made into a “standard” shape and size. It has a constant circular cross
section with enlarged ends, so that failure will not occur at the grips.
d0 ! 0.5 in.
To perform a tension
or compression
test apunch
specimen
material
Before
testing, two small
marksof
arethe
placed
alongisthe specimen’s
uniform
length.
Measurements
are
taken of
both the
specimen’s initial
made
into
a
“standard”
shape
and
size.
It
has
a
constant
circular
cross
L0 ! 2 in.
A0, and
L0 between the
area,
gauge-length
section with enlargedcross-sectional
ends, so that
failure
willthenot
occur at distance
the grips.
punchpunch
marks.marks
For example,
when along
a metalthe
specimen
is used in a tension
Fig. 3–1 testing, two small
Before
are placed
specimen’s
test it generally has an initial diameter of d0 = 0.5 in. (13 mm) and a
uniform length. Measurements
are taken of both the specimen’s initial
L0 ! 2 in. Bölüm 3. Malzemelerin Mekanik
gauge length
of L0 = 2 in. (50 mm), Fig. 3–1. In order to apply an axial
Özellikleri
A
,
L0 ends
cross-sectional
area,
and
the
gauge-length
distance
between
the seated into
ofATthe
the
are usually
8
2
C
HAPTER 3
M
E C H A N I Cload
A 0L Pwith
R O P Eno
R T bending
IES OF M
E R Ispecimen,
ALS
punch
marks.
For
example,
when
a
metal
specimen
is
used
in
a
tension
Fig. 3–1 3 Çekme ve Basınç
ball-and-socket joints. A testing machine like the one shown in Fig. 3–2 is
Testleri
82
C H Atest
P T E it
R 3
M E C H Ahas
N I C Aan
L Pinitial
RO
P E R to
Tdiameter
I Estretch
S O F Mthe
AT
R I0A L
d
=S 0.5atin.a very
generally
ofEspecimen
(13 mm)
and a rate until it
then
used
slow, constant
d0 ! 0.5
in.
Tomachine
perform
a tension
or
compression
testaxial
a to
specimen
the material is
2 in.The
gauge
length
of L0 = fails.
(50
mm), Fig.
3–1. Intoorder
to load
apply
an
is designed
read
the
required
maintainof
this
made
into a “standard”
shape
and seated
size. It into
has a constant circular cross
uniform
stretching.
load
with
no
bending
of
the
specimen,
the
ends
are
usually
d0 ! 0.5 in.
To perform
a tension
orthe
compression
test a specimen
of the material is
frequent
intervals
during
recorded
of not
the applied
section
with
enlarged
so data
that
will
occur at the grips.
ball-and-socket joints. AAt
testing
machine
like
theends,
onetest,
shown
inisfailure
Fig.
3–2
made
into
a the
“standard”
shape
and
size.
It
hasa is
a constant
circular cross
readout.
load
P,
as
read
on
dial
of
the
machine
or
taken
from
digital
Before testing,
two small
punch marks
are
placed along the specimen’s
then used to stretch thesection
specimen
a very
slow,
constant
until
withatenlarged
ends,
so
that rate
failure
willitmarks
not occur
at the grips.
Also,uniform
the elongation
d
=
L
L
between
the
punch
on
the
Typical steel specimen with attached
0
length.
Measurements
are
taken ofthis
both the specimen’s initial
fails.
totesting,
read
the
load
required
to
maintain
L0 !The
2 in. machine is designed
strain
gauge.
Before
two
small
punch
marks
are
placed
along
the
specimen may be measured using either a caliper or a mechanical or specimen’s
!
cross-sectional area, A0, and the gauge-length distance L0 between the
uniform stretching. optical
devicelength.
called an
extensometer. This
of dof(delta)
then
used
uniform
Measurements
arevalue
taken
both isthe
specimen’s
initial
L0Fig.
!
in.
punch the
marks.
For
example,
whenof
a the
metal
specimen
is used in a tension
At23–1
frequent intervals
during
test,
data
is ,recorded
the
applied
to
calculate
the
average
normal
strain
in
specimen.
Sometimes,
A
L
cross-sectional
area,
and
the
gauge-length
distance
between
the
Test sonrası (deformasyon yapmış durum)
noktalar
arası
0uzaklık L kumpas ile ölçülerek 0
in.read
test
it machine
generally
has
an
initial
of dpossible
(13 mm) and a
readout.
load P, as read on the however,
dialpunch
of the
orexample,
taken
from
a diameter
digital
0 = 0.5 to
this
measurement
is
not
taken,
since
it
is
also
marks.
For
when
a
metal
specimen
is
used
in
a
tension
Fig. 3–1
L =an
2electrical-resistance
in.
gauge
length
(50 mm),
Fig.on3–1.
Ingauge,
orderwhich
to apply an axial
Also, the elongation
=
Lit-directly
L0 hesaplanır.
between
the
punch
marks
strain
byofusing
strain
steel specimen with
attachedUzama/kısalma
bulunur.
miktarıthe
! d test
Bu
işlem
yerine
strain-gage
(birim
dthe
generally
has0 an
initial
diameter
of
(13 mm) and a
0 = 0.5 in.
with
no
bending
specimen,
the ends
aregauge
usually
looksload
likeusing
the one
shown
in of
Fig.the
3–3.
The
operation
of
this
is seated into
uge.
specimen
may
be
measured
either
a
caliper
or
a
mechanical
or
3
L0electrical
= 2 in. resistance
gauge
length of in
(50 mm),of
Fig.
3–1.thin
In order
to
apply an axial
based
on the change
a very
piece in Fig. 3–2 is
ball-and-socket
joints.
Aoftesting
machine
like
thewire
oneorshown
device called an
extensometer.
This
value
d
(delta)
is
then
used
deformasyon optical
ölçer) kullanılabilir.
load with
no bending
of the specimen,
the isends
are usually
into
of metal
foil
under
strain.
Essentially
the Sometimes,
gauge
cemented
to the seated
3
then
used
to stretch
thespecimen.
specimen
at a very
slow, constant
rate until it
to calculate the average
normal
strain
in the
ball-and-socket
joints.
A
testing
machine
like
the
one
shown
in
Fig.
3–2
is
specimen
along
its length.isIf designed
the cementtoisread
very strong
in comparisonto
tomaintain this
fails.
The
machine
the
however, this measurement
isused
not taken,
sincethe
it is
also possible
to load
read required
specimen
at a very
thethen
gauge,
thento
thestretch
gauge is
in effect
an integral
part slow,
of theconstant
specimen,rate until it
uniform
stretching.
the strain directly by so
using
electrical-resistance
strain
gauge,
which
that
when
the
specimen
is strained
the
direction
ofrequired
the gauge,
fails.an
The
machine
is designed
toinread
the load
tothe
maintain this
At
frequent
intervals
during
the
test,
data
ismeasuring
recordedthe
of the applied
looks like the one shown
in
Fig.
3–3.
The
operation
of
this
gauge
wire
and specimen
will experience the same strain. Byis
uniform
stretching.
load resistance
P, asresistance
readof
onthe
the
dial
of the
machine
or
taken from
a digital readout.
based on the change in
electrical
of
a
very
thin
wire
or
piece
electrical
wire,
the
gauge
may
be
calibrated
to
read
At frequent intervals during the test, data is recorded of the applied
! Typical steel specimen
Also,normal
the elongation
d = L cemented
- L0 between
punch marks on the
attachedstrain.
values
strain
directly.
of metal with
foil under
gauge
totaken
thethefrom
loadofEssentially
P, as read
onthe
the
dial of is
the machine
or
a digital readout.
strain gauge. specimen along its length.specimen
may is
bevery
measured
using
either a caliper
or a mechanical or
If
the
cement
strong
in
comparison
to
3.2
T
HE STRESS
–S
TRAIN
D
IAGRAM
83
Also,
the
elongation
d
=
L
L
between
the
punch
marks on the
Typical
steel
specimen
with
attached
0
movable
optical
devicean
called an extensometer.
This value of d (delta) is then used
the gauge, then the gauge
is in effect
partusing
of the
specimen,
strain
gauge.
upper
specimen
may beintegral
measured
either
a caliper or a mechanical or
to iscalculate
the
average
normal
inthe
the specimen. Sometimes,
so that when the specimen
strained
in the
direction
of thestrain
gauge,
optical
device called
an
extensometer.
This
value
of d (delta) is then used
load
however, this
measurement
is
not
taken, since
it is also possible to read
wire and specimen will
experience
the
same
strain.
By
measuring
the
to
calculate
the
average
normal
strain
in
the
specimen. Sometimes,
dial
The Stress–Strain
Diagram
the
strain
directly
by
using
an
electrical-resistance
strain gauge, which
electrical
resistance
of
the
wire,
the
gauge
may
be
calibrated
to
read
tension
however, this measurement is not taken, since it is also possible to read
specimen
looks like the one shown in Fig. 3–3. The operation of this gauge is
values of normal strain directly.
the strain directly by using an electrical-resistance strain gauge, which
L0, change in electrical resistance of a very thin wire or piece
t is not feasible to prepare a test specimen to match the motor
size,
A0 and
based
on the
andlooks
load like the one shown in Fig. 3–3. The operation of this gauge is
f each structural member. Rather, the test results mustcontrols
beofreported
so under strain. Essentially the gauge is cemented to the
metal foil
based on the change in electrical resistance of a very thin wire or piece
hey apply to a member of any size. To achieve this, the
load
and
specimen along its length. If the cement is very strong in comparison to
of metal foil under strain. Essentially the gauge is cemented to the
orresponding deformation data are used to calculate various
values then
of the gauge is in effect an integral part of the specimen,
the gauge,
specimen along its length. If the cement is very strong in comparison to
load
he stress and corresponding straindial
in the specimen. A plot so
of that
the results
when the specimen is strained in the direction of the gauge, the
the Tgauge,
then
the gauge
is in effect an
part of the specimen,
HE STRESS
–STRAIN
Dwill
IAGRAM
83integral
Electrical–resistance
roduces a curve called the stress–strain diagram. There3.2
are
two
ways
in
wire
and
specimen
experience
the
same strain. By measuring the
so that when strain
the specimen
is strained in the direction of the gauge, the
gauge
! described.
electrical
resistance
ofexperience
the wire, the
beBy
calibrated
to read
which it is normally
wire
and specimen
will
thegauge
same may
strain.
motor
3measuring the
Fig. 3–3
Fig. 3–2
values
of
normal
strain
directly.
and load
electrical resistance of the wire, the gauge may be calibrated to read
controls
he
Stress–Strain
Diagram
Conventional
Stress–Strain
Diagram.
We can values
determine
the strain directly.
of normal
C H A P T E R 3 movable
M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S
ominal or engineering
stress
by
dividing
the
applied
load
P
by
Gerilme-Birim
Deformasyon Diyagramıthe
upper
ble
to prepare
a test
specimen
to
match
size,calculation
A0 and L0,assumes that
pecimen’s
original
cross-sectional
area
Athe
movable
crosshead
0. This
s reported so
upper
true fracture stress
tural
member.
Rather,
the the
test cross
resultssection
must be
he stress
is constant
over
and
throughoutload
the gauge
s¿f
crosshead
o a member
load and dial
ength.
We haveof any size. To achieve this, theElectrical–resistance
load
tension
strain gauge
g deformation data
are used to calculate various values
ofdial
specimen
ultimate
tensionin the specimen. A plot of the results motor
corresponding
strain
Fig. 3–3
su
Fig. 3–2
stress
fracture
P
specimen
rve called the stress–strain diagram.
in and load
s = There are two ways
stress
proportional
limit(3–1)
motor
controls
s
A0
f
rmally described.
elastic limit
crosshead
3.2
and load
yield controls
stress
sY
spl
3
nal Stress–Strain
can determine
the
Likewise,
the nominal Diagram.
or engineeringWe
strain
is found directly
from the
engineering
stress byordividing
the the
applied
load
the
train gauge reading,
by dividing
change
in P
thebyspecimen’s
gauge
Electrical–resistance
iginal
cross-sectional
area
A
.
This
calculation
assumes
that
0
ength, d, by the specimen’s original gauge length L0. Here the strain is
strain gauge
constant
over
the crossthroughout
section andthe
throughout
the gauge
ssumed to
be constant
region between
the gauge points. Electrical–resistance
strain gauge
strain
necking
Fig. 3–2 elastic yielding
ve
Thus,
region
hardening
Fig. 3–2
!
s =
P
A0
!
P =
d
L0
!
elastic
behavior
plastic behavior
P
Fig. 3–3
Fig. 3–3
(3–1) and true
(3–2)
Conventional
stress-strain diagrams
for ductile material (steel) (not to scale)
Fig. 3–4
nominal
or engineering
strain
is and
found
directly
from
the the vertical
If the corresponding
values
of s
P are
plotted
so that
reading,
or
by
dividing
the
change
in
the
specimen’s
gauge
xis is the stress and the horizontal axis is the strain, the resulting curve is
the
original
gauge length
the strain
is
L0. Here
Elastic
Behavior.
Elastic
behavior
of the material occurs when
alledspecimen’s
a conventional
stress–strain
diagram.
Realize,
however,
that
two
etress–strain
constant throughout
the
region
between
the
gauge
points.
thematerial
strains inwill
thebespecimen
are within
diagrams for a particular
quite similar,
but the light orange region shown in
3–4. Here
the curve
is actually
will never be exactly the same. This isFig.
because
the results
actually
depend a straight line throughout most of
this region,
so that the stress
is proportional to the strain. The material
n variables such as the material’s
composition,
microscopic
in
this
region
is
said
to
be
linear
elastic. The upper stress limit to this
d
mperfections, the way it is manufactured, the rate of loading, and the
P =
(3–2)
linear
relationship
is
called
the
proportional
limit, spl. If the stress
emperature during the time
L0 of the test.
slightly exceeds the proportional limit, the curve tends to bend and
!30
A0 is
a decreasing
load,
since
constant
when
calculating
Whensupports
this diagram
is plotted
it has
a form
shown
by the
light-blue
3.3 . However,
STRESS–STRAIN
BEHAVIOR
OF DUCTILE AND BRITTLE MATERIALS
87
s = the
P>A
s
–
P
from
the
true
diagram,
s
–
P
curveengineering
in Fig. 3–4. stress,
Note that
conventional
and
true
diagrams
0
the actual area
A3.3
within
the–S
necking
is always
decreasing until
are practically
coincident
when
the
strainregion
is small.
The differences
STRESS
TRAIN BEHAVIOR OF DUCTILE AND BRITTLE MATERIALS
87
3.3 Sin
TRESS
TRAIN
BEHAVIOR
OF Drange,
UCTILEstress,
AND BRITTLE MATERIALS
87
fracture,
s¿f, and begin
so thetomaterial
actually
sustains
increasing
between
the diagrams
appear
the–Sstrain-hardening
Stress–Strain
of more
Ductile
since
s = P>A. Behavior
where
the magnitude
of strain becomes
significant. In particular,
there
a large divergence
within the necking region. Here it can be
andisBrittle
Materials
Stress–Strain
Behavior
of
s –Ductile
P diagram
seenStress–Strain
from the
conventional
the specimen actually
Behavior
of that
Ductile
and
Brittle
Materials
A
supports
a
decreasing
load,
since
is
constant
when on
calculating
0 brittle, depending
Materials canand
be classified
as either
being ductile or
Brittle
Materials
s
=
P>A
.
s
–
P
engineering
stress,
However,
from
the
true
diagram,
This steel specimen clearly shows the necking
0
their stress–strain characteristics.
3.3
STRESS
–STRAIN
BEHAVIOR
OF DUCTILE AND BRITTLE MATERIALS
8 7 that occurred just before the specimen failed.
ls can be classified
as either
ductile
or brittle,
depending
on decreasing until
the actual
area Abeing
within
the
necking
region
is always
Materials
can besclassified
either
being ductile
or brittle,
depending
on stress,
This resulted in the formation of a “cup-cone”
ess–strain
characteristics.
¿f, and
fracture,
soasmaterial
the
material
actually
sustains
increasing
Ductile
Materials.
Any
that can
be subjected
to
large
shape at the fracture location, which is
their stress–strain
characteristics.
since s
= P>A. is called a ductile material. Mild steel, as
strains
before
it fractures
characteristic of ductile materials.
Stress–Strain
Behavior
of
Ductile
ediscussed
Materials.
Anyismaterial
that
can be
subjected
to choose
large Failure
previously,
a
typical
example.
Engineers
often
ductile
of
a
Necking
3
and
Materials
Ductile
Materials.
Any
material
that can
besteel,
subjected
to large
before
itBrittle
fractures
called
ductile
material.
ductile material
materials
for
design
becausea these
materials
areMild
capable
ofas
absorbing
! itis fractures
!
strains
before
is
called
a
ductile
material.
Mild
steel,
as
(b)
dsshock
previously,
is aastypical
example.
Engineers
often they
choose
(a)
3
energy,
and
if being
they
become
overloaded,
usually
exhibit
can
be or
classified
either
ductile
brittle, depending
onwillductile
discussed
previously,
is amaterials
typicalorexample.
Engineers
often
choose
ductile
s
for
design
because
these
are
capable
of
absorbing
3
This steel specimen clearly shows the necking
ess–strain
characteristics.
large deformation
before failing.
Fig.
3–5capable of absorbing
materials
for
design
because
these they
materials
are
that occurred just before the specimen failed.
energy,
and
if
they
become
overloaded,
will
usually
exhibit
One way to specify the ductility of a material is to report its percent
This
resulted in the formation
of a “cup-cone”
shock before
or energy,
and if(Düktil)
they
overloaded,
they will usuallyGerilme-Birim
exhibit
Sünek
Gevrek
Deformasyon
İlişkisi
eformation
Materials.
Any
material
that become
can ve
be subjected
toMalzemelerin
large
failing.
elongation
or percent
reduction
in area Mild
at the
time
of fracture. The
shape at the fracture location, which is
before
itspecify
fractures
is ductility
called
a ductile
material.
steel, its
as percent
large
deformation
before
failing.
way
to
the
of
a
material
is
to
report
elongation
is the
specimen’s
fracture
strain
expressed
as a
characteristic
of ductile materials.
Sünek
malzeme:
Kırılma
öncesi
büyük
birimitsdeformasyon
yapabilen
dpercent
previously,
isway
a typical
example.
Engineers
often
ductile
One
to
specify
the
ductility
of time
achoose
material
is to report
percent
3 malzeme
on
or
percent
reduction
in
area
at
the
of
fracture.
The
percent.
Thus,
if
the
specimen’s
original
gauge
length
is
L
and
its
length
Failure
of
a
Necking are capable of absorbing
s for design because these materials
0
elongation
or percent
reduction
in strain
area atexpressed
the timeductile
of afracture.
The
elongation
isLthe
specimen’s
fracture
as
material
energy,
and ifisthey
overloaded,
they
usually
exhibit
Yumuşak
(mild)
çelikwill
için
düktilitenin
belirlenmesi:
at fracture
then
f,become
percent
elongation
is
the
specimen’s
fracture
strain
expressed
as
a
(b)
(a)
Thus,
if
the
specimen’s
original
gauge
length
is
L
and
its
length
formation before failing.
0
Thus,
if
the
specimen’s
original
gauge
length
is
L
and
its
length
way
the
ductility
of
a
material
is
to
report
its
percent
0
re istopercent.
,
then
Lspecify
- L3–5
Lf Fig.
f
0
on oratpercent
reduction
in area
at the time
The
fracture
is Percent
Lf, then
(3–3)
1100%2
elongation
= of fracture.
L0
elongation is the ! specimen’s fracture
as a
L0 expressed
Lf -strain
Thus, if the
specimen’s
original gauge
L0 L
and
itsL
length
(3–3)
Percent
elongation
= length is1100%2
f 0
L0 this
re
is seen
Lf, then
(3–3)
1100%2
Percent
= value would
As
in Fig. 3–6,
be 38%
for a mild
= 0.380,
Pf elongation
Lf: since
Kırılma/kopma
anındaki
uzunluk
Lo:Başlangıç
uzunluğu
L0
steel specimen.
- L0
Lf this
in Fig.
since
would
be 38%
fordeğer)
a mild
= 0.380,
The3–6,
percent
reduction
areavalue
is another
to
specify
ductility. It is
(3–3)
1100%2
Percent
elongation
= in (yumuşak
!Pf 3–6,
çelikway
için
tipik
As
seen
in
Fig.
since
P
L
0f = 0.380, this value would be 38% for a mild
ecimen.
defined within the region of necking as follows:
steel
specimen.
ercent
reduction
in area is another
way to specify
ductility.
It is kesit Af ve başlangıçtaki kesit Ao a göre de
Düktilite
çekmewould
deneyinde
kopma
anındaki
in Fig. 3–6,
since
38% forway
a mild
0.380, this value
Pf = reduction
The
percent
infollows:
area is be
another
to specify ductility. It is
A
A
within
the
region
of
necking
as
0
f
cimen.
(3–4)
Percent
area =as follows: 1100%2
defined within
thereduction
region of of
necking
ercent reduction in incelenebilir.
area is another way to specify ductility.
A0 It is
A
A
f
within the region of necking as follows: 0
(3–4)
Percent reduction of area =
1100%2
A0 - Af
A
Here A0 is the specimen’s
original
area and
Af is the area
0
(3–4)
Percent reduction
of
area
=
1100%2
A0 - Across-sectional
f
A(3–4)
0 of 60%.
Percent
= steel has1100%2
of the
neckreduction
at fracture.
Mild
a typical value
! of area
A0
is the
specimen’s
original
cross-sectional
areamolybdenum,
and Af is the area
Besides
steel, other
metals
such as brass,
and zinc may s (ksi)
Here
is
the
specimen’s
original
cross-sectional
area
and
the area
Awhereby
A
60
eck
at
fracture.
Mild
steel
has
a
typical
value
of
60%.
0
f issonuç
Yukarıdaki
formülün
yumuşak
çelik
için
tipik
değeri
%60 tır.
also
ductile
stress–strain
characteristics
similar
is
the exhibit
specimen’s
original
cross-sectional
area and Af is the
area to steel,
s (ksi)
sYS ! 51
the
neck
at
fracture.
Mild
steel
has
a
typical
value
of
60%.
esthey
steel,
other
metals
such
as
brass,
molybdenum,
and
zinc
may
eck
atof
fracture.
Mild
steel
has
a
typical
value
of
60%.
undergo elastic stress–strain behavior, yielding at sconstant
stress, 50 s (ksi)
(ksi) and 60
Besides
steel,
other
metals
such similar
asuntil
brass,
molybdenum,
zinc may
es
steel,
other
metals
such
asfinally
brass,
molybdenum,
and to
zinc
may whereby
ibit
ductile
stress–strain
characteristics
steel,
strain
hardening,
and
necking
fracture.
In60 most
metals,
sYS ! 5160
ibit ductile
stress–strain
characteristics
similar
to steel,
whereby
also
exhibit
ductile
stress–strain
characteristics
similar
to
steel,
whereby
dergo
elastic
stress–strain
behavior,
yielding
at
constant
stress,
40
sYS !50
51One
however,
constant
yielding
will
not
occur
beyond
the
elastic
range.
sYS ! 51
dergothey
elastic
stress–strain
behavior,
yielding behavior,
at constant yielding
stress, 50
undergo
elastic
stress–strain
at
constant
stress,
ardening,
and
finally
necking
until
fracture.
In
most
metals,
50
metal
for
which
this
is
the
case
is
aluminum.
Actually,
this
metal
often
ardening, and finally necking until fracture. In most metals,
strain
hardening,
and occur
finally
necking
fracture.
In
40 metals,30
r,
constant
yielding
will
theuntil
elastic
range.
notyielding
have
a will
well-defined
yieldbeyond
point,
and
consequently
it ismost
standard
40 One
r,does
constant
not not
occur beyond
the elastic
range.
One
however,
constant
yielding
will
not
occur
beyond
the elastic
range.
One 40
which
this
is
the
is strength
aluminum.
Actually,
thisoften
metal
oftencalled
rrpractice
which
this
the
case
is aluminum.
Actually,
metal
toisdefine
acase
yield
using
athis
graphical
procedure
20
30 the
30
metal
for which
this
is
the
is
aluminum.
Actually,
thisand
metal
often
t have
have
well-defined
yield
and
consequently
it)isisstandard
aawell-defined
yield
point,
and case
consequently
it isin.>in.
standard
offset
method.
Normally
apoint,
0.2%
strain
(0.002
chosen,
from
30
to
define
anot
strength
using
a graphical
called
thecalled
does
have
well-defined
yieldprocedure
point,
and
consequently
standard10
20 theit is
to
define
ayield
yield
strength
a graphical
procedure
20
this
point
on
the
ausing
line
parallel
to the
initial
straight-line
portion
Pa axis,
ethod.
Normally
0.2%
strain
(0.002
) using
is chosen,
and fromand
in.>in.in.>in.
practice
toadefine
adiagram
yield
strength
called
ethod.
Normally
a 0.2%
strain
(0.002
)aThe
isgraphical
chosen,
from this
the
is drawn.
point procedure
where
linethe 20
10
P (in./in.)
ntof
on offset
the P stress–strain
axis,
a
line
parallel
to
the
initial
straight-line
portion
10
method.
Normally
a
0.2%
strain
(
)
is
chosen,
and
from
0.002
in.>in.
nt
on
the
axis,
a
line
parallel
to
the
initial
straight-line
portion
P
0.005
0.010
intersects the
curve
the point
yieldwhere
strength.
An example of the
stress–strain
diagram
is defines
drawn. The
this line
0.002
P
(in./in.)
10
point
on determining
the
axis,
astrength.
line
parallel
to the
initial
straight-line
portion
stress–strain
diagram
is drawn.
The
point
where
line alloy
0.005
construction
for
the yield
for
aluminum
is (0.2%0.010
ts
thethis
curve
defines
theP yield
Anstrength
example
of an
thethis
P (in./in.)
offset) Yield strength
0.002
for an aluminum alloy
0.005
0.010
of
the
stress–strain
diagram
is
drawn.
The
point
where
this
line
tstion
the
curve
defines
the
yield
strength.
An
example
of
the
for determining
theFrom
yield strength
for anthe
aluminum
alloy
is (0.2%
P (in./in.)
offset)
shown
in Fig. 3–7.
the graph,
yield 8strength
isC s
51
ksi
8
H AYS
P T E=
R 3
M
E
C
H
A
N
I
C
A
L
P
R
O
P
E
R
T
I
E
S
O
F
M
AT E R I A L S
0.002
Yield strength for an aluminum alloy
0.005
0.010
Aluminyum
alaşım
diyagram:
! example
curve
defines
the için
yield
strength.
of offset)
the
n(352
Fig.intersects
3–7.
From the
the
graph,
the
yield
strength
sYS
= 51 ksi An
ction
for
determining
the yield
strength
forisan
aluminum
alloy
is (0.2%
3–7
MPa).
Yield0.002
strength for Fig.
an aluminum
alloy
Fig.is3–7 (0.2% offset)
construction
the yield
strength
aluminum
alloy
na).Fig.
3–7. From for
thedetermining
graph, the yield
strength
is
sfor
= 51
ksi
YS an
Yield strength for an aluminum alloy
s (ksi)
strength
is
s
=
51
ksi
Fig. 3–7
Pa). shown in Fig. 3–7. From the graph, the yield
YS
2.0
Fig. 3–7
(352 MPa).
3.3
3.3
1.5
s (ksi)
1.0
3
sf " 22
20
B
!0.06 !0.05 !0.04 !0.03 !0.02 !0.01 A
0.5
0.01
!20
Doğal lastik için diyagram: !
2
4
6
8
s–P diagram for natural rubber
10
P (in./in.)
!40
!60
Fig. 3–8
!80
!31
C
!100
!120
s–P diagram for gray cast iron
L
PROPERTIES
M AT E R I A L S
OF
Gevrek Malzemeler
Gevrek malzemeler kırılma öncesi çok az akar (yield) veya hiç akmaz.
Gri dökme demir örneği:
s (ksi)
sf " 22
20
B
!0.06 !0.05 !0.04 !0.03 !0.02 !0.01 A
0.01
P (in./in.)
!20
10
P (in./in.)
!40
!60
!80
!100
!120
C
s–P diagram for gray cast iron
!
Fig. 3–9
3.3
STRESS
–STRAIN BEHAVIOR OF
STRESS–STRAIN BEHAVIOR OF DUCTILE AND3.3
BRITTLE
MATERIALS
8 9DUCTILE A
Bir gevrek malzemenin çekme altındaki kırılması/kopması
s (ksi)
(st)max " 0.4
2
!0.0030 !0.0025!0.0020!0.0015!0.0010!0.0005 !0.0030 !0.0025!0.0020!0.0015!
P (in./in.)
Realize that the yield strength is not a physical property of the
0 0.0005
material, since it is a stress that causes a specified permanent strain in the
!2
material. In this text, however, we will assume that the yield strength,
yield point, elastic limit, and proportional limit all coincide unless
! An exception would be natural rubber,
Basınç
altında
otherwise stated.
which
in fact dışarı doğru şişme:!
!4
Compression
causes
Compression causes
Tension failure of
Tension failure of
(sc)max " 5
material
to bulgeare
outa brittle
does not even have a proportional
limit, since stress
and strain
not material
material to bulge out
a brittle material
3
(b) which 89
(a)UCTILE
linearly
Instead,
as shown
in AND
Fig. B3–8,
material,
is (a)
(b)
3.3 Srelated.
TRESS–STRAIN
BEHAVIOR
OF D
RITTLEthis
MATERIALS
!6
known as a polymer,
exhibits
nonlinear
elastic behavior.
Betonarme
karışım
için
Fig.tipik
3–10 gerilme-birim deformasyon diyagramı
Fig. 3–10
Wood is a material that is often moderately ductile, and as a result it is
s!P diagram for typical concrete mix
s!P diagra
s (ksi)
usually designed to respond only to elastic loadings. The strength
Fig. 3–11
F
characteristics of wood vary greatly from one species to
2 another, and for
(st)max " 0.4
each species they depend on the moisture content, age, and the size and
Materials
exhibit
little
or no
before that exhibit little or no yielding before
!0.0025!0.0020!0.0015!0.0010!0.0005
Brittle
Materials.
arrangement !0.0030
ofBrittle
knots
inMaterials.
the wood. Since
wood isthat
a fibrous
material,
itsyieldingMaterials
P (in./in.)
failure
are
referred
to
as
brittle
materials.
Gray
cast
iron
is
an
example,
0 0.0005
referred to as brittle materials. Gray cast iron is an example,
tensile or compressive characteristics will differ greatly failure
when
itare
is loaded
a stress–strain
tensionhaving
as shown
by
portion AB
of thein tension as shown by portion AB of the
a stress–strain
diagram
either parallel having
or perpendicular
to diagram
its grain.inSpecifically,
wood
splits
!2
curve
in
Fig.
3–9.
Here
fracture
at
(152
MPa)
took
place at sf = 22 ksi (152 MPa) took place
s
=
22
ksi
curve
Fig. and
3–9. Here fracture
easily when it is loaded in tension perpendicularf to
its in
grain,
initially
at
an
imperfection
or
microscopic
crack
and
then
spread
rapidly
initially
an imperfection or microscopic crack and then spread rapidly
consequently tensile loads are almost always intended
to beatapplied
acrossofthe
specimen,
causing complete fracture.
Since
the appearance
of
!4 the
across
specimen,
causing complete
fracture. Since the appearance of
parallel to
the grain
wood
members.
Compression
causes
brittle
materials
do
not
have
(s
)
"
5
material to bulge out initial cracks in a specimen is quite random,
c max in a specimen
initial cracks
is quite random, brittle materials do not have
3
a well-defined tensile fracture stress. Instead
the average
fracture
stressstress. Instead the average fracture stress
(b)
a well-defined
tensile
fracture
!6
from a set of observed tests is generally
reported.
A
typical
failed
from a set of observed tests is generally reported. A typical failed
specimen is shown in Fig. 3–10a.
specimen
is shown in Fig. 3–10a. Steel rapidly loses its strength when
Steel
s!P
diagram
for
typical
concrete
mix
! Compared with their behavior in tension,
heated.
For this
reason
engineers such
often as
brittle
as in
heate
Compared materials,
with their such
behavior
tension,
brittle
materials,
require main structural members to be
Fig. 3–11
gray cast iron, exhibit a much
higher resistance
toiron,
axialexhibit
compression,
requi
gray az
castise
a gelir.
much as
higher
resistance
to
axial
compression,
as
Çelikteki
karbon
miktarı
fazla
ise
gevrek,
sünek
hale
insulated in case of fire.
evidenced by portion AC of the curve in evidenced
Fig. 3–9. For
this
case
any
cracks
by portion AC of the curve in Fig. 3–9. For this case any cracks insula
or
imperfections
in
the
specimen
tend
to
close
up, and
the
load sıcaklıklarda
at exhibit little orDüşük
no yielding
before
sıcaklıklarda malzemeler dahaorsert
ve gevrek
olurlar.
Yüksek
ise up,
daha
imperfections
inasthe
specimen
tend to close
and as the load
increases
the material will generally bulge
or
become
barrel
shaped
as s (ksi)
erials. Gray cast iron
is an example,
increases the material will generally
bulge or become barrel shaped as s (ksi)
the
strains
Fig. 3–10b. the strains become larger, Fig. 3–10b.
40# F
vebecome
sünek
olurlar.
nsion as shown byyumuşak
portion
AB
of
the larger,
9
Like took
gray place
cast iron, concrete is classified
as
a
brittle
material,
and
it
sf = 22 ksi (152 MPa)
Like gray cast iron, concrete is classified as a brittle material, and it 9
also
has
a
low
strength
capacity
in
tension.
Thea characteristics
of its 8in tension. The characteristics of its 8
scopic crack and then spread rapidly
also has
low strength capacity
stress–strain
diagram
depend
primarily
on
the
mix
of
concrete
(water,
7
te fracture. Since the appearance of
stress–strain diagram depend primarily
on the mix of concrete (water, 7
sand, gravel,
and cement) and the time and temperature of curing. A
!32
andom, brittle materials
do not have
110# F
sand, gravel, and cement) and the6 time and temperature
of curing. A
6
typical
example
of
a
“complete”
stress–strain
diagram
for
concrete
is
. Instead the average fracture stress
typical example of a “complete” 5stress–strain diagram for concrete is
given
in
Fig.
3–11.
By
inspection,
its
maximum
compressive
strength
enerally reported. A typical failed
given in Fig. 3–11. By inspection, its maximum compressive strength 5
is almost 12.5 times greater than its tensile strength, 1sc2max = 5 ksi 4
s = EP
(3–5)
s = EP
(3–5)
the previous section, the stress–strain
diagrams for most
Here E exhibit
represents
the constant
of proportionality,
materials
a linear
relationship
between stress which
and is called the
modulus
of elasticity
Young’s
modulus,
after
Thomas
Young,
he
elastic
region.
Consequently,
an of
increase
innamed
stress causes
Here
E represents
theorconstant
proportionality,
which
is called
the
who
published
an
account
of
it
in
1807.
ate
increaseofinelasticity
strain. This
fact wasmodulus,
discovered
by Robert
modulus
or Young’s
named
after Thomas Young,
Equation
3–5and
actually
represents
the equation of the initial straight76who
using
springs
is known
published
an account
of itasin Hooke’s
1807. law. It may be
lined
portion
of
the
stress–strain
diagram
up to
the proportional
limit.
thematically
Hook,
Equationas
3–5Hooke
actually Kanunu:
represents (Robert
the equation
of the1676)
initial straightFurthermore, the modulus of elasticity represents the slope of this line.
lined portion of the stress–strain diagram up to the proportional limit.
Since strain is dimensionless,
Eq. 3–5, E will have
the same units as
= EP offrom
(3–5)
Furthermore, thesmodulus
elasticity represents the
slope of this line.
stress, such as ! psi, ksi, or pascals. As an example of its calculation,
Since strain is dimensionless, from Eq. 3–5, E will have the same units as
consider constant
the stress–strain
diagram for
steelis shown
in Fig. 3–6. Here
esents
proportionality,
which
calledModülü
the
stress,the
such as E:
psi,of
ksi,
or pascals.
As
an example
of
its calculation,
Elastisite
Modülü
veya
Young
s
=
35
ksi
and
P
=
0.0012
in.>in.,
so
that
pl
pl
lasticity
orthe
Young’s
modulus,diagram
named after
Thomas
Young,
consider
stress–strain
for steel
shown
in Fig. 3–6. Here
ds
an
itGerilme-birim
in
= 35 ksiofand
Ppl1807.
=s0.0012
in.>in.,
deformasyon
ilişkisindeki orantılılık (proportionality) sınırlarını kullanarak:
pl account
pl
35
ksiso that
3
E = the equation
=
= 29110
2 ksi
–5 actually represents
of the initial
straightPpl
0.0012
in.>in.
s
35 the
ksi proportional3 limit.
of the stress–strain diagram up to
E =
=
= 29110 2 ksi
theAs
modulus
the slope
of this
Ppl represents
in.>in.
shownof
in elasticity
Fig. 3–13,
the0.0012
proportional
limit
for aline.
particular type of
!
dimensionless,
from
Eq.
3–5,
E
will
have
the
same
units
asgrades of steel,
steel alloy depends on its carbon content; however,
most
As ksi,
shown
in
Fig. 3–13,
the example
proportional
limit
for a particular type of
s from
psi,
or
pascals.
As
an
of
its
calculation,
için
farklı
karbon
içerikleri
elastisite
the softestÇelik
rolled
steel
to the
hardest
tool steel,
have modülünde
about the önemli değişikliğe sebep olmaz:
steel alloy depends
onfor
its carbon
content;
stress–strain
diagram
steel shown
in however,
Fig. 3–6. most
Heregrades of steel,
s (ksi)sosteel
from
softestin.>in.,
rolled
and
Ppl the
= 0.0012
that to the hardest tool steel, have about the
E =
spl
Ppl
=
s (ksi)
35 ksi
180
0.0012 in.>in.
180
160
= 2911032 spring
ksi steel
(1% carbon)
spring steel
(1%
carbon)
a particular
n Fig. 3–13, the proportional limit for
type of
160
140
pends on its carbon
content; however, most grades of steel,
est rolled steel to140
the hardest toolhard
steel,
steelhave about the
120
s (ksi)
180
160
140
120
100
80
60
40
20
(0.6% carbon)
heat treated
hard
steel
(0.6% carbon)
heatsteel
treated
machine
100
80
spring steel
(0.6% carbon)
(1% carbon)
machine steel
80
60
structural
steel
(0.6% carbon)
(0.2% carbon)
60
40
structural
soft steel steel
(0.2%
(0.1% carbon)
carbon)
40
20hard steel soft steel
(0.6% carbon)
(0.1% carbon)
20heat treated
0.002 0.004 0.006 0.008 0.01
!
machine steel
Fig. 3–13
(0.6% carbon)
0.002 0.004 0.006
0.008 0.01
120
100
structural steel
(0.2% carbon)
soft steel
(0.1% carbon)
0.002 0.004 0.006 0.008 0.01
P (in./in.)
P (in./in.)
Fig. 3–13
P (in./in.)
Fig. 3–13
!33
3.5 Strain Energy
As a material is deformed by an external loading, it tends to store energy
internally throughout its volume. Since this energy is related to the
strains in the material, it is referred to as strain energy. To obtain this
strain energy consider a volume element of material from a tension
Birim Deformasyon Enerjisi
test specimen. It is subjected to uniaxial stress as shown in Fig. 3–15.
s
This stress develops a force ¢F = s ¢A = s1¢x ¢y2 on the top and
bottom faces of the element after the element of length ¢z undergoes a
3
vertical displacement P ¢z. By definition, work is determined by the
product of the force and displacement in the direction of the force. Since
the force is increased uniformly from zero to its final magnitude ¢F
!z
when the displacement P ¢z is attained, the work done on the element
by the force is equal to the average force magnitude 1¢F>22 times the
92
C H A P T E R 3 M E C H A N I C A L P R O P E Rdisplacement
T I E S O F M AT E R
L S This “external work” on the element is equivalent to
PI A¢z.
!x
!y
the
“internal
work”
or strain energy stored in the element—assuming
ARPOTPEERR 3
M
E
C
H
A
N
I
C
A
L
P
R
O
P
E
R
T
I
E
S
O
F
M
AT
E
R
I
A
L
S
T I E S O F M AT E R I A L S
L P R O P E R T I E S O F M AT E R I A L S
92energy is lost
C H Ain
P Tthe
ER 3
Mof
E C heat.
H A N I CConsequently,
A L P R O P E R T I E Sthe
O Fstrain
M AT Eenergy
RIALS
that
no
form
s
TF
E RM3AT! EM
EC
ATis
E R¢U
I A L S = 11 ¢F2 P ¢z = 11 s ¢x ¢y2 P ¢z. Since the volume of the
O P E RC
T IHEA
S PO
RIA
L SH A N I C A L P R O P E R T I E S O F M
¢U
2
2
Strain Energy
3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S
1
Fig. 3–15
element
is ¢V
=H A¢x
¢y
¢z,
then
9
2
C
H
A
P
T
E
R
3
M
E
C
N
I
C
A
L
P
R
O
P
Ebütün
R T I¢U
E S O=
F 2
MsP
AT E¢V.
Rboyunca
IALS
Strain
Energy
Strain
Energy
Bir
cisim
dış
yük
altında
deformasyon
yaptığında
hacmi
bir iç enerji
O F M AT E R I A L S
M
E C H A N I C A LStrain
P E R T I E S O F M AT E R I A L S
it
is
sometimes
convenient
specify
the Energy
strain
HAPTER 3
MPE RCO
H A N I C AEnergy
L P R O P E R T I E S O F M As
AT Ea
R Imaterial
A L SFor applications,
is deformed by an external loading, it tends totostore
energy
Strain
Strain
Energy
energy
per
unit
volume
of
material.
This
is
called
the
strain-energy
Strain
Energy
depolar.
Bu
enerji
içindeki
birim
deformasyonlara
bağlı
olduğundan
As
aan
material
iscisim
deformed
external
loading,
it tends
toHstore
energy
internally
throughout
Since
energy
is related tobuna
the birim
As a material
is deformed
by
external
loading,
it by
tends
to
storeits
energy
9an
2 and
C Hvolume.
APTER 3
M E Cthis
A N I C A L P R O P E R T I E S O F M AT E R I A L S
Strain
Energy
density,
itstore
canit
expressed
As a material
is deformed
by an external
loading,
tends
toSince
energy
throughout
its
volume.
this
energytoisasasrelated
to As
the
strains
init
the
isthe
referred
strain
energy.
To
obtain is
this
internally
throughout
itsinternally
volume.
Since this
energy
is material,
related
tobe
Strain
Energy
a
material
deformed by an external load
Strain
Energy
aSince
material
is
deformed
byto
an
external
loading,To
it tends
tothis
store
energy
deformasyon
enerjisi
denmektedir.
Eksenel
bir
çubuk
için birim
Energy
As
ainternally
material
deformed
by
external
loading,
itenergy
tends
to
store
energy
strains
inAs
the
itenergy
is referred
as
strain
energy.
obtain
throughout
its
volume.
thisEnergy
is
related
the çekmeye
consider
a to
volume
element ofmaruz
material
from
a
tension
Strain
strains
in theis
material,
itStrain
is an
referred
tomaterial,
as strain
strain
energy.
To
obtain
this
internally
throughout its volume. Since thi
As aits
material
is
deformed
by
an aexternal
loading,
itthe
tends
to this
store¢U
energya1 istension
internally
throughout
its
volume.
Since
energy
related
to the
nternally
throughout
volume.
Since
energy
is related
to
energy
consider
volume
element
of
material
from
strains
in the
material,
itvolume
is referred
tothis
as
energy.
To
this
strain
energy
consider
astrain
element
ofstrain
material
from
aobtain
tension
test
specimen.
ItYukarıdaki
is subjected
to As
uniaxial
stressisas
shown
inby
Fig.
a= material
deformed
an3–15.
external
loading,
it tends to
to as
stors
strains
in
the
material,
italt
is referred
deformasyon
enerjisini
belirleyelim.
şekilde,
normal
gerilme
elemanın
üst Energy
ve
u
=
sP
(3–6)
rial
is
deformed
by
an
external
loading,
it
tends
to
store
energy
strains
in
the
material,
it
is
referred
to
as
strain
energy.
To
obtain
this
Strain
internally
throughout
its
volume.
Since
this
energy
is
related
to
the
As
ais material
is
by
ansubjected
loading,
tends
to as
store
energy
trains
in the
material,
its is
to
asis
strain
energy.
Toexternal
obtain
this
test
specimen.
It
is
to
uniaxial
stress
shown
Fig.
3–15.
As
adeformed
material
deformed
by
an
loading,
it tends
to
store
teststrain
specimen.
Itconsider
subjected
to
uniaxial
stress
as shown
in
Fig.
3–15.
This
stress
develops
ait
¢F
=
s ¢A
=in
s1¢x
¢y2
on
the
top
and
energy
areferred
volume
element
ofexternal
material
from
aforce
tension
¢V
2 energy
internally
throughout
its
volume.
Since
this
energy
is
related
strain
energy consider a volume element o
strain
energy
consider
athe
volume
element
oftomaterial
from
a tension
throughout
its
volume.
Since
this
energy
tothis
strains
in
thestress
material,
it of
isisreferred
to
as
strain
energy.
To¢y2
obtain
this top
internally
throughout
its
Since
energy
isafter
related
the
train
energy
consider
aforce
volume
element
material
from
as
tension
s stress
This
develops
arelated
force
¢F
=the
¢A
=
s1¢x
the
internally
throughout
its
volume.
Since
energy
ison
to and
the
This
develops
¢F to
= uniaxial
s
¢Avolume.
= stress
s1¢x
¢y2
on
top this
and
strains
inrelated
the
material,
itundergoes
is referred
as strain to
energy.
To ob
bottom
faces
of
the
element
the element
of
length
¢z
a istosubjected
test
specimen.
It isa subjected
as
shown
in
Fig.
3–15.
yüzeyinde
!
kuvvetini
oluşturmaktadır.
Burada
!
kenarı
test
specimen.
Itdeformed
uniaxial
st
test
specimen.
It
is
subjected
to
uniaxial
stress
as
shown
in
Fig.
3–15. isuzayacaktır
As
a
material
by an external
load
the
material, itIt
isisstrain
referred
tomaterial,
asconsider
strain
To
obtain
this
a referred
volume
element
ofelement
material
from
a ¢z
tension
3 specimen.
strains
in energy
the
it
isthe
toafter
as¢z
strain
energy.
obtain
this
est
to
uniaxial
stress
aslength
shown
in
Fig.
3–15.
bottom
faces
ofenergy.
element
the
ofTo
length
undergoes
a a volume
strains
in
the
material,
itIf
is
referred
to
as
strain
energy.
To
obtain
this
bottom
faces of
thesubjected
element
after
the
element
of
undergoes
a
strain
energy
consider
element
of
material
from
a
vertical
displacement
P
¢z.
By
definition,
work
is
determined
by
the
This
stress
develops
a
force
¢F
=
s
¢A
=
s1¢x
¢y2
on
the
top
and
the
material
behavior
is
linear
elastic,
then
Hooke’s
law
applies,
s
This
stress
develops
a
force
¢F
=
s
¢A
=
s1¢x
¢y2
on
the
top
and
internally
throughout
its
volume.
Since
thi
s
This
stress
develops
a
force
¢F
=
s
¢A
=
testaenergy
specimen.
It
is material
subjected
uniaxial
as
shown
Fig.
3–15.a It
rgy
consider
astrain
volume
element
ofs
from
adetermined
tension
consider
a consider
volume
element
ofstress
material
from
adetermined
tension
his stress
develops
force
¢F
=
¢A
=product
s1¢x
¢y2
onforce
the
topby
and
vertical
displacement
Pto¢z.
By
definition,
work
isingöre
byisthe
strain
energy
a
volume
element
of
material
from
tension
test
specimen.
subjected
to
uniaxial
stress
as
shown
in
F
vertical
displacement
P
¢z.
By
definition,
work
is
the
of
the
and
displacement
in
the
direction
of
the
force.
Since
ve
yapacağı
deplasman
!
olacaktır.
Tanıma
iş,
kuvvet
ile
kuvvetin
etkime
bottom
faces ofThis
the
after
theaaselement
ofin
¢z=therefore
undergoes
aon
s
=
EP,
and
we
can
express
the elastic
strain-energy
density
bottom
faces
of
the
after
the
element
of
length
¢zstrains
undergoes
a of
in
the
material,
itinis referred
as s
bottom
faces
the
element
after the to
elemen
stress
develops
force
¢F
=length
selement
¢A
s1¢x
¢y2
the
top
and
men.
isofsubjected
toelement
uniaxial
stress
Fig.
3–15.
specimen.
Itspecimen.
is
subjected
to
uniaxial
stress
as
shown
in
Fig.
3–15.
ottomItfaces
oftest
the element
after
the
element
of
length
¢z
undergoes
adirection
product
of
the
and
displacement
in
the
of
the
force.
test
Itshown
isdirection
subjected
to
uniaxial
stress
as
shown
in
Fig.
3–15.
This
stress
develops
amagnitude
force ¢F ¢F
= s ¢A = s1¢x ¢y2 on the
product
the
force
and
displacement
inforce
the
the
force.
Since
the
force
issof
increased
uniformly
from
zero
to
its Since
final
3
terms
of
the
uniaxial
stress
as
vertical
displacement
P
¢z.
By
definition,
work
is
determined
by
the
vertical
displacement
P
¢z.
By
definition,
work
is
determined
by
the
strain
energy
consider
a
volume
element
o
bottom
faces
of
element
the
element
length
¢z
undergoes
aoftop
vertical
displacement
definition,
sertical
develops
a increased
force
¢F
=the
s
¢A
=the
s1¢x
¢y2
on
the
top
stress
develops
a deplasmanın
force
¢F
=
s
¢A
=and
onattained,
the
topthe
and
force
is
increased
from
zero
to
its
final
magnitude
¢F
P uniformly
¢z.
By
definition,
work
isforce
determined
by
the
s displacement
This
stress
develops
auniformly
¢F
=s1¢x
sof
¢A
=elde
¢y2
on
the
and
çarpımı
ile
edilir.
Kuvvet
sıfırdan
başlayıp
! P ¢z.ofBy
!This
z doğrultusundaki
bottom
faces
the
element
after
the element
length
¢z und
the
force
isof
from
zero
toafter
its
final
magnitude
¢F
when
the
displacement
P¢y2
¢z
iss1¢x
work
done
onspecimen.
the
element
product
of
the
force
and
displacement
in
the
direction
of
the
force.
Since
product
the
force
and
displacement
in
the
direction
of
the
force.
Since
test
It
is
subjected
to
uniaxial
st
3
vertical
displacement
Pthe
¢z.
By
definition,
is determined
by
the
ces
of the
element
after
the
element
of
¢zthe
awork
product
thedefinition,
force and displacement
in the
bottom
ofisthe
element
after
element
of element
length
¢z
a element
displacement
Pundergoes
¢z
attained,
the
work
done on
the
roduct
of the
force
andfaces
displacement
in length
the
direction
ofisafter
the
force.
Since
bottom
faces
of
the
element
the
of undergoes
length
¢z
undergoes
aP ¢z.of
vertical
displacement
By
work is determined
when
the
displacement
Pwhen
¢z
attained,
done
on
the
element
by
the
force
is
equal
to
thefrom
average
force
magnitude
1¢F>22
times
the
2
the
force
iswork
increased
uniformly
zero
to
its
final
magnitude
¢F
sthe
This
stress
develops
a uniformly
force ¢F =
s ¢A
=
the
force
is¢z.
increased
uniformly
from
zero
to
its
final
magnitude
¢F
stimes
1üniform
product
of
the
and
displacement
in
the
direction
of
force.
Since
splacement
P
By
definition,
work
is
determined
by
the
deplasmanına
karşılık
gelen
ve
son
şiddeti
olan
!
değerine
bir
artış
gösterdiğinden,
vertical
displacement
P
¢z.
By
definition,
work
is
determined
by
the
by
the
force
is
equal
to
the
average
force
magnitude
1¢F>22
the
he
force
is
increased
uniformly
from
zero
to
its
final
magnitude
¢F
the
force
is
increased
from
zero
vertical
displacement
P
¢z.
By
definition,
work
is
determined
by
the
product
of
the
force
and
displacement
in
the
direction
of
the
for
by!zthe force is equal to the average
force
magnitude
1¢F>22
times
the the
(3–7) after the elemen
u on
= the
displacement
P ¢z.
This
“external
work”
element
is element
equivalent
to
when
the
displacement
P
¢z
is
attained,
work
done
on
the
bottom
faces
of
the
element
!
x
when
the
displacement
P
¢z
is
attained,
the
work
done
on
the
element
!
z
the
force
is
increased
uniformly
from
zero
to
its
final
magnitude
¢F
displacement
Pdisplacement
¢z.
This
“external
work”
on
the
element
isthe
equivalent
to uniformly
the force
and product
displacement
inforce
the work”
direction
of
the
Since
of“external
the
and
in on
the
direction
the
force.
when
the displacement
¢z
is
attained,
the
work
done
the
element
2isforce.
E
product
of
the
force
and
displacement
theof
direction
ofSince
Since
when
the displacement
is attained,
the
y P
the
force
from zeroP ¢z
to its
final magnit
displacement
onthe
the
element
is the
equivalent
to
3force.
!xs P ¢z.!This
“internal
work”
orin
strain
energy
stored
inincreased
the
byto
the
force
is
equal
to
average
force
magnitude
1¢F>22
times
the
vertical
displacement
¢z. By definition,
!y by the force
iseleman
equal
to
the
average
force
1¢F>22
times
the
when
the
displacement
Pmagnitude
¢z
is
attained,
work
done
on
the
element
üzerinde
gerçekleştirilen
iş the
ortalama
kuvvet
şiddeti
! element—assuming
ile
deplasman
! P the
the
“internal
work”
or
strain
energy
stored
in
the
element—assuming
!magnitude
zin
the
force
is
increased
from
zero
to
its
final
magnitude
¢F
y the
force isuniformly
equal
to
average
force
1¢F>22
times
the
is
increased
from
zero
its
final
magnitude
¢F
the
force
isuniformly
increased
uniformly
from
zero
to
its
final
magnitude
¢Fthe
when
the
displacement
P
¢z
is
attained,
the
work
done
on
the
by
force
is
equal
to
average
force
m
the
“internal
work”
or the
strain
energy
stored
the
element—assuming
that no
energy
is“external
lost in thework”
form ofon
heat.
Consequently,
the strain
energy
displacement
Pthe
¢z.
This
the
element
is equivalent
to force and displacement in the
product
of the
s
that
no
energy
is
lost
in
form
of
heat.
Consequently,
the
strain
energy
by
force
is
equal
to
the
average
force
magnitude
1¢F>22
times
the
!
x the
displacement
P
¢z.
This
“external
work”
on
the
element
is
equivalent
to
when
displacement
P
¢z
is
attained,
the
work
done
on
the
element
isplacement
P
¢z.
This
“external
work”
on
the
element
is
equivalent
to
displacement
P
¢z
is
attained,
the
work
done
on
the
element
when
the
displacement
P
¢z
attained,
the
work
done
on
element
1
1
by
the
force
is
equal
to
the
average
force
magnitude
1¢F>22
t
displacement
P
¢z.
This
“external
work”
on
that
no
energy
is
lost
in
the
form
of
heat.
Consequently,
the
strain
energy
y
!
Modulus
of
In the
particular,
when
thekaybı
stress
s reaches
s
¢UPiswork”
¢U= =1or
¢F2
P Resilience.
¢z
=¢z.
12 stored
s
¢x biçiminde
¢y2
Pvolume
¢z.element—assuming
Since
the
volume
of
the
1 eşittir.
112 strain
the=This
“internal
energy
the
force
is
increased
uniformly from zero
!the
xin
değerinin
çarpımına
Sistemde
ısıya
dönüşme
bir
enerji
olmadığını
1 work”
1P
¢U
is
¢U
1
¢F2
¢z
s
¢x
¢y2
P
Since
of
the
displacement
¢z.
“external
work”
on
the
element
is
equivalent
to
by
the
force
is
equal
to
the
average
force
magnitude
1¢F>22
times
the
he
“internal
work”
or
strain
energy
stored
in
the
element—assuming
y
!
by
the
force
is
equal
to
the
average
force
magnitude
1¢F>22
times
the
“internal
or
strain
energy
stored
in
the
element—assuming
ce
is
equal
to
the
average
force
magnitude
1¢F>22
times
the
s
displacement
P
¢z.
This
“external
work”
on
the
element
is
equiv
1
2
2
pl
¢Uthe
is
¢U
=
1
¢F2
P
¢z
=
1
s
¢x
¢y2
P
¢z.
Since
the
volume
of
the
!x
work”
or3–6
strain
stored
the
proportional
the strain-energy
density,
as“internal
calculated
by Eq.
!¢y
z limit,
!
xform
2
2
1¢z,
Fig. 3–15
element
¢x
then
¢U = 2 sP ¢V.
that
no=energy
is!¢z,
lost
in =
the
of
heat.
Consequently,
the the
strain
energy
when
the
displacement
P ¢z isenergy
attained,
the
yis ¢V
Fig.
3–15
element
¢x
¢y
then
¢U
=1 the
sPelement—assuming
¢V.
s isdisplacement
the
work”
or
stored
in
Pthen
¢z.
This
“external
work”
on
the
element
is
equivalent
to resilience,
hat
no
energy
in“internal
the
form
ofisheat.
Consequently,
the
strain
energy
displacement
P 1strain
¢z.
This
“external
work”
on
the
element
is equivalent
to i.e.,
ent
P ¢z.
work”
on
the
element
isenergy
equivalent
to
noisThis
energy
is
lost
in
the
form
of¢V
heat.
Consequently,
the
strain
energy
the
“internal
work”
or
strain
energy
stored
inform
the of
element—a
2to
1
element
¢V
=lost
¢x
¢y
¢z,
¢U
=
sP
¢V.
!
x!ythat
!x1“external
or
3–7,
is
referred
as
the
modulus
of
that
no
energy
is
lost
in
the
heat.
Con
varsayarsak,
eleman
üzerindeki
dış
iş,
iç
işe
veya
eleman
içinde
depolanmış
olan
birim
2
¢U
is
¢U
=
1
¢F2
P
¢z
=
1
s
¢x
¢y2
P
¢z.
Since
the
volume
of
the
For
applications,
it
is
sometimes
convenient
to
specify
the
strain
by
the
force
is
equal
to
the
average
force
m
1
1that
1is
2 of
2
s the
For
it
sometimes
convenient
to
specify
theis strain
the
work”
oris
strain
energy
stored
inthe
element—assuming
¢U
iswork”
¢U
=or12=
¢F2
Penergy
=sometimes
1stored
s1applications,
¢x
¢y2
Pthe
¢z.
Since
the
volume
ofstrain
the
no
in
form
heat.
Consequently,
the
“internal
work”
or
strain
energy
stored
inthe
the
element—assuming
1 heat. Consequently,
1
nalFor
strain
in
the
element—assuming
that
no energy
energy
lost
in is
the¢U
form
of
the
strai
1strain
¢U
is
¢U
Pis energy
¢z
slost
¢x
¢y2
P¢V
¢z.
Since
the
volume
of
the
2=“internal
applications,
it¢z
convenient
to
specify
2 element
¢U
=
1
¢F2
P
¢z
=
1
s
¢x
¢y2
P
¢
s
Fig.
3–15 12 ¢F2
is
=
¢x
¢y
¢z,
then
¢U
=
sP
¢V.
displacement
P
¢z.
This
“external
work”
on
energy
per
unit
volume
of
material.
This
is
called
the
strain-energy
1
1
1
2
2 !the
1=¢V.
energy
per
unit
volume
of form
material.
This
isthe
called
that
no
energy
lost
the
of
heat.
Consequently,
the
strain
energy
lement
isper
¢V
=that
¢x
¢y
¢z,
then
¢U
=the
sP
energy
isConsequently,
lost
in
form
heat.
Consequently,
strain
¢U
¢U
=material.
1then
¢F2
PThis
1in
s
¢x
¢y2
PSonuç
¢z.
Since
volume
ofstrain-energy
the
¢U
isxenergy
¢U
= 112 ¢F2
P ¢z = 112 s 2¢x ¢y2 P ¢z. Since
the volum
ergy
iss lost
in
the
form
heat.
the
strain
energy
1
2¢z
2enerjisine
2of
!ythe
element
isunit
¢V
=no
¢x
¢z,
¢U
=is
sP
¢V.
energy
volume
is
called
the
strain-energy
deformasyon
eşit
olacaktır.
olarak:!
uisof
r ¢yof
For
applications,
it
is
sometimes
convenient
to
specify
the
strain
2
Fig.
3–15
the
“internal
work”
or
strain
energy
stored
element
is
¢V
=
¢x
¢y
¢z,
then
¢U
=
s
1
1
1
1
density,
and
it
can
be
expressed
as
1
1
2
density,
and
it
can
be
expressed
as2¢z.
¢U
isPas
¢U
=
1convenient
Pvolume
¢z
=3–15
1specify
¢x
¢y2
Pthe
¢z.volume
Since
the
volume
of2 the
applications,
isit¢x
sometimes
to
specify
theSince
strain
¢U
is
=is
1¢V
Pconvenient
¢z
12 s
¢x
¢y2
of is
the
=
¢x
¢y
¢z,
then
¢U
=
sP
¢V.
Fig.
element
¢Vthe
¢y ¢z, then ¢U = 2 sP ¢V.
=For
112For
¢F2
P it
¢zcan
=element
1it12 ¢U
s
¢y2
¢z.
Since
of
the
2=¢F2
2Ps
2 ¢F2
density,
and
be
expressed
applications,
is
sometimes
to
the
pl
energy
perthe
unit
volume
of material.
is called
strain-energy
1= s¢x
that
noapplications,
energy is lost itin is
thesometimes
form of heat.
Con
1 strainThis 1
1
For
conven
1 ¢x
Fig.
3–15
element
is
¢V
=
¢x
¢y
¢z,
then
¢U
=
sP
¢V.
s
nergy
per
unit
volume
of
material.
This
is
called
the
strain-energy
element
is
¢V
=
¢y
¢z,
then
¢U
=
sP
¢V.
For
applications,
it
is
sometimes
convenient
to
specify
the
strain
it is sometimes
convenient
ur olacaktır.
=For sapplications,
(3–8)
1 to specify th
2
¢V
= ¢xper
¢y unit
¢z,
then
¢U
= 2material.
sP
¢V. This
2the ! strain-energy
plPpl =
Hacim
! of
olduğuna
energy
volume
called
density,
and
itis can
begöre
expressed
as
¢U is ¢U
=unit
112 ¢F2
P ¢z of
= 1material.
¢y2 P ¢
2specify
Eenergy
permaterial.
volume
2 s ¢x
¢U
1strain
For
it is the
sometimes
to
the2 strain
ensity, anditit is
cansometimes
be
expressed
as applications,
For
applications,
it is to
sometimes
convenient
the
¢U is convenient
1to specify
energy
perconvenient
unit
volume
of
material.
This
called
the
strain-energy
energy
per
unit
volume
of
This
is
called
the This
strai
plications,
specify
strain
u =3–15 = sP
(3–6)
¢U
1
density, and it can be expressed
as
Fig.
element
is
¢V
=
¢x
¢y
¢z,
then
¢U
= 12 s
u
=
=
sP
(3–6)
density,
and
it
can
be
expressed
as
energy
per
unit
volume
of
material.
This
is
called
the
strain-energy
energy
per
unit
volume
of
material.
This
is
called
the
strain-energy
¢V
2
u
=
=
sP
(3–6)
density,
and
it
can
be
expressed
as
Uygulamalarda
birim
hacimdeki
birim
deformasyon
enerjisinin
belirlenmesi
kimi
zaman
density,
and
it
can
be
expressed
as
P
¢V
2
r unit volume of Pmaterial.
This¢V
is called
strain-energy
For applications, it is sometimes conven
pl
2 the expressed
and
as u = ¢U = 1 sP
density,as
anddensity,
it can
be
expressed
¢U
1it can beas
(3–6)
d it can be expressed
energy
per
unit
volume
This
From
the
elastic
region
of
the
stress–strain
diagram,
Fig.
3–16a,
notice
u
=
=
sP
(3–6)
Modulus
of
resilience
u
¢V
2
daha uygun
olmaktadır.
birim
deformasyon
enerjisi
yoğunluğu
denmektedir
veof material.
¢U
1 If Buna
r
¢U
1
¢U
1expressed
the
material
behavior
is
linear
elastic,
then
Hooke’s
law
applies,
¢U
1
¢V
2= behavior
If the
material
is
linear
elastic,
then
Hooke’s
law
applies,
u
=
sP
(3–6)
density,
and
it
can
be
as
that
u
is
equivalent
to
the
shaded
triangular
area
under
the
diagram.
r
u
=
=
sP
u
=
=
sP
If the material behavior
is
linear
elastic,
then
Hooke’s
law
applies,
u
=
=
sP
(3–6)
(a)
¢U
1
¢V
2
1express
s = ¢U
EP,
and
therefore
weelastic
canresilience
express
therepresents
elastic
strain-energy
density
in 2 to
s = 1EP, and therefore
we
can
the
strain-energy
density
in
¢V
¢V
2
¢V
2 a material’s
¢U
Physically
the
ability
of
the
material
u
=
=
sP
(3–6)
u
=
=
sP
(3–6)
s = EP, and therefore
we
can
express
the
elastic
strain-energy
density
in
Ifuniaxial
the material
behavior
linear
elastic, then Hooke’s law applies,
u = Fig.
=
sPthe
(3–6)
¢V isstress
2 as any
of
stress
as
terms
of
the
¢V
2uniaxial
If theofmaterial
behavior
is
elastic,
then
Hooke’s
law
applies,
3–16
absorb
energy
damage todensity
¢U
1
¢Vterms
terms
the uniaxial
stress
as 2linear
s =şeklinde
EP, and therefore
we canwithout
express
thepermanent
elasticlineer
strain-energy
in
!
ifade
edilmektedir.
Malzeme
elastik the
ise, material.
Hooke
kanunu,
u =Hooke’s
= law
sP
If the
is linear
elastic,
then
Hooke’s
law
= EP,
and material
therefore
we
express
theof
elastic
strain-energy
in
If the
material behavior
is linear elastic,
then
Ifbehavior
thecan
material
behavior
linear
elastic,
thenapplies,
Hooke’s
law
applies,
If the material
behavior
is linear
elastic,
¢V
2
terms
theisuniaxial
stress density
as
2
2 and
the
material
behavior
isthe
linear
elastic,
then
Hooke’s
law
applies,we can express the elastic strain-energy d
erms
the uniaxial
stress
as If
EP,
s of
= EP,
and therefore
we
can
express
the
elastic
strain-energy
density
in s
1s
material
behavior
iscan
linear
then
Hooke’s
law
applies,
1= s
=linear
EP,
therefore
express
strain-energy
density
intherefore
2we
s = EP, and
therefore
we can express the ela
material behaviorIf
elastic,
then
Hooke’s
lawelastic,
applies,
! sisthe
,sand
geçerli
ve
elastik
birim
deformasyon
enerjisi
yoğunluğu
eksenel
gerilme
stherefore
1olur
(3–7)
u =elastic
u strain-energy
= density
=
EP,
and
we
can
express
the
elastic
density
in
terms
of
the
uniaxial
stress as (3–7)
suniaxial
=terms
EP, and
therefore
we
can
express
the
elastic
strain-energy
in
terms
of
the
stress
as
(3–7)
u
=
2
E
of
the
uniaxial
stress
as
2
2
E
terms
of
the
uniaxial
stress
asis linear elastic,
If
the
material
behavior
nd therefore
we can express
the elastic
strain-energy
density in
1s
2 2E
of the
uniaxial
s
terms of theterms
uniaxial
stress
as stress as
(3–7)
u
=
s
1
s
=
EP,
and
therefore
we
can
express the ela
he uniaxial stress as
(3–7)2 E
u =
2
2In particular, when the stress s reaches
2
s
1
2
E
terms
of
the
uniaxial
stress
as
Modulus
of
Resilience.
s Resilience.
1of
1 solacaktır.
Modulus
In particular, when the stress s reaches
2
cinsinden: ! Inuparticular,
u =
1 s2
2 stress s
1s
Modulus
of Resilience.
when
the
reaches
(3–7)
u =
(3–7)
=
s
1
2
E
the proportional
limit,
the
strain-energy
density,
as
calculated
by
Eq.
3–6
u
=
spl
2
(3–7)s reaches
=
the
density,
calculated
by Eq. 3–6
2 Eu limit,
2 density,
E theof
1s
(3–7)theasstress
uproportional
=Resilience.
s
Instrain-energy
particular,
the proportional limit,u the
calculated
by
Eq.
3–6
2 E2
2 resilience,
E
3–7,InisModulus
referred
towhen
as as
the
modulus
of
i.e.,of when
(3–7)
=orstrain-energy
2referred
Estress
1s
Modulus
of
Resilience.
particular,
the
s
reaches
s
or
3–7,
is
to
as
the
modulus
resilience,
i.e.,
proportional
limit, the strain-energy density, as calculated by Eq. 3–6
or 3–7, is referred to as the2modulus
resilience, i.e.,
E theof
uthe
= stress s
Modulus
of
Resilience.
In
particular,
when
he proportional limit,
the
strain-energy
density,
as
calculated
by
Eq.
3–6
2
E
or
3–7, is referred
assthe
modulus
ofstress
resilience,
i.e.,
of Resilience.
In to
particular,
when
the
s reaches
Modulus Modulus
oftouModulus
Resilience.
Inof
when
the
stress
s reaches
sparticular,
Modulus
of Resilience.
In when
particular,
when
stress s reaches
plresilience,
Modulus
of
Resilience.
In particular
thesthe
proportional
limit,
the strain-energy
density, as calculated
by
r 3–7, is referred
as
the
modulus
i.e.,
of
Resilience.
In
particular,
the
stress
reaches
r
2
the
proportional
limit,
the
strain-energy
density,
calculated
by Eq. 3–6
theResilience.
proportional
limit,
the
density,
assstrain-energy
calculated
byasdensity,
Eq.
2 by Eq.
s of
In particular,
when
the
reaches
s
thestrain-energy
proportional
limit,
the
calculated
3–6
pl as
1pl
1 s3–6
orby
3–7,
is13–6
referred
to
asproportional
the modulus limit,
of resilience,
i.e.,
2stress
the
the
strain-energy
dens
s
theor
proportional
limit,
the
strain-energy
density,
as
calculated
Eq.
pl
ur
1
s
urby
=
s3–6
(3–8)
3–7, isthe
referred
to as
the
modulus
ofEq.
resilience,
i.e.,
Modulus of
Resilience. In particular
pl as the
1 is referred
1 to
plPplof=resilience,
3–7,
modulus
or 3–7,
is referred
modulus
of
resilience,
i.e.,
tional
limit,
the
density,
as
ur E
= spli.e.,
Ppls2=
(3–8) to as the modulus of resili
2
or
3–7,
is referred
uor
s
Pplthe
= calculated
(3–8)
orstrain-energy
3–7,toisas
referred
toplas
modulus
of
resilience,
i.e.,
s2
r =
pl
pl 2 E
1
1
the
proportional
limit, the strain-energy dens
2
2
2
2 E
eferred to as the modulus of resilience,
i.e.,
ur = splPpl =
(3–8)
1
1 spl ur
P
or
3–7,
is
referred
to as the
2 modulus of resili
2
2
E
ur
ur = splPpl =P
(3–8)
1
1 spl
Ppl
2
2 the elastic2 region
E 21 of the stress–strain
s2pl ur s
1
u
=
s
P
=
2
From
diagram,
Fig.
3–16a,
notice
r
pl pl
1s
1 pl
P
s of Presilience ur
1 2
uFrom
su
(3–8)
2 E
plP
1=spl the
1s
elastic
region
of
diagram,
Fig. 3–16a, 2notice
ur the stress–strain
r1diagram,
plP
pl
pl the elastic
==
(3–8)
1
1s
From
region
of uthe
3–16a,
Modulus
of resilience
u stress–strain
pl =notice
equivalent
triangular
area under
the
diagram.
(3–8)
2plE
s2plthe
Pplr shaded
=Fig.
(3–8)
r u=rr1isss
plplPplur= = to
2
2
E
1 that
u
=
s
P
=
(a)
r
pl
pl
From
the
elastic
region
of
the
stress–strain
diagram,
Fig.
3–16a,
notice
that
u
is
equivalent
to
the
shaded
triangular
area
under
the
diagram.
2
2
E
2
2
E
1
1s
that
ur isofequivalent
to
area
under
the diagram.
Modulus
resilience
r
P ability of the material to
ur = ur(a)
splthe
Ppl shaded
=
(3–8)
Physically
atriangular
material’s
represents
the
Ppl resilience
ur = 2 splPpl = 2
rom
the
elastic
region
of
the
stress–strain
diagram,
Fig.
3–16a,
notice
P
2
2
E
that
u
is
equivalent
to
the
shaded
triangular
area
under
the
diagram.
P
Physically
a
material’s
resilience
represents
the
ability
of
the
material
to
r
Physically
a material’s resilience
represents
the ability
of the material
to to the material.
2
2
Fig.
3–16 P(a)
absorb
energy
without
anyunder
permanent
damage
From
the
elastic
region
of thetostress–strain diagram,
Fig. 3–16
hat ur is equivalent
to3–16
the
shaded
triangular
area
the udiagram.
Modulus
of resilience
Physically
athe
material’s
resilience
the
ability
the
material
r any
Fig.
absorb
energy
without
permanent
damage
toPofthe
material.
absorb
energy
without
any
permanent
damage
to
the
material.
Ppl represents
From
the
elastic
region
of
stress–strain
diagram,
Fig.
3–16a,
notice
From
the
elastic
region
of
the
stress–strain
diagram,
Fig.
3–16a,
notice
ence
u
that unotice
is the
equivalent
to the shaded triangular area under the d
lus
of resilience
urFrom the
r
P
rto
hysically
a material’s
resilience
represents
the ability
of the
to
3–16
region
ofenergy
stress–strain
diagram,
Fig.
3–16a,
absorb
without
anymaterial
permanent
damage
material.
(a)
Parea
the
elastic
region
of
the
stress–strain
diagram,
Fig.
3–16a,
notice
e ur FromFig.
pl
that
ur elastic
isthat
equivalent
to
the shaded
triangular
under
theunder
diagram.
From
the elastic
region of
stress–strain
ur is
equivalent
to
the
shaded
triangular
area
the
diagram.
Physically
a
material’s
resilience
represents
thethe
ability
of the ma
Modulus
of
resilience
u
(a) energy
r
elastic
region
of
the
stress–strain
diagram,
Fig.
3–16a,
notice
bsorb
without
any
permanent
damage
to
the
material.
thatPhysically
ur is to
equivalent
to triangular
the
shaded
triangular
area
under
the
diagram.
From
the
elastic region
of
the
stress–strain
that ur is equivalent
the
shaded
area
under
the
diagram.
3
!
4
Modulus
ofof
resilience
uof
a material’s
resilience
represents
the
ability
the
material
Physically
a material’s
resilience
represents
the
ability
the to
material
toany
r energy
that
u
is
equivalent
to
the
shaded
triangul
Fig.
3–16
absorb
without
permanent
damage
to
the
material.
r
equivalent to the
shaded
triangular
area
under
the
diagram.
(a)
a material’s
resilience
represents
the material to
that ur is equivalent to the shaded triangul
Physically
a Physically
material’s
resilience
represents
the
ability
of the
the ability
material
to
Fig.
3–16
absorb energy
any
permanent
damage
to
theofmaterial.
absorbwithout
energy
without
any
permanent
damage
to the material.
(a)
Physically
a
material’s
resilience
represents
t
a material’s resilience
represents
the ability
of the material
toto the material.
absorb
energy
anydamage
permanent
damage
Physically a material’s resilience represents t
absorb energy
without
any without
permanent
to the
material.
Fig.
3–16
absorb
energy
without
any
permanent
dama
ergy without any permanent damage to the material.
Fig. 3–16
absorb energy without any permanent dama
3.5
3.5
3.5
3.5
3.5
3.5 3.5
3.5
3.5
3.5
3.5
3.5
3
MECHANICAL PROPERTIES
OF
M AT E R I A L S
T I E S O F M AT E R I A L S
A N I C A L P R O P E R T I E S O F M AT E R I A L S
OHE
PAC
EPR
E
M
IA
LI ESASL OPFR O
R SI C
3O
C
N
E RETRI EI A
S LO
AAPNTIECRA L3 P RC
M
HTTAEI N
AF
LM
P ERAT
OH
PEA
ERR
TI C
MPAT
SF
A L P R O P E R T I E S O F M AT E R I A L S
CHAPTER 3
96
CHAPTER 3
MECHANICAL PROPERTIES
OF
M AT E R I A L S
M AT E R I A L S
HAPTER 3
M E C H A N I C A L P R O P E R T I E SEXAMPLE
O F M AT E R I A L S
M E C9H6A N I C A L P RC
O P E R T I E S O F M AT E R I A L S
3.3
An aluminum rod shown in Fig. 3–20a has a circular cross section and is
XAMPLE
3.3
3.3
An aluminum rod shown in Fig. 3
subjected to an axial load of 10 kN. If a portion of the
stress–strain
s (MPa)
minum rod
in Fig.
3–20a
hasinain
circular
cross
section
andcross
is approximate
subjected to an axial load of 10
Anshown
aluminum
rod
shown
Fig.
3–20a
hasdetermine
a circular
section and is
diagram
is
shown
Fig.
3–20b,
the
elongation
EXAMPLE
3.3
An
aluminum
rod
in
3–20a
has
a circular
An
aluminum
rod
shown
in
Fig.shown
3–20a
hasFig.
a circular
cross
sectioncross
and issection and is
! oftothe
LE
3.3
An
aluminum
shown
in
Fig.
3–20a
has
a
circular
cross
section
and
is
ed sto(MPa)
ansubjected
axial
load
ofanrod
10
kN.
If
a
portion
of
the
stress–strain
diagram is shown in Fig. 3–20b,
axial
10 kN.
a portion
stress–strain
rod
when
theof
load
is
applied.
Take
Eof
=the
70
GPa.
An aluminum rod shown
inload
Fig.
3–20a
has
aIf
circular
cross
section
and
is If
al
subjected
to
an
axial
load
of
10
kN.
portion
of theCstress–strain
subjected
to
an10
axial
load
of
10
kN.
If
a stress–strain
portion
of athe
stress–strain
subjected
to
an
axial
load
of
kN.
If
a
portion
of
the
m isFshown
in
Fig.
3–20b,
determine
the
approximate
elongation
96
H
A
P
T
E
R
3
E Crod
Hsection
AN
I C Aand
Lthe
P is
Rload
O P Eis
RT
IES
60
ofM
the
when
applie
An
aluminum
rod
shown
in
Fig.
3–20a
has
a
circular
cross
diagram
in
Fig.of3–20b,
determine
the
approximate
elongation
56.6
subjected
to is
anshown
axial
load
10
kN.
If
aFig.
portion
of
the
stress–strain
s diagram
(MPa)
Dairesel
kesitli
bir
aluminyum
çubuk
103–20b,
kNhas
eksenel
yüke
maruzdur.
için
şekilde
F Aluminyum
isshown
shown
in
Fig.
determine
approximate
elongation
is
shown
in
3–20b,
determine
the
elongation
An
rod
in70
Fig.
3–20a
aapproximate
circular
cross
section
and
is a portion of
50axial the
diagram
shown
in
Fig.aluminum
3–20b,
determine
the
approximate
elongation
od whenof
the
load
iswhen
applied.
Take
E
=diagram
70
GPa.
al
20
mm
subjected
to
an
load
of
10
kN.
If
the
stress–strain
the
rod
the
load
is
applied.
Take
E
=
GPa.
15
diagram is shown in of
Fig.
3–20b,
determine
the
approximate
elongation
oftothe
rod
when
is
applied.
Take
Ealof=the
70 stress–strain
GPa.
the
rodiswhen
the
load
isalload
applied.
Take
Emm
70
GPa.
sYportion
! 40
56.6 60
al =
subjected
anTake
axial
ofload
10
kN.
If
2
of the rodverilen
when
load
E
=the
GPa.
F the
al
diagram
is aCshown
ingöre,
Fig. 3–20b,
determine
thealtında
approximate
elongation
A applied.
B70
gerilme-birim
deformasyon
ilişkisi
verildiğine
verilen
yükleme
meydana
50of theFrod when
the
load
is
applied.
Take
E
=
70
GPa.
al
M EPC H!A N
I C A L P R O P E R T I20
E Smm
Odiagram
F M AT E Ris
I A15
Lshown
S mm in Fig. 3–20b, determine the30approximate elongation
A
0.0450
20
mm
60
of
the
rod
when
the
load
is
applied.
Take
E
=
70
GPa.
15
mm
Y ! 40BC
10
kN
56.6
al
20 mm
20 mm
20 10 kN
15 mm
EXAMPLE
3.3
15 mm
F
PBC ! 0.0450
20 mm
A
10 kN
15 mm Take
50
ofB the rod
when
the
is
applied.
70 GPa. değerini kullanınız.
C
30
A yaklaşık
B load
20
mm
CB 3
gelecek
uzama
belirleyiniz.
! Eal =10
B
15 A
mm
A miktarını
C
F
C
s ! 40
10 kN 400
B
.0450
600 mm
20 mm
04 20 0.06 P10
C mm
15 mm
0.0450
10 kN O
! 0.0450AY PA
BC !
BCkN
10BkN
10 kN
10
kN
10
kN
C
0.02
0.04
0.06
30
0.045010
20 mm
10 kN
10 kNA15 mm
s
(MPa)
B
(a)
C
10 kN
10 kN
600 mm 20
O 0.04
600400
mmmm
400
mm
0.0450
600kN
mm
400
mm (b)
mm
0.02
0.04
0.06
02
0.06
APBC !600
B400 mm
10
C
3
10
600
mm in Fig. Fig.
400 mm
An aluminum
shown
3–20a
has a circular cross section and is
6
(a) 600rod
P(b)
(a) 400 mm3–20 (a)
mm
BC ! 0.0450
(a)
10
kN
10 kN
(b)
subjectedO to an 0.02
axial load
of
10
kN.
If a portion of the stress–strain
600 mm
400 mm
0.04
0.06
(a)
10 kN
(a)
600 mm
An aluminu
subjected to
diagram isFigs
of the rod w
Fig.
56.6 60 (a)
Fig.3–20b,
3–20 determine
Fig.400
3–20
diagram
is 3–20
shown in(b)
Fig.
the approximate
600Fig.
mm3–20
mmelongation
0.06
F
SOLUTION
50
SOLUTION
Fig.
3–20
of the rod when theFig.
load
is applied.
Take Eal = 70 GPa.
3–20
(a)
For
the
analysis
we
will
neglect
the
localized
deformations
at
the
point
(b)
F
sY ! 40Fig. 3–20
For the analysis we will neglect t
ION SOLUTION
SOLUTION
of loadSOLUTION
application and where20the
rod’s
cross-sectional
area
suddenly
30
mm
15 mm
analysis
we the
will analysis
neglect the
localized
deformations
atdeformations
thethe
point
For
the
analysis
we
will
neglect
deformations
Forwill
the neglect
analysis
we
will
neglect
localized
deformations
at the point at the point of load application and where t
SOLUTION
Fig.
3–20
For
we
the
localized
atthe
thelocalized
point
SOLUTION
changes.
(These
effects
will
be
discussed
in
Sections
4.1
and
4.7.)20
( These effects will be
Bdeformations
SOLUTION
ofthe
load
application
and
where
cross-sectional
area suddenly changes.
application
and
where
rod’s
cross-sectional
area
suddenly
of
and
where
the
rod’s
cross-sectional
area suddenly
PBC ! 0.0450
Cthe
Forload
the application
analysis
weload
willAapplication
neglect
the
localized
at
therod’s
point
of
and
where
rod’s
cross-sectional
suddenly
10 kN of
the
analysis
we the
will
neglect
the
localized
deformations
atarea
the
point
3
Throughout
the
midsection
of
each
segment
the
normal
stress
and
Throughout
thepoint
midsection
! 0.0450
PBCFor
10
10
kNbe
kN in
changes.
( These
effects
will
bein
discussed
Sections
4.1 and 4.7.)
For
the
analysis
we10will
neglect
localized
deformations at the
changes.
( These
effects
willinand
be
discussed
Sections
4.1
and the
4.7.)
s. (These
discussed
in
Sections
4.1
4.7.)
ofeffects
load
application
and
where
the
rod’s
cross-sectional
area
suddenly
changes.
(will
These
effects
will
be
discussed
Sections
4.1
and
4.7.)
of load
application
and
where
the
rod’s
cross-sectional
area
suddenly
SOLUTION
deformation
are uniform.
deformation
are uniform.
Throughout
the
midsection
ofstress
each
segment
theOnormal
stress and
the
midsection
each
segment
the4.7.)
normal
stress
and
of
load
application
and where
the
rod’s
area suddenly
houtchanges.
theThroughout
midsection
of Throughout
each
segment
the
normal
stress
and
changes.
(These
effects
will
be
discussed
inof
Sections
4.1
and
the
midsection
of
each
the
normal
and
0.02cross-sectional
0.04 In order
0.06
( These
will
beanalysis
discussed
in Sections
4.1we
and
4.7.)
For
the
wesegment
will
neglect
the
localized
deformations
at the point
600
mm
400
mm
0.06
to
the elongatio
Ineffects
order
to
find
the
elongation
of
the
rod,
must
first
obtain
the
!
!
deformation
are
uniform.
are
uniform.
(stress
These and
effects will be discussed in Sections 4.1 andfind
4.7.)
ationThroughout
aredeformation
uniform.
Throughout
thedeformation
midsection
of each
segment
thechanges.
normal
are
uniform.
the
midsection
of application
each
segment
the
normal
stress
and
of
load
and
where
the
rod’s
cross-sectional
area
suddenly
strain.
This
is
done
by calculating
(b)
strain.
This
isorder
done
byfind
calculating
the
stress,
then
using
the
stress–strain
(a)
In
order
to
find
the
elongation
of
the
rod,
we
must
first
obtain
the
In
to
the
elongation
of
the
rod,
we
must
first
obtain
the
Throughout
the the
midsection of each segment the normal stress and
rder deformation
to find
elongation
of
the
rod, we must
first
obtain
the first obtain
Inthe
order
touniform.
finduniform.
the
elongation
of
the
rod,
we must
deformation
are
are
changes.
(These
effects
will
be
discussed
in
Sections
4.1
and
4.7.)
diagram.
The
normal
stress within
Çözüm:
Çubuk
farklı
iki
kesit
bölgesinde
oluşmaktadır.
Herbir
bölge
için
önce
gerilmeleri
ve
diagram.The
normal
stress
within
each
segment
is
strain.
This
is
done
by
calculating
the
stress,
then
using
the
stress–strain
strain.
This
is
done
by
calculating
the
stress,
then
using
the
stress–strain
deformation
are uniform.
This is done
calculating
the
stress,
then
using
the3–20
stress–strain
This
istodone
by
calculating
the
then
the
stress–strain
Inbyorder
find
the
elongation
of stress,
the
rod,
weusing
must
first
obtain
the
Fig.
Instrain.
order
to find
the
elongation
of
the
rod,
we
must
first
obtain
the
Throughout
the
midsection
of
each
segment
the
normal
stress
and
The normal
stress
withinto
each
is
diagram.
The diagram.
normal
stress
within each
is segment
Insegment
order
the elongation
of the rod, we must first obtain the
m.The
normal
stress
within
each
segment
is each
diagram.
stress
within
segment
is using
strain.
This
isnormal
done
by
calculating
the
stress,
the
stress–strain
buna
bağlı
olarak
ve
diyagramı
kullanarak
birimfind
deformasyonları
hesaplamalıyız. Buradan
strain.
This
isThe
done
by
calculating
the
stress,
then3then
using
the stress–strain
deformation
are
uniform.
101
strain.
This
is
done
by
calculating
the
stress, then using the stress–strain P
2 Nis
10110
P each
diagram.The
normal
within
segment
SOLUTION
diagram.
The
normal
stressstress
within
each
segment
is
sABSOLUTION
=
=
InsAB
order
to =
find the elongation
of
the
rod,
we
must
first
obtain
the
3normal
3 31.83
=
=
MPa
diagram.
The
stress
within
each
segment
is
2Pdeformations
N 10110 2 N
10110
deplasmana
A
2
3
p10.
3 p10.01
analysis
we This
will
neglect
the
localized
at the
point
A
2strain.
NPgeçebiliriz.
10110
2=N P
10110
PFor the
sm2
= 31.83
=
isABdone
by=calculating
the
using
theMPa
stress–strain
s
=stress,
31.83then
MPa
AB =
2
2
For
the
ana
sAB = of =loadsapplication
=
31.83
MPa
A
= rod’s
31.83
MPa
m2 suddenly
3
p10.01
m2 p10.01area
the
cross-sectional
AB = 2 = and
3 where
2 N2A
10110
diagram.The
normal
stress
2N
P 10110
A
A
3
p10.01Pm2
p10.01
m2
10
3 within each segment is
P
changes.
( These
be
discussed
in
Sections
4.1 and P4.7.) 10110 2 N
10110
2
N
== effects
=
31.83
MPa
=
Pwill
sAB s=AB
=
31.83
MPa
sBCof
= load
= app
3
2
3
2
s
=
= 31.83 MPa
=
A
10110
2
N
p10.01
m2
10110
2
N
s
=
=
=
56.59
MPa
AB
A
P
p10.01
m2
A
P
2
BC
p10.
Throughout
midsection
segment =the =normal
stress
and p10.01
changes. (T
=A56.59
MPa m2
10110
2of
N each
101103the
2N
sABC 3=
= sBCm2=2
56.59 MPa
p10.0075
P
P
3 2
2
2
N
10110
A
A
p10.0075
m2
P
p10.0075
m2
=
=
56.59
MPa
sBC = deformation
= sBC =are uniform.
=
56.59
MPa
3
Throughou
3
10110
2m2
N=2 of the
2 NsAB
= 31.83
MPa the 1011032 N
= rod, we must
A
P 210110
A In p10.0075
p10.0075
Pto find
m2
2
order
the
elongation
first
obtain
P
From
the
stress–strain
diagr
A
s
=
=
=
56.59
MPa
p10.01 m2
sBC =BC =
= 56.59 MPa
deformatio
2 m22
sstress–strain
= is
= 56.59 MPa
From
stress–strain
the material
inthe
segment
BC = AB
A
A
strain.
Thisthe
is done
byp10.0075
calculating
the
stress,
then
using
p10.0075
m2diagram,
2 AB is strained elastically since s
From
the
stress–strain
diagram,
the
material
in
segment
From
the
stress–strain
diagram,
the
material
in
segment
AB
is
A
AB 6
p10.0075 m2
! strainedThe
In order
elastically
since
law, Using
6 each
sY =segment
40
MPa.
s
3 Using Hooke’s
normal
stress
within
is
AB
Fromdiagram.
thediagram,
stress–strain
diagram,
in
segment
AB
is
m the stress–strain
the elastically
material
inthe
segment
AB
is
strained
elastically
since
Hooke’s
law,
6
s
=
40
MPa.
s
10110
2
N
strained
since
Using
Hooke’s
law,
6
s
=
40
MPa.
smaterial
P
AB
Y
AB
Y
strain. This
sBC
= in
= AB
56.59
MPa
strained
Using
Hooke’s
6 Using
sthe
40= MPa.
sMPa.
d elastically
since
Hooke’s
law,
6 sYsince
= 40
AB
Y =
2 law,
AB stress–strain
From
diagram,
the
material
inFrom
segment
iskullanarak:
From
the elastically
stress–strain
diagram,
material
segment
AB
is
!sthe
olduğuna
göre,
Hooke
kanununu
sAB
A
p10.0075
m2
the
stress–strain
diagram,
the
material
in
segment
AB
is 31.8311
6
3 2 Pa 6
6 law,
PAB = diagram.Th
=
2Using
NsAB
10110
sAB
strained
elastically
Using
Hooke’s
6 =s
=
40
MPa.
strained
elastically
sincesince
Hooke’s
law,
sY
40
MPa.
sAB s6
P
31.83110
2
Pa
31.83110
2
Pa
AB
Y
s31.83110
strained
elastically
since sAB 6 sY = 40 MPa. Using Hooke’s law,Eal
AB
701109
=
=
0.0004547
mm>mm
=
=
31.83
MPa
=
6
P
=
=
=
0.0004547
mm>mm
P
=
=
=
0.0004547
mm>mm
AB
6 PABs=
AB
9
AB A 2 Pa
2 9
9
31.83110
s
31.83110
2
Pa
E
sAB
70110
Pa Ediagram,
p10.01270110
m2
AB
al
Pa
al2 Pa 70110
al
the E
stress–strain
the 2material
in segment AB is
=From
= 0.0004547
mm>mm
PAB =
= PAB = 9
= 0.0004547
6 9 6 mm>mm
6
31.83110
2
Pa
EPa
31.83110
2
Pa
sAB 2s
70110
2
Pa
Eal
al
70110
31.83110law,
2 Pa
AB
The material within segme
s
strained
elastically
since
Using
Hooke’s
6
s
=
40
MPa.
s
3
AB
AB
Y
PAB! P=AB = = =
0.0004547
2 N mm>mm
= 0.0004547
mm>mm
P 9 = 10110
P
=
=
= 0.0004547 mm>mm
9
AB
material
within
segment
BCE is
strained
9plastically, since sBC 7 sY = 40 MPa. From the
Eal material
material
segment
is MPa
strained
plastically,
since
The
within
segment
BC
is= BC
strained
plastically,
since
EThe
sBC
=2 PaThe
56.59
70110
2=within
Pa
al 70110
70110 2 Pa
al
2
m2
From
the
graph,
for PsMPa,
sA
7p10.0075
splastically,
= graph,
40
MPa.
= Pdeğeri
56.59
From
the
graph,
sisYsegment
=
40
MPa.
=MPa,
56.59
From
the
for
swithin
=7 40
MPa.
sBCfor
= !s
56.59
BC
Y
BC L
BC
BC since
BC L MPa,
The material
BC
is
strained
BC 7 s
Y BC
BC
6plastically,
material within
strained
since
! ssegment
olduğuna
göre,
grafikten
için! PBC L ! 0.045 mm>mm. The approxima
31.83110
2
Pa
s
AB
0.045
mm>mm
.
The
approximate
elongation
of
the
rod
is
therefore
0.045
mm>mm
.
The
approximate
elongation
of
the
rod
is
therefore
0.045
. The
approximate
elongation
ofMPa,
the
rod
isLtherefore
From
graph,
for
PBC
7 From
smaterial
= within
40mm>mm
MPa.
sBCPplastically,
= 56.59
the
graph,
for
sY = The
40sMPa.
s
= BC
56.59
Lplastically,
BC
Y
Pthe
= since
0.0004547
mm>mm
material
segment
BC
is strained
BC
BC9The
AB
The
within
segment
is=MPa,
strained
since
material
within
segment
BC
is strained plastically, since
E
70110
2 rod
Pa isinPbölgedeki
Çubuktaki
uzama
miktarı
herbir
uzama
miktarlarının
toplamı olarak
d = ©PL = 0.0004547
al is
From
theFrom
stress–strain
diagram,
the
material
segment
AB
is
. The
approximate
elongation
of56.59
the
therefore
mm>mm
The
elongation
of
the
rod
therefore
the
graph,
for
sBC. 0.045
7BCsapproximate
=solacaktır.
40
MPa.
s
=
MPa,
L
From
the
graph,
for
P
s
7mm>mm
=
40
MPa.
s
=
56.59
MPa,
L
Y
BC
BC
From
the
graph,
PBC L
s
7
s
=
40
MPa.
Y
BC= 0.00045471600
BC
BCmm2 Y
=Y ©PL
mm2
+ mm2
0.04501400 mm2 for sBC = 56.59 MPa,
dsince
==©PL
=6d
0.00045471600
+ Hooke’s
0.04501400
strained
elastically
Using
law,
s
=
40
MPa.
s
AB
d
=
©PL
0.00045471600
mm2
+
0.04501400
mm2
= 18.3 mm
0.0450.045
mm>mm
.
The
approximate
elongation
of
the
rod
is
therefore
mm>mm
.
The
approximate
elongation
of
the
rod
is
therefore
0.045
mm>mm
.
The
approximate
elongation
of
the
rod
is
therefore
aşağıdaki
gibi
hesaplanır.
=mm2
18.3
mm
Ans.
=
18.3 mm
d = ©PL ==mm2
0.00045471600
mm2
+ 0.04501400
mm2
The
material
within
segment
BC
is strained plastically,
since
d = ©PL = 0.00045471600
0.04501400
18.3 +
mm
Ans. Ans.
6
31.83110
Pa
From
the
graph,
sBC
40 MPa.
sAns.
= 0.00045471600
56.59 MPa, PBCmm2
L + 0.04501400 mm2
s7ABsY =
== 18.3
mm
d = d©PL
= 0.00045471600
mm2
+ 20.04501400
mm2
BC =
From the
d for
= ©PL
©PL
mm2
+ 0.04501400
mm2
= 18.3 mm
Ans.
PAB=0.045
=0.00045471600
=
= 0.0004547
mm>mm
9
mm>mm
.
The
approximate
elongation
of
the
rod
is
therefore
E
70110
2
Pa
= 18.3
mm
Ans.
strained
ela
al
=
18.3
mm
Ans.
= 18.3 mm
Ans.
0.02
0.04
!
d = segment
©PL = 0.00045471600
mm2
+ 0.04501400
The material within
BC is strained
plastically,
sincemm2
= 18.3the
mm
graph, for sBC = 56.59 MPa, PBC L
sBC 7 sY = 40 MPa. From
0.045 mm>mm. The approximate elongation of the rod is therefore
d = ©PL = 0.00045471600 mm2 + 0.04501400 mm2
= 18.3 mm
Ans.
PA
The mat
sBC 7 sY
0.045 mm>m
Ans.
d =
=
!35
amount d, and its radius contracts by an amount d¿. Strains in the
3 lateral or radial direction are,
longitudinal or axial direction and in the
respectively,
102
C H A P T E R 3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S
d
d¿
and Plat =
Plong =
r
L
Consider a bar having an original radius r and length
to the tensile force P in Fig. 3–21. This force elongate
amount d, and its radius contracts by an amount d¿
longitudinal or axial direction and in the lateral or rad
respectively,
3.6that within
Poisson’s
Ratio
In the early 1800s,
the French
scientist S. D. Poisson realized
the
Poisson
Oranı
Plong =
d
L
and Plat =
d¿
r
elastic range the ratio of these strains is a constant, since the deformations
In the earlyto1800s,
the French
S. D.
Poisson realize
a deformable
an axial
tensile scientist
force, not
only
as Poisson’s
ratio, body is subjected
d and d¿ are proportional. This constant is referred toWhen
elastic
range
the
ratio
of
these
strains
is
a
constant,
since t
it elongate
but it also contracts laterally. For example, if a rubber
particular
material
n (nu), and it has a numerical value that is unique for adoes
and
are
proportional.
This
constant
is
referred
to as
d
d¿
band is stretched,
it can be noted that both the thickness and width of the
that is both homogeneous and isotropic. Stated mathematically
it is
(nu),
and
it
has
a
numerical
value
that
is
unique
for
a pa
n
band are decreased. Likewise, a compressive force acting on a body causes
that
is
both
homogeneous
and
isotropic.
Stated
mathema
it to contract in the direction of the force and yet its sides expand laterally.
Plat
(3–9)
n = Consider a bar
having an original radius r and length L and subjected
Plong
Plat
to the tensile force P in Fig. 3–21. This force elongates the
n =bar
- by an
3
amount d,(positive
and its radius contracts by an amount d¿. Strains Pinlongthe
The negative sign is included here since longitudinal elongation
longitudinal
or axial direction and in the lateral or radial direction are,
strain) causes lateral contraction (negative strain), and
vice versa. Notice
The negative sign is included here since longitudinal elon
respectively,
that these strains are caused only by the axial or longitudinal force P; i.e.,
strain) causes lateral contraction (negative strain), and v
no force or stress acts in a lateral direction in order to strain the material
d
d¿
and strains
Plat =are caused only by the axial or longitud
Plong =that these
in this direction.!
!
r in a lateral direction in order to st
noLforce or stress acts
Poisson’s ratio is a dimensionless quantity, and for most nonporous
in
this
direction.
1 altında
1 thebir
In
early
1800s,
the
French uzama/kısalmanın
scientist S. D. Poisson realized
thatenine
within the
Eksenel
veya
basınç
cisimde
eksenel
yanında
When 4 the
block
is compressed
solids it has a value
that isçekme
generally
between
and rubber
values
of
3 . Typical
Poisson’s
ratio is asince
dimensionless
quantity, and for
elastic
range
the
ratio
of
these
strains
is
a
constant,
the
deformations
(negative
strain)
its
sides
will
expand
1
1
n for common engineering materials are listed on the inside back cover.
solids
it
has
a
value
that
is
generally
between
4 and 3 . T
doğrultuda
daralma/genişleme
meydana
gelecektir.
(positive
strain).
The
ratio
of
these
strains
and
are
proportional.
This
constant
is
referred
to
as
Poisson’s
ratio,
d
d¿
For an “ideal material” having no lateral deformation when it is stretched
for common
engineering
materials
are listed on the in
remains constant.
and
it has ainnumerical nvalue
that is unique
for a particular
material
nit(nu),
or compressed Poisson’s
ratio
will
be
0.
Furthermore,
will
be
shown
For
an
“ideal
material”
having
no
lateral
Orijinal
yarıçapı r ve uzunluğu Lthat
olan
ve eksenel
çekme etkisine
maruz
birmathematically
çubuk gözönüne
isATboth
isotropic.
Stated
it is deformation wh
ERST
IO
E SFCO
M
FA
AT
MTEthat
AT
E3
LRSIthe
A LM
102
CS HEmaximum
A
R IC
3 A L MP
ERCOHPAENRITCIAE
LS PORFO M
Pfor
E AT
R T IPoisson’s
OLFS M
E
R I A Lhomogeneous
Sis 0.5. C H A P Tand
102
ER 3
MECHAN
I C A L P R O P E R T I E S O F M AT E R I A L S
Sec.
10.6
possible
value
ratio
H
P
ERRI A
CPHTAEN
EERSI A
or compressed
Poisson’s ratio will be 0. Furthermore, it
Therefore 0 … nalalım.
… 0.5.
Sec. 10.6 Pthat
the maximum possible value for Poisso
lat
O F M AT E R I A L S
(3–9)
n = - 0 … n … 0.5.
Therefore
d/2
Plong
Poisson’s
Poisson’s
Ratio
Ratio
6
3.6 Poisson’s
3.6 Poisson’s
Ratio Ratio
3.6 Poisson’s
Ratio
d/2
P
Ltensile
The
negative
included
here
since
longitudinal
(positive
ndeformable
a deformable
body
body
is subjected
is subjected
toWhen
an
to axial
an aaxial
tensile
force,
notnot
only
d/2force,
When
a deformable
bodysign
is subjected
to tensile
an
axial
tensile
force,
not elongation
onlydeformable
deformable
body
isonly
subjected
toisan
axial
force,
not
only
oisson’s
Ratio
When
body is subjected
P avice
strain)
causes
lateral
contraction
(negative
strain),
and
versa.
Notice
it
longate
elongate
butbut
it also
it also
contracts
contracts
laterally.
laterally.
For
example,
example,
ifit aalso
ifrubber
a but
rubber
does
itbut
elongate
it also contracts
For example,
ifdoes
a rubber
L
does
itFor
elongate
contracts
laterally. laterally.
For example,
if a rubber
it
elongate
but
it also d/2
contracts
that
these
strains
arethat
caused
by
the
axial
longitudinal
force P; i.e.,
band
isitand
stretched,
it
be noted
bothonly
the thickness
andofor
width
of the
tretched,
is stretched,
it can
it can
besubjected
noted
be
noted
that
that
both
thethe
thickness
thickness
and
width
width
of
the
ofcan
the
band
istensile
stretched,
can
be
noted
that
both
the
thickness
and
width
the
Original
Shape
ormable
body
is
to
anboth
axial
force,
not
only
Final
Shape
band
is
stretched,
it
can
be
noted
that
no
force
or
stress aacts
in a lateral
direction
to strain the material
band
decreased.
compressive
force
onin
aorder
body causes
are
decreased.
decreased.
Likewise,
a compressive
a compressive
force
force
acting
acting
onare
body
arubber
body
causes
causes
band
are
decreased.
aLikewise,
compressive
force
acting
onacting
a body
causes
gate
but it Likewise,
also
contracts
laterally.
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example,
ifaon
aLikewise,
Original
Shape
band
are
decreased.
Likewise,
a
compr
Final
inof
this
direction.
toexpand
contract
in
the
direction
ofand
the force
yetexpand
its sides laterally.
expand laterally.
ontract
ract
initthe
in
the
direction
the
of the
force
force
and
and
yet
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expand
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laterally.
it to
contract
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the
direction
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the force
yet itsand
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ched,
can
bedirection
notedofthat
both
the
thickness
width
the
it to contract
in
the
direction
of
the
for
When the rubber block is compressed
PPoisson’s ratio is a dimensionless quantity, and
for most nonporous
Tension
Consider
a bar
having anradius
original
radius
r andLlength
L and subjected
d¿subjected
aexpand
bar
having
an
original
r and
length
and subjected
der
nsider
a bar
a Likewise,
bar
having
having
ana
original
an original
radius
radius
rConsider
and
r and
length
length
and
L and
subjected
creased.
aLbody
causes
r an original r
! compressive
(negative
strain) itsforce
sidesacting
will on
1 Consider
1
a
bar
having
P
solids
itby
has
value
that
is
generally
between
andby3.an
Typical values
of
to
the
tensile
force
Pan
ina Fig.
3–21.
This
force elongates
to
the
tensile
force
Pbar
inbar
Fig.
This
force
elongates
the
bar the
by 4bar
an
etnsile
tensile
force
force
P (positive
in
Pof
Fig.
inthe
Fig.
3–21.
3–21.
This
This
force
force
elongates
the
the
by
an3–21.
in
the
direction
force
and
yet
itsofelongates
sides
expand
laterally.
Tension
strain).
The
ratio
these
strains
d¿ T
3
to the
tensileback
force
P in Fig. 3–21.
n
for
common
engineering
materials
are
listed
on
the
inside
cover.
amount
,
and
its
radius
contracts
by
an
amount
Strains
in
the
d
d¿.
amount
,
and
its
radius
contracts
by
an
amount
Strains
in
the
d
d¿.
nt
,
and
,
and
its
its
radius
radius
contracts
contracts
by
by
an
an
amount
amount
Strains
Strains
in
the
in
the
d
d
d¿.
d¿.
Fig.
3–21
remains
constant.
a bar having an
original
radius
r and
lengthaltında
L and subjected
Çubuk,
eksenel
kuvvet
! kadar
kadar kısalacaktır.
Buna
göre
3 uzayacak ve yarıçapı !
amount
its radius contracts
d, and
longitudinal
or an
axial
direction
in
theorlateral
ordirection
radial
direction
are,
For
“ideal
material”
having
no lateral
deformation
when
it is stretched
orthe
axial
and
in theand
lateral
radial
are,
nal
udinal
or axial
orPaxial
direction
and
and
in the
in longitudinal
the
lateral
lateral
or radial
or
radial
direction
are,
are,
le
force
indirection
Fig.
3–21.
This
force
elongates
bardirection
by
an
Fig.
3–21direction and in
longitudinal
or
axial
respectively,
boyuna ve
birim d¿.
deformasyonlar
ctively,
ely, its radius contracts
and
by enine
an respectively,
amount
Strains or
in compressed
the sırasıylaPoisson’s ratio will be 0. Furthermore, it will be shown in
respectively,
Sec.are,
10.6 that the dmaximum possible
value for
Poisson’s ratio is 0.5.
or axial direction andd indthe lateral or
direction
d¿
d
d¿
d¿ radial
d¿
and
Plat =
and
=
Plong = 0Plong
Therefore
… n= …Plat
0.5.
= = andandPlat P=lat =
PlongPlong
d
r
L
r
L
r olacaktır.
r
L L ve !
and
Plong =
!
L
d/2
d
d¿
In
thethat
early
1800s,
the
FrenchS.scientist
S. D.realized
Poisson that
realized
thatthe
within the
early
1800s,
thethat
French
scientist
D. Poisson
within
rly
early
1800s,
1800s,
the
the
French
French
scientist
S.Plat
D.S.In
Poisson
D.the
Poisson
realized
realized
within
within
the
the
= scientist
and
=
Plong
1800
lü
yılların
Fransız
bilimadamı
S.D.
Poisson,
elastik
bölgede
bu1800s,
birim
r başlarında
L
elastic
range
the
ratio
of
these
strains
is
a
constant,
since
the
deformations
In
the early
the French scientist S
elastic
range
the
ratio
of
these
strains
is
a
constant,
since
the
deformations
cnge
range
thethe
ratio
ratio
of these
of these
strains
strains
is aisconstant,
a constant,
since
since
thethe
deformations
deformations
P
L
and
are
proportional.
This
constant
is
referred
to
as
Poisson’s
ratio,
d
d¿
elastic
range
the
ratio
of these strains i
and
are
proportional.
This
constant
is
referred
to
as
Poisson’s
ratio,
d
d¿
d/2
are
are
proportional.
proportional.
This
This
constant
constant
is
referred
is
referred
to
as
to
Poisson’s
as
Poisson’s
ratio,
ratio,
d¿
1800s, the French scientist
S. D. Poisson
realized that
within
the
deformasyonlar
oranının
sabit
olduklarının
farkına
varmıştır.
Çünkü
boyuna
ve enine
(nu),
and
it
has
a
numerical
value
that
is
unique
for
a
particular
material
n
(nu),
and
it
has
a
numerical
value
that
is
unique
for
a
particular
material
n
and
are
proportional.
This consta
d
d¿
nd
),the
and
it has
it
has
a
numerical
a
numerical
value
value
that
that
is
unique
is
unique
for
for
a
particular
a
particular
material
material
ratio of these strains is a constant, since the deformations
that
is both ithomogeneous
andOriginal
isotropic.
Stated mathematically
it(nu),
is and it has a numerical value tha
that
is
both
homogeneous
and
isotropic.
Stated
mathematically
it
is
n
Shape
oth
s proportional.
both
homogeneous
homogeneous
and
and
isotropic.
isotropic.
Stated
Stated
mathematically
mathematically
is
it
is
This
constant
is
referred
to
as
Poisson’s
ratio,
Final
Shape
deformasyonlar orantılıdır. Bu sabit Poisson oranı olarak adlandırılmış ve !
sembolü
ile
that is both homogeneous and isotrop
t has a numerical value that is unique for a particular material
Plat
r
Plat PHomojen
ve izotropik
malzeme
buPlatoran:
lat mathematically
homogeneous andgösterilmiştir.
isotropic.
it is bir
n = (3–9) (3–9)
n = - için
(3–9)
(3–9)
n =n -=Stated
P
Plong
P
Tension
long
d¿
PlongPlong
n = Plat
(3–9)
n = The
negative
sign ishere
included here since longitudinal
elongation
(positive
The
negative
sign
is included
(positive
Fig. 3–21
Psince
ative
egative
signsign
is included
is included
here
here
longitudinal
longitudinal
elongation
elongation
(positive
(positivesince longitudinal elongation
longsince
strain)
causes
lateral
contraction
(negative
strain),
and
vice
versa.
Notice
strain)
causes
lateral
contraction
(negative
strain),
and
vice
versa.
Notice
!
The negative sign is included here sin
auses
) causes
lateral
lateral
contraction
contraction
(negative
(negative
strain),
strain),
andand
vicevice
versa.
versa.
Notice
Notice
that these
strains are
caused
only
by or
thelongitudinal
axial or longitudinal
force
P; i.e.,
that
these
strains
are
caused
only
by
the
axial
force
P;
i.e.,
e strains
signstrains
is included
here
since
longitudinal
elongation
(positive
hese
areare
caused
caused
only
only
by the
by
the
axial
axial
or longitudinal
or
longitudinal
force
force
P; i.e.,
P; i.e.,
strain) causes lateral contraction (neg
no force
or
stress
acts indirection
a lateral direction
in strain
order to
strain
the material
no
force
or
stress
acts
in
a
lateral
in
order
to
the
material
Buradaki
eksi
işareti,
boyuna
ve
enine
deformasyonların
birbirlerinin
aksi
biçimde,
biri uzama
es
contraction
(negative
strain),
vice
rce
orlateral
stress
or stress
acts
acts
in ainlateral
a lateral
direction
direction
in order
inand
order
to
strain
toversa.
strain
theNotice
the
material
material
that these strains are caused only by th
in this direction.
in this direction.
rains
are
caused
byblock
the axial
or longitudinal force P; i.e.,
she
rection.
direction.
no force
or stress acts in a lateral direc
When
the only
rubber
is compressed
rubber
block
is
(+)compressed
iken
diğerinin
kısalma
(-)is olmasından
kaynaklanmaktadır.
deformasyon
Poisson’s
ratio is a dimensionless
andBurada
for
mostenine
nonporous
Poisson’s
ratio
a dimensionless
quantity, quantity,
and for most
nonporous
stress
acts
in
a
lateral
direction
in
order
toand
strain
the
material
(negative
strain)
its
sides
will
expand
n’s
sson’s
ratio
ratio
is
a
is
dimensionless
a
dimensionless
quantity,
quantity,
and
for
for
most
most
nonporous
nonporous
e strain) its sides will expand
1
1
1 1
in
this
direction.
a value
that is generally
between
and 3. values
Typical of
values of
solids
that
is generally
between
and 3.isTypical
1it has
1 solids
1a value
1 it has
4 compressed
(positive
strain).
ratio of
these
strains
When
the
rubber 4 block
strain).
ratio
of
these
strains
tion.
has
it has
a value
aThe
value
that
that
isoluşurken,
generally
isThe
generally
between
between
Typical
Typical
values
of
of
Poisson’s
ratio is a dimensionless
enine
bir values
kuvvetin
etkimediğine
dikkati
çekmemiz
gereklidir.
4 and
4 and
3 .common
n3.for
engineering
materials
listed
on
theback
inside
back
cover. Poisson
n fordoğrultuda
common
engineering
materials
are
listedare
on
the inside
cover.
remains
constant.
constant.
(negative
strain)
its
sides
will
expand
ratio engineering
is aengineering
dimensionless
quantity,
and
nonporous
mmon
common
materials
materials
areare
listed
listed
onfor
the
onmost
the
inside
inside
back
back
cover.
cover.
solids
it
has
a value that is generally
For
an
“ideal
material”
having
no
lateral
deformation
when
it
is
stretched
For1 an “ideal
material” having(positive
no lateral
deformation
it isstrains
stretched
strain).
The 1/4
ratio when
of
these
oranı
çoğu
boşluksuz
için
ile
1/3
arasında
bir
değerdir.
a“ideal
value
that ishaving
generally
between
andve13. Typical
values
of malzeme
ndeal
material”
material”
having
no boyutsuzdur
lateral
no
lateral
deformation
deformation
when
when
it
is
it
stretched
is
stretched
4
n
for
common
engineering materials
or
compressed
Poisson’s
ratio
will
be
0.
Furthermore,
it
will
be
shown
in
or compressed Poisson’s ratioremains
will beconstant.
0. Furthermore, it will be shown in
on engineering
materials
are
listed
on the inside
back
cover.
ressed
mpressed
Poisson’s
Poisson’s
ratio
ratio
willwill
be 0.
be
Furthermore,
0.
Furthermore,
it will
it
will
be
shown
be shown
in
For isan0.5.
“ideal material” having no later
Sec.
10.6
that
the in
maximum
possible
for Poisson’s
Sec.
10.6 that
the
maximum
possible
value
forvalue
Poisson’s
ratio is ratio
0.5.
l that
material”
having
no lateral
deformation
when
it is stretched
10.6
that
thethe
maximum
maximum
possible
possible
value
value
for for
Poisson’s
Poisson’s
0.5.
or compressed Poisson’s ratio will be
0ratio
… isn 0.5.
…is 0.5.
Therefore
0Therefore
… n … ratio
0.5.
eed
fore
0 Poisson’s
…0 n……n 0.5.
…ratio
0.5. will be 0. Furthermore, it will be shown in
Sec. 10.6 that the maximum possibl
d/2
d/2
at the maximum possible value for Poisson’s ratio is 0.5.
Therefore 0 … n … 0.5.
d/2 d/2
… n … 0.5.
P
P
d/2
L
L
!36
P P d/2
d/2
d/2
L
P
Original
Original
Shape
Shape
L
L
d/2
d/2 d/2
Original Shape
Original Shape
FinalFinal
Shape
Shape
Final Shape
Final Shape
P
3.6
NICAL
PROPERTIES
OF
EXAMPLE
3.4
Kitabınızın arka kapağında
yaygın mühendislik
malzemeleri için bu oranın tipik değerleri
aa
3.6
es of Typical Engineering
Materialsa Ratio
Poisson’s
I Units)
POISSON’S RA
M AT E R I A L S
AverageMechanical
MechanicalProperties
PropertiesofofTypical
TypicalEngineering
EngineeringMaterials
Materials
Average
verilmiştir.
(SIUnits)
Units) shown in Fig. 3–22. If an
A bar made of A-36 steel has the (SI
dimensions
POISSON’S RATIO
When a deformable body is subjected to
an force
axial tensile
only to the bar, determine the3.6
axial
of P = force,
80 kNnot
is applied
change
Strength does
(MPa) it elongate
Ultimate Strength
(MPa)
Coef.
of
Therm.
Modulus
of
Modulus
of
Yield
Strength
(MPa)
Ultimate
Strength
(MPa)
Modulus
of
Modulus
of
Yield
Strength
(MPa)
Ultimate
Strength
(MPa)
but it also contracts laterally.
For and
example,
if a rubber
3.6 Pafter
OISSON’S RATIO
inPoisson’s
its length
the change
in the dimensions
of its cross section
sY
su
% Elongation in Density
Expansion
ss
ss
Materials
%%Elongation
Materials
Densityrr Elasticity
ElasticityEEa Rigidity
RigidityGG
Elongationinin
YY
uu
band is stretched,
itb can
be noted
that both
the3 thickness
and
width
of the
-6
b
bb
bb
3
applying
the
load.
The
material
behaves
elastically.
(10
)>°C
Comp.
Shear Tens. Comp.
Shear 50 mm specimen (Mg>m
(GPa)
(GPa)
Tens.
Mg>mn) )
(GPa)
(GPa)
Tens. Comp.
Comp. Shear
Shear Tens.
Tens. Comp.
Comp. Shear
Shear 5050mm
mmspecimen
specimen
band are decreased. Likewise,
a compressive
force acting on a body causes
EXAMPLE
3.4(Ratio
Metallic
Metallic of the force
it to contract in the direction
sides expand laterally.
EXAMPLE
3.4and yetPits
! 80 kN
Consider
a
bar
having
an
original
radius
r
and
length
L and subjected
A
bar
made
of
A-36
steel
has
the
shown
Fig. 3–22.
an
414
172
469
469
290 2014-T6
0.35
23 dimensions
2.79
73.1
2727
414
414
172
469
290
1010
2014-T6 10
2.79
73.1
414 in
414
172 If469
469
469
290
Aluminum
Aluminum
to
the
tensile
force
P
in
Fig.
3–21.
This
force
elongates
the
bar
by
an
axial
force
ofA-36
P = steel
80
is the
applied
to the
determine
the131
change
Wrought
Alloys
A
bar
made
has
shown
If an290
255
131
290 Wrought
290 Alloys
186
12 of
0.35
24dimensions
6061-T6
2.71
68.9
2626 bar,
255
255
290
186
1212
6061-T6
2.71kN
68.9
255 in Fig.
255 3–22.
131
290
290
186
amount d, and its radius
contracts
anchange
amount
Strains
in determine
the
d¿. dimensions
axial
of and
P by
=the
80
kN
is applied
to the bar,
change
in itsforce
length
in the
of its crossthe
section
after
–
–
179 Cast
– ASTM
0.28
12
Gray
7.19
67.0
2727
––
––
179
669
––
0.6
Gray
ASTM20
200.6 in the lateral
7.19
67.0
179
669
0.6
Iron
Cast669
Iron direction
longitudinal
or
axial
and
or
radial
direction
y– –after
in
its length
and
the The
change
in the
dimensions
ofare,
its
cross–section
applying
the
load.
material
behaves
elastically.
Alloys
Alloys
–
–
276
572
–
5 A-197
0.28
12
Malleable
ASTM
7.28
172
6868
––
––
276
572
––
55
Malleable
ASTM
A-197
7.28
172
–
276
572
respectively,
applying the load. The material behaves elastically.
703
924
0.35
0.35
0.28
0.28
0.28
0.28
70.0
70.0
––
241
241x
241
241
––
3535
0.35
0.35
345
345
––
655
655
655
655
––
2020
0.34
0.34
Magnesium
Magnesium
– early
276 1800s,
276the French
152
1
0.30
26
[Am
1.83
44.7
1818
152
[Am1004-T61]
1004-T61]
1.83
44.7
152
In the
scientist
S. D. Poisson
realized
that within
the
Alloy
Alloy
152
152
––
276
276
152
152
11
0.30
0.30
–
207
0.35
0.35
345
345
–
345
250
Poisson’s
Poisson’s
Ratio
Ration n
70.0
1.5 m 70.0
70.0
152
50 mm
10
1 0 3C
241 Copper
241
Copper
Alloys
655 Alloys
655
Red
Red–Brass
BrassC83400
C83400
dP35
! 80 kN
0.35
8.74
8.74
=P !2080and
P
– long
Bronze
C86100
Bronze
C86100
kN
Plat
=
0.34
8.83
8.83
L
d¿
r
18
101
101
3737
17
103
103
3838
elastic range the ratio of these strains is a constant, since the deformations
y
– d¿ 400
400 Structural
–
30
0.32
12
y 250
A36
7.85
200
7575 ratio,
250
Structural
A36 constant
7.85
200
250
250
are proportional.
This
is
referred
to
as
Poisson’s
d and
Steel
Steel
50 mm
–
517 itAlloys
–
40 that is unique
0.27
17
Stainless
304
7.86
7575
207
Stainless
304value
7.86
193
2073–22207
and
has517a numerical
for a193
particular
material
n (nu),
Fig.
Alloys
x207
50
mm
1.5 200
m
800 homogeneous
800 Tool
–L2
22
0.32
12
8.16
75
703
Tool
L2 and isotropic.
8.16 mathematically
200
75 is
703 x 703
703
that– is both
Stated
it
1.5 m
–
1,000
Titanium
Titanium
1,000
Alloy
Alloy
–
[Ti-6Al-4V]
[Ti-6Al-4V]
16
n = -
Plat
Plong
0.36
4.43
4.43
9.4
120
120
4444
924
924
100 mm
100 mm
SOLUTION
100 mm
The normal stress in the bar is
The negative sign is included here since longitudinal elongation (positive
z
400
400
400
400
––
3030
0.32
0.32
––
517
517
517
517
––
4040
0.27
0.27
––
800
800
800
800
––
2222
0.32
0.32
1,000
1,000
1,000
1,000
––
1616
0.36
0.36
P924
! 80 kN– –
924
z
276
276
––
(3–9) P ! 80 kN
Nonmetallic
Nonmetallic
P ! 80 kN
z
Fig. 3–22 ––
12
–
–
– Strength
0.15
11 3–22
Low
2.38
22.1
––
––
1212
––
––
––
––
0.15
Low
Strength –
2.38
22.1
0.15
Fig.
Concrete
Concrete
strain)
causes
lateral
contraction
(negative
strain),
and
vice versa.
Notice
3
–
38
–
–
– Strength
0.15
11
High
2.38
29.0
–
–
–
38
–
–
–
–
0.15
High
Strength –
2.38
29.0
–
–
–
38
–
–
–
–
0.15
80110
2
N
P
that these strains are caused only by the axial or longitudinal
sz = force
= P; i.e.,
= 16.011062 Pa
–
–
717
483acts 20.3
0.34
–
Athe–material
Plastic
49
131
––
–m2
––
717
483
20.3
2.8
0.34
Plastic
Kevlar
49 2.8
1.45order to
131
– 10.1 m210.05
–
717
483
20.3
2.8
0.34
no force
or stress
inKevlar
a lateral
direction 1.45
in
strain
–
–
90
131
–
–
0.34
–
Reinforced
30%
Glass
1.45
72.4
–
–
–
–
90
131
–
–
0.34
Reinforced
30%
Glass
1.45
72.4
–
–
–
–
90
131
–
–
0.34
in this direction.
SOLUTION
essed
From the
table
the inside
back cover for A-36 steel Est = 200 GPa,
Poisson’sc ratio
is aSOLUTION
dimensionless
quantity,
and
foron
most
nonporous
Wood
Wood
d
e
c c
dd
dd
e e
pand–
The
normal
stress
in0.47
the
is
1 bar
1 – in the z––direction
26d
6.2
–Fir
0.29
–
2.1
2626
6.2
––
0.29
Douglas
Fir
13.1
– – is
––
––
2.1
6.2
0.29
Douglas
0.47
13.1
2.1
and
so
the
strain
The
normal
stress
in
the
bar
is
solids
it
has
a
value
that
is
generally
between
and
Typical
values
of
.
Select
Structural
SelectdStructurald
3
e 4
c c
dd
dd
e e
rains–
–
2.5c
36
6.7
–
0.31
–
White
Spruce
3.60
9.65
–
–
–
–
2.5
36
6.7
–
0.31
White
Spruce
3.60
9.65
–
–
–
–
2.5
36
6.7
–
0.31
Grade
Grade
n for common engineering
materials are listed on the inside
back cover.
3 32 N
!
! PP ! 80110
80110
6
2 sN
16.0110
2 Pa
For an “ideal material” having no lateral
when
itz is stretched
6 62 Pa
= ==
16.0110
szsz=deformation
= =16.0110
2 Pa
P
=
=
=mechanical
80110-6working
2working
mm>mm
aa
z
A
10.1
m210.05
m2
9
tion, mechanical
working
of
the
specimen,
or
heat
treatment.
For
a
more
exact
value
Specific
values
may
vary
for
a
particular
material
due
to
alloy
or
mineral
composition,
ofofthe
ororheat
or compressed
Poisson’s
ratio
will
be
0.
Furthermore,
it
will
be
shown
in
Specific
values
may
vary
for
a
particular
material
due
to
alloy
or
mineral
composition,
mechanical
thespecimen,
specimen,
heattreatment.
treatment.For
Fora amore
moreexa
ex
A
10.1
m210.05
m2
Poisson
oranının
mümkün
olan
maksimum
değeri
0.5’tir
(Bkz.
hacim
modülü
hesabı).
Böylece
E
200110
2
Pa
st
reference
books
for
the
material
should
be
consulted.
reference
books
for
the
material
should
be
consulted.
3.6 POISSON’S RATIO
103
Sec. 10.6 thatb the maximum possible value for Poisson’s ratio is 0.5.
bThe yield and ultimate strengths for ductile materials can be assumed equal for both tension and compression.
oth tension and compression.
The
yield
and
ultimate
strengths
for
ductile
materials
can
be assumed
equalsteel
for
both
and
compression.
From
thetable
tableonon
the
inside
back
cover
A-36
steel
200
GPa,
From
the
the
inside
back
cover
forfor
A-36
EstEtension
= =200
GPa,
st
Therefore ! 0 …
n …
0.5.
olmaktadır.
cc
The
axial
elongation
of the bar is therefore
Measured
perpendicular
to
the
Measuredand
perpendicular
tostrain
thegrain.
grain.
and
so
the
in
the
z
direction
is
so
the
strain
in
the
z
direction
is
dd
Measured
grain.
Measuredparallel
parallelto
tothe
thed/2
grain.
3.6 POISSON’S RATIO
103
-6
EXAMPLE
3.4 perpendiculartotothe
Ans.
= load
Pload
=
[80110
2]11.5
m2 = 120 mm
zLiszisapplied
6dz6the
grain
when
along
the
grain.
the grain.
Deformationmeasured
measuredperpendicular
the
grain
when
the
applied
along
the
grain.
! eeDeformation
szsz 16.0110
16.01102 Pa
2 Pa
P
-6 -6
PzPz= L
= =80110
2 mm>mm
= ==
80110
2 mm>mm
9
d/2
9
E
Using
Eq.
3–9,
where
from
the imal
insideedilmiş
back bir
n
A
bar
made
of
A-36
steel
has
the
dimensions
shown
in
Fig.
3–22.as
Iffound
an
E
200110
2
Pa
200110
2
Pa
st
st
st
A36 çeliğinden (Tablodan ! =!
ve ! = 0.32
değerleri
alınır)
–
EXAMPLE
3.4
the to
lateral
contraction
strains
in both the x and y directions are
axial force of P = 80 kN iscover,
applied
the bar,
determine
the change
Original
Shape
Final
Shape
The
axial
elongation
of
the
bar
is is
therefore
in
its
length
and
the
change
in
the
dimensions
of
its
cross
section
after
çubuk
şekilde
gösterildiği
gibi
80
kN
eksenel
çekme
kuvvetine
maruzdur.
Yük uygulandıktan
The
axial
elongation
of
the
bar
therefore
A bar made of A-36 steel has the dimensions shown in Fig. 3–22. If an
-6
Px =elastically.
Py = -nstPz = -0.32[80110 2] = -25.6 mm>m
applying
load.isThe
material
behaves
r the
-6-6 the change
axial force
P = the
80 boyundaki
kN
applied
to
bar,
determine
Ans.
dzdve
=
P
L
=
[80110
2]11.5
m2m2= =120
mm
sonraof çubuk
çubuk
kesiti
değişimleri
hesaplayınız. Malzeme
z
z
Ans.
=
P
L
=
[80110
2]11.5
120
mm
P boyutlarındaki
z
z z
Tension
d¿
in its length and the change in the
dimensions
of
its
cross
section
after of the cross section are
Thus the changes in the
dimensions
! 80 material
kN
applying
the load.Pelastiktir.
The
elastically.
Using
3–9,
the
nstn = =0.32
davranışı
UsingEq.
Eq.behaves
3–9,where
where
foundfrom
from
theinside
insideback
back
0.32asasfound
st
-6
Fig. 3–21
cover,
the
lateral
contraction
strains
in
both
the
x
and
y
directions
Ans.
d
=
P
L
=
-[25.6110
2]10.1
m2 = are
-2.56
x in
x both the x and y directions
cover, the lateral contractionxstrains
are mm
P ! 80 kN
-6
d = PyLy = -6
-[25.6110
2]10.05
m2 = -1.28 mm
PxP = =PyP = =- n-n
2]
mm>m
-6 = - 25.6
stPzP = =y- 0.32[80110
-0.32[80110
2] = -25.6
mm>m
y
x
st z
y
3
mmcross section are
Thus the changes in the dimensions of50the
Thus the changes in the dimensions
of the cross
section are
x
y
1.5 m
-6
dx = PxLx = - [25.6110
2]10.1 m2 = - 2.56 mm
-6
50 mm
dx = PxLx = -[25.6110
2]10.1 m2 = -2.56 mm
x
! 80 kN
-6
1.5 m
dy = PyLy = - [25.6110 -6
2]10.05 Pm2
= - 1.28 mm
dy = PyLy = -[25.6110 2]10.05 m2 = -1.28 mm
P 100
! 80mm
kN
Fig. 3–22
100 mm
!
Ans.
Ans.
Ans.
Ans.
z
z
Fig. 3–22
SOLUTION
The normal stress in the bar is
SOLUTION
The normal stress in the bar isP
8011032 N
sz =
=
= 16.011062 Pa
A 310.1 m210.05 m2
3
Ans.
!37
PP!!80
80kN
kN
100 mm
100
mm
100
mm
zz z
100
100mm
mm
zz
Fig.Fig.
3–223–22
Fig. 3–22
Fig.
Fig.3–22
3–22
SOLUTION
SOLUTION
SOLUTIONÇözüm:
SOLUTION
The normal
stress in the
bar is
SOLUTION
The
Thenormal
normalstress
stressininthe
thebar
barisis
The
in
the
bar
is
Thenormal
normalstress
stress
in
the
bar
is
Eksenel çekme gerilmesi:
3 3 32 N
80110
N
P=PP = 80110
80110
33 2 2
6 6 62 Pa
s
= 16.0110
22NN N = =16.0110
szsz=PzP
=A= 80110
PaPa
80110
=
16.0110
66 2 2
10.1
m210.05
m2
ssz == A=A
== 16.0110
10.1
m210.05
m2
10.1
m210.05
m2
= 10.1
16.011022Pa
Pa
z
A
m210.05
m2
A
10.1 m210.05 m2
! table
From
the
on the
inside
back
cover
for A-36
steel
= 200
GPa,
From
the
table
on
the
inside
back
cover
for
steel
Est stE=st=200
GPa,
From
the
table
on the
inside
back
cover
forA-36
A-36
steel
200
GPa,
From
the
table
on
the
inside
back
cover
for
A-36
steel
EEst E
== 200
GPa,
and
so
the
strain
in
the
z
direction
is
From
the
table
on
the
inside
back
cover
for
A-36
steel
200
GPa,
and
ininthe
z zdirection
isisdeformasyonu: st
andso
sothe
thestrain
strain
the
direction
Eksenel
uzama
birim
and
andso
sothe
thestrain
strainin
inthe
thezzdirection
directionisis
6
sz 16.0110
16.0110
2 Pa
6
s=s
PaPa
-6
6 262
16.0110
z z 16.0110
6
P
=
= 80110
2 mm>mm
-6
s
2
Pa
z
z
9
s
16.0110
2
Pa
P
=
=
=80110
2 2mm>mm
-6 -6
P
=
=
=80110
80110
mm>mm
z
Est
200110
9 9 2=Pa
-62 mm>mm
PPz z==z EE
=
=
=
80110
2
mm>mm
200110
2
Pa
9
st
200110
2
Pa
! z EEstst st 200110
200110922Pa
Pa
The
axial
elongation
of
the
bar
is therefore
The
isistherefore
Eksenelofthe
uzama
deformasyonu:
Theaxial
axialelongation
elongation
ofthe
thebar
bar
therefore
The
Theaxial
axialelongation
elongationof
of thebar
barisistherefore
therefore
-6
Ans.
dz P=LPzL
= [80110
2]11.5
m2 120
= 120
mm
z [80110
-6-6-6
Ans.
2]11.5
mm
Ans.
= [80110
2]11.5
m2
= 120
mm
zP
z z=[80110
-62]11.5
Ans.
ddzdz=d=z=PP=
=
m2m2
= =120
mm
zL
zL
z
Ans.
zLz = [80110 2]11.5 m2 = 120 mm
! z
Using Eq. 3–9, where nst = 0.32 as found from the inside back
Using
Eq.
where
asasfound
from
back
Using
Eq.3–9,
3–9,
where
found
fromthe
theinside
inside
back
nst=0.32
=0.32
0.32
Using
Eq.
where
as
from
back
nnstnst=
Using
Eq.
3–9,
where
as found
found
from
the
inside
backenine
0.32
cover,
the3–9,
lateral
contraction
strains
in both
the xthe
andinside
y directions
are
st = hesaplayarak
Poisson
etkisini
çubuk
enkesitindeki
cover,
the
lateral
contraction
strains
in
the
x xand
y ydirections
are
cover,
the
lateral
contraction
strains
inboth
both
the
and
directions
are
cover,
the
lateral
contraction
strains
in
both
the
x
and
y
directions
are
cover, the lateral contraction strains in both the x and y directions are
-6
Px = Py = -nstPz = -0.32[80110
2] = -25.6 mm>m
-6 -6-6
-0.32[80110
-25.6
mm>m
=
P=y=-=n-n
-0.32[80110
-25.6
mm>m
PPxPx=P=x! =PP
PstPzPst=
0.32[80110
=2]
mm>m
-62] 2]
zP
z=-=
yPy=
st-n
n
=
0.32[80110
2]
==-=
-25.6
25.6
mm>m
x
y
st z
Thus the changes in the dimensions of the cross section are
Thus
the
changes
inthe
the
dimensions
of
the
cross
section
are
Thus
the
changes
in
the
dimensions
of
the
cross
section
are
Thus
the
changes
dimensions
the
cross
section
are
Çubuk
enkesitindeki
deformasyonlar:
Thus
the
changes
inin
the
dimensions
ofof
the
cross
section
are
-6
Ans.
dx = PxLx = -[25.6110
-6-6-6 2]10.1 m2 = -2.56 mm
ddxdx=d=x=PP
LxPxLxL
=x=-=[25.6110
m2
===-=
2.56
mm
-62]10.1
Ans.
=
-[25.6110
2]10.1
m2
-2.56
mm Ans.
Ans.
-[25.6110
2]10.1
m2
-2.56
mm
xP
Ans.
m2
2.56
mm
x
xL
x x= - [25.6110 2]10.1
-6
d L= P=yL-y [25.6110
Ans.
= -[25.6110
-6-6-6 2]10.05 m2 = -1.28 mm
ddydy=d=y=PPy=
m2
===-=
mm
-62]10.05
Ans.
=-[25.6110
-[25.6110
2]10.05
m2
-1.28
mm Ans.
Ans.
2]10.05
m2
-1.28
mm
yPL
yL
yL
yP
y=y= Ans.
[25.6110
2]10.05
m2
-1.28
1.28
mm
y
!
y
birim deformasyonları belirleyelim:
y
Çubuğun x ekseni doğrultusundaki kesit boyutu 100 mm iken deformasyon sonrası
(100-0.00256) mm olacaktır.
Çubuğun y ekseni doğrultusundaki kesit boyutu 50 mm iken deformasyon sonrası
(50-0.00128) mm olacaktır.
!38
homogeneous and isot
M pure
AT E R I A
LS
to
shear,
equilibrium requires that equal shear stresses
y
element uniformly, Fig.
eveloped on four faces of the element. These stresses txy must
3
gxy
gxy measures the angul
d toward or away from diagonally opposite corners of the
2
originally along the x an
as
shown inStress–Strain
Fig. 3–23a. Furthermore,
if the material is
e Shear
Diagram
The behavior of a ma
ous and isotropic, then this shear stress will distort the
laboratory
using specim
g
xy
niformly,
Fig. 3–23b.
As mentioned
in Sec. 2.2,
shear strain
was shown
that when
a small element
of the
material
is
them to a torsional loa
2
res the
angular
distortion
of
the that
element
relative
to
the sides
pure
shear,
equilibrium
requires
equal
shear
stresses
x
Kayma
Gerilme
- Birim
Deformasyon
Diyagramı
torque and the resultin
p g
along
the
x
and
y
axes.
oped on four faces of the element. These stresses txy must
! xy
explained in Chapter 5
2
1can
0 4 be
H Ain
P TaE R bulunması
3 M E C H A N durumu)
ICAL PROPE
R T I E S O F M ATkayma
E R I A L S gerilmeleri
avior ofora away
material
subjected
pure
shear
studied
Basit
kaymato(sadece
kayma
gerilmelerinin
durumunda
oward
from
diagonally
opposite
corners
of Cthe
stress and shear strain,
(b)
using specimens
in theFurthermore,
shape of thiniftubes
and subjecting
shown
in Fig. 3–23a.
the material
is
example of such a diag
aşağıda
şekilde
torsional
loading.
If measurements
are
made
of they applied
and isotropic,
then
thisgösterildiği
shear stress
will yönlenmişlerdir.
distort
the
Fig. 3–23
Like the tension test, th
d the Fig.
resulting
of twist, inthen
the methods
to be
The Shear
Stress–Strain
Diaga
ormly,
3–23b.angle
As mentioned
Sec.by
2.2,
shear strain
linear-elastic behavior
Malzeme
homojen
ve the
izotropik
ise kayma gerilmesi, elemanda üniform
biçimde
aşağıda
in Chapter
5, the dataofcan
used to
determine
sheartxy
the
angular distortion
the be
element
relative
to thethe
sides
Also, strain hardening
and
a shear stress–strain
diagram
plotted. sebep
An
In Sec. 1.5basit
it was
showntesti
thatverileri
when a And
smallfinally,
elemen
ngshear
the xstrain,
and y axes.
gösterildiği
gibi bir biçim
değişimine
olacaktır. Laboratuvarda
kayma
reached.
th
f such
a diagram
for a ductile
material
is shown
in Fig.
subjected to pure shear, equilibrium
requires
that
equw
or
of a material
subjected
to pure
shear can
be studied
in 3–24.
a
until it reaches a point
kullanılarak
birsubjecting
kayma
gerilme-birim deformasyonu
edilebilir.
ension
test, thisinmaterial
when
subjected
toand
shear
will exhibit
must be elde
developed
on four faces of
themost
element.
These
ing specimens
the shape
ofaşağıdaki
thin tubesgibi
For
engineering
x
tic
behavior
andIf itmeasurements
will have a defined
proportional
limit tpl.
be
directed
toward
or
away
from
diagonally
opposit
sional
loading.
are made
of the applied
t
behavior is linear, and s
(a)
in hardening
will occur
untilthen
an by
ultimate
shear stress
tu is
element, as shown in Fig. 3–23a. Furthermore, if
he
resulting angle
of twist,
the methods
to be
104
C H Ato
P Tto
E R lose
3 M
H A N I Cstrength
A L P R O P E R T I E S O F M AT E R I A L S
And
finally,
the data
material
begin
itsE Cshear
homogeneous and isotropic, then this shear stress
Chapter
5, the
can will
be used
determine
the shear
y
tu
element uniformly, Fig. 3–23b. As mentioned in Sec. 2.2
chesstrain,
a point
where
it fractures,
tf.
ear
and
a shear
stress–strain
diagram plotted. An
3
tf
gxyelastic
gxy measures the angular distortion of the element rela
t engineering
likematerial
the
one
just
described,
the
uch
a diagram materials,
for a ductile
is
shown
in
Fig.
3–24.
y
2
originally
along the x andDiagram
y axes.
son
linear,
Hooke’s
law subjected
for shear can
be written
as
The tShear
Stress–Strain
test, and
this so
material
when
to shear
will exhibit
pl
Here Gtoispure
called
thecas
The behavior of a material subjected
shear
txy
behavior and it will have a defined proportional
limit tpl.
rigidity.
Its
value
repre
laboratory
using
specimens
in the
of thin
that when
a small
element
of shape
material
is tube
gxy 1.5 it was shown
hardening will occur until an ultimate shear stress tu isIn Sec.
(3–10) 2
t = Gg
is, Gstresses
Typi
= tpl>g
pl. mad
to a torsional
loading.
If that
measurements
are
to pureG shear,them
equilibrium
requires
that equal
shear
finally, the material will begin to lose its shear strengthsubjected
x
listed
on
the
inside
back
torque
theelement.
resulting
angle
of twist,
then by the
be developed on four
facesand
of the
These
txy must
p must
s a point where it fractures, tf.
g stresses
x
! gxy
gpl explained gin
gr 5, the data
u Chapter
Gcorners
will
be
the
as
can
be
used
to dete
2
be
directed
toward
or
away
from
diagonally
opposite
of
thesame
! one
! elastic of
!
ngineering
like the
just described,
s called thematerials,
shear modulus
of elasticity
or thethe
modulus
radians,
a
dimensionless
and shear strain, and a shear stress–strain diag
(b)
(a)can be written as
element,
as shown in stress
Fig. 3–23a.
near,
andrepresents
so Hooke’sthe
law for shear
Fig. 3–24Furthermore, if the material is
s value
of the
line on the t–g diagram,
It will
be
shown
Se
example
of such
diagram
for will
a ductile
material
is sho
Lineerslope
elastik
malzemede
kayma gerilmeleri
için
Hooke
kanunu
şu ashear
şekilde
yazılabilir:
homogeneous
and
isotropic,
then
this
stress
distort
the in
engineering materials
are3–23
= tpl>gpl. Typical values for common
Fig.
y
G
are
actually
related
Like the
test,inthis
material
toby
s
uniformly, Fig. 3–23b.
Astension
mentioned
Sec.
2.2, the when
shear subjected
strain
he inside back cover.
that
the units of measurementelement
for
3 t =Notice
g
behavior
and it
will have
a defined
propo
xy
(3–10)gxy measures the angularlinear-elastic
Gg
distortion
of
the
element
relative
to
the
sides
the same as those
in
! for t (Pa 2or psi), since g is measuredoriginally
strain hardening will occur until an ultimate s
along the x andAlso,
y axes.
dimensionless quantity.
reached.
And finally,
the
material
begin
The
behavior
of
a
material
subjected
to pure
shear
can diğer
bewill
studied
in to
a lose it
Burada
G three
ye
elastik
kayma
modülü
veya
Kayma
modülünü
elled
shown
Sec. 10.6
that the
material
constants,
E, n,
and rijitlik modülü denmektedir.
the in
shear
modulus
of
elasticity
or
the
modulus
of
untilin
it reaches
a point
where
it fractures,
tf.
laboratory
using
specimens
the
shape
of
thin
tubes
and
subjecting
g
ally related
by the
alue
represents
theequation
slope of the line on2xythe t–g diagram,them to a torsional loading.
ForIfmost
materials,
one just desc
measurements
are
made oflike
thethe
applied
malzeme
sabitleri cinsinden aşağıdaki gibi yazabiliriz (bkz. Bölüm
10.6engineering
)
Typical
values
for
common
engineering
materials
are
>g
.
x
t
behavior
is
linear,
and
so
Hooke’s
law
for
shear
be w
pl pl
torque and the resulting angle of twist, then by the Provided
methods E
to and
be can
p g
G are
xy
nside back cover. Notice that the
of measurement forexplained in Chapter 5, the data can be used to determine
!units
the shear
E 2
from this
equation rat
same as those forGt=(Pa or psi), (b)
since g is measured(3–11)
instress and shear strain, and a shear stress–strain diagram
plotted.
An in the
tu
211
+
n2
For
example,
t = Gg
ensionless quantity.
tf
example
of
such
a
diagram
for
a
ductile
material
is
shown
in
Fig.
3–24.
! the three material
Gst = 11.011032 ksi, so t
Fig. 3–23
own in Sec. 10.6 that
constants, E, n, andLike the tension test, this material when subjected to shear
will exhibit
y related by the equation
tG
pl ve E bilinirse
linear-elastic
behavior
and
it
will
have
a
defined
proportional
tpl.
Buradan
görülüyor
ki,
Poisson
oranı
formülden
hesaplanabilir
ve
deneysel
Here G is called the shear modulus limit
of elasticity
or
E and G are known, the value of n can then be determined
Also,
strain
hardening
will
occur
until
an
ultimate
shear
stress
t
rigidity. Its value represents the slope of the
line on t
u is
equation rather than through experimental measurement.
olarak ölçülmesine gerek kalmaz.
reached. And finally, thethat
material
begin
to lose
its shear
strengthengineer
is, G =will
values
for common
tpl>g
mple, in the case of EA-36 steel, Est =G 2911032 ksi and
pl. Typical
3
until
it
reaches
a
point
where
it
fractures,
.
t
(3–11)
G from
=
listed on the insidef back cover. Notice that the units of m
3–11, nst = 0.32.
110 2 ksi, so that,
g
211Eq.
+ n2
gu most engineering
gpl
gr
For
materials,
one
described,
theor
elastic
G will be like
the the
same
asjust
those
for t (Pa
psi), since g
t
behavior is linear, and soradians,
Hooke’sa law
for
shear
can
be
written
as
dimensionless quantity.
Fig. 3–24
It will be shown in Sec. 10.6 that the three material co
nd G are known, the value of n can then be determined
G
are actually related by the equation
uation rather thantuthrough experimental measurement.
(3–10)
t = Gg
3
t
, in the case of f A-36 steel, Est = 29110 2 ksi and
3
2 ksi, so that, from Eq. 3–11, nst = 0.32.
E
tpl
G the
= modulus of
Here G is called the shear modulus of elasticity or
211 + n2
rigidity. Its value represents the slope of the line on the t–g diagram,
G
that is, G = tpl>gpl. Typical values for common engineering materials are
listed on the inside back Provided
cover. Notice
thatGthe
of measurement
E and
areunits
known,
the value of for
n can then
g
gu
gpl
gr
G will be the same as those
for
(Pa
or
psi),
since
measured
in
t
g
from this equation rather thanisthrough
experimenta
radians, a dimensionless For
quantity.
example, in the case of A-36 steel, Est =
Fig. 3–24
It will be shown in Sec.G10.6= that
the 3three
constants,
E, n,nand
so that, from
Eq. 3–11,
11.0110
2 ksi, material
st
st = 0.32.
G are actually related by the equation
3.7
3.7
G =
E
211 + n2
(3–11)
Provided E and G are known, the value of n can then be determined
from this equation rather than through experimental measurement.
For example, in the case of A-36 steel, Est = 29110!3392 ksi and
Gst = 11.011032 ksi, so that, from Eq. 3–11, nst = 0.32.
CHA1
P T0E6
R 3
M E CCHHAANPITCEARL 3P R OMP EE CR THIAE N
S IO
F LMPAT
106
CA
R OEPREI A
R LT SI E S
C
H
A
P
T
E
R
3
M
E
C
H
A
N
I
C
A
L
P
R
O
P
E
R
T
I
E
S
O
F
M
AT
E
R
I
A
L
S
1
0
6
C H A P T E R 3 M E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S
OF
M AT E R I A L S
106
A
TAE
C HEA
R IEARLISA L S
CN
H IACNAILC APLR O
P PR EORPTEI RE TS I EOSF OMF AT
MEAT
106
E106
E RC
I AHLEC
SRPH
ESR TO
I EFS M
OAT
F M
AT
IA
LPSRT E3R 3M EM
M E C H A NM
I C A L P R O P E R T I EEXAMPLE
S O F M AT E R I A L S EXAMPLE
3.6
PTER 3
E C H A N I C A L P R O P E R T I E S O F M 3.6
AT E R I A L S
M
E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S
3.6
T I E S O F M AT E R I A L S EXAMPLE
EXAMPLE
C H A P T E R 3!
M3.6
E C H A N I C A L P R O P E R T I E S O F M AT E R I A L S
An aluminum specimen
shownspecimen
in Fig. 3–26
diameter
of a d
An aluminum
shownhas
in aFig.
3–26 has
An
aluminum
specimen
shown
in
Fig.
3–26
has
a
diameter
of
and
a
gauge
length
of
If
a force
d
=
25
mm
L
=
250
mm.
and
a
gauge
length
of
If
a
force
of
165
kN
d
=
25
mm
L
=
250
mm.
An aluminyum
aluminum specimen
shown
0
0
Şekilde verilen
numune
için in
! 0 Fig. 3–26 has
ve ! a0 diameter ofolduğu
and
a
gauge
length
of
If
a
force
of
165
kN
d
=
25
mm
L
=
250
mm.
0
0
elongates
the
gauge
length
1.20
mm,
determine
elongates
the
gauge
length
1.20
mm,
determine
AMPLE
3.6
and
a of
gauge
length
force
of
165
daluminum
=diameter
mmspecimen
LFig.
250
minum
specimen
shown
in Fig.
3–263–26
has
a0 aluminum
of
aluminum
specimen
shown
in Fig.
has
a25 diameter
An
shown
inofinFig.
3–26
hashas
aIf aadiameter
of ofkN the modulus of the m
0 =
165 165
kN kN
An
specimen
shown
3–26mm.
diameter
elongates
the
gauge
length
1.20
mm,
determine
the
modulus
of causes th
An
aluminum
specimen
shown
in
Fig.
3–26
has
a
diameter
of
elasticity.
Also,
determine
by
how
much
force
bilinmektedir.
165
kN
kuvvet
ölçüm
uzunluğunu
1.20
mm
uzattığına
görethe diameter
elasticity.
Also,
by
much
the
force
the
gauge
length
1.20
determine
theof modulus
ofcauses
and and
a gauge
length
of Lof
If
force
of
165
kN
=specimen
aAn
gauge
length
If
force
of
165
kN
=mm
L250
250specimen
and
a3–26
gauge
length
of
Ifofahow
165
kN
dmm.
25
mm
L0 L
=0mm,
250
mm.
An
shown
in
Fig.
3–26
has
adetermine
diameter
6525
kNmm
and
aofgauge
length
of
Ifforce
a force
of
165
kN
d=a0elongates
=aa25
mm
=ofby
250
mm.
0aluminum
0 = mm.
aluminum
shown
in
Fig.
has
aforce
diameter
0 has
uminum
specimen
shown
in
Fig.
3–26
diameter
elasticity.
Also,
determine
how
much
the
force
causes
the
diameter
and
a
gauge
length
of
If
a
of
165
kN
d
=
25
mm
L
=
250
mm.
An
aluminum
specimen
shown
in
Fig.
3–26
has
a
diameter
of
165
kN
of
the
specimen
to
contract.
Take
and sY =
G
=
26
GPa
0
0
elasticity.
Also,
determine
by1.20
how
much
the
force
causes
the
diameter
the
specimen
to
contract.
Take
andalsY = 440 MPa.
Galmodulus
= 26
GPa
s the the
gauge
length
determine
the
modulus
of
gates
gauge
length
1.20
mm,
determine
the
modulus
ofIf
elongates
the
gauge
1.20
mm,
determine
the
modulus
of
and
a gauge
length
ofofgauge
If
akuvvet
force
of
165
kN
d0of
=mm,
L
250
mm.
elongates
the
length
mm,
determine
the
of
and
a=mm
gauge
length
of
aAyrıca
force
of
165
kN
delongates
251.20
mm
L0mm,
=of
250
mm.
0 =length
0 =length
If
a
force
165
kN
5 mm and a gauge
L25
250
mm.
the
specimen
to
contract.
Take
and
G
=
26
GPa
=
440
MPa.
s
elastisite
modülünü
belirleyiniz.
etkisi
ile
numunenin
çapındaki
the
length
1.20
determine
the
modulus
of
0gauge
al
Y
and
a
gauge
length
of
If
a
force
of
165
kN
d
=
25
mm
L
=
250
mm.
0
0 Take Gal = 26 GPa and sY = 440 MPa.
of
the
specimen
to contract.
y.ticity.
Also,Also,
determine
by how
much
thethe
force
causes
the
diameter
determine
byelongates
how
much
the
force
causes
the
diameter
elasticity.
Also,
determine
by
how
much
the
force
causes
thethe
diameter
gauge
length
1.20
mm,
the
modulus
ofcauses
elasticity.
Also,
determine
by
how
much
force
diameter
the
gauge
length
mm,
the
modulus
ofthe
tes the gaugeelongates
length 1.20
mm,
determine
the
modulus
of determine
elongates
the
gauge
length
1.20
mm,
determine
the modulus of
elasticity.
Also,
determine
by 1.20
how
much
the
force
causes
the diameter
SOLUTION
pecimen
to contract.
Take
and
G
=
26
GPa
=
440
MPa.
s
he specimen
toelasticity.
contract.
Take
and
G
=
26
GPa
=
440
MPa.
s
al
Y
of
the
specimen
to
contract.
Take
and
G
=
26
GPa
=
440
MPa.
s
elasticity.
Also,
determine
by
how
much
the
force
causes
the
diameter
al
Y
SOLUTION
of
the
specimen
to
contract.
Take
and
G
=
26
GPa
=
440
MPa. olarak
s
Also,
determine
by
how
much
the
force
causes
the
diameter
al
Y
al
Y
ty. Also, determine
by specimen
how much
causes
diameter
büzülme
!
elasticity.
Also,
determine
by how
the diameter
of the
tothe
contract.
Take the
and much
G
26
GPa
440force
MPa.causes ve
sY =! the
SOLUTION
3forcemiktarını
al = hesaplayınız.
3 specimen
SOLUTION
Modulus
of
Elasticity.
The
average
normal
stress isin the
of
the
specimen
to
contract.
Take
and
G
=
26
GPa
=
440
MPa.
s
of
the
to
contract.
Take
and
G
=
26
GPa
=
440
MPa.
s
al
Y
al
Y
specimen to contract.
Take
and
G
=
26
GPa
=
440
MPa.
s
3
Elasticity.
The
normal stress
in the
specimen
of the specimen
Take
andaverage
= 26
GPaThe
440
MPa.
sY =average
al
Y to contract.Modulus
al of
3
Modulus
ofGElasticity.
normal
stress in the specimen is
ON
verilmektedir.
UTION
Modulus
of
Elasticity.
The
average
normal
stress
in
the
specimen
is
SOLUTION
SOLUTION
16511032 N
3 P
SOLUTION
2 N=
3
P 165110165110
SOLUTION
= 336.1 MPa
s
=
s3 of Elasticity.
The The
average
normal
stress
in
the
specimen
is
2
N
SOLUTION
dulus
of Elasticity.
average
normal
stress
in
the
specimen
is
3
P
TION
SOLUTION
of of
Elasticity.
The
average
normal
in in
thethe
specimen
Modulus
Elasticity.
The
average
specimen
is MPa
=stress
=isMPa
336.1
2=Nis stress
Çözüm: Modulus
A
Modulus of Elasticity.
The
average
normal
stress
specimen
m22
=
=m2
336.1
= s normal
2 1p>4210.025
L0 P in thes165110
A
2
1p>4210.025
L
d
=
=
336.1
MPa
s
=
Modulus
of
Elasticity.
The
average
normal
stress
in
the
specimen
is
A
0
3
Modulus
Elasticity.
The
average
stress
in
specimen
isspecimen
3 of
1p>4210.025
m2
us of Elasticity.
average
normal
stress
in0 the
specimen
isnormal
Modulus
of
Elasticity.
The
average
normal
is
L0
2in the
3 the
3 stress
165110
165110
P P The
1p>4210.025
d0 2 N d0
L02 N
3
2N
165110
2and
N m2
P PA 165110
165110
2
N
the
average
normal
strain is
=
336.1
MPa
s = s == d0 =
=
336.1
MPa
P
3
= average
= 336.1
MPa
s =
=MPa
s=32=the
the32average
normal
isMPa
2N
AP A1p>4210.025
336.1
sN 2=m2
16511032m2
P 2 = 165110
165110
Nand
165110
Nnormal
strain
is336.1
2 2 =strain
1p>4210.025
P2and
P
A
A
1p>4210.025
m2
1p>4210.025
m2
L
L
and
the
average
normal
strain
is
A
=
=
336.1
MPa
s
=
0
0
1p>4210.025
=s
336.1
s = d0 d=0
= 336.1 MPa
= s =m22 =
= 336.1 MPa
= MPa
2
d
1.20 mm
! m2A
1p>4210.025
m2A
A L0 1p>4210.025
1p>4210.025
m22
A
1p>4210.025
m22
L0dnormal
d mm
1.20 mm
P =
=
= 0.00480 mm>mm
average
strain
is
the average
normal
strain
is
d
1.20
0
and
the
average
normal
strain
is
and
the
average
normal
strain
is
L
250mm>mm
mm
P
=
=
=
0.00480
d
1.20
mm
P =
=
= 0.00480 mm>mm
and the average normal strain is
P
=
=
=
0.00480
mm>mm
L
250
mm
L
250
mm
Boyuna
birim
deformasyon:
and
the
average
normal
strain
is
and the
average
normal
e average normal strain and
is the average
normal
strain
is strain isL
250 mm
d d1.20 1.20
mm mm
d d 1.20
mm
1.20
mm Since s 6 sY = 440 MPa, the material behaves elast
P = P == =
= 0.00480
d mm>mm
1.20
mm
= 0.00480
mm>mm
P
=
=
=Y 0.00480
mm>mm
P
=
=
==MPa,
0.00480
mm>mm
the
material
behaves
elastically.
The The
s mm
6 ssY 6
= s
440
dSince
1.20
Since
the material
behaves
elastically.
440 MPa,
L d L250
mm
= 0.00480
mm>mm
250
mm P = d =1.20 dmm 1.20
1.20
mm
modulus
of
elasticity
is therefore
mm
L
250
mm
Since
the
material
behaves
elastically.
The
s
6
s
=
440
MPa,
L
250
mm
P
=
=
=
0.00480
mm>mm
L
250
mm
P
=
=
=
0.00480
mm>mm
P =
=
= 0.00480 mm>mm
Y = 0.00480
P
=
=
mm>mm
modulus
of
elasticity
is
therefore
modulus
of elasticity is therefore
L
250
mm
L
250modulus
mm
L
250 mm
elasticity
L
250ofmm
!
the the
material
behaves
elastically.
The The is therefore
e6 ssY6 =sY440
material
behaves
elastically.
= MPa,
440 MPa,
336.111062 Pa
s The
6 elastically.
Since
thethematerial
behaves
elastically.
6
MPa,
Since
material
behaves
The
ss
6Ys=Y 440
=behaves
440
MPa,
6=
Since s 6 sY = 440 MPa,
thes material
elastically.
The
336.1110
2
Pa
E
=
= 70.0 GPa
s6
336.1110
2 Pa
isSince
therefore
al The
dulus
of
is therefore
Since
the elastically.
material
behaves
s the
6 s
= 440 MPa,
s elastically.
material
behaves
elastically.
The
sof6elasticity
sYelasticity
= 440
MPa,
material
behaves
sthe
6ofselasticity
MPa,
Y
Y =
336.1110
2=Pa
P =GPa
0.00480
s
Ans. Ans.
Eolarak
= The
= 70.0
of elasticity
is
therefore
Since
the
material
behaves
elastically.
The
s 440
6 sis
=modulus
440
MPa,
modulus
of
elasticity
is
therefore
modulus
therefore
al
E
=
=
70.0
GPa
Y
!
yani
malzeme
elastik
davranmaktadır
al
kN Eal =
modulus
of elasticity165
is therefore
P
Ans.
=
70.0 GPa
us of elasticity modulus
is therefore
of
elasticity
therefore
P=0.00480
0.00480
6
6 is
165
kN 165 kN is therefore
modulus
of
P
0.00480
336.1110
2 Pa
336.1110
2 elasticity
Pa
s s165
6 Contraction
6
6
kN
of
Diameter.
First we will determine Pois
336.1110
Pa
6
336.1110
2 Pa
s
s s 336.1110
Ans.
Eal =
= 70.0
GPa
Eal == 336.1110
=
= 70.0
GPa 2Fig.
6
336.1110
Pa 2 Pa
3–26
s=EAns.
2EPa
Contraction
of2 Diameter.
First
we
will
determine
Poisson’s
ratio ratio
336.111062 Pa
6=
s
P s P 0.00480
Ans.
=
=
70.0
GPa
0.00480
Ans.
E
=
=
70.0
GPa
Ans.
=
=
=
70.0
GPa
Contraction
of
Diameter.
First
we
will
determine
Poisson’s
al
al
for
the
material
using
Eq.
3–11.
al
Fig. =
3–26
Ans.
=Diameter.
=Ans.
70.0
GPa
2 Pa
Contraction
of70.0
First
we
will
determine Poisson’s
ratio
s E336.1110
al = =
Ans.
Eal = Fig.
= 3–26
=Fig.
GPa
P70.0
E
=3–26
GPa
0.00480
Pmaterial
0.00480
P
0.00480
al
for
the
using
Eq.
3–11.
P
0.00480
Eal for
=0.00480
= 70.0
GPa using Eq.Ans.
P 165
kN kN 0.00480
for Eq.
the
material
3–11.
Hooke Pkanunu
ile,=! material
165 kN165
using
3–11.
PthePoisson’s
0.00480
tion
of Diameter.
FirstFirst
we will
determine
Poisson’s
ratioratio
traction
of Diameter.
we will
determine
E
65 kN
Contraction of Diameter.
First
we of
will
determine
Poisson’s
ratio
Contraction
Diameter.
First
wewe
will
determine
Poisson’s
ratio
Contraction
of
Diameter.
First
will
determine
Poisson’s
ratioG =
E ratio
Contraction
of
Firstbu
weoran
will
determine
Poisson’s
he material
using
Eq.
3–11.
material
using
Eq.
3–11.
action
of
Diameter.
First
weDiameter.
will
determine
Poisson’s
ratio
Fig.
3–26
Contraction
of
First
weDiameter.
will determine
Poisson’s
ratio
Şimdi
de
Poisson
oranını
hesaplayalım,
çünkü
ile
büzülme
miktarı
hesaplanabilir.
Fig.
3–26
E
Fig.
3–26
211
+ n2
G
=
E
for the material
using Eq.
3–11.
for
thethe
material
using
Eq.
3–11.
Contraction
of
Diameter.
First
we
will
determine
ratio 211
for
material
using
Eq.
3–11. G Poisson’s
the
material
using
Eq.
3–11.
G =+ n2
=
material
Eq.the
3–11.
material usingfor
Eq.
3–11.
g.
3–26 usingfor
211
+
n2
211 + n2
for the material
E E using Eq. 3–11.
70.0 GPa
E
G =G = E
26 GPa =
E
E
E GPa = 70.0 GPa
E
G =
211
+
n2
211
+
n2
70.0
GPa
26
211 + n2
G
=
70.0
G
=
G =
G =
G = 211 + n2 E
211 =+ n2
26 GPa
+211
n2
26211
GPa
=211
+ n2
+
n2
211 + n2
211
+
n2
G =
211 + n2 n = 0.347
211 + n2
70.0 70.0
GPaGPa
70.0 GPa211 + n2 70.0 GPa
n = 0.347
26 GPa
= GPa
26 GPa
= 70.0
70.0
GPa
70.0
GPa
= GPa
+ n2+ n226 GPa =70.0 GPa 26 GPa
n = 0.347
n = 0.347
26 26
GPa
26 GPa =211 211
211= +=
n2211
211 + n270.0 GPa
Since
Pn2
long = 0.00480 mm>mm, then by Eq. 3–9,
211
+ n2
+
211 + n2 26 GPa =26211
Since
then
by
Eq. 3–9,
P
=
0.00480
mm>mm,
+
n2
GPa =
long
= 0.347
n = n0.347
!
Since
by Eq.
3–9, then by Eq. 3–9, Plat
Plong
= 0.00480
211Since
+mm>mm,
n2=
Plong
= then
0.00480
mm>mm,
n
0.347
n
=
0.347
n =n 0.347
n = 0.347
= 0.347
Plat
n = 0.347
n = -P
n = -P
long
engP=
then
by 3–9,
Eq. 3–9,
= 0.00480
mm>mm,
then
by deformasyonu
Eq.
mm>mm,
P3–9,
Plat
n
=
0.347
long0.00480
Enine
birim
hesaplayalım:
lat
Since
then
by
Eq.
P
=
0.00480
mm>mm,
long
Since
then
by
Eq.
3–9,
P
=
0.00480
mm>mm,
long
nthen
=then
-by
longthen by Eq. 3–9,
Plong = 0.00480Since
mm>mm,
Since
Eq.Eq.
3–9,
Plong
= 0.00480
mm>mm,
Since
by
3–9,n = - Plong
Plong
= 0.00480
mm>mm,
P
Plat
then
by
Eq.
3–9,
Plong
=
0.00480
mm>mm,
long
Plat Plat
Plat 0.347 = Plat
Plat then by Eq. 3–9,
= P-latPPlong = 0.00480 mm>mm,
n = n-Since
0.347
=
0.00480 mm>mm
Plat
PlatPlat
n = -P
Plong long
Plat
n = -Plat
0.00480 mm>mm
n = long
n0.347
=
- P=- n =
n = - PPlong
0.347 = Plong
P
P
0.00480
mm>mm
lat
longlong
P = - 0.00166 mm>mm
long
0.00480 mm>mm
Plat Plat
Plat
n = -Plat
Plat = - 0.00166 mm>mmlat
0.3470.347
= -= P
Plat 0.347 = 0.347
=
long
P
P
Plat
lat latmm>mm
0.00480
mm>mm
P = 0.00166
0.00480
mm>mm
0.347 = 0.00480
Plat = -0.00166
mm>mmis therefore
0.347
= lat=
- - mm>mm
The contraction
of the diameter
0.347
0.347 = - 0.00480 mm>mm
0.00480 mm>mm
Plat
The contraction
of0.00480
the diameter
is therefore
0.00480
mm>mm
0.00480
mm>mm
mm>mm
0.347
=
P
=
-0.00166
mm>mm
P! lat =lat-0.00166 mm>mm The contractionPlat
=
-diameter
0.00166 mm>mm
of The
the
is therefore
d¿ = 10.001662125 mm2
P = -0.00166 mm>mm
0.00480
mm>mm
contraction
ofd¿the
is therefore
Plat = - 0.00166 mm>mm
= diameter
10.001662125
mm2
Platlat= -0.00166 mm>mm PlatPlat
= =
-0.00166
mm>mm
-0.00166
mm>mm
= 0.0416 mm
= 10.001662125
mm2 mm
contraction
of diameter
the
diameter
is therefore
traction
of the
is therefore
The contraction
of= the
diameter
isd¿
therefore
= 0.0416
Ans.
-0.00166
mm>mm
Çap
boyunca
enine
deformasyonu
(büzülmeyi)
hesaplayalım:
d¿
= 10.001662125 mm2
The
contraction
of
the diameter
isPlat
therefore
ntraction of the
diameter
is therefore
The
contraction
of the diameter
iscontraction
therefore
The
contraction
of
the
diameter
is
therefore
The
of
the
diameter
is
therefore
=
0.0416
mm
Ans.
= 10.001662125
d¿ =d¿10.001662125
mm2mm2
d¿ = 10.001662125 mm2
= 0.0416 mm
Ans.
d¿
mm2
d¿ = 10.001662125
mm2
The contraction
of =the10.001662125
diameter ismm2
therefore
= 10.001662125
mm2
d¿
=mm
10.001662125
mm2
= d¿
0.0416
Ans.
= 0.0416
Ans.
= 0.0416
mm mm d¿ = 10.001662125Ans.
= 0.0416 mm Ans.
Ans.
d¿ mm
= 10.001662125 mm2
= 0.0416
Ans.
! = 0.0416 mm
= 0.0416
mm
Ans.
= 0.0416
mm
Ans.
EXAMPLE
EXAMPLE3.6
3.6
165 kN
3.6
165 kN
165 kN
165 kN
= 0.0416 mm
Ans.
!40
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