. 2–1b. The change
inislength
ed to as shear strain. This
angle is
denoted
by =g becomes
(gamma)
-6
1mm>m2,
where
1 mm
10
m. In
theis Foot-Pound-Second
system,
Fig.
2–1
ancak
#
gibi
birimlerle
gösterilmesi
uygundur.
of the
is therefore
- ¢s.
we define the average normal strain
s measured in radians (rad),
are dimensionless.
For
example,
strainwhich
is often
stated in units
ofline
inches
per inch¢s¿
( in.>in.
). If
Sometimes
the symbol
der the line segments AB andœ AC originating
fromusing
the same
point Pavg (epsilon), then
p
- ¢s
p gnt =
lim
u¿¢s¿
-and
¢s
-= ¢s
body, and directedPalong
the
tA
axes,
Fig.
2–2a.
= nu¿
lim
- Plim
gnt perpendicular
(2–3)(2–2)
2
BA
:
along
n(2–1)
avg =
B
:
along
n
(2–3)
¢s
2
B : A alongCn : A along t
¢s¿ - ¢s
¢s lines
deformation, the ends of both
are
displaced,
and
the
lines
P =
lim
C : A along t
#
#
B
:
A
along
n
¢s
elves become curves, such that the angle between them at A is u¿,
œ
¢s
¢s
B¿
–2b. Hence the shear
strain
at point
A associated
with the n and tbirim deformasyon ise kısalmayı gösterir.
Pozitif
birim
deformasyon
uzamayı,negatif
Pavg =
(2–1)
ecomes
¢s dik doğru parçaları
!s¿
Kayma
birim
deformasyonu:
Şekil
değiştirmeden
önce
birbirine
Hence,
when
P
(or
P
)
is
positive
the
initial
line
will
elongate,
whereas
if
avg A, the length of the line
nt B is chosen closer and closer to point
Notice
that
if u¿
smaller
than
p>2 the
shear
strain olarak
is positive,
whereas ifBirimi radyandır. Deformasyon
A¿ is açıda
is
negative
the ¢s
line
contracts.
meydana
değişim
tanımlanır.
that
u¿Parasındaki
smaller
p>2
strain
is positive,
if
meNotice
shorter
andif shorter,
such than
that
:the
0. shear
Also,gelen
this
causes
B¿ to whereas
Hence, when
P (or Pavg) is positive the initial line will elongate, wh
u¿
is
larger
than
p>2
the
shear
strain
is
negative.
Note
that
normal
strain
is
a
dimensionless
quantity,
since
is a ratio
oftaktirde
u¿ issuch
larger
than
p>2
theConsequently,
shear
strainbüyük
isinnegative.
sonrası
dik
açıdan
açıthe
meydana
aksithe
pozitiftir.
h A¿,
that
¢s¿
: 0.
thebir
limit
normal geliyorsaPnegatif,
isitnegative
line contracts.
two
lengths.
Although
this
is
the
case,
it
is
sometimes
stated
in
terms
of
a
point A and in the direction
of
n
is
Note
that
normal
strain
is the
a dimensionless
since it is a r
As point B is chosen closer and closer to point A,
length of thequantity,
line
p
lim u¿
ratio
If the SI system is used, then the basictwo
unit
for length
= length
- units.
gnt of
lengths.
Although
this
is
the
case,
it
is
sometimes
stated
in ter
will become(2–3)
shorter and shorter, such that ¢s : 0. Also, this causes B¿ to
2 (m).
B : A along n
is theDeformed
meter
Ordinarily, for most engineering applications
Plength
will beunits. If the SI system is used, then the basic unit for
Cbody
: A along t
ratio
of
approach A¿, such that ¢s¿ : 0. Consequently, in the limit the normal
#
very
small,
meter
(b)so measurements of strain are in micrometers
is theper
meter
(m). Ordinarily, for most engineering applications P
-6 strain at point A and in the direction of n is
1mm>m2, where
1
mm
=
10
m.
In
the
Foot-Pound-Second
system,
¢s¿
¢s
very
small,
so
measurements of strain are in micrometers per
Fig. 2–1
P strain
=
lim
(2–2)
is
often
stated
in
units
of
inches
per
inch
(
in.>in.
).
Sometimes
1mm>m2,
where
1 mm = 10-6 m. In the Foot-Pound-Second s
B : A along n
¢s
strain is often stated in units of inches per inch ( in.>in.). Som
e that if u¿ is smaller than p>2 the shear strain
is positive, whereas if
n
u¿
n
t
u¿
¢s¿ - ¢s
t p>2 the shear strain is negative.
arger than
B¿ P =
lim
(2–2)
C¿ B¿
C¿
B : A along n
¢s
p
hen P (or Pavg) is positive
the initial line
p
B will elongate, whereas if
2
A¿
tive the line contracts.
2 C B
C
A¿
hat normal strain is a dimensionless
quantity, since it is a ratio of
A
A
hs. Although this is the case, it is sometimes statedHence,
in terms
of aP (or Pavg) is positive the initial line will elongate, whereas if
when
ength units. If the SI system is used, then the basic unit
lengththe line contracts.
P is for
negative
# for most engineering applications
#
ter (m). Ordinarily,
P will
be
Note
that
normal
strain is a dimensionless quantity, since it is a ratio of
Deformed
body
Undeformed body
Deformed body
Undeformed body
all, so measurements ofn strain
are
in
micrometers
per
meter
two
lengths.
Although
this is the case, it is sometimes stated in terms of a
(b)
(a)
u¿
(b)
t
(a)m. In the Foot-Pound-Second
where 1 mm = 10-6
system,
ratio
of
length
units.
If
the
SI system is used, then the basic unit for length
B¿
often stated in units of inches per inchC¿( in.>in.
).
Sometimes
Fig. 2–2
is
the
meter
(m).
Ordinarily,
for most engineering applications P will be
Fig. 2–2
p
very
small,
so
measurements
of strain are in micrometers per meter
B
2
C
A¿ 1mm>m2, where 1 mm = 10-6 m. In the Foot-Pound-Second system,
#23
A
strain is often stated in units of inches per inch ( in.>in.). Sometimes
2.2
2.2
2.2
deformation of the body in Fig. 2–3a. ToCartesian
do so, imagine
the Components.
body is
Strain
subdivided
into
small
elements
such
as
the
one
shown
in
Fig.
now show how
Cartesian Cartesian
Strain Components.
Strain Components.
Using the definitions
Using theofdefinitions
normal of normal and shear strain, we will2–3b.
¢x,
¢y, and
This
element
is
rectangular,
has
undeformed
dimensions
Cartesian Strain
Components.
Using
the
definitions
of
normal
deformation
of
the
body
in Fig. 2–3a
and
shear
strain,
and
we
shear
will
strain,
now
show
we
will
how
now
they
show
can
how
be
used
they
to
can
describe
be
used
the
to
describe
the
z
RAIN
68
C H A P68
TER 2
S T R ACI H
N A P T E R 2¢z,S T
and
is To
located
in imagine
the
neighborhood
point in the
body,
Fig.elements
2–3a.
2deformation
and shear strain,
we will now
how they
canbody
be used
to
the
into
small
such
deformation
ofshow
the body
of
in the
Fig.
2–3a.
in
ToFig.
dodescribe
2–3a.
so, imagine
do the
so,
body
is the body of
is asubdivided
If the
element’s
dimensions
are
very 2–3b.
small,This
thenelement
its deformed
shape will has undef
deformation
of
the
body
in
Fig.
2–3a.
To
do
so,
imagine
the
body
is
is
rectangular,
subdivided
into
subdivided
small
elements
into
small
such
elements
as
the
such
one
shown
as
the
in
one
Fig.
shown
2–3b.
in
Fig.
2 STRAIN
bein
aundeformed
parallelepiped,
Fig.
since
very
small
will
remain
Cartesian
Strain
Components.
Using
the
definitions of
Cartesian
Components.
Using
the segments
definitions
ofthe
normal
subdivided
into
elements
such asisthe
one
shown
Fig. 2–3b.
¢z, line
and
is located
in
neighborhood
¢x,
¢y,2–3c,
¢y,
Thissmall
element
This
is rectangular,
element
rectangular,
hasUsing
undeformed
has
dimensions
dimensions
and¢x,
and
2
Cartesian
Strain
Components.
the definitions
of Strain
normal
approximately
straight
after
the
body
is
deformed.
In
order
to
achieve
Birim
Deformasyon
Kartezyen
Eksenlerdeki
Bileşenleri
and
shear
strain,
we
will
now
show
how
they
can
be
used
descr
y
n Components.
Using
definitions
shear
strain,
we
will
now
show
they
be used todimensions
describe the
¢x,
This
element
isstrain,
rectangular,
has
undeformed
dimensions
and
If can
the element’s
areto
very
sm
¢z,the
and we
is
located
and
in
is
thelocated
neighborhood
innormal
the
neighborhood
of
point
in
of
the
a body,
point
in
the
2–3a.
body,
Fig.how
2–3a.
and
shear¢z,
will
now
show
howof
they
can
be aand
used
to¢y,
describe
theFig.
this
deformed
shape,
we
will
first
consider
how
the
normal
strain
sian
Strain
Components.
Using
the
definitions
of
normal
deformation
of
the
body
in
Fig.
2–3a.
To
do
so,
imagine
the
z
Cartesian
Strain
Components.
Using
the
definitions
of
normal
e will
now
show
how
they
can
be
used
to
describe
the
deformation
of
the
body
in
Fig.
2–3a.
To
do
so,
imagine
the
body
is
¢z,
and
is
located
in
the
neighborhood
of
a
point
in
the
body,
Fig.
2–3a.
be
a
parallelepiped,
Fig.
2–3c,
since
ver
If
the
element’s
If
the
dimensions
element’s
are
dimensions
very
small,
are
then
very
its
small,
deformed
then
its
shape
deformed
will
shape
will
z
deformation of the body in Fig. 2–3a. To do so, imagine the body is
changes
the
lengths
ofremain
the sides
of
theaselements
rectangular
element,
and
then
ar
strain,
weFig.
will
now
show
how
they
can
be
used
to2–3c,
describe
the
subdivided
into
small
asstraight
theFig.
oneafter
shown
in Fig
and
shear
we
will
now
show
how
they
can
be
used
to
describe
the
e the
body
in
2–3a.
To
do
so,
imagine
the
is small
subdivided
into
small
elements
such
the
one such
shown
in
2–3b.
If
element’s
are
very
small,
then
its
deformed
shape
will
approximately
the body
bedimensions
ainto
parallelepiped,
be
astrain,
parallelepiped,
Fig.
2–3c,
since
Fig.
very
since
line
very
segments
small
line
will
segments
will
remain
subdivided
small
elements
such
as body
the
one
shown
in
Fig.
2–3b.
y
¢x
how
the
shear
strain
changes
the
angles
of
each
side.
For
example,
ation
of
the
body
in
Fig.
2–3a.
To
do
so,
imagine
the
body
is
¢x, c¢
This
element
is
rectangular,
has
undeformed
dimensions
of
the
body
in
Fig.
To
do
so,
imagine
theorder
body
mall
elements
such
as
the
one
shown
instraight
Fig.
2–3b.
¢x, ¢y,weand
Thisbody
element
is and
rectangular,
has
undeformed
be
a parallelepiped,
Fig.
2–3c,
since
very
small
line
segments
will
remain
deformed shape,
will first
approximately
approximately
straight
after
the
body
after
is2–3a.
deformed.
the
isIn
deformed.
order
to achieve
In
tois
achieve this dimensions
¢x,
¢y,
This
rectangular,
has
undeformed
dimensions
x is
y element
ydeformation
P
¢x
¢x
+
P
¢x
elongates
,
so
its
new
length
is
.
Therefore,
the
ded
into
small
elements
such
as
the
one
shown
in
Fig.
2–3b.
¢z,
and
is
located
in
the
neighborhood
of
a
point
in
the
body,
Fi
2deformed
x shown
xthe
subdivided
into
small
elements
such
as
theachieve
inthe
Fig.
2–3b. strain
¢x,
ctangular,
has
undeformed
dimensions
and
¢z,
and
isone
located
in the
neighborhood
of achanges
point in
thelengths
body, Fig.
approximately
straight
after
the
body
is deformed.
In
order
to
2is
of 2–3a.
the sides of th
this
deformed
this
shape,
we
will
shape,
first
we
consider
will
first
how
consider
the
normal
how
strain
normal
¢z, and
located
in
the
neighborhood
of¢y,
a point
in
the
body,
Fig.
2–3a.
(a)
approximate
lengths
of
the
three
sides
of
the
parallelepiped
are
¢x,
¢y,
ment
is
rectangular,
has
undeformed
dimensions
and
If
the
element’s
dimensions
are
very
small,
then
its
deformed
sha
¢x,
¢y,
This
isthe
rectangular,
has
dimensions
in
ofthe
aelement
point
body,
Fig.how
2–3a.
Ifnormal
the
dimensions
areand
very
thenthe
its deformed
shape
will the angle
histhe
deformed
shape,
we
will in
first
consider
the
shear strain
changes
changes
changes
lengths
of
the
the
lengths
sides
of
theundeformed
rectangular
ofelement’s
thestrain
element,
rectangular
element,
then
and small,
then how
If neighborhood
the element’s
dimensions
are
very
small,
then
itssides
deformed
shape
willand
x a point in the
is located
inlengths
the
neighborhood
ofdeformed
achanges
point
inshape
thechanges
body,
Fig.
2–3a.
be
a
parallelepiped,
Fig.
2–3c,
since
very
small
line
segments
will
¢z,
and
is
located
in
the
neighborhood
of
body,
Fig.
2–3a.
mensions
are
very
small,
then
its
will
be
a
parallelepiped,
Fig.
2–3c,
since
very
small
line
segments
will
remain
changes
the
of
the
sides
of
the
rectangular
element,
and
then
P
¢x
elongates
,
so
its
new length
¢x
¢x
how
the
shear
how
strain
the
shear
strain
the
angles
of
the
each
angles
side.
of
For
each
example,
side.
For
example,
be a parallelepiped, Fig. 2–3c, since very small line segments will remain
x
ement’s
dimensions
are
very
small,
then
its
deformed
shape
will
approximately
straight
after
the
body
is
deformed.
In
order
to
Ifstraight
the
dimensions
are
very
small,
deformed
d, Fig.approximately
2–3c,
since
very
small
line
segments
remain
yto
approximately
straight
after
the
body
is
to of
achieve
how
the
shear
strain
changes
angles
ofwill
example,
approximate
lengths
the three sides
Pelement’s
¢x
¢x
+ then
Px ¢x
¢x
elongates
,the
so
itsPbody
new
, iseach
so
length
its
new
isFor
is its
.¢x
Therefore,
.shape
the
Therefore,
yside.
after
the
deformed.
Inlength
order
achieve
11
++ PPx(a)
2¢x
¢x
11will
+ Py2 the
¢ydeformed.
11 + In
Pz2order
¢z
xelongates
x ¢x
x
allelepiped,
Fig.
2–3c,
since
very
small
line
segments
will
remain
this
deformed
shape,
we
will
first
consider
how
the
normal
adeformed.
parallelepiped,
Fig.
2–3c,
small
line
segments
will will
remain
aight
afterdeformed
body
isits
In
to
achieve
this
deformed
shape,
first consider how the normal strain
(a)
Pthe
¢x
+sides
Pthe
elongates
, sobeshape,
new
is
.very
Therefore,
the
approximate
approximate
lengths
oforder
the
lengths
three
ofsince
ofthree
the
parallelepiped
sides
of
the
parallelepiped
are we
are
this
we length
will
first
consider
how
the
normal
strain
x ¢x
x ¢x
mately
straight
after
the
body
is sides
deformed.
In
order
to changes
achieve
changes
the
lengths
of rectangular
the sides of element,
the rectangular
element, an
approximately
straight
after
the
body
is
deformed.
In
order
to
achieve
ape,
we
will
first
consider
how
the
normal
strain
y
the
lengths
of
the
sides
of the
and then
approximate
lengths
of
the
three
of
the
parallelepiped
are
changes the lengths of the sides of
rectangular element, and then
11
+
P
2
¢x
11 + Py2
ormed
shape,
we
will
first
consider
how
the
normal
strain
x
how
the
shear
strain
changes
the
angles
of
each
side.
this
deformed
shape,
weand
will
first
consider
how strain
the
strain
hs ofhow
the the
sidesshear
of the
rectangular
element,
then
how
theexample,
shear
changes
the angles
of sides
each are
side. For example, ¢x For exam
¢x+normal
strain
changes
angles
of
each
side.
For
And
approximate
angles
between
these
xthe
x sides of
11
+
P
2
¢x
11
+
11
P
+
2
¢x
P
2
¢y
11
+
11
P
+
2
¢y
P
2
¢z
11
P
2
¢z
the
lengths
of
the
the
rectangular
element,
and
then
x
x
y
y
z
z
P
¢x
¢x
+
P
¢x
elongates
,
so
its
new
length
is
.
changes
lengths
the
sides
the
ain changes
the Pangles
of each
side.
For of
example,
Px ¢xelement,
+ Px ¢x. Therefore, xthe Therefo
elongates
, the
so
its and
new
length
is ¢x
uzama
miktarı:
# xthen
Böylece
deformasyon
¢x+# ¢x
+Pof2Pboyundaki
¢x. rectangular
elongates
, ¢x
so
itsthe
new
length
is 11
Therefore,
x" ¢x
x¢z
11
+
P
2
11
+
P
2
¢y
¢x
esoshear
strain
changes
the
angles
of
each
side.
For
example,
x
y
z
(a)
approximate
lengths
of
the
three
sides of the
parallelepiped are
how
strain
changes
of each
For
example,
¢x
+ Pthree
its
new length
is the
. sides
Therefore,
theangles
(a)
approximate
lengths
the three¢x
are
x ¢x
p of
psides of thepparallelepiped
approximate
lengths
of shear
the
of thethe
parallelepiped
are side.
And
the
approximate
angles between t
g
g
g
P
¢x
¢x
+
P
¢x
es
,
so
its
new
length
is
.
Therefore,
the
xy
yz
xz sırasıyla
x new
Px ¢x# , so its
Px ¢x. 2Therefore,
elongates
length Herbir
is ¢x +eksendeki
the
hs ofxthe three sides
of the
parallelepiped
are
sonrası
uzunluk
olacaktır.
deformasyon
sonrası
2
2boylar
And
approximate
And
the
approximate
angles
between
angles
these
between
sides
are
these
sides
are
mate lengths of
thethe
three
sides
of
the
parallelepiped
are
approximate lengths of the three sides of the parallelepiped are 11 + P 2 ¢x
11 ++ PPzy22 ¢z
¢y p 11 + Pz2 ¢z
p
11 + Px2 ¢x
11 + Pyx2 ¢y
11
And the approximate
between
11 +angles
Px2 ¢x
11 these
+ Py2 sides
¢y are11 + Pz2 ¢z olacaktır.
#
Kenarlar
arasındaki
yaklaşık
açılar
da - gxy
- gy
Px2 ¢x
11 + Py2 ¢y
11 +pPz2 ¢z
p
p
p
p normal strains cause a change in volume of the2 element,
2
Notice
that
the
g
g
g
g
g
11 + Px2 ¢x p 11 + Py2 ¢y
11 + P 2 ¢z xy
yz
yz
xz11 + P 2 ¢zxz
+ Pywhereas
2 ¢y
p11
2 -+gPxxy2 ¢xp22z - 11
2
2 z strains cause a change in its shape. Of course, both of
the shear
- gxy
gxz
yz
And
the approximate
angles
between
the
approximate
angles
between
these
are these sides are
2
2 between these
2 sides areAnd
And the approximate
angles
these
effects
occur
simultaneously
during
thesides
deformation.
# these sides are
olacaktır. Görülmektedir
ki normal
birim
deformasyonlar
Notice
that the normal strains cause a
ate angles between
Ininsummary,
then,
the state
ofelement,
strain atwhereas
a pointthe
in shear
a body
requires
strains
cause a chan
Notice
that
the
Notice
normal
that
strains
the
normal
cause
strains
a
change
cause
volume
a
change
of
in
the
volume
element,
of
the
e approximate angles
between
these sides
are between
p
p
And
these sidesthree
are normal
p deformasyonları
pPx , p
p
pthe approximate
p angles
p olurken
P
,
P
,
gxzxy ,
specifying
strains,
and
three
shear
strains,
g
g
g
hacimsel
değişikliğine
sebep
kayma
birim
biçim
y of
z xy
yz
g
g
g
Notice
that
the
normal
strains
cause
a
change
in
volume
of
the
element,
these
effects
occur
simultaneously
durin
whereas
the
shear
whereas
strains
the
shear
cause
strains
a
change
cause
in
its
a
change
shape.
Of
in
its
course,
shape.
both
Of
of
course,
both
- gyz
- gxz
xy
p
p
p gxy
2 xz deformation
2
2
2 2 yz
2 the
gcourse,
, during
. These
gxzboth
strains
completely
describe
ofstate
a of strai
2pcause
2p occur
2 Ofthe
yzdeformation.
- gxythepshear
- strains
geffects
- a
geffects
yz
xz
whereas
change
in
its
shape.
of
In
summary,
then,
the
these
these
occur
simultaneously
simultaneously
during
the
deformation.
p -olmaktadır.
p Bir
p herhangi
2
2- gdeğişikliğine
2- gyz sebep
cismin
bir
noktasındaki
birimlocated
deformasyon
element
of requires
material
thenormal
point strains,
and P , P ,
ggxzxy
-the
-atrectangular
gyz
-atavolume
gaxzbody
hese effects2occur
during
deformation.
specifying at
three
Inxysimultaneously
summary,
then,
summary,
the
then,
of strain
the2state
aofpoint
strain
in
pointrequires
in a body
x
y
2 In
22 state
2 so
oriented
that
its
sides
are
originally
parallel
to
the
x,
y,
z
axes.
Notice
that
the
normal
strains
cause
a
change
in
volume
of the ed
Notice
that
the normal
strains
cause a
change
volume
of the
element,
In summary,
then,
state
of strain
atiçin#
achange
point
body
. kayma
These
strains
completely
Px , Pystrains,
,inPvolume
Px ,three
Prequires
,birim
Pzshear
, and
gxy , strains,
gxy
, # gyz , ingxz
three
specifying
normal
three
strains,
and
strains,
three shear
Notice specifying
that
the the
normal
strains
cause
anormal
in
of
element,
za, normal
ythe
durumunu
belirlemek
deformasyonları
ile
#
Provided
these
are
defined
at
points
the
body,
whereas
strains
cause
ainchange
in itsthen
shape.
course,
ormal
strains
volume
of
thestrains
element,
whereas
the
strains
cause
a change
its shape.
Ofvolume
course,
boththe
ofOf of
Pcompletely
gshear
, strains
specifying
three
strains,
shear
strains,
element
mate
gcause
,shear
. strains
gyzin
, cause
gaxzchange
These
strains
gPxz
These
describe
completely
thedescribe
deformation
the deformation
ofthe
a shear
of allainrectangular
whereas
the
inthree
its shape.
Of
course,
both
of
x ,. Pa
y ,change
z , and
xy
yznormal
deformed
shape
of
the
body
can
be
determined.
e
that
the
normal
strains
cause
a
change
in
volume
of
the
element,
these
effects
occur
simultaneously
during
the
deformation.
Notice
that
the
normal
strains
cause
a
change
in
volume
of
the
element,
cause
a
change
in
its
shape.
Of
course,
both
of
these
effects
occur
simultaneously
during
the
deformation.
gstrains
,
.
g
These
strains
completely
describe
the
deformation
of
a
oriented
so
that
its
sides
are
origina
rectangular
rectangular
volume
element
volume
of
element
material
of
located
material
at
the
located
point
at
and
the
point
and
effects birim
occur simultaneously
during bilinmesi
the deformation.
deformasyonlarının
gereklidir.
yz these
xz
srectangular
the shear strains
cause
a change
its that
shape.
Of course,
both
ofitstoshape.
In
then,
theatProvided
state
of strain
at
a point
a bodyatr
whereas
thestate
shear
cause
a at
change
Ofy,
course,
both
simultaneously
during
the
deformation.
Inain
summary,
then,
the
state
strain
a point
in
a body
requires
volume
element
of
material
the
point
these
strains
areindefined
oriented
so
oriented
that
itsin
sides
so
arelocated
its
originally
areparallel
the
parallel
x,
to
zsummary,
axes.
the
x, of
y,of
z axes.
In summary,
then,
the
ofstrains
strain
at sides
a point
inoriginally
body and
requires
fects
occur
simultaneously
during
the
deformation.
P
,
P
,
P
,
specifying
three
normal
strains,
and
three
shear
effects
occur
simultaneously
during
en, the
state
of
strain
atthese
aProvided
point
in these
body
x shape
y
z of
Deformasyon
öncesi
eleman:
Pz ,deformed
gxy ,can bestrai
specifying
three
strains,
and
three
shear
strains,
oriented
so that
itsthese
sides
are
originally
the
x, the
y,at
axes.
the body
det
Provided
strains
strains
are
attothree
all
defined
points
inzdeformation.
all
the
points
body,
then
the
the
body, Pthen
Paare
, Prequires
gnormal
specifying
three
normal
strains,
shear
strains,
x , Py ,the
x , Pdefined
yparallel
z , and
xy ,in
( p $ gxy)
mmary,
then,
the
state
of
strain
at
a
point
in
a
body
requires
g
,
.
g
These
strains
completely
describe
the
deformatio
In
summary,
then,
the
state
of
strain
at
a
point
in
a
body
requires
P
,
P
,
P
,
g
,
ormalg strains,
and
three
shear
strains,
yz
xz
g
,
.
g
These
strains
completely
describe
the
deformation
of
a
x
y
z
xy
2
Provided
these
strains
are
defined
at
all
points
in
the
body,
then
the
deformed
shape
deformed
of
the
shape
body
can
of
the
be
determined.
body
can
be
determined.
deformation
of a
xz
p
yz , gxz . These strains completely describe the yz
Px , can
Pythree
, Pbe
gxyand
, three
ng three
normal
strains,
and
three
shear
rectangular
element
of material
locatedand
at the poi
z ,deformation
Pastrains,
gofxy ,material
specifying
normal
strains,
shear
strains,volume
strains
completely
describe
the
of
volume
element
located
at the point
x , Prectangular
y ,atPz ,the
deformed
shape of
the
body
determined.
2 point
rectangular
volume
element
of
material
located
and
.
These
strains
completely
describe
the
deformation
of
a
oriented
so
that
its
sides
are
originally
parallel
to
z
gyz ,itsgxz
. These
strains
completely
describe
deformation
a
me
element
located
the point
and to
oriented
that
its sides areoforiginally
parallel to the x, y, z axes.the x, y,
orientedofso material
that
sides
areatoriginally
parallel
the x, so
y,the
z axes.
p
p
p
(1
"
#
)!z
ular
volume
element
of
material
located
at
the
point
and
z
Provided
these
strains
are
defined
at all
points in the body, th
element
of material
at
point
and at all points
its sides
are originally
parallel
to the x,
axes.
( $!z
( strains
defined
in the
body,
) these
) are
gxylocated
$ gpthe
Provided
theserectangular
strains
are volume
defined
at y,
allzpoints
inProvided
the
body,
then
p then the
2axes.
2 the2xy
2
p
(
p
p
drains
so that
its
sides
are
originally
parallel
to
the
x,
y,
z
$ gxz)
deformed
shape
of
the
body
can
be
determined.
so in
that
itsbody,
sides
are
parallel
to of
the x, y, z axes.
are defined
atoriented
all points
thebe
( then
)originally
$
gxythe
deformed
shape
2
deformed
shape
determined.
! x the body can be determined.
p
2the body,
2of the body
2can
p
dthe
these
strains
are
defined
at
all
points
in
then
the
Provided these strains are defined at all 2points
body can be determined.
(1 " #p
(1 " #z
x)!x
!yin the body, then the ( p $ g )
2 of the body can be determined.
#
yz
ed shape
(1 "!z#y)!y
2
deformed shape of the body(1can
" #zbe
(1 " #z)!z
)!zdetermined.
2
z
p
!zp
! x( p $ gxy)
( pDeformed
Undeformed
$pgxy)
( p $eleman:
( p $ gxz)
( p $Deformasyon
(1
" #z)!z
gxz)p
gxy)
2
2
2
sonrası
2
p
!z
2
2
!y
2
p element 2
element
x
!x
p
( $!
gp
xy)
( p $ gxz)
2
p
2
p
p
p
(1
"
#
(1
"
#
)!x
)!x
2
2 ( ( $$
(!y$ gxy)
x
x
2
!y
2
(c)
( $ gyz)
p
g (b)
g) )
2
2
2#y)!y
2 yz xy (1 " #y)!y
Undeformed
(1 (1
""
#x)!x
p
!y
2
( p $ gyz)
(1element
" #z)!z
(1 "p#z)!z
(1 " #y)!y
2 (1 " #z)!z
!z
2
p
Deformed
Deformed
Undeformed
Undeformed
p
!z
Fig.
2–3
!z
p
p
(1 " # )!z
2
element
(b) ( $ gxz)
element
( $ gxz) element
2
p
2
Deformed
Undeformed z element
!x
p
2
p
2
!
x
(1
"
#
)!z
p
!
x
z
(1 " #z)!z (c)
p p
2
element
(c) 2
(b)
(b)
xz)
p ( 2 $ gelement
!y
!z
p
p
#x)!x
( p $ g(1yz"
x
p
p
(1 " #x!y
)!x
)
2
(
(
2 2 !y
)
$
g
)
$
g
(
( $ gxz)
xz
(1 " #y)!y
gyz) (c) 2
yz
2(1 "$
(1 " #y)!y2
2
p
#
)!x
(1
"
#
)!y
! x(b)
2
2
x
y
!
x
( $ gyz)p
Fig.
2–3
Fig.
2–3
(1
"
#
)!y
Deformed
p
Undeformed
2
(1
"
#
)!x
y
!y
Deformed
Deformed
Undeformed 2 ( !y
$ gyz)
(x p $ gyz) Undeformed (1 " #x)!x
element
element
(1 " #y)!y Fig. 2–3
2
(1element
" #y)!y
element
2
Deformed
element
element
#
2.2
S
TRAIN
69
(c)
(b)
element
Deformed
Undeformed
(c)
Undeformed
(c)
2.2Deformed
S(b)
TRAIN
69
(b)
element
element
element
(c) Deformasyon Analizi:
element
Küçük Birim
Fig. 2–3
(c)
(b)
(c)
(b)
Fig. 2–3
!z
p
p
2
! x2
Fig. 2–3
engineering design involves
Fig.
2–3 tasarımları
alysis.
Most engineering
design
involves
Small
Straindeformasyonlara
Analysis. Most
Most
engineering
design
involves
Small
Analysis.
design
için Strain
küçük
izin engineering
verilir.
Bu derste
bir involves
cisimde
tions are allowed.Çoğu
In thismühendislik
text,
Fig. text,
2–3applications
Fig.
2–3 small
ly small deformations are allowed. In this
applicationsfor
forwhich
which
only
smalldeformations
deformationsare
areallowed.
allowed.InInthis
thistext,
text,
only
ations that take place
withingelen
a
meydana
deformasyonların
hemen
sonsuz
oldukları
varsayılmaktadır.
that the deformations
that take place
withintherefore,
atherefore,
wewill
willhemen
assumethat
thatthe
theküçük
deformations
that
takeplace
placewithin
withina a
we
assume
deformations
that
take
r, the normal strains occurring
mal. In particular, the normal strains occurring
bodyare
arealmost
almostinfinitesimal.
infinitesimal.InInparticular,
particular,the
thenormal
normalstrains
strainsoccurring
occurring
body
red to 1, so that # P V 1. This
1 denare
çok
çoksmall
küçük)
ery small compared to 1, (normal
so that P birim
This
withinthe
thematerial
material
are
very
small
comparedtoto1,1,sosothat
thatP PV
This
V 1. deformasyonlar
V1.1.This
within
very
compared
in engineering, and it is often
2
ctical application in engineering, and it is often
assumptionhas
haswide
widepractical
practicalapplication
applicationinin
engineering,and
andititisisoften
often
2engineering,
assumption
varsayım
can be used, forBu
example,
to mühendislikte çok yaygındır ve genellikle küçük birim deformasyon analizi
rain analysis. It can be used, for example, to
referredtotoasasa asmall
smallstrain
strainanalysis.
analysis.ItItcan
canbe
beused,
used,for
forexample,
example,toto
referred
u = u, provided u is very small.
os u = 1, and tan uolarak
= u, provided
u
is
very
small.
approximate
sin
u
=
u,
cos
u
=
1
,
and
tan
u
=
u
,
provided
u
is
verysmall.
small.
approximate
sin
u
=
u,
cos
u
=
1
,
and
tan
u
=
u
,
provided
u
is
very
bilinir. # çok küçük olduğunda #
ve #
olarak yaklaşımı
kullanılabilir.
e rubber bearing support under this
Theisrubber
bearing
ncrete bridge girder
subjected
to support under this
concrete
bridge
girder is subjected to
h normal and shear strain. The
bothbynormal
and shear strain. The
rmal strain is caused
the weight
normal strain is caused by the weight
Therubber
rubberbearing
bearingsupport
supportunder
underthis
this
The
concretebridge
bridgegirder
girderisissubjected
subjectedtoto
concrete
bothnormal
normaland
andshear
shearstrain.
strain.The
The
both
normalstrain
strainisiscaused
causedbybythe
theweight
weight
normal
#24
(p $
2
(p $ g
2
(1 " #x)!
2.2 S
2.2
od shown in Fig. 2–4 is subjected to an increase of
70
CC
HA
TPFig.
ETnormal
RE R22–4
RTstrain
A
IANI N
70rod creates
HP
Aa
2 S TSis
Rsubjected
slender
shown
in
to an
of
long The
its axis,
which
in the
rodincrease
of
slender
rodrod
shown
in in
Fig.
2–42–4
is is
subjected
to
anan
increase
of the
he
shown
Fig.
subjected
to
increase
of rod of
1>2slender
temperature
along
its
axis,
which
creates
a
normal
strain
in
zitsis-3
measured
inin meters.
the
zmperature
, where
The
slender
rod which
shown
Fig.
2–4 Determine
isstrain
subjected
to(a)
an
increase of
1>2which
perature
along
axis,
creates
a normal
in in
the
rod
of of
along
its
creates
a normal
the
rod
P-3
z iswhich
measured
in strain
meters.
Determine
(a)rod
theof
= 1>2
40110
2zaxis,
, where
ztemperature
along
its
axis,
creates
a
normal
strain
in
the
-3
1>2
of40110
end
the
rod
due
to
temperature
increase,
zthe
is
inthe
meters.
Determine
(a)(a)
thetheincrease,
2z 2zB
, where
where
z measured
isend
measured
in
meters.
Determine
= the
40110
,of
EXAMPLE
2.1
EXAMPLE
2.1
displacement
of
B
of
the
rod
due
to
the
temperature
-3
1>2
Pzof=of
z due
is to
measured
in meters.
Determine
40110
, where
# end
acement
the
end
B2zof
the
rodrod
due
the
temperature
increase,
erage
normal
strain
in
the
rod.
splacement
the
B of
the
to
the
temperature
increase, (a) the
and
(b)
the
average
normal
strain
in
the
rod.
displacement
of
the
end
B
of
the
rod
due
to
the
temperature
increase,
(b)
thethe
average
normal
strain
in in
thethe
rod.
d (b)
average
normal
strain
rod.
The
ininFig.
2–4
ananincrease
Theslender
slender
rodshown
shown
Fig.artışına
2–4isissubjected
subjectedtotoBu
increaseofof
Şekildeki
narin çubuk
ekseni rod
boyunca
sıcaklık
maruzdur.
and (b) the average normal strain
in the rod.
A
22
z
dz
temperature
temperaturealong
alongitsitsaxis,
axis,which
whichcreates
createsa anormal
normalstrain
strainininthe
therod
rodofof
-3-3 1>2
1>2
P
where
z
is
measured
in
meters.
Determine
(a)
the
=
40110
2z
,
P
where
z
is
measured
in
meters.
Determine
(a)
the
=
40110
2z
,
z
z
değişim
çubukta #
normal birim deformasyonu meydana
A
displacement
displacementofofthe
theend
endBBofofthe
therod
roddue
duetotothe
thetemperature
temperatureincrease,
increase,
z
average
ininthe
and(b)
(b)the
the
averagenormal
normal
strain
therod.
rod.
getirmektedir. (a) and
Sıcaklık
artışından
dolayıstrain
B
ucunun
deplasmanını
A
A A
z z
dz dz
z
dz
hesaplayınız. (b) çubuktaki ortalama normal birim
deformasyonu
AA
200200
mmmm
dz
200 mm
200 mm
200 mm hesaplayınız.
B B
Fig.Fig.
2–42–4
zz
dzdz
B
200
200mm
mm
B 2–4
Fig.
B
Fig. 2–4
SOLUTION
UTION
OLUTION
Part
(a). Since Fig.
the normal
strain is reported at each point along the
SOLUTION
2–4is reported
BB
(a).(a).rod,
Since
the
normal
strain
at at
each
point
along
thethe
art
Since
the
normal
strain
isdz,
reported
each
point
along
Çözüm
a
differential
segment
located
at
position
z, Fig.
2–4,
has the
a
Part
(a).
Since
the
normal
strain
is
reported
at
each
point
along
a
differential
segment
dz,
located
at
position
z,
Fig.
2–4,
has
a
d, a differential
segment
at position
z, Eq.
Fig.2–1;
2–4,that
has a
deformed
length
thatdz,
canlocated
be determined
from
Fig.
Fig.2–4
2–4
rod,
athat
differential
segment
dz,
located
at
position
z, Fig.is,2–4, has a
Deformasyon
sonrası
dz
uzunluğu
dz'
rmed
length
cancan
bebe
determined
from
Eq.Eq.
2–1;
that
is,
formed
length
that
determined
from
2–1;
that
is, olacaktır.
length that
can=be
Eq.
2–1; that
dzdetermined
+ at
Pz dz
SOLUTION
SOLUTION
ce the deformed
normal dz¿
strain
is +dz¿
reported
eachfrom
point
along
theis,
= dz
P+z Pdz
dz¿
= dz
dz
z
-3 1>2
Part
Since
Part(a).
(a). a
Sincethe
thenormal
normalstrain
strainisisreported
reportedatateach
eachpoint
pointalong
alongthe
the
dz¿=at=C 1position
dz
Pz dz
ntial segment dz, located
z,
has
dz¿
++
40110
2z Fig.
D dz2–4,
-3 -31>2 1>2
dz¿dz¿
= =
1
+
40110
2z
dz
C
1 + 40110 2zD D dz-3 1>2 rod,
C
a
differential
segment
dz,
located
at
position
z,
Fig.
2–4,
has
aa
rod,
a
differential
segment
dz,
located
at
position
z,
Fig.
2–4,
has
th that
determined
2–1;yields
is, deformed length
dz¿ from
= C 1 +Eq.
40110
2zthat
dz
D the
Thecan
sumbe
of# these
segments
along
the
axis
deformed
length
that
can
be
determined
from
Eq.
2–1;
that
is,
deformed
length
that
can
be
determined
from
Eq.
2–1;
that
is,
sum
ofofof
these
segments
along
thethe
axis
yields
thethe
deformed
length
he
sum
these
segments
along
axis
yields
deformed
length
the
rod,
i.e.,these
The
along the axis yields the deformed length
Deformasyon
dz¿sum
= of
dz + Psegments
ethe
rod,
i.e.,
rod,
i.e.,
z dz sonrası çubuk boyu:
dz¿
dz¿==dzdz++PzPdz
z dz
0.2 m
of the rod, i.e.,
m
0.2 m
-3-3 1>2
1>2
z¿
= -32z
1 + 1>2
40110-32z1>2 D dz
dz¿
2z2z1>2D dz
dz¿==C 1C 1++40110
40110
D dz
dz¿ = C 1 +0.240110
0.2 Cm
-3 D-3dz
z¿ z¿
= =
2z 2z1>2
C 1 C+1L040110
D dzD dz-3 1>2
+ 40110
dz sum
C 1 + 40110 2z The
D The
L0 L0 z¿ =
sumofofthese
thesesegments
segmentsalong
alongthe
theaxis
axisyields
yieldsthe
thedeformed
deformedlength
length
2 3>2
ese segments along the= axis
deformed
length
-3the
0.2 m
zL
+0 2yields
40110
2
z
C -3
D
ƒ
of
the
rod,
i.e.,
0
of
the
rod,
i.e.,
2 3>2
0.2 m
0.23m
= =
C z C+z 40110
+ 401102-332z3>2
3 z D ƒ 0 D ƒ 0-3 2 3>2 0.2 m
0.20.2
mm
=
z
+
40110
2 3 z D ƒ0
C
= 0.20239 m
-3-3 1>2
0.2
m
=
0.20239
m
z¿
=
2z2z1>2D dz
z¿
=
40110
=
0.20239
m
C 1C 1++40110
D dz
#
0.20239
m
0
L
The displacement of the=end
of
the
rod
is
therefore
0
L
-3 1>2
z¿ = of of
+ of
40110
2z
C 1end
D dz
displacement
thethe
end
of
the
rodrod
is therefore
he
displacement
the
is therefore
ucunun
deplasmanı:
-3-3 2 2 3>2
mm
0.2
The displacement
of the
end of the rod is therefore
L0¢B
Ans. ==C zC z++40110
2 23 z3 z3>2D ƒ 0.2
B = 0.20239 m - 0.2 m = 0.00239 m = 2.39 mm T
40110
0D ƒ 0
¢ B¢=B 0.20239
mm
- 0.2
mm
= 0.00239
mm
= 2.39
mm
T T Ans.
Ans.
= 0.20239
- 0.2
= 0.00239
= 2.39
mm
Ans.
¢ =average
0.20239
- 0.2
2.39 mm T from
Part (b). #The
strain
the rodmis=determined
2m3>2
-3 normal
0.2m
m =in0.00239
==0.20239
0.20239mm
=average
zwhich
+ B40110
2that
C average
ƒrod
(b).
The
normal
strain
in Din
the
rod
is determined
from
0 the
art
(b).
The
normal
strain
rod
is determined
from
3 z the
Eq.
2–1,
assumes
or
“line
segment”
has
an
original
Part
(b).
The
average
normal
strain
in
the
rod
is
determined
from
2–1,
which
assumes
that
thethe
rod
or or
“line
segment”
hashas
an
original
q. 2–1,
which
assumes
that
rod
“line
segment”
an
original
The
displacement
The
displacementofofthe
theend
endofofthe
therod
rodisistherefore
therefore
Çubuktaki
normal
deformasyon:
length
of
200
mm
and
a ortalama
change
in length
of birim
2.39
mm.
Hence,
Eq.
2–1,
which
assumes
that
the
rod
or
“line
segment”
has an original
=
0.20239
m
h
of
200
mm
and
a
change
in
length
of
2.39
mm.
Hence,
ngth of 200 mm and a change in length of 2.39 mm. Hence,
¢¢
Ans.
0.20239mm--0.2
0.2mm==0.00239
0.00239mm==2.39
2.39mm
mmT T Ans.
length of 200 mm¢s¿
and-a2.2
change
in length
B B==0.20239
¢s STRAIN
2.39
mm of 2.39
71 mm. Hence,
P
Ans.
=
=
=
0.0119
mm>mm
2.2
S
TRAIN
71
¢s¿
¢s
2.39
mm
avg
¢s¿
¢s
2.39
mm
ent ofPthe
end of the=rod
is therefore
mm
Part
(b).
Part
(b). The
Theaverage
averagenormal
normalstrain
strainininthe
therod
rodisisdetermined
determinedfrom
from
Ans.
= = #
0.0119
mm>mm
Pavg
Ans.
=¢s - ¢s =200
=
0.0119
mm>mm
¢s¿
2.39
mm
avg
¢s
200200
mm
mm=
P¢s
Ans. that
= 0.0119Eq.
mm>mm
avg =
2–1,
2–1,which
whichassumes
assumes
thatthe
therod
rodoror“line
“linesegment”
segment”has
hasananoriginal
original
¢s
200=
mm
Ans.
= 0.20239 m - 0.2 m = 0.00239
m
2.39 mmEq.
T
length
of
200
mm
and
a
change
in
length
of
2.39
mm.
Hence,
length
of
200
mm
and
a
change
in
length
of
2.39
mm.
Hence,
EXAMPLE 2.2
#
average
normal strain in the rod is determined from
¢s¿
¢s¿--¢s
¢s 2.39
2.39mm
mm
ne Fig.
2–5a, the
Pavg
=in
=
==0.0119
P
=
=
0.0119mm>mm
mm>mm
When
force
P
is
applied
to
the
rigid
lever
arm
ABC
Fig.
2–5a,
the
avg
assumes
that the rod or “line segment” has an original
angle
of 0.05°.
¢s
200
¢s
200mm
mm
arm rotates Dcounterclockwise about pin A through an angle of 0.05°.
mm and a changeDetermine
in lengththeofnormal
2.39 mm.
D
strainHence,
developed
2 in wire BD.
P
¢s¿ - ¢s SOLUTION
2.39 mmI 300 mm
P
Ans.
=
=
= 0.0119 mm>mm
tg rotates about
¢s
200 mm The orientation of the lever arm after it rotates about
Geometry.
s figure,
point
A isBshown in Fig.
A From the geometry of this figure,
C EXAMPLE
2.22–5b.
B
C
400 mm
#
Şekildeki
sistemde ABC kaldıraç koluna P yükü
400
mm
° = 36.92°
0 mm
2.2
300 mm
SOLUTION
f = 90° - Ia + 0.05° = 90° - 53.1301° + 0.05° = 36.92°
Geometry. The orientation of the lever arm after it rotates about
point A
is shown
in Fig. 2–5b.theorem
From thegives
geometry of this figure,
For triangle
ABD
the Pythagorean
400 mm
LAD = 21300 mm2 -+1 1400
mm2 = 500 mm
b = 53.1301°
a = tan a
300 mm
2
2
71
400 mm
BD
D
2
300 mm
P
C
2
STRAIN
A
When force P
arm ABC in Fig. 2–5a, the
tan - 1 a to the rigid
b =lever
53.1301°
a is
= applied
300
mm
arm
rotates
counterclockwise
about
pin
A
through an angle of 0.05°. dönmektedir.
uygulandığında,
(a) kol saat yönünün tersi yönde A noktası etrafında #
(a)
Determine
the
normal
strain
developed
in
wire
BD.
Then
halatında meydana gelen normal birim deformasyonu belirleyiniz.
Ans.
Ans.
B
#25
400 mm
A
400 mm
01°
(a)
0.05° = 36.92°
2.2
2.2
STRAIN
2.2
2.2
EXAMPLE 2.2
2.2 STRAIN
Çözüm - 1 2.2
EXAMPLE
EXAMPLE
2.2
EXAMPLE
2.2
= 500
mm
2.2in Fig.
STRAIN
71
When force P is applied to the rigid lever arm ABC
2–5a, the
Kosinüs teoremi
ile
halatın
dönme sonrasıabout
uzamış
boyunun
hesabı:
arm
rotates
counterclockwise
pin
A
through
an
angle
of
0.05°.
When
force
P
is
applied
to
the
rigid
lever
arm
ABC
in
Fig.
2–5a,
the
force
applied
to thearm
rigidABC
leverinarm
ABC
Fig. 2–5a, the
EXAMPLE
2.2 Ptoisthe
When
force When
P is applied
rigid lever
Fig.
2–5a,in
the
o triangle
AB!D,
Determine
strain
developed
in wireof
arm rotates
rotates counterclockwise
counterclockwise
about pin
pin
A through
through
an angle
angle
ofBD.
0.05°.
400 mm the normal
arm
about
A
an
0.05°.
2.2 STRAIN D
arm rotates
counterclockwise about pin A through an angle of 0.05°.
XAMPLE
2.2
Determine
the
normal
strain
developed
in
wire
BD.
D
When force
P is applied
to the
rigiddeveloped
lever arm in
ABC
Fig. 2–5a, the
Determine
the normal
strain
wireinBD.
STRAIN
71
71
STRAIN
TRAIN
S
71
71
71
D
D
Determine the normal strain
developedI in wire BD.
300 mm
2
2
D SOLUTION
2.2 STRAIN P
arm rotates counterclockwise
about pin A through an angle of 0.05°.
2 71
When force P is applied
to
the
rigid
lever
arm
ABC
in
Fig.
2–5a,
the
300 mm
mm
P D
SOLUTION
I
300
P
The orientation
it rotates
Determine
the normal
strain developed
in wire BD. of the lever arm after
SOLUTION
I Geometry.
300 about
mm
P
EXAMPLE
SOLUTION
I 2.2
rm
rotates
about
pin A through
an angle of 0.05°.
21400
mm2counterclockwise
cos 36.92°
a
2
300lever
mm
point
A
is
shown
in
Fig.
2–5b.
From
the
geometry
of
this
figure,
Geometry.
The
orientation
of
the
arm
after
it
rotates
about
D
Geometry.
The of
orientation
of
theafter
leverit arm
after
it rotates about
Determine
the normal
strain
developed
in wire
BD.
Geometry.
The
orientation
the
lever
arm
rotates
about
300
mm
P
B
C
EXAMPLE
2.2
SOLUTION
IA
point
A$is
is
shown
in0.05#
Fig.
2–5b.
From
the
geometry
of this
this
figure,
When
forcepoint
P isP applied
to the
rigid2–5b.
leverFrom
arm the
ABC
in Fig. 2–5a,
the
2
shown
Fig.
of
figure,
LBD
"
fthe geometry
point A is shown
in
Fig.
2–5b.u in
From
ofgeometry
this figure,
400 mm
A
B
C
400
mm
A
B
arm
rotates
counterclockwise
about
pin
A
through
an
angle
of
0.05°.
B
300Bmm C
- 1 it rotates about
P
orientation of the lever
arm
after
A
OLUTION Geometry.
IWhen force PThe
C
tanABC
a in Fig. 2–5a,
b = 53.1301°
a =arm
is
applied
to
the
rigid
lever
the
400
mm
D
400
mm
400 mm
A 400
Determine
normalinstrain
developed
in
wire
BD. of this
300figure,
mm
point Athe
is shown
Fig. 2–5b.
the
geometry
400 mm
400
mm
tan
apin
b =
= 53.1301°
53.1301°
a
= From
eometry. arm
Therotates
orientation
of the
arm-- 11after
itmm
rotates
about
counterclockwise
about
A
through
an angle of 0.05°.
- 1 lever
tan
a
b
a
=
2 (a)
B¿
A
B
C
tan
a
b
=
53.1301°
a
=
300 mm
mm
400 mm
C
300
Then
D
oint A is shown
in# Fig.
2–5b.
From
the
geometry
this
figure,
300
mm of
Determine
the normal
strain
developed
in wire
BD.
300 mm
P
400
mm
SOLUTION
I
(a)
A (a)
B
2
C
ThenAns. a = tan - 1 a 400 mm b = 53.1301°
= 0.00116
mm>mm
(a)
Then
Then
Geometry.
The
orientation
of
the
lever
arm
after
it
rotates
about
300
mm
(b)
400
mm
SOLUTION I - 1 400 mm f = 90° - a + 0.05° = 90° - 53.1301° + 0.05°P = 36.92° 300 mm
b+ =
53.1301°
a = tan
point A is shown
infFig.
2–5b.
From
the
geometry
of
this figure,
(a)
= a90°
90°
-mm
a +
0.05°
= lever
90° -arm
53.1301°
+ rotates
0.05° =
=about
36.92°
Then
300 a
0.05°
90°
53.1301°
0.05°
36.92°
Geometry.
orientation
of
the=
after
it+
A
f = # 90° - afThe
+= 0.05°
=
90°
53.1301°
+
0.05°
=
36.92°
B
Fig.
2–5
C
For triangle ABD the Pythagorean theorem gives
(a)
point
A
is
shown
in
Fig.
2–5b.
From
the
geometry
of
this
figure,
400
mm
hen
400
mm
1the
For
triangle
ABD
the
Pythagorean
theorem
gives = 36.92°
A
B
f =triangle
90°
aABD
+ -0.05°
=theorem
90°b-=53.1301°
+ 0.05°
theorem
gives
C
tan
a Pythagorean
53.1301°
a -=
Forby
triangle
ABD
the
Pythagorean
btained
approximating
# For
300 mmLAD gives
= 21300 mm22 + 1400 mm22 = 500 mm
400
mm
mm
- 1 400 +
= 90° - a + 0.05° = La90°= 53.1301°
0.05°
=53.1301°
36.92°
2–5b. f
Here,
22 b+ =1400
(a)
=tan
21300
mm2
mm222 =
= 500
500 mm
mm
AD
2 a
ABD
the
Pythagorean
theorem
gives
LAD
=
21300
mm2
+2 1400
mm2
ThenFor triangle
LAD =
21300
mm2
+
1400
mm2
=
500
300
mm
Using this result and applyingmm
the law of cosines to triangle AB!D,
#
400 mm
(a)
or triangle ABD
the
Pythagorean
theorem
gives
2
2cosines to triangle AB!D,
Using
this
result
and
applying
the law
law
of
=0.05°
21300
mm2
+cosines
1400
mm2
= 500
mm
Using
and
applying
the
cosines
to
triangle AB!D,
m2 = Using
0.3491Then
mm
f =result
90° and
-Lthis
a applying
+result
=
90°
-of53.1301°
+ of
0.05°
= AB!D,
36.92°
400
mm
AD
this
the
law
to
triangle
400 mm
Kosinüs teoremi ile:
400 mm
D
2LB¿D = 2L2AD
2 + L2AB¿ - 21LAD21LAB¿) cos f
LAD = f21300
mm2
+
1400
mm2
=
500
mm
=result
90°
-Pythagorean
+ 0.05°
=
90°
+ triangle
0.05° = AB!D,
36.92°
2
2a applying
Using this
and
the
law-of53.1301°
cosines
to
For triangle
ABD
the
theorem
gives
D
2
2
L
=
2L
+
L
21L
21L
)
cos
f
D 400 mm
2AB¿) cos f
AB¿ 21500
B¿D = 22LAD
AD + LAB¿
AD
2 B¿D
D
AD
mm2
+ 1400 mm22 - 21500 mm21400
mm2 cos
36.92°
LB¿D = 2LL
21L-AB¿21L
) cos
f21LAB¿
a
AD + LAB¿ - 21LAD=
2 cosines to triangle
2
Using this result
and
applying
the
law
of
AB!D,
2
2
For triangle
ABD
the
Pythagorean
theorem
gives
=
21500
mm2
+
1400
mm2
21500
mm21400
mm2
cos
36.92°
2+
2400
mm
a
2
2
=
21500
mm2
1400
mm2
21500
mm21400
mm2
cos
36.92°
P
300 mm
mm
mm2
+ -1400
mm2
= 500 mm2
mm cos 36.92°
2
2= 21300
a
= 300.3491
mm
AD
=B¿D
21500
mm2
+ 1400
mm2
21500
mm21400
D
300
a
L
= L2L
$300
LBDmm u " 0.05#
AD + LAB¿ - 21LAD21LAB¿) cos f
f
P
6 mm>mm 2
Ans.
=
300.3491
mm
B
P $L
2
2
=L21L
300.3491
mm) mm2
" 0.05#
0.05#f
P
BD
= 221300
mm2
= 500AB!D,
mm cos 36.92°
# 2AB¿
D
300.3491
mm
$LBD
uu "
2L=
+ =result
L
fof2+cosines
AD
21500
mm2
+ AB¿
1400
mm2
-1400
21500
mm2
B¿D = Using
AD
AD21L
this
and
applying
thecos
law
to mm21400
triangle
$LBD
u " 0.05#
a f
A
300 mm
B
400 f
mm
Normal Strain.
B
B
2
2
B¿
P
A
= 21500
mm2
+
1400
mm2
- 21500
mm21400
mm2 to
costriangle
36.92° AB!D,
400 mm
=BD
300.3491
mm
halatındaki
normal
birim
deformasyon:
C
au " 0.05#300
A
Normal
Strain.
Using
this
result
and
applying
the
law of
cosines
$LBD
Normal
Strain.
A mm
mm
Normal
Strain.
B¿ 400 f
2
2
LB¿D -) cos
LBD
300.3491
mm
300
mm
B¿
D
400
mm
B
C
L
=
2L
+
L
21L
21L
f
400 mm
B¿ C
AB¿
B¿D
AD
AB¿
P
= 300.3491
mm
PBD AD
=
=
=u "0.00116
Ans.
400 mm
0.05#f mm>mm
BD
LB¿D
- L
LBD
300.3491
mm - 300
300 mm
mm
300 mm $LC
B¿D
BD
A
L
(b)
300.3491
mm
2 300.3491
2 -LBD
Normal
Strain.
2
2
L
L
mm
300
mm
P
=
=
=
0.00116
mm>mm
Ans.
B
D
= B¿D
21500
mm2
+ 1400 mm2
21500
mm2=cos
36.92°mm>mm Ans.
BD
BD
2L
) mm21400
cos
f0.00116
=
= AD-21L
0.00116
a
AD= +LLAB¿ - 21L
AB¿
B¿
300
mm
PBD = LB¿D =PBD
=
mm>mm
Ans.
300
mm
400 mm (b)
(b)
BD
C
LBD
L# BD
300 mm 2 300 mm
Normal Strain.
(b)A
2 300.3491
Fig. 2–5
P
= 300.3491
mm
Lmm2
mm -- 300
mmmm21400 mm2 cos 36.92°
=LB¿D
21500
+ 1400 mm2
21500
BD
B¿ $LBD 400 mm
a
u
"
0.05#
300 mm
f
PBD =
=SOLUTION II
= 0.00116C mm>mm Ans.
Fig.
2–5
Fig.
2–5
B
LBD - 2mm
(b)
Fig. 2–5
LB¿D - LBD=
Çözüm
300.3491
mm - 300 300
mmmm
P
300.3491
$LBD by approximating
small, this
same result
u " 0.05#fA
SOLUTION
II Since the strain=is 0.00116
=
mm>mm
Ans.can be obtained
BD = Normal Strain.
SOLUTION
II
SOLUTION
II
LBD Since
300
B
B¿(b)Here,
the
elongation
of wire
BD
as ¢L
, shown in
Fig.
2–5b.
the
strain
ismm
small, this
this same
result
can
beBD
obtained
by
approximating
Küçük
birim
deformasyon
analizi
ile can
400 mmFig. 2–5
Sinceis the
strain
is
small,
be
by
C approximating
Since the strain
small,
this
same
result same
can
beresult
obtained
byobtained
approximating
A
Normal
Strain.
the
elongation
of
wire
BD
as
¢L
,
shown
in
Fig.
2–5b.
Here,
LB¿Dthe
- elongation
L II
300.3491
BD, shown in Fig. 2–5b. Here,
SOLUTION
of ¢L
wiremm
BD-as300
¢Lmm
BD
Fig.B¿2–5
the
of BD
wire=BD
as
2–5b.
Here,
BD, shown in Fig. =
PBDelongation
=
0.00116
mm>mm
Ans.
400 mm
0.05°
C
Since the
same
can
by approximating
LBDstrain is small, this
300
mm=result
¢L
b1p rad2
uL-AB
= mm
cbe
a obtained
d1400 mm2 = 0.3491(b)
mm
OLUTION II
BD mm
LB¿D - LBD
300.3491
300
0.05°
180°
the
ofcinsinden
wire
BD as
, shown
in# Fig.
2–5b.
Here,
0.05°
PBDelongation
=uL
=
0.00116
BD
¢L
b1p rad2
= 0.05°
=¢L
rad2
d1400
mm2
=mm>mm
0.3491 mm
mmAns.
ince the strain
is=Radyan
small,¢L
thisBD
same
result
can
bemiktarı:
obtained
by
approximating
BD
AB dönme
b1p
=
uL
=
ccrad2
aa300
mm2
=
0.3491
mm
AB
(b)
¢LBD = uLL
b1p
d1400
mm2 d1400
= 0.3491
mm
BD= c a
180°
Fig. 2–5
AB
180°
he elongation of wire BD as ¢LBD
, shown in Fig. 2–5b. Here,
180°
Therefore,
0.05°
SOLUTION II
¢L
b1p rad2 d1400 mm2 = 0.3491 mm
Fig. 2–5
BD = uLAB = c a
Therefore,
Therefore,
Since
the
strain
is
small,
this
same
180°result can be obtained by approximating
Therefore,
0.05°
SOLUTION
II
¢L
0.3491
mm
BD
¢L
b1p
= uLAB =ofc wire
a
d1400
mm2
0.3491
mm
the
BD
asrad2
¢LBD
, shown
2–5b.
Here,
BDelongation
=in =Fig.
= obtained
PBD
= 0.00116 mm>mm
Ans.
180°
Since the strain
is small, ¢L
this
same
result
can
be
¢LBD
0.3491
mm
L
300 mmby approximating
BD
Therefore,
0.3491
mm
BD
¢L
0.3491
mm
=
=
mm>mm
P
=
0.00116
Ans.
BD
BD =
mm>mmAns.
PBD
= 0.00116
Ans.
the elongation
BDLas ¢L=BD=, 300
shown
Fig.
2–5b. Here,
= of wire
=0.05°
P#
0.00116
mmin mm>mm
BD
LBD
300
mm
herefore, ¢LBD BD
mm
= uLABLBD
= ¢L
c a 300b1p
rad2
d1400
mm2
=
0.3491
mm
0.3491 mm
BD
180°
=0.05°
PBD =
= 0.00116 mm>mm
Ans.
LBD
mm
¢L
= uL
= mm
c a 300b1p
rad2 d1400 mm2 = 0.3491 mm
¢LBD
AB
0.3491
BD
180°
=
PBD = #
= 0.00116 mm>mm
Ans.
Therefore,
LBD
300 mm
Therefore, ¢L
0.3491 mm
BD
=
PBD =
= 0.00116 mm>mm
Ans.
LBD
300 mm
#
¢LBD
0.3491 mm
=
PBD =
= 0.00116 mm>mm
Ans.
LBD
300 mm
#26
A
300 m
CHAPTER 2 STRAIN
72
Due
a loading,
the plate is deformed into the dashed shape shown
C H Ato
PTER 2
S T72
RAIN
C H Aaverage
PTER 2
S T R A I Nstrain along the side
in Fig. 2–6a. Determine (a) the
normal
AB, andEXAMPLE
(b) the average2.3
shear strain in the plate at A relative to the
x
and
y
axes.
PLE 2.3
# EXAMPLE 2.3
Due to a loading, the plate is deformed into the dashed shape shown
in
2–6a.
Determine
(a)the
thedashed
average
normal
strain along the side
Due
to
a loading,
theFig.
plate
is
deformed
into
shown
ythe plate
Due to a loading,
shape
shown
C
H A Pis
T Edeformed
R 2
S T R Ainto
I N the dashed
72
Due
to
ayükleme
loading,
the
platedeforme
isshape
deformed
into thedeformasyon
dashed shape shown
Bir
levha
altında
olmuş
AB,
and
(b)
the
average
shear
strain
in
the
plate
at Ave
relative to the
Fig.
2–6a.
Determine
theFig.
average
normal
along
the side
in Fig. 2–6a. Determinein(a)
the
average
normal(a)
strain
along
theDetermine
side strain
in
2–6a.
(a)
the
average
normal
strain
along the side
2
x and
y shear
axes.
AB,
andstrain
(b) the
in
the
plate
at A
relative
to in
the
AB, and (b) the average
shear
in average
the
plate
atAB,
A strain
relative
to
the
and
(b)
the
average
shear
strain
the
plate
at
A
relative
to the
sonrası
hali
şekilde
kesikli
çizgi
ile
gösterilmiştir.
(a)
AB
x and2.3
y axes. 3 mm
x and y 2axes.EXAMPLE
x and y axes.
kenarı boyuncay ortalama normal birim deformasyonu
B
y
2 mm
y
2
250 mm
Due to a loading, the plate is deformed into the dashed shape shown
y
hesaplayınız.
Levhanın
noktasında
x ve ystrain
eksenlerine
in Fig. 2–6a. (b)
Determine
(a)Athe
average normal
along the side
3 mm
AB,ortalama
and (b) the
average
shear
strain in the platehesaplayınız.
at A relative to the
göre
kayma
birim
deformasyonunu
3 mm
B
x and y axes.
B
Due to a loading, the plate
is deformed into
dashed shape shown
B
to a loading,
the plate
is average
deformed
into the
dashed
shape
shown
in Due
Fig. 2–6a.
Determine
(a) the
normal
strain
along
the
side
y
x strain along the side
in and
Fig. (b)
2–6a.
(a) the
average
A
C normal
300 mm
AB,
theDetermine
average
shear
strain
in the
plate2at
A relative
to
the
2
mm
mm
250 mm to the
AB,yand
strain in the plate at A relative
x and
axes.(b) the average shear
(a)
250
mm
x and y axes. 250 mm
250 mm
3 mm
3 mm
B
the
Çözüm:
y
Fig. 2–6
2 mm
2 mm
3 mm
B
y
x
A
300 mm
x
hesaplayalım:
C
A(a)
(a) AB kenarının
deformasyon
sonrası
boyunu
A
SOLUTION
A
C
300 mm
300 mm
3 mm
(a) after
(a)y axis, becomes
Part (a). Line AB, coincident
with the
3 mm
3 mm line AB¿
B
deformation,
as
shown
in
Fig.
2–6b.
The
length
of
is
AB¿
3 mm
B
250 mmFig. 2–6
C
300 mm
x
C 2 mm
x
(a)
3 mmFig. 2–6
Fig. 2–6
Fig. 2–6
2 mm
2
2
SOLUTION
AB¿ =2 mm
21250 mmB¿- 2 Bmm2 + 13 mm2
=2 mm
248.018 mm
x
2 mm
mmbecomesCline AB¿
SOLUTION 2 mm Part (a). SOLUTION
B¿
Line AB, coincident Awith the y300
axis,
SOLUTION
250 mm
B¿
250 mm
250 mm
average normal
strain
for
AB(a).
is therefore
(a) after
deformation,
as shown
Fig.becomes
2–6b. The
length
of AB¿ is
Part
Line AB,
coincident with
the yin
axis,
line
AB¿
B
B
after
The Part (a). Line AB, coincident with the y axis, becomes line AB¿ after
Part (a). Line AB, coincident with the y axis, becomes line AB¿ after
50 mm deformation, as shown in
as shown
2–6b.
of AB¿
is 2–6b. The length of AB¿ is
3 mm
Fig.
2–6b. The
length in
of Fig.
is The length
AB¿
250deformation,
mm
deformation,
as shown
in Fig.
2
2–6 2 = 248.018 mm
248.018 mm - 250
mm
AB¿ - AB
AB¿
=
21250
mm
2
mm2
+ Fig.
13 mm2
x
=
1PAB2avg =
AAB B 300
2 21250
2 - x2 mm22 + 13 mm22 = 248.018 mm2
mm
mm2=mm
mm
AB¿# = 21250
mm - 2# AB¿
mm2
+ 250
13 Cmm2
= 248.018
mm= 21250 mm - 2 mm2 + 13 mm22 = 248.018 mm
A
C
A
AB¿
300 mm
The
average
normal strain for AB is therefore
(a)
SOLUTION
B¿
-3
(b)
A
Ans.
= - 7.93110
2 mm>mm
(a)
Aaverage
Thefor
strain for
AB
is therefore
The
normal strain
AB
is normal
therefore
Ortalama
normal
birim
deformasyon:
Part
(a).
Line
AB, coincident
withisthe
y axis, becomes line AB¿ after
The
average
normal
strain for AB
therefore
(b) average
Fig. 2–6
248.018
mmlength
- 250ofmm
AB¿
- AB
250 mm (b) Fig. 2–6
deformation,
as
shown
in
Fig.
2–6b.
The
AB¿ is
=
1P
2avg =
The negative sign
strain causes a contraction
of
AB.
AB
y indicates the
mm AB¿
- 250-mm
AB¿ -- 250
ABmm248.018AB
250
mm
248.018 mm
AB¿ - AB
248.018 mm - 250 mm
AB
=
1P
=AB2avg =
1P 2 =
2avg
2
AB
250=mm
SOLUTION AB avgy
AB¿1P=AB21250
mm AB
- 2 mm2=2 + 13 mm2
= 248.018 mm
AB
250 mm
-3
250 mm
Ans.
= - 7.93110 2 mm>mm
SOLUTION
Part (b).
As noted in Fig. 2–6c, the once 90° angle BAC between the
(a).plate
Line
AB,
coincident
with
ythe
axis,
becomes
line
-3 AB¿ after
-3 the
A
Ans.
=
7.93110
2
mm>mm
-3
sidesPart
of
the
at
A
changes
to
due
to
displacement
of
B
to
u¿
Ans.
- 7.93110 with
2 mm>mm
Part
AB,=coincident
y axis,
becomes
line AB¿
after strain
mm
Ans.
= - 7.93110
2 mm(a). # Line
The
average
normal
for AB 2ismm>mm
therefore
as 3shown
Fig.g2–6b.
Thethe
length
of
AB¿
The
negative
sign
indicates the strain causes a contraction of AB.
(b)2–6b.
Since
then
is the
angle
shown
inisthe
figure.
B¿. deformation,
gxy =B p>2
-shown
u¿, in
xy
deformation,
as
in
Fig.
The
length
of
is
AB¿
B¿
3 mm
3 mm
2 mm
The
negative
sign
indicates
the strain
contraction
of
AB. 248.018
Thus,The negative
(b)
Ortalama
kayma
birim
deformasyonu
hesabı
sign
indicates
the
strain
causes
a contraction
of causes
AB.signa indicates
negative
strain
causes amm
contraction
of AB.
- 250 mm
AB¿ the
- AB
B¿
AB¿ = 21250B mm
- 2 mm22 +2 13 mm22 =2 The
248.018
mm
B¿
=angle BAC between the
1P
2
=
AB
avg
Part
(b).
As
noted
in
Fig.
2–6c,
the
once
90°
AB¿ = 21250
mm - 2 mm2 + 13 mm2 = 248.018 mm
y
AB
250 mm
3 mm
gxy -1
sides
of the
plate
at
A changes
toBAC
to the the
displacement of B to
u¿ due
(b).
As
noted
inangle
Fig.
2–6c,
the
once
90°
between
gxy As
Ans.
= tan
ain Part
b =
0.0121
rad
Part
(b).
noted
Fig.
2–6c,
the
once
90°
BAC
between
theangle
250
mm
Part
(b).
As
noted
in
Fig.
2–6c,
the
once
90°
angle BAC between the
The
average
normal
strain
for
AB
is
therefore
gxy
250
mmof-AB
2 mm
Since
then
is the-3angle
shown
in the figure.
B¿.
gxy to
= u¿
p>2
- to
u¿,to
gxy for
The average
normal
therefore
=displacement
-gchanges
7.93110
2 of
mm>mm
sides
the
plate
at
changes
due
the
B
toto the
xy
sides
of the plate
at Astrain
changes
to u¿is
due
to A
the
displacement
of
B
sides
of
the
plate
at
A
to
due
displacement of Ans.
B to
u¿
250 mm
Thus,
Since
then
is
the
angle
shown
in
the
figure.
B¿.
g
=
p>2
u¿,
g
u¿= p>2 AB¿
then
is
the
angle
shown
in
the
figure.
B¿. Since gxy
u¿,
g
xy
xy
250Since
mm gxy = p>2 - u¿, then gxy is the angle shown in the figure.
- AB xy 248.018 mm - B¿.
3 mm
2 mm
= x248.018 mm The
1PAB2avg = Thus,
- 250
mm sign indicates the strain causes a contraction of AB.
AB¿ - AB
negative
Thus,
u¿
B B¿u¿ AB
250 mmThus,
=
3 mm
A 1PAB2avg =
C
AB
250 mm gxy = tan-1 a
Ans.
b = 0.0121 rad
x
x
3 mm
250 mm - 2 mm
(c)
-3 3 mm
C
-1 x
3
mm
-1
Ans.
=
7.93110
2
mm>mm
A
C
Part
(b).
As
noted
in
Fig.
2–6c,
the
once
90°
angle
BAC
betweenAns.
the
Ans. b = 0.0121 rad
b==tan
0.0121
-1
-3 gxy = tan
gxy = tan
b = a0.0121 rad
gAns.
a rad
Ans.
= -ga7.93110
mm>mm
xy
250
mm
2 plate
mm
(c)
xy250 mm2 2 mm
250
mm
2
mm
(c)
sides
of
the
at
A
changes
to
due
to
the
displacement
of
B
to
u¿
250 mm
The negative sign indicates the strain causes a contraction
of AB.
B¿. Since
gxy = p>2 - u¿, then gxy is the angle shown in the figure.
The negative sign indicates the strain causes a contraction of AB.
Thus,
u¿
Part (b). As noted in Fig. 2–6c, the once 90° angle
x BAC between the
3 mm
Part
in Fig. 2–6c,
once
90° displacement
angle
BAC between
the
Cto the
# As noted
sides
of(b).
the plate
atAA changes
to u¿the
due
of B
gxyto= tan-1 a
Ans.
b = 0.0121 rad
ofgthe =plate
dueangle
to theshown
displacement
of B to
u¿the
250 mm - 2 mm
(c) g to is
Since
then
in the figure.
B¿.sides
p>2at- Au¿,changes
xy
xy
then gxy cinsinden)
is the angle shown in the figure.
B¿. Since # gxy = p>2 - u¿, (radyan
Thus,
Thus,
x
3 mm
x
gxy = tan-1 a-1
Ans.
b = 0.0121 rad
3 mm
250
mm
- 2 mm b = 0.0121 rad
g
Ans.
=
tan
a
xy
#
250 mm - 2 mm
#27
2.2
2.2
STRAIN
2.2
73
EXAMPLE 2.4
2.4
# EXAMPLE
The plate shown in Fig. 2–7a is fixed connected along AB and held in
73
STRAIN
73
STRAIN
x
y
the horizontal guides at its top and bottom, AD and BC. If its right
x
y
plate shown in Fig. 2–7a
is fixed connected along AB and held in
x
D
side CD is given a uniform horizontal displacement of 2 mm,
A
the horizontal guides at its top and bottom, AD and BC. If its right
2
determine (a) the average normal strain along the diagonal AC, and
D
side CD is givenD a uniform horizontal displacement
of 2.4
2 mm,
EXAMPLE
A
A
(b) the shear strain at E relative to2the x, y axes.
2
determine (a) the average normal strain
along the diagonal AC, and
150 mm
E
(b) the shear strain at E relative to the x, y axes.The plate shown in Fig. 2–7a is fixed connected along AB and held in
150 mm
2.2150Smm
TRAIN
73
STRAIN
73
the horizontal guides at2.2its top
and bottom,
AD
and BC. If its right
E
SOLUTIONE
B
C
side CD is given a uniform horizontal
displacement
of 2 mm,
Part (a). When the plate is deformed, the determine
diagonal AC
150along
mm the 2diagonal
mm
SOLUTION
(a) becomes
the average normal
strain
AC, and
B
C
B Fig. 2–7b. The
C length of diagonals AC and AC¿ can be found
EXAMPLE Part
2.4
AC¿,
(a)axes.
EXAMPLE
2.4
(b)
the
shear
strain
at
E
relative
to
the
x,
y
(a). When
the2 plate
150 mm
2 mm
AC becomes
mm
mm is deformed, the diagonal AC becomes
from the150
Pythagorean
theorem. We have
# AC¿,
Fig. 2–7b.
The length of diagonals AC and AC¿ can bey found
(a)
an be The
found
(a)
x
plate shown
shown in
in Fig.
Fig.2–7a
2–7a isis fixed
fixed connected
connected along
along AB
AB and
and held
held in
in
76 mm
x
y
The plate
A 76 mm
from the
Pythagorean
theorem.
We have
Şekildeki
levha
AB
kenarı
boyunca
ankastre
mesnetlidir.
Ayrıca
AD
ve
BC
kenarları
boyuncaD¿
the
horizontal
guides
at
its
top
and
bottom,
AD
and
BC.
If
its
right
2 and BC. If 2SOLUTION
the horizontal guides76at
its
top
and
bottom,
AD
its
right
AC
=horizontal
210.150 m2
+ 10.150 m2
0.21213
D 76 mm73 76 mm D¿
76 mm
AD
2.2 STRAIN75 mm
side CD
CD isis given
given
uniform
displacement
of 2=
2 mm,
mm, m
A a mm
D¿
A
side
a uniform
horizontal
displacement
of
düşey
hareket
engellenerek
yatay
verilmiştir.
CD
kenarının
yatay
olarakthe
2
2 harekete izin
2 Part
A
(a).
When
the
plate
is
deformed,
diagonal
AC becomes
u¿ 2 mm
determine (a)
(a) the
the average
averageAC
normal
strain along
along
the
diagonal
AC,
and mm
== 210.150
m2
m2
== 0.21213
2
3 m determine
AC¿
210.150
m22 +
+ 10.150
10.152
m22AC,
0.21355
normal
strain
the
diagonal
and
75
mm
AC¿,
AC¿
Fig.
2–7b.
The
length
of
diagonals
AC
and
can
be found
75
mm
2.2 STRAIN
73
(b) the
the shear
shear
strain at
atdeplasman
E relative
relative
to
the x,
x,yy axes.
axes.
u¿ birim
üniform
yapması
durumunda
(a)2AC
diyagonali
boyunca
ortalama
normal
u¿
(b)
strain
E
to
the
2
E¿
150
mm
from
the
Pythagorean
theorem.
We
have
AC¿
=
210.150
m2
+
10.152
m2
=
0.21355
m
EXAMPLE
2.4 Therefore the average normal strain along the diagonal is 150 mm
5m
E 75 mm
E
B and held in
C. If its right
t of 2 mm,
onal AC, and
yThe
E¿
deformasyonuE¿ve (b) E noktasında, x ve y eksenlerine göre kayma birim deformasyonunu
75 mm x
mm
Therefore
the
average
normalalong
strainAB
along
diagonal
isy
C¿
shown in75
Fig.
2–7a is
fixed
connected
andthe
held
in
EXAMPLE
2.4
B
s The plate
SOLUTION
AC = B210.150 m22 C
+ 10.150 m22 = 0.21213
m
SOLUTION
AC¿
AC
0.21355
m
0.21213
m
the horizontal guides
at its top and bottom, AD and BC. If its right
hesaplayınız.
(b)
B
C
C¿
Part
(a).
When
the
plate
is
deformed,
the
diagonal
AC
becomes
1P
2
=
=
B
150 mm
2
mm
C¿
AC
avg
B
D 2
(a).
When
the plate
is deformed,
the diagonal
becomes
150 mm
2 mm xm22 = 0.21355 m
y
side Part
CD
is
given
a uniform
horizontal
displacement
of AC
2 0.21213
mm,
The
plate
shown
Fig. 2–7a
is fixed
connected
along
AB
and
held
AC
m in m
AC¿
m2 + 10.152
AC¿
-ACAC
0.21355
m
-found
0.21213
A = 210.150
AC¿,
AC¿
Fig.
2–7b.
Theinlength
length
of diagonals
diagonals
and AC¿
can
be
13 m AC¿,
(a)
(b)of
Fig.(b)
2–7 2
Çözüm:
Fig.
2–7b.
The
AC
and
can
be
found
1P
2
=
=
(a)
determine
(a)
the average
normal
along
the diagonal
thethe
horizontal
guides
atAC
itsstrain
top
and
bottom,
AD andAC,
BC.and
If its right
avg
from
Pythagorean
theorem.
We
have
AC
0.21213
m
=
0.00669
mm>mm
Ans.
D
from
the
Pythagorean
theorem.
We
have
(b) the shear
strainisatgiven
E relative
to the x,horizontal
y axes.
side CD
a uniform
displacement Therefore
of 2 mm,the averageAnormal strain along
the diagonal is
Fig. 2–7
Fig.
2–7
(a) AC diyagonal
uzunluğunu
deformasyon
öncesi ve150
sonrası
belirleyelim
mm
mm
76 mm
2
STRAIN
A 76mm
= 0.00669
mm>mm
Ans.
76 mm D¿
determine (a) the2.2
average
normal
strain73
along the diagonal AC, and
Ans.
A 76
E
D¿
2 the shear strain
2
Part
(b).
To
find
at
E
relative
to
the
x
and
y
axes,
it
AC
=
210.150
m2
+
10.150
m2
=
0.21213
m
2
(b) the shear
at E relative
the x,m2
y axes.
AC strain
= 210.150
m22 + to
10.150
= 0.21213 m
0.21355 m - 0.21213 m
75
mm
150
mm AC¿ - AC
is first necessary to find the angle u¿ after deformation, Fig.
2–7b.
mmWe
SOLUTION
1PAC
2avg
=it
=
u¿E
Part
(b).
To
find
the
shear
strain
yB75axes,
2
2 at E relative to the x and
and y axes, it
C
u¿
AC
0.21213 m
AC¿
210.150 m2
m22 ++ 10.152
10.152 m2
m22 = 0.21355 m
m
have
AC¿
==plate
210.150
is first
necessary
to findthe
the diagonal
angle =u¿0.21355
after
deformation, Fig. 2–7b. We
Part
(a).WeWhen
the
is deformed,
AC becomes
150 mm
2 mm
Fig.
2–7b.
#
E¿
SOLUTION
0.00669
Ans.
have
E¿ mm>mm
AC¿, Fig.
2–7b. yThe
length of diagonals xAC and AC¿ can be found
75 mm= (a)
B
C
Therefore
the
average
normal
strain
along
the
diagonal
is
and held
in
u¿
76
mm
75
mm
the
average
normal
strain
along
the
diagonal
is
fromTherefore
thePart
Pythagorean
theorem.
We
have
AC
diyagonali
boyunca
ortalama
normal
birim
deformasyon:
(a).
When
the
plate
is
deformed,
the
diagonal
AC
becomes
150 mm
2 mm
tan a b =
If its right
75mm
mm AC and AC¿ can be found
C¿
u¿2 Dof diagonals
76
B shear
AC¿, Fig. 2–7b. The length
(a)
76
mm
76
mm
Part
(b).
To
find
the
strain
at
E
relative to the x and y axes, it
C¿
A
B
of 2 mm,
D¿
tan
a - bAC
=
A
AC¿
0.21355
m2-- 0.21213
0.21213 m
mis first necessary to find the (b)
from the1P
Pythagorean
theorem.
We
have
2
75
mm
u¿
angle
after
deformation, Fig. 2–7b. We
2
2
AC¿
AC
0.21355
m
p
2
=
=
(b)
al AC, and AC1P=AC210.150
m2 +u¿10.150
m2 = 0.21213
m
avg
a
b190.759°2
= 1.58404
rad
AC2avg =
AC = =90.759° =0.21213
0.21213
m
75 mm
have
76 mm
76 mm
AC
m
A
180°
D¿
p
u¿Fig. 2–7
2
150=mm
Fig.
2–7
2
= 90.759°
= 0.21355
am22 =b190.759°2
= Ans.
1.58404 rad
AC¿
m22 +u¿mm>mm
10.152
m2
m
404 rad
0.00669
AC =
== 0.00669
210.150
m2
+
10.150
0.21213
m
E
# 210.150
180°
mm>mm
Ans.
u¿ 75 mm
76 mm
Applying Eq. 2–3, the shear strain at E is therefore
u¿
tan
a b = E¿
2
2
75 mm
Therefore the average
normal
strain
along
the
diagonal
is
AC¿
=
210.150
m2
+
10.152
m2
=
0.21355
m
2
75 mm
(b)
Deformasyon
sonrası
levhanın
biçim
değişimini
çizelim.
Deformasyon
öncesi x ve y
Part
(b).
To
find
the
shear
strain
at
E
relative
to
the
x
and
y
axes,
it
Applying
Eq.
2–3,
the
shear
strain
at
E
is
therefore
B the shear strain
C
Part (b). To find
at E
p relative to the x and y axes, it
E¿ C¿
u¿
is
first
necessary
to
find
the
angle
after
deformation,
Fig.
2–7b.
We
C becomes
150 the
mm angle
mm
B 75Ans.
g 2u¿
- deformation,
1.58404 rad =Fig.
- 0.0132
rad
= after
p
is first
necessary
find
2–7b. We
Therefore
the to
average
normalxy
strain
the
diagonal
=mm
90.759°
a
b190.759°2 = 1.58404 rad
eksenleri
açı p
# 2 along
idi.
E =noktası
deformasyon
AC¿arasındaki
-(a)AC
0.21355
m -Deformasyon
0.21213
m issonrası açı ise # u¿ radyan.
have
n be found
(b)
180°
have
=xy =
g
- 1.58404 rad = - 0.0132 rad
Ans.
avg =
Ans.1PAC2sonrası
C¿
E'AC
ne kayacaktır.2 0.21213 m
B
Fig. 2–7
u¿
The76
negative
sign
indicates
that
the
angle
is
greater
than
90°.
AC¿
AC
0.21355
m
0.21213
m
u¿
76
mm
(b) at E is therefore
mmmm76 mm
Applying Eq. 2–3, the shear strain
1P=AAC
2 76
= mm>mmD¿ =
0.00669
Ans.
tanaau¿
tan
bb negative
==avg75 mmsign
AC
0.21213
m
2
u¿
The
indicates
that
the
angle
is
greater
than
90°.
n
m90°.
2
75 mm
NOTE: If the x and y axes were horizontal and vertical at point E,p Fig. 2–7
75 mm
0.00669
Ans.
u¿
- 1.58404 rad = - 0.0132 rad
=
Ans.
Part (b). To find the
shear=strain
at Emm>mm
relative to
the axes
x andwould
y axes,not
it change
xy the
then
90°
angle
these
due gto
pp
u¿ ==the
90.759°
=
a between
b190.759°2
= 1.58404
1.58404and
rad vertical
2
NOTE:
If
the
x
and
y
axes
were
horizontal
at
point
E,
almisatfirst
point
E,
u¿
90.759°
=
a
b190.759°2
=
rad
u¿ after
necessary to deformation,
find the angleand
= 0 at pointFig.
so180°
gxydeformation,
E. 2–7b. We
E¿ angle 180°
thethe
90°
between
axes
notychange
ehave
due toPart
the (b). then
To find
shear strain
at E these
relative
to would
the x and
axes, it due to the
75 mm
=
0
deformation,
and
so
g
at
point
E.
The
negative
xy
Applying
Eq.2–3,
2–3,the
the
shear
strain
at E
E
therefore
u¿isisafter
is first necessary
toshear
find the
angle
deformation, Fig. 2–7b.
We sign indicates that the angle u¿ is greater than 90°.
Applying
Eq.
strain
at
therefore
C¿
B
76 mm
have u¿ #
tan a b =
m
NOTE: If the x and y axes were horizontal and vertical at point E,
p(b)
2
75
gxymm
1.58404 rad
rad == --0.0132
0.0132 rad
rad
= p -- 1.58404
Ans.
thenAns.
the 90° angle between these axes would not change due to the
gu¿
=
xy
276 mm
2
tan a b Fig.
= 2–7 p
deformation,
and so gxy = 0 at point E.
2
75
u¿ = 90.759°
= mm
a
b190.759°2 = 1.58404 rad
Ans.
180°
u¿ isis greater
The negative
negative sign
sign indicates
indicates that
that the
the angle
angle u¿
greater than
than 90°.
90°.
The
p
u¿ = 90.759° = a
b190.759°2 = 1.58404 rad
ndApplying
y axes, itEq. 2–3, the shear strain at E is 180°
therefore
#
NOTE:
If
the
x
and
y
axes
were
horizontal
and vertical
vertical at
at point
point E,
E,
g. 2–7b.
We
NOTE:
If the x and y axes were horizontal and
then
the
90°
angle
between
these
axes
would
not
change
due
to
the
pbetween
then
the 90° Eq.
angle
these
axes
not change due to the
Applying
2–3,
the shear
strain
at would
E is therefore
gxyand
rad
= E.
- 0.0132 rad
= so Ans.
= 00 at
deformation,
gxy1.58404
at point
point
=
deformation,
and
so
g
E.
2
xy
#
p
gxy =
- 1.58404 rad = - 0.0132 rad
Ans.
The negative sign indicates that2the angle u¿ is greater than 90°.
4NOTE:
rad
The
negative
indicates
that the angle
is greater
than 90°.
If the
x andsign
y axes
were horizontal
andu¿vertical
at point
E,
then the 90° angle between these axes would not change due to the
0 aty point
deformation,
andIfsothe
gxyx=and
E. horizontal and vertical at point E,
NOTE:
axes were
then the 90° angle between these axes would not change due to the
#28
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