to restrict
To properly
design
a structural
it isnecessary
stress
that
the applied
load to oneelement
thatinisthe
less
than the load
the the stress in the material to a level that wi
accidental
loadings
can
occurrestricts
thatmember
may
notorbemechanical
accounted
for
Todoing
ensurethis.
thisFor
safety, it is therefore necessary to choose an
necessary
to restrict
the stress
indecay,
thesupport.
material
to a level
that will
be
safe.
member
canTo
fully
There
are
many
reasons
for
design.
Atmospheric
corrosion,
or
weathering
tend
to
cause
properly design a structural member or mechanical element it is
stress
that
restricts the applied load to one that is less than the
To ensure
this safety,
it
is the
therefore
necessary
tosome
choose
an
allowable
example,
load
for
which
the
member
is
designed
may
be
different
materials
to deteriorate
during
service.
And
lastly,
materials,
such
necessary to restrict the stress in the material to a level that will be safe.
member can of
fully support. There are many reasons for doing
thatconcrete,
restricts
the
applied
load to one
thaton
is lessThe
thanintended
the load
the
from
loadings
measurements
asstress
wood,
or actual
fiber-reinforced
composites,
show
high
To
ensure placed
this
safety,it.
it can
is therefore
necessary
to choose aan allowable
example,
theorload
for which the member is designed may be
member in
can
fullystructure
support.
are may
manynot
reasons
for due
doing
this. For
orThere
machine
be
exact,
to
errors
in
fabrication
inthe
variability
mechanical
properties.
stress that restricts the applied load to one that is less than
load the
from actual
loadings
placed on it. The intended measurem
example,
the of
load
which
the
is designed
may
be
different
theforassembly
of member
its component
parts.
Unknown
vibrations,
impact,
or
One method
specifying
themember
allowable
load
for
a
member
is
to
use
a
can fully support. There are many reasons for doing this. For
orinmachine
may not be exact, due to errors in fabrica
from actual
loadings
placed
onThe
it. The
intended
measurements
of
astructure
accidental
loadings
can
occur
that
may
not
be
accounted
for
the
number
called Emniyet
the
factor
ofGerilmesi
safety.
factor
of
safety
(F.S.)
is
a
ratio
of
example, the load for which the member is designed may be different
assembly
of its component parts. Unknown vibrations, i
structure
or machine
mayallowable
not be exact,
due
to. Here
errors
or inthetend
Atmospheric
corrosion,
decay,
tomeasurements
cause
the
failure load
Ffaildesign.
to the
Fallow
Fin
isor
found
failfabrication
from load
actual
loadings
placed
onweathering
it.from
The intended
of a
accidental
loadings
can occur
that may
not be accounted f
the assembly
of
its
component
parts.
Unknown
vibrations,
impact,
or
materials
to
deteriorate
during
service.
And
lastly,
some
materials,
such
experimental
testing
of
the
material,
and
the
factor
of
safety
is
selected
Bir elemanın üzerine
gerçek
sırasında
alınan
yüklerden
farklı
structure gelen
or machine
mayyükler
not betasarım
exact, due
to errors dikkate
in fabrication
or in
Atmospheric
corrosion, decay, or weathering tend
accidental
loadings
can
occur
that
may
not
be accounted
for inare
thedesign.
as wood,
concrete,
or
fiber-reinforced
composites,
can
show
high
based
on experience
so that
the
above
mentioned
uncertainties
the assembly of its component parts. Unknown vibrations, impact, or
materials to deteriorate
during
service.
And lastly, some mate
imalatı
sırasında
ölçüsel
gibi elemanın
yerine
montajı
design. Atmospheric
corrosion,
decay,
or
weathering
tendhatalar
to cause
variability
inaccidental
mechanical
properties.
accounted
for olabilir.
when
theElemanın
member
is
usedloadings
under
similar
conditions
of yapılabildiği
can occur
that may
not be accounted for in the
as wood,
concrete,
or fiber-reinforced composites, can sh
materials
deteriorate
during
service.
And lastly,
some materials,
such
One
method
of
specifying
the
allowable
load
for
a
member
is
to
use
a
loading
and to
geometry.
Stated
mathematically,
design. Atmospheric
corrosion, decay, Elemanlar
or weathering
tend to
cause umulmayan
sırasında
hatalar
söz ofcomposites,
konus
tasarım
sırasında
in
mechanical
properties.
as wood, concrete,
or da
fiber-reinforced
can show
highvariability
number
called
the factor
safety. olabilmektedir.
The
factor
of safety
(F.S.)
is asome
ratio
of
materials
to deteriorate
during
service.
And
lastly,
materials,
such
One
method
of
specifying
the allowable load for a member
variability in mechanical
properties.
the
failure
load
F
to
the
allowable
load
F
.
Here
F
is
found
from
fail
allow
fail
yüklere maruzaskalabilirler.
Elemanların
kullanım
sırasında
paslanma,
wood,
concrete,
or fiber-reinforced
composites,
can showçürüme
high vb.
1.7etkilere
DESIGNofOFsafety
SIMPLE(F.S.)
CONNEC
called
the factor of safety. The
factor
is
Ffail
One method ofexperimental
specifying the
allowable
load
for a member
isfactor
to useofanumber
testing
of
the
material,
and
the
safety
is
selected
variability
in
mechanical
properties.
F.S.
=
(1–8)
the
failure
load
F
to
the
allowable
load
F
.
Here
F
is
fo
number called
the
factor
of
safety.
The
factor
of
safety
(F.S.)
is
a
ratio
of
based
on
experience
so
that
the
above
mentioned
uncertainties
are
fail
allow
fail
F
maruz kalabilirler.
Ayrıcaof elemanların
üretiminde
kullanılan
allow
One method
specifying the allowable
load for
a member beton
is to use ve
a ahşap gibi
testing
ofbethe
material, and the factor of safety i
the failure load Faccounted
. Here
Fused
is
found from
for
whenload
theFmember
under
conditions
ofmust
In
ofisthese
equations,
theexperimental
factor
of safety
greater
fail to the allowable
allow
number
called
theany
factor
offailsafety.
The similar
factor
of
safety
(F.S.)
is a ratio
of than 1 in
based
on experience
so gösterebilir.
that
the on
above
mentioned uncerta
mekanikand
özellikleri
/oftaşıma
büyük
değişimler
Tüm
experimentalmalzemelerin
testing
of the
the factor
safety
iskapasiteleri
selected
loading
andmaterial,
geometry.
order
to avoid
the potential
for
failure.
Specific
values
depend
the
the
failure Stated
load
Fmathematically,
Ffail is found
from
fail to the allowable load F
allow . Here for
1.7
DESIGN
OF
SIMPLE CONNECTIONS
47
accounted
when
the of
member
is used under similar con
based
on
experience
that
the above
mentioned
uncertainties
are
If the
load
applied
tosothe
member
is
linearly
related
to
the
stress
types
of
materials
to
be
used
and
the
intended
purpose
the
structure
experimental
testing
of theyükü
material,
and the
factorondan
of safety
is selected
bunlar
alınarak
yapıya
yüklemek
yerine
daha
düşük
izin verilebilir
loading
geometry.
Stated
mathematically,
accountedwithin
for
when
the member
used
under
similar
conditions
developed
the dikkate
member,
as inison
the
case
ofgöçme
using
s example,
=
P>A
and
or
machine.
theof
F.S.
used and
in the
design of
aircraft
based
experience
soFor
that
the
above
mentioned
uncertainties
are or spaceF
loading
andthen
geometry.
mathematically,
tavg
canStated
also express
factor
ofolacaktır.
safety
asfail
a Genellikle
ratio
of
the
vehicle
components
may
close
to 1 similar
insırasında
orderconditions
to reduce
the
weight ofizin
P theuygun
than
1=inV>A,
birweyükleme
yapmak
tasarım
göçmeof
yükünün
accounted
for
when
is be
used
under
F.S. the
= member
1.7 power
D(1–8)
ESIGN OF SIMPLE CONNECTIONS
47
failure
stress
s
(or
t
)
to
the
allowable
stress
s
(or
t
);∗
that
is,
the
vehicle.
Or,
in
the
case
of
a
nuclear
plant,
the
factor
safety
1
F
nd on the
fail OF SIMPLE
fail CONNECTIONS
allow Stated
allow
allow
1.7 DESIGN OF SIMPLEof
CONNECTIONS
47
loading and geometry.
mathematically,
1.7
DESIGN
47
F
1.7
D
ESIGN OF SIMPLE CONNECTIONS
47
fail
for bilinen
some ofbir
itsgüvenlik
components
may /bekatsayısı
as high belirlemek
as 3 due touygun
uncertainties
verilebilir yüke oranı olarak
faktörü
structure
F.S. =olacaktır.
Ffailthe
Fallow
in loading
orsafety
material
behavior.
In than
many1 cases,
the factor of safety for
aP
or spaceIn any of theseF.S.
equations,
factor of
must
be
greater
in
= equations,
(1–8)
(!b)allow
P
In
any
of
these
the
factor
of
safety
must
be
greater
than
1
in
P
s
eater
than
1
in
If
the
load
applied
to
the
member
is
linearly
related
to
the
stress
F
specific
case
can
be
found
in
design
codes
and
engineering
handbooks.
fail
weight
1
allowfor failure. Specific values
order to avoid F.S.
the Ppotential
depend
on
the
han 1 inof
F
fail
= the potential for 1failure.F.S.
order to avoid
Specific
values
depend
on
the
=(1–9)
(1–8) public and Assumed uniform 1
within
member,
asintended
in are
the intended
case
of using
sstructure
=aP>A
and
These
values
to the
form
balance
of ensuring
sthe
of
of materials
to be
used
and 1the
purpose
of
depend
onsafety
theon the typesdeveloped
allow
F
normal
stress to
allow
types
of materials
bealso
used
and the
intended
purpose
the
structure
Ifof
load
applied
to the solution
member to
is linearly
related
the structure or machine.
tavg
= V>A,
then weto
can
express
the
factorand
of
as
athe
ratio
of the
safety
providing
aspacereasonable
economic
ertainties
For example,
theenvironmental
F.S.
used
in the
design
of safety
aircraft
or
tructure
# in
distribution
B the
or
machine.
For
example,
F.S.
used
the
design
of
aircraft
or
spacedeveloped
within
applied
to
the smember
isdesign.
linearly
totothe
orIfspacefailure
stress
(or tbe
)close
to
the
sstress
tweight
that
is, the member, as in the case of using s =
fety
for
athe load
failmay
fail
allow (or
allow);∗ of
vehicle
components
toallowable
1 related
in orderstress
reduce
the
rraft
spaceP
ordeveloped
(!b)vehicle
allow
components
may
be
close
to
1
in
order
to
reduce
the
weight
of
t
=
V>A,
then
we can also
express
the factor of
within
the
member,
as
in
the
case
of
using
s
=
P>A
and
the
weight
of
avg
A "safety as a ra
ndbooks.
the
vehicle.
Or,deplasman
in the
caseload
of a applied
nuclearoran
power
theve
factor
of safety
Eğer
yük ve
arasındaki
# plant,
# linearly
eşitliklerinde
olduğu
gibi
eight of
If the
to the
member
is
related
to the stress
!
)allowt
(
the
vehicle.
Or,
in
the
case
of
a
nuclear
power
plant,
the
factor
of
safety
b
failure stress sfail (or tfail) to the allowable stress sallow (or
tof
= V>A,
thensome
we can
express the factor
of safety
a ratio
of to
theuncertainties
actor
safety
Assumed
uniform
allow
avg
ublic
and
for
of also
its components
may be
asmember,
highasas
3asdue
of
safety
developed
within
the
in
the
case
of
using s = P>A and
B
normal
stress
for
some
of
its
components
may
be
as
high
as
3
due
to
uncertainties
s
t
failure
stress
s
(or
t
)
to
the
allowable
stress
s
(or
t
);∗
that
is,
uncertainties
fail
B
fail
fail
fail
allow
allow
olution
in loading or material
InF.S.
the (1–10)
factor
of safety
for a(1–9)
rtaintiesto
Bmany
tor
V>A,behavior.
then
we
can
also express
the factor
factor
of safety
as
distribution
= cases,
==behavior.
avgmaterial
B
(!bthe
)allow
in loadingF.S.
In
many
cases,
of
safety
fora aratio of
Design
ofthe
Simple
Connections
of safety
The area of the column
tstress
s
(!b)allow
allowin s
case
can
be
found
design
codes
and
engineering
handbooks.
allow
ety
for a for a specific
(
!
)
failure
(or
t
)
to
the
allowable
stress
s
(or
t
);∗
that
is, katsayısı
b
allow
doğrusal
ise:
#
ve
yazılabilir.
Bu
denklemlerde
FS from the allowable bea
fail # in design
fail
allow
allow emniyet
Pallow
(
!
)
sfail
specific
case
can
be
found
codes
and
engineering
handbooks.
b
g
handbooks.
Assumed
uniform
AThese
"
values are intended to form a balance of ensuring public and
dbooks.
F.S.
=
Assumed
uniform
These
values are intended
to formsimplifying
a balance assumptions
of ensuring public
andnormal
!b)allow
(uniform
Assumed
uniform
stress
sproviding
regarding
thebüyük
behavior
of sthe
ng
failBy making
allow
Assumed
environmental
safety
and
a
reasonable
economic
solution
to
blicpublic
and and
nin
1’den
büyük
olması
gerektiği
aşikardır,
aksi
taktirde
göçme
yükünden
bir
yüke
izin
normal
stress
F.S.
=
(1–9)
orenvironmental
stress
safety
and
providing
a reasonable
economic
solution
todistribution
normalnormal
stress
s
=
P>A
t
=
V>A
material,
the
equations
and
can
often
be
used
to
ic
solution
to
s
avg
distribution
allow
design.
ution
distribution
s
*Into
some cases,
such
as
columns,
the
applied
load
is
not
linearly
related
to
stress
and
fail
distributiondesign.
P
analyze
or See
design
or mechanical
In
F.S.a 13.
=simple connection
verilmiş
olunur.
FS
değerleri
tasarım
alanına
ve kullanılacak
malzemeye
göre
The
area
thebe
column
platenin
B the
is özel
determined
P
therefore only
Eq.
1–8ofcan
used
determine
factor
of safety.
Chapter
A " (1–9)element.
or
P tobase
s
allow
P
A
"a section, its
from the allowable
bearing stress for the concrete.
)allow
(!bat
t
particular,
if
a
member
is
subjected
to
normal
force
A
"
fail
A"
(!b)allow
or
F.S.area
=uygun
(1–10)
değişecektir.
değerine
bakılabilir.
(!b)allow(!b)allow İlgili yönetmeliklerden
required
theFS
section
is determined
from
r of the
tat
allow
tfail B is determined
or
e used to
Design
of Simple
Connections
The area of the column
base plate
F.S.
=
Design
of
Simple
Connections
The
area
of
the
column
base plate B is determined
sment. In The areaTheofarea
tfail
of the column
baseBplate
B is determined
from the allowable bearingtstress
allow for the concrete.
the column
base F.S.
plate
P
from the allowable bearing stress for the concrete.
= is determined
(1–10)
from
the
allowable
bearing
stress
for
the
concrete.
A = of the
(1–11)
tassumptions
allowable
bearing
stress for the
concrete.
ction, its from the
By
making
simplifying
regarding
thelinearly
allow
tbehavior
Basit
Bağlantıların
*In
some
cases, such
as Tasarımı
columns,
the applied
load
is not
stress
fail
sto
By
making
simplifying
assumptions
regarding
therelated
behavior
ofand
the
allow
havior
F.S.
=
(1–10)
Assumed unif
s
=
P>A
t
=
V>A
the
equations
and
can
often
be
used
to
of theof the material,
therefore only Eq. 1–8 can be used to determine
safety.
See Chapter 13.
tallow
tallow
= ofV>A
material, the equations s = P>A avg
andthet factor
can
often be used to
en
be to
used to analyze
or design a simple connection or avg
mechanical element.
In
used
*In some cases, such as columns, the applied load is not linearly related t
analyze or design a On
simple
connection
or
mechanical
element.
In
element.
In
the linearly
other
hand,
iftoforce
the
section
is subjected
to can
an average
force,
therefore
onlyits
Eq. 1–8
be used to shear
determine
the factor of safety.
particular,
if a member
isload
subjected
to normal
section,
ment. *In
In some cases,
P See Chap
such particular,
as columns,
the
is not
related
stress at
anda
l " —————
if applied
a member
subjected
toarea
normal
force
at a section, its
P is the
a (1–11)
section,
required
tallowpd
required
area
attothe
sectiontheisthen
determined
from
tion,
its its
therefore
only Eq.
1–8 can be
used
determine
factor
of safety.
See
Chapterat
13.the section is
required area
the section is determined from
*Inat
some
Assumed uniform shear
stresscases, such as columns, the applied load is not linearly related to stress and
tallow therefore only Eq. 1–8 can be used to determine the factor of safety.
V See Chapter 13.
A =
(1–12)
P
P
t(1–11)
ear force,
P
A P=
allow
P
P
=
(1–11)
sA
l " —————
(1–11)
P
allow
t" allowpd
(1–11)
Assumed uniform shear stress
"
sallow
Thestress
embedded leng
Assumed
uniform shear
tallow
Assumed
uniform
shear stressAs discussed in Sec. 1.6, the allowable stress used in each
of these
Assumed
uniform
shear stress
tallow
can be determined
tallow
t
equations
is determined
eithershear
by applying
of safety to the
allow hand, if the section
stress of th
On the other
is subjected
to an average
force, a factor
d hand, if the section
P——
(1–12)
On the other
is subjected
to anfailure
averagestress
shearorforce,
———
l "finding
e shear
P
P
material’s
normal
or shear
by
these
stresses
the
required
area
at
the
section
is
P
ar
force, force, then
pd
t
————
allowl " —
——— the required area at the section is
P
l—
"———
then
tallowpd
l " ———
directly from an appropriate design code.
tallowpdtallowpd
P
The embedded length l of this rod in concrete
Three V
examples of where the above equations apply are shown in
of these
d
can be determined using the allowable shear
V
A
=
(1–12)
d
P
ty to the
tA
dglue. Fig. 1–25.
=
(1–12)
stress of the bonding
allow
(1–12)
d
P
tallow
P
e (1–12)
stresses
P
"
"
(Burada üniform kayma gerilmesi varsayımı yapılmıştır.)
The embedded length l of this rod in concrete
As discussedlength
in Sec.
1.6,rodthe
allowable stress used in each of these
The embedded
length
l of this rod
in concrete
ofin
this
in
concrete
can be determined
using
the allowable
shear
PThe embedded
As
discussed
1.6,
the allowable stress used in each of theseAssumed
embedded
length
l of thisl rod
inSec.
concrete
uniform
each
ofintheseThe equations
can
be determined
using
the allowable shear
shown
is
determined
either
by
applying a factor of safety to the
of
these
can
be
determined
using
the
allowable
shear
stress
of
the
bonding
glue.
can be determined
using the
allowable sheareither by applying a factor of safety to the shear stress stress of the bonding glue.
equations
is determined
of the
bonding
tallow
normal
or
shearglue.
failure
ysafety
to theto the material’s
stress ofstress
the bonding
glue.
V " Pstress or by finding these stresses
material’s
normal
or shear
failure stress or by finding theseP stresses
these
stresses
directly from an appropriate design code.
stresses
P
directly from an appropriate design code.
P A"
The area of the
Three
examples of where the above equations apply are shown in
P tallow
P
are shown
in PFig. 1–25.Three examples of where the above equations apply are shown in
is determined f
m
hown
in
which is larges
Fig. 1–25.
1.7
1.7 1.7
"
w
niform
ess
w
P
" t
allow
P
P
"
The area of the bolt for this lap joint
is determined from the shear stress,
V " is
P largest between the plates.
which
V"P
P
P
P
Fig. 1–25
The
area
of
the
bolt
for
this
lap joint
The area of the bolt for this lap joint
is determined
from
thestress,
shear stress,
is determined
from the
shear
which is largest between the plates.
Assumed uniform
shearAssumed
stress uniform
tallowshear stress
tallow
P
P
A" t
P
allow
A" t
allow
Fi
#17
P
P
The area of the bolt for this lap joint
The
area
of
the
bolt
for
this
lap
joint
is determined from the shear stress,
determined
from
shear stress,
which isislargest
between
thethe
plates.
which is largest between the plates.
1.7
DESIGN
SIMPLEOFCS
ONNECTIONS
1.7 OF
DESIGN
IMPLE CONNECTIONS
49
49 1.7
DESIGN OF SIMPLE
DESIGN OF SIMPLE CONNECTIONS
49
1.7
D
ESIGN OF SIMPLE CO
1.7 DESIGN OF SIMPLE CONNECTIONS
49
1
EXAMPLE 1.13
1
1.7 DESIGN OF SIMPLE CONNECTIONS
49
1.13
he control
arm isEXAMPLE
subjected
to the
loading
shown in
Fig. in
1–26a.
DESIGN OF SIMPLE
CO
The control
is subjected
to the loading
shown
Fig.arm
1–26a.
1
The control
is subjected to the loading shown in Fig.1.71–26a.
" arm
1.7
XAMPLE
1.13 1.13
EXAMPLE
1
EXAMPLE
1.13
1
etermine
to the nearest
diameter
of the steel
pinsteel
EXAMPLE
1.13
in. the
Determine
to the
nearest
required
diameter
of the
4 in. the4 required
Determine
to thepin
nearest 14 in. the required diameter of the steel pin
The
control
arm
is
subjected
to
the
loading
shown
in Fig. 1–26a.
=
8
ksi.
C if the
allowable
shear
stress
for
the
steel
is
t
allow is tat
= the
8 ksi.
at C if the allowable shear stress for the steel
= 8Fig.
ksi.1–26a.
C
if
allowable
shear stress
for
the
steel
is
tallow
1
Çelik
için
kayma
emniyet
gerilmesi
# allow
ise isCshown
mesnedindeki
perçinin
çapını
belirleyiniz.
EXAMPLE
1.13
The
control
arm
subjected
to pin
the
loading
shown
in
in.
Determine
to the
nearest
the
required
diameter
of
The control
arm
is subjected
to
the
loading
insteel
Fig.
1–26a.
1
EXAMPLE 1.13 the
4
1
1
in.
Determine
to
the
nearest
the
required
diameter
of
the
steel pin
at CDetermine
if the allowable
stress
forthe
therequired
steel is tdiameter
4 the steel pin
in.
to theshear
nearest
of
allow = 8 ksi.
4
A to the
FAB
The control arm Bisatsubjected
inthe
Fig.
1–26a.
A loading shown
FAB stress forAthe steel is tallow = 8 ksi.
at
C for
ifcontrol
the
allowable
shear
F
=
C ifB the allowable
shear stress
steel
tallow
The
armis is
subjected
B 8 ksi.to the loading shown in Fig. 1–26a. AB
Determine to the nearest 14 in. the required diameter of the steel pin
1
Determine
diameter of the steel pin
A = 8 ksi.to the nearest 4 in. the Frequired
AB
CESIGN
if theOFallowable
shear stress for theBsteel
1.7 at D
SIMPLE CONNECTIONS
49 is tallow
at C if the allowable shear stress for
A the steel is tallow = 8 ksi.
FAB
A
B
8 in.
B
1.7
1–26a.
teel pin
8 in.
1
DESIGN OF SIMPLE CONNECTIONS
8 in.
C
C
C
49
8 in.
8 in.
C
FAB
B
8 in.
FAB 8 in.
8 in.
B
A
Cx
C Cx
8 in.
1
FAB
A
8 in.
8 in.
8 in.
8 in.
C
C
Cx
C
5
5
8 in.
3
3
8
in.
5
4
3
3 in.
2 in.
4
3
C3 in. Cx 2 in. C
3 in.
3Cin.
2 3in.in.
5 kip
2 4in. 5 kip
C 4
3 in.
2 in. 5 kip4
5
Cx
C
C
hown in Fig. 1–26a.
5
kip
3
C
C
C
C
3
kip
5
kip
5
x
y
Cy
3 kip
F
Cy
3 kip
3k
4
3 kip
" 2 in. 3 4 5
5
meter ofABthe steel " pin
3 kip3 in. 3 52 in.
C
3 in.
35 kip
(b)
Fig.
1–26
(b)
3
in.
2(b
C
(a)
Fig. 1–26
Cx35 kip C
C
(a)
4
33 in.
2 in.
4Fig. 1–26
2Cin.
allow = 8 ksi.
(a) 3 in.
kip
C
4
C
y 5
in.
2 in.
x
5C
kip
5
kip
Çözüm: Serbest cisim diyagramı: 5 33 kip
5 kip
3
Cy
3 kip
3 in.
2 in. 3 kip 4 Cy 3(b)5 3 kip
(a) 3 4
3 kip Fig. 1–26
OLUTION
3
in.
3
in.
2
in.
5 kip
SOLUTION F
(b) 2
Fig.4 1–26
SOLUTION
(a)
3 in.
2 in.
(b)
Fig. 1–26
AB
(a) 5 kip
8 in.
C
3
kip
5
kip
y
nternalInternal
Shear Force.
A free-body
diagram
of the arm
is shown
3diagram
kip
Cy
Shear Force.
A free-body
ofInternal
the
arm Shear
is in
shown
in
3 kip
Force.
A free-body 3diagram
of the arm is shown in
kip
SOLUTION
(b)
g. 1–26b.
For
equilibrium
we
have
Fig.
1–26
(a)
Fig. 1–26b. For equilibrium we have
(b)
Fig.
1–26b.
For
equilibrium
we
have
Fig.
1–26
SOLUTION
(a) in
SOLUTION
Internal
Shear Force. A free-body diagram of the arm is shown
8 in.
3
5C
5
1
C
+ ©MCd +=©M
0; C = 0;
FAB18 in.2
3in.2
kip -133in.2
kip in.2
0 Force.
A 35 d+
FAB-18For
kip -135we
in.2
5B 15
kip
A 35Shear
B=15
Internal
diagram
armA 3is
shown in
©M
= in.2
0; =of0the
FABA
18free-body
in.2
- 3 kip
5 kip
Fig. Internal
1–26b.
equilibrium
have
Shear
Force.
A
free-body
arm
is shown
in 13 in.2of-the
Cdiagram
5 B 15 in.2 = 0
SOLUTION
8 in.
Cx
C
Fig.
1–26b.
For
equilibrium
we
have
SOLUTION
= F
3For
kipequilibrium
Fig.F1–26b.
we
have
3
AB
3 kip
AB =
d+ ©M
=3 0;
FAB
18 in.2 kiparm
13 in.2
- 5 kip
= 0= 3 kip
5
C A
Internal Shear Force.
free-body
diagram
of 3the
is shown
in A 5 B 15 in.2FAB
4
+
Internal
Shear
Force.
A
the arm
shown
3 free-body
4
d+
©M
=
0;
F
18
in.2
3 kip diagram
13 in.2 - of5 kip
= 0 in
A 35 B 15isin.2
+
4
: ©F
=
0;
-3
kip
C
+
5
kip
=
0
C
=
1
kip
C
AB
3
in.
2
in.
A
B
d+-3
©M
0;
in.2
kip
- 5kip
in.2
+3 3kip
x:
xC
x0 A 5 B 15
©Fx For
= 0;equilibrium
kip
-5have
C + 5F
kip
==
Cxx 13
== 1in.2
Fig.
1–26b.
we=
ABA18
:
©F
0;kip
3 kip
- have
Cx += 50kip A 45 B = 0
Cx = 1 kip
5 BAB
F
kipx 5
Fig.
1–26b.
For
equilibrium
we
FAB
= 3 kip
c ©F
=c ©F
0; y=Cy=0;0; C3+
3C
kip
- 35kip
= in.2
0 A 3 B -=C
=c=6A 3C
kip
A 35 5B 13
4=in.2
3 15
kip
ykip
y0
-0;
kip
--3
kip
60;kip
d +y+©M
F
18y=in.2
3 kip
5+
kip
6.082
kip
AB
3
C
ABx
+5©M
©F
=0;=
3 kip
kip
kipin.2
C
= 6 kip
©F
kip
-5 CxF
kip
0= 0 FCxC18
=y 6.082
1 kip
A3B = 0
A
B
5 By
y=
5
d+
in.2
3 -kip5 13
C:
4 5 - 5 kip A 5 By15 in.2 = 0
x(b) C
C
AB
+ ©F = 0; 4
he pinThe
at Cpin
resists
the
resultant
force
at
C,
which
is
:
3
kip
C
+
5
kip
= 0
Cx = 1 kip
+
A
B
x
x
at C resists
the
resultant
force
at
C,
which
is
5
5
:
©F
=
0;
-3
kip
C
+
5
kip
=
0
C
=
1
kip
3
A
B
The
pin at C5 resistsCthe=resultant
force at C, which
is
x
x6 kip
kip
5
+ c ©Fy = x0;FAB = 3 C
6.082 kip
y3 - 3 kip - 5 kip A 5 B = 0
y
FAB = 3 kip3.041
3
3 kip
23 in.
4 2
2kip2
2in. 2 = 6.082
3
c
FC = 211
kip2
+
16
kip
+
©F
=
0;
C
3
kip
5
kip
=
0
Cy kip
= 6 kip
3.041
kip
4
A
B
4 +
2
2
y
y
FC -3
=
+ the
16
kip2
kip
c211
54 = 6.082
+
©F
=
C
-== 306.082
kip
-C
kip1which
= is
0 = 211
Cy kip2
= 6 kip
A5B F
6.082
:
©Fx = 0; The
kip
C0;
kip
kip
+ 16 kip2
kip
pin
at C
resists
resultant
force
at
C,
A 5y5B kip
y-kip2
x + 5
x5 =
+ ©F
C 5 kip
:
=
0;
3
kip
C
+
5
kip
=
0
C
= 1 kip
A
B
3.041
kip
x
x
x
C
3
kip
5
3.041
kip
y
nce the
pin the
is subjected
to double
shear,
ashear,
shear
3.041
Theof
pin
at2 at
C
resists
the
force at C, which is
Since
pin is subjected
to Cdouble
aforce
force
ofkip
3.041
kipisresultant
3 resultant
The
pin
at
resists
the
force
C,
which
2 shear
Since
the
pin
is subjected
to double
a shear force of 3.041
3.041kipkip
Pin atkip
C shear,
c
+
©F
=
0;
C
3
kip
5
kip
=
0
C
=
6
kip
F
=
211
kip2
+
16
kip2
=
6.082
kip
A
B
3
6.082
Pin
at
C
hown
in
y
y
y
C
5
(b) between
cts
over
its over
cross-sectional
area between
the armthe
andarm
each
supporting
cand
Fig.
1–26
+2acts
©F
0;
Cy -kip2
3 kip
-between
5kip2
kip 2A 5the
=
0 and
Ceach
kip
2
B
acts
its" cross-sectional
area
each
supporting
y = its
y = 6supporting
over
cross-sectional
area
arm
2F
=
211
+
16
=
6.082
kip
"
(c)
3.041 kip 3.041 kip
(c)
FC =at 211
kip2 is +shear,
16 kip2
=C 6.082
The
C resists
the1–26c.
resultant
force
C,towhich
Since
the
pin is subjected
double
a shear
forcekip
of 3.041 kip
af for
thepin
pin,
Fig.
1–26c.
leaf
foratthe
pin,
Fig.
The
C resists
the
resultant
force
at C,awhich
is
leafpin
foratpin
the
pin,
Fig.
1–26c.
Pin at
Since
the
is
subjected
to
double
shear,
shear
force
ofC3.041
kip
3.041 kip
actsSince
over its
cross-sectional
area
the arm
each
supporting
the
tobetween
double
shear,
a and
shear
force
of 3.041 kip
2
equired
Area. Area.
We
3.041 kip
.2 = 0Required
FChave
= We
211have
kip2pin
+ is16subjected
kip22 = 6.082
kip
(c)
2
2
Pin
at
C
Required
Area.
We
have
acts
over
its
cross-sectional
area
between
the
arm
and
each
supporting
F
=
211
kip2
+
16
kip2
=
6.082
kip
for the
pin,
1–26c.
" leafVacts
C
over
its Fig.
cross-sectional
area between the arm
and each supporting3.041 kip
3.041
kip
3.041
kip a shear
V double
(c)
2leafforce
pinAisin
subjected
of pin,
3.041
kip1–26c. V
forthe
Fig.
f theSince
arm isthe
shown
3.041
kip shear force
2the
=Required
=forto
=shear,
0.3802
in
leaf
the =
pin,2We
Fig.
1–26c.
Since
pin
is
subjected
to double
shear,
Area.
have
A
=
=
0.3802
in
Pin at Ca
A = kesilmeye
=
= 0.3802 in2 of 3.041 kip
tallow tallow
2 tesirlidir.
8area
kip>in
acts over its cross-sectional
between
the
armRequired
and each
supporting
C
mesnedindeki
perçin
çift
Yani
iki
noktadan
çalışılmaktadır.
2
8
kip>in
t
Area.
We haveallow
8 kip>in
acts over
cross-sectional
area between
(c)the arm and each supporting
kip
Area. WeVhave 3.041
kip its
2
leaf for the pin, Fig. Required
1–26c.
2 =
d 2
A
= 2 leaf for2the
= pin,
0.3802
in1–26c.
2 kip
d
3.041
Fig.
2
V
5 kip A 35 B 15 in.2 = 0
2
3.041 kip
tallow Vin 8 kip>in
pa b pa
= 0.3802
in
b =A
0.3802
kip Required Area.
A = in2 pa
=d
= 0.3802
6.082We
kip have
= 0.3802
in2 in
=
=
= 0.3802
2
2
tallow 28bkip>in
2
2
Area.
We
have
2Required
tallow
8 kip>in 2
d
3.041
kip
Vd = 0.696
d =in.
0.696
in.
2
d = kip
0.696 in.
pa
b =in20.3802
in
V d 23.041
A =
=
=
0.3802
2
2 d
=b = 0.3802
= 0.3802
in2
Use C
a xpin
diameter
of 3.0418of
= having
1 kip
2 A =
pa
allow
kip>in2
kip
Use
a pin ahaving
a tdiameter
pad =b 0.696
=
0.3802
in
Use ain.
pin having a diameter
tallow 2 8ofkip>in2 in
3
2
30.750 in.
d =3.041
Ans.
= 2kip
d=
Cy = 6 kip
dkip
in.a diameter
= 0.7502in.
4 in.
041 kip
in. in.
6.082
d d=d342 =in.0.696
Ans.
= 0.750
4 =
d = 0.696 in. Ans.
Use
a
pin
having
of
pa
b
0.3802
in
Pin at C
"
"
"
pa of
b = 0.3802 in2
2
Use
a
pin
having
a
diameter
hporting
is
3
2
Use
a
pin
having
a
diameter
of
(c)
Ans.
d = 0.696din.= 4 in. = 0.750 in.
d == 0.696
d = 34 in.
Ans.
0.750in.
in.
3.041dkip
82 kipUse a pin having a diameter of
= 34 in. = 0.750 in.
Ans.
Use a pin having a diameter of
3.041 kip
ar force of 3.041 kip
d = 34 in.Pin=at0.750
Ans.
in.
C
d = 34 in. = 0.750 in.
Ans.
and each supporting
(c)
02 in2
Ans.
Ans.
#18
6.08
6.082 k
6.082 k
disk as shown in Fig. 1–27a. If the rod passes through a 40-mm-diameter
hole,
determine
the minimum
required diameter of the rod and
The suspender rod is supported at
its end
by a fixed-connected
circular
The suspender rod is supported
at its end thickness
by a fixed-connected
circular to support the 20-kN load.
the
minimum
of
the
disk
needed
50
C H Adisk
P T E as
R 1shown
S T R EinS SFig. 1–27a. If the rod passes through a 40-mm-diameter
TER 1
STRESS
disk
as
shown
in
Fig.
1–27a.
If
the
rod
passes
through
a
40-mm-diameter
50
CHAPTER 1 STRESS
allowable
normal
for the
STRESS
hole, determine the minimum The
required
diameter
ofstress
the rod
androd is sallow = 60 MPa, and the
the minimum
required
diameter
of disk
the rod
and= 35 MPa.
allowable
shear
stress
for
the
is tallow
50
C H A P Tthe
E R 1 hole,
S
T R Edetermine
SS
minimum thickness of the disk needed to support the 20-kN load.
the minimum thickness of the disk needed to support the 20-kN load.
EXAMPLE The
1.14
allowable normal stress for the rod is sallow = 60 MPa, and the
1.14
1EXAMPLE 1.14
= 60 MPa,t and the
The allowable normal stress for the rod is sallow
40 mm
40 mm
1
allowable shear stress for the disk is tallow = 35 MPa.
=
35
MPa.
allowable
shear
stress
for
the
disk
is
t
50 C H A P T E R C
H
A
P
T
E
R
1
S
T
R
E
S
S
1 S EXAMPLE
RESS
allow
1.14
The
rod is supported
at its end by a fixed-connected circular
1 "
The suspender rod is supported at its
end
bysuspender
a fixed-connected
circular
The
suspender
supported
atthe
itsrod
endpasses
by a fixed-connected
circular
t rodinis
40
mm
The suspender
rod
is
supported
at
its
end
by
a
fixed-connected
circular
disk
as
shown
Fig.
1–27a.
If
through a 40-mm-diameter
tallow
40 mm
disk as shown in Fig. 1–27a. If the roddisk
passes
through
at 40-mm-diameter
40
mm
A
40amm
as
shown
in
Fig.
1–27a.
If
the
rod
passes
through
a
40-mm-diameter
The
suspender
rod
is
supported
at
its
end
by
fixed-connected
circular
Şekildeki
askı
çubuğu
ucundaki
dairesel
disk
vasıtasıyla
mesnetlidir.
Eğer
çubuk
40
mm
disk as shown
in
Fig.
1–27a.
If
the
rod
passes
through
a
40-mm-diameter
hole,
determine
the
minimum
required
diameter
of
the
rod
and
hole, determine the minimum required
diameter the
of the
rod and
EXAMPLE
1.14
PLE
1.14 hole,
hole,
determine
minimum
required
the
rod
and
disk
as
shown
inen
Fig.
1–27a.
If
rod
passes
through
a of
40-mm-diameter
determine
the thickness
minimum
diameter
of
the
rod
and çapı
the
minimum
thickness
ofthe
theve
disk
needed
to support
theiçin
20-kN
load.
çapında
bir delikten
geçiyorsa
için
gerekli
küçük
20
kNdiameter
yükü
taşımak
the minimum
ofrequired
the çubuk
disk the
needed
to
support
the
20-kN
load.
tallow
minimum
thickness
of
the
disk
needed
to
support
the
20-kN
load.
hole,
determine
the
minimum
required
diameter
of
the
rod
andthe
A
t
thediskin
minimum
thickness
of
the
disk
needed
to
support
the
20-kN
load.
s
=
60
MPa,
The
allowable
normal
stress
for
the
rod
is
and
allow
allow
A
send
= its
MPa,
The
normal
stress
the
and
the
enallowable
küçük
kalınlığını
belirleyiniz.
Çubuk
için
normal
emniyet
gerilmesi
# = 60the
allow
The suspender
rod
isrod
supported
at
end
by
athe
fixed-connected
The suspender
rod
is for
supported
atisits
by
a60and
fixed-connected
circular
scircular
MPa,
allowable
normal
stress
for
the
rod
is to
andload.
the
the
minimum
thickness
offor
disk
needed
support
20-kN
allow
s
= 35
60 MPa.
MPa,
The allowable
normal
stress
for
the
rod
isisThe
the
allowable
shear
stress
the
disk
allow
dis tallow = 35 MPa.
=
allowable
shear
stress
for
the
disk
t
20 kN
allow
disk
as
shown
in
Fig.
1–27a.
If
the
rod
passes
through
a
40-mm-diameter
ve
disk
için
kayma
emniyet
gerilmesi
#
olarak
verilmektedir.
disk
as
shown
in
Fig.
1–27a.
If
the
rod
passes
through
a
40-mm-diameter
=
35
MPa.
allowable
shear
stress
for
the
disk
is
t
s
=
60
MPa,
The
allowable
normal
stress
for
the
rod
is
and
the
allow
allow
allowable shear stress for the disk is tallow = 35 MPa.
hole,
determine
the
minimum
required
diameter
of
the
rod
and
hole, determine the minimum
required diameter
of the
and= 35 MPa.
the
disk rod
is ttallow
40 mm
40 mm
t allowable shear stress for
40 mm
(b)
40 to
mmsupport
t
40 mm
thickness
the disktoneeded
the
20-kN
d ofneeded
the40minimum
thickness
of the disk
support
the
20-kN
load.
40 mm
20 kN load.
t
20
kN
mm the minimum
d
40 mm
20 kN
sallow
=t and
60 MPa,
The allowable
normal
foristhe
rod
40
mm
sallow
=is 60
MPa,
The allowable
normal stress
forstress
the rod
the and the
40 mm
tallow
(b)
(a)
=
35
MPa.
allowable
shear
stress
for
the
disk
is
t
t
MPa.
allowable shear stress for the disk is tallow = 35allow
A
allow
(b)
A
t
20 kN
20 kN A
t
t
40 mm (a)40 mm
(a)
end by a fixed-connected circular
passes through a 40-mm-diameter
quired diameter of the rod and
needed to support the 20-kN load.
" MPa, and the
e rod is sallow = 60
tallow = 35 MPa.Çözüm:
d
tallow
40 mm
Fig. 1–27
40 mm
allow
A
tallow
A
SOLUTION
Fig. 1–27
20 kN
Fig. 1–27 d 20 dkN
tallow
tallow
A
20 the
kN rod is 20 kN.
Diameter
20 of
kN Rod.A By inspection, the axial force in
d
(b)kN
Thus the required
(b) cross-sectional area of the rod is 20
d
SOLUTION
SOLUTION
20 kN
(b)
20 kN
(b)
kN in the rod is 20 kN.
Diameter
of Rod. By inspection, the axial20force
20 kN
3
(b)
Diameter of Rod. By inspection,
the
axial
force
in
the
rod
is
20
kN.
20110 2 N
P the 20
p 2
(a)
kNis
Thus the
cross-sectional
rod
(a)required
d Aarea
;
= of
=
d
d cross-sectional
20
kN
Thus
the required
area
of
the
rod
is
sallow(a)
420 kN 6011062 N>m2
(a)
Fig. 1–27
3
(a)
Fig.
1–27
20110 2 N 3
P Fig. 1–27
p 2
Fig.
(b) 1–27 (b)
20110
2
N
P
p
;
A =
d = so
2 that
2
; 20 kN 4 20 kNd60110
A
=
= 62 N>m
Fig. 1–27
sallow
40 mm
s
4
SOLUTION
6011062 N>m2
allow
SOLUTION
SOLUTION
(a)
(a)
SOLUTION
ofinRod.
By
inspection,
axial
in mm
the rod is 20 kN. Ans.
= the
0.0206
m force
= 20.6
so that By inspection,Diameter
Diameter of Rod.
theDiameter
axial force
theBy
rodinspection,
is 20
kN. dthe
SOLUTION
ofrequired
Rod.
axial
force
in is
the rod is 20 kN.
so
that
t
allow By inspection, the axial force
Diameter
of the
Rod.
in
the
rod
is
20
kN.
Thus
the
cross-sectional
area
of
the
rod
Fig.
1–27
A Thus
Fig.
1–27
required cross-sectional area
ofthe
therequired
rod is cross-sectional area of the rod is
Thus
Diameter
of Rod. Byofinspection,
the axial force
the free-body
rod is 20 kN.
Thus the required cross-sectional area of the
rod
Thickness
on inthe
diagram in
d = is0.0206
m = 20.6 mmDisk. As shown
3 Ans.
Ans.
d
=
0.0206
m
=
20.6
mm
Thus
required
cross-sectional
area
of
the
rod
is
3 the
20110
2
N
P
p
Fig.
1–27b,
the
material
at
the
sectioned
area
of
the
disk
must resist
20110
2
N
3
PSOLUTION
p 2 3
SOLUTION
;
d2 =20110 26N
pprevent
20110
;
A =
d = 2 NA 6=A =P s2 ;shear
2the disk through the hole. If this
24
P
p 2
stress
to
movement
of
60110
2 N>m
=rodin
dthe
allow on the
3rod
Thickness
Disk.
shown
theaxial
free-body
diagram
in kN.
4 of of
60110
2As
N>m
Diameter
Rod.
By
inspection,
force
; sallow
A =
dThickness
6 kN.
20110
2 diagram
Nis2 20
Diameter
of=Rod.
inspection,
the
axial force
isthe
20
PAs
sallow
4passumed
Disk.
shown
onisin
the
in
6Byof
2
60110
2 N>m
2 free-body
shear
stress
to be
uniformly
distributed
over the sectioned
sallow 20 kN
4Thus
60110
2
N>m
;
A
=
=
d
Fig.
1–27b,
the
material
at
the
sectioned
area
of
the
disk
must
resist
the
required
cross-sectional
area
of
the
rod
is
Thus
the
required
cross-sectional
area
of
the
rod
is
6
2
Fig.
1–27b,
the
material
at
the
sectioned
area
of
the
disk
must
resist
s
4 V = 60110
#
2
N>m
we
have
20
kN,
allowarea, then, since
so
that
shear stress to prevent movement of the disk through the hole. If this
so that
so that movement
shear stress to prevent
of the
3 disk through the hole. If this
so that (b)
3
shear
stressPis assumed
to be20110
uniformly
distributed
over the sectioned
20110
2distributed
N
2
N
p
P shear
passumed
2
stress
is
to
be
uniformly
overmthe
2 that d =
so
; V =dm
Ans.
d
=
0.0206
= sectioned
20.6 mm
; = dsince
A = area,Athen,
==kN,
1.7
D
ESIGN
OF
SIMPLE
CONNECTIONS
we
have
20
2011032 N 51
6
2
Ans.
=
0.0206
20.6
mm
6 we
2V 2 N>m
sthen,
4
60110
sallow
4
area,
since
have
V
=
20
kN,
Ans.
d
=
0.0206
m
=
20.6
mm
allow
60110
2
N>m
A
=
;
2p10.02
m21t2
=
Ans.
d
=
0.0206
m
=
20.6
mm
Çubuk çapı hesabı: #
#
tallow # d = 0.0206 m = 20.6 mm
3511062 N>m2
Ans. in
Thickness
of
Disk.
As
shown
on
the
free-body diagram
3
Thickness
of Disk.
As shownThickness
on the free-body
in on the free-body
1–27
so that
20110diagram
2As
N shown
so that
3
V
of
Disk.
diagram
20110
2
N
Thickness
of Disk.
As
onsectioned
the free-body
diagram
in1.7atresist
Fig.m21t2
1–27b,
the disk
material
the
sectioned area of the disk must in
resist
EXAMPLE
= shown
;atVthe
2p10.02
=the
Fig.1.15
1–27b,
theAmaterial
area
of
D
ESIGN OF SIMPLE CONNECTIONS
1 in51
6 must
2shown
A
=
;
2p10.02
m21t2
=
Fig.
1–27b,
the
material
at
the
sectioned
area
of through
the4.55
diskmm
must
resist
t
-3disk
Thickness
of
Disk.
As
on
the
free-body
diagram
allow
35110
2
N>m
6
2 4.55110
Fig. 1–27b,
the
material
at
the
sectioned
area
of
the
disk
must
resist
shear
stress
to
prevent
movement
of
the
the
hole.
If
this Ans.
t
=
2
m
=
t
allow # d of
stresshesabı:
to prevent
movement
the stress
disk
the
hole.
IfN>m
35110
2mm
Diskshear
kalınlığı
#
# this
Ans.
d =
0.0206
m
=
20.6
Ans.
=shear
0.0206
mthrough
=the20.6
mm
to
prevent
movement
of the
disk
through
hole.
If
this
Fig.
1–27b,
material
attothe
sectioned
area
of
the the
disk
must
resist
to
prevent
movement
of
the
disk
through
the
hole.
If
this
shear
stress
is
assumed
be
uniformly
distributed
over
the
sectioned
The shear
shaft stress
shown
in
Fig.
1–28a
is
supported
by
the
collar
at
C,
which
is
shear stress is assumed to be uniformly
distributed
over thetosectioned
shear
stress
isthe
assumed
bekN,
uniformly
distributed
overthe
thehole.
sectioned
shear
stress
prevent
the
disk through
If this
the axial force
in theto
rod
kN.
shear
stress
is20
assumed
toVbe
uniformly
distributed
over
sectioned
area,
then,
since
weofhave
V
20
-3to
attached
theisshaft
and
located
on
the
right
side
of
the
bearing
B. =movement
area,
then,
since
we
have
=
20
kN,
EXAMPLE
1.15
t
=
4.55110
Ans. inover the sectioned
2
m
=at=free-body
4.55
mm
-3
Thickness
of
Disk.
As
shown
on20
the
free-body
diagram
Thickness
of
Disk.
As
shown
on
the
diagram
in
area,
then,
since
we
have
V
kN,
1
shear
stress
is
assumed
to
be
uniformly
distributed
#
t
=
4.55110
Ans.
2
m
=
4.55
mm
ea of the rod
is
area, then,
have
V value
= 20 kN,
Determine
thesince
largest
of Pwe
for
the axial forces at E and F so
Fig.
1–27b,
the at
material
at the
sectioned
area
of must
the
disk
must resist
Fig.
1–27b,
the
material
the
sectioned
area
of
the
disk
resist
area,
then,
since
we
have
V
=
20
kN,
3
that the
bearing
stress
oninthe
collar
does
not exceed
allowable
3an
The shaft
shown
Fig.
1–28a
is to
supported
byofthe
C,disk
which
is
20110
2N
Şekildeki
şaft
B mesnedinin
sağ
tarafında
bulunan
noktasında
şafta
bitişik
yaka
ile3 mesnetlidir.
E ve
VC
shear
stress
prevent
movement
ofat
the
through
the
hole.
If this
shear
stress
to
prevent
movement
disk
through
the
hole. If
this
20110
2the
Ncollar
11032 N stress of
V75
3 A stress
20110
2
N
1s
2
=
MPa
and
the
average
normal
in
the
shaft
=
;
2p10.02
m21t2
=
V
b
allow
attached
toshear
the
shaft
and
located
on
the
right
side
of
the
bearing
at
B.
20110
2
N
A
=
;
2p10.02
m21t2
=
6
2
V
F
noktalarında
etki
eden
taşınabilecek
en
büyük
P
kuvvetini
hesaplayınız.
Yakadaki
emniyet
gerilmesi
shear
stress
is
assumed
to
be
uniformly
distributed
over
the
sectioned
t
stress
is assumed
to be
uniformly
over
the sectioned
6 allow
2
35110
2 N>m
A 35110
=55 MPa.
; distributed
2p10.02
m21t2
=
tallow
allowable
stress
of =1s
= exceed
; an
2p10.02
m21t2
6 32 N 2
062 N>m2 doesAnot
20110
t2
allow6 =
tallow
V2 N>m
the
largest
value
P
the
axial
forces
at E and F so
2kN,
35110
2 N>m
area,
then,
have
VA
=we
75 tDetermine
MPa
emniyet
gerilmesi
55
olarak
area,
then,
since
V of
=since
20for
kN,
allow ve şafttaki
35110
2MPa
N>m
=20have
;weverilmektedir.
2p10.02 m21t2 =
that the bearing stress on the collar doestnot
allow exceed an allowable
3511062 N>m2
-3
-3
t
=
4.55110
Ans.
2
m = 4.55 mm
stress of 1sAb2allow = 75tMPa
and
normal stress
in the 3shaft
= -3
4.55110
Ans.
2average
m =204.55
60
mmtheB
-3
3 t 20110
mm mm
=
4.55110
Ans.
2
m
=
4.55
mm
2
N
20110
2
N
V
V
P
P
t
=
4.55110
Ans.
2
m
=
4.55
mm
2allow
= 55 MPa.
does not exceed
stress
of 1s
t80
A; =
;
2p10.02
-3
3P
mm m21t2 = t = 4.55110
A = an allowable
2P
Ans.
2 62 N>m2 2 m = 4.55 mm
Ans.
m = 20.6 mm
tallow2p10.02 m21t2 = 35110622PN>m
t
35110
allow
F
E
C
#
A(a)
on the free-body diagram in
Axialmust 2resist
P
oned area of the disk
Force
the disk through the
hole. If thisF
mly distributed over the sectioned
3P
e
Axial
2P
20110 2 N
3511062 N>m2
3P
2P
B
20 mm
m
=
= 80
4.55
mm
mm
=-3 4.55110-32
t 2m
t = 4.55110
E
(b)
4.55
2 Pmm
C
Ans.
Ans.P
3P
(b)
(a)
Force
3
P
60 mm
Position
(c)
Fig. 1–28
Position
SOLUTION
(c)
Ans.we will determine P for each possible failure
m = 4.55 mm
To solve the problem
Fig. 1–28
condition. Then we will choose the smallest value. Why?
Normal Stress.
Using the method of sections, the axial load within
SOLUTION
region FETo
of solve
the shaft
2P, whereas
the determine
largest axialPforce,
3P, occurs
the isproblem
we will
for each
possible failure
within region
EC, Fig.
1–28b.
Thechoose
variation
the internal
loading is
condition.
Then
we will
the of
smallest
value. Why?
clearly shown on the normal-force diagram, Fig. 1–28c. Since the cross-
#19
80 mm
2P
P
B 160
20mm
mm B
80 mm
20 mm
2P
1
2
P
E
B
60
mm
C
B
P
60
mm
A
P F 20 mm
P
A
20 mm
(b)
3P
80
mm
E
P
P
2
P
the collar at C, which is 2 P
80
mm
C
P
2
P
2
P
B
60
mm
3P
80 mm byisthe
A shown2 P
2 Pcollar at
80 C,
mmwhich is
The shaft
in FFig.
1–28a
is supported
by the
collar
at
2P
2 P20 mm
The
shaft
in Fig.
1–28a
is
(a)supported
E
Fshown
C C, which
E
side of the bearing
at PB.
P
C
FP
FE and
3P(b)at B.
80
mm
CtheEright
Axial
(a)the Cbearing
2of
P the on
(b)
2P
attached
to
the
shaft
and
located
on
the
right
side
bearing
at
B.
attached
to
the
shaft
located
side
of
3P
al forces atFE and F so
(b)
(b)
Axial
E Determine
Cthe
(a)
Determine
the largest
value ofForce
P the
for(a)
axial
forces
E and
F
so forces at E and F so
largest
value
of at
P for
the
axial
ot exceed an
allowable
(a)
Force
(b)
(a)
Axial
Axial
(b)
that
theshaft
bearing
stressthat
on Force
collar
does
noton
exceed
an allowable
the bearing
stress
the collar
does not exceed an allowable
Axial
Axial
Force
normal stress
in the
(a)
3P
Force
Force
1s
2
=
75
MPa
stress
of
and
the
average
normal
stress
in
the
shaft
1s
2
=
75
MPa
stress
of
and
the
average
normal stress in the shaft
Axial
b allow
Çözüm:
2P b allow
w = 55 MPa.
3P
Force does not exceed an allowable
=
55
MPa.
stress
of 1s
2
does not
exceed
ant2allowable
stress
of
1s
allow
2P
t allow = 55 MPa.
Position
3P
3P
2P
3P
Serbest cisim diyagramı:
Position
2P
3P
(c)
2P
B
2P
20
Position
(c) Fig. 1–28
3P mm
Position
B
60 mmA
P
60 mm B
Position
A
20 mm
2P
Position
20 mm
3P (c)
80 mm
(c) P
2P
P Fig. 1–28
P
P
(c)
3P
Position
80
mm
80 mm
(c)
2P
# 2P
2P
C
Fig. 1–282 P Fig. 1–28
SOLUTION
E F
(b)
(c)F
EFig. 1–28
C
C Fig. 1–28
SOLUTION
(b)
To
solve the problem
we will determine P for each possible
failure
(b)
Normal
kuvvet
diyagramı:
Fig.
1–28
SOLUTION SOLUTION
To
solve
the
problem
we
will
determine
P for each possible failure
(a)
(a)
condition.
Then
we
will
choose
the
smallest
value.
Why?
SOLUTION
SOLUTION
1.7 P
DESIGN
OF SIMPLE
CONNECTIONS
51 Why?
ToAxial
solve the To
problem
weAxial
will
determine
forThen
each
possible
failure
condition.
we
the smallest
value.
solve the
problem
we
will determine
Pwill
forchoose
each
possible
failure
To
solve the problem
we
will
determine
P
for
each
possible
failure
Normal
Stress.
Using
the
method
of
sections,
the
axial
load
within
OLUTION
To
solve
the
problem
we
will
determine
P
for
each
possible
failure
Force
Forcewethe
condition. Then
we will Then
choose
smallest
Why? value. Why?
condition.
will
choosevalue.
the smallest
3P
2P
EXAMPLE 1.15 EXAMPLE
1.1560 mm
A
F A
P
3P
P
(b) 3P
3P
Normal
Stress.
Using
theaxial
method
of3P,
sections,
Then
we will
choose
the
smallest
value.
region
ofeach
theThen
shaft
is 2P,
whereas
largest
force,
occursthe axial load within
o
solve the problemcondition.
we will determine
PFEfor
possible
failure
condition.
we
will
chooseWhy?
the
smallest
value.
Why?
ure
Normal
Stress.
Using
the
method
of
sections,
the
axial
load
within
region
FE
of
the
shaft
is
2P,
whereas
the
largest
axial force, 3P, occurs
Normal
Stress.
Using
the
method
of
sections,
the
axial
load
within
within
region
EC,
Fig. of
1–28b.
The
variation
ofsections,
the
internal
loading
is
ndition.
Then we will
chooseStress.
the smallest
value.
LE 1.15
Normal
Using
theWhy?
method
sections,
the
axialof
load
within
Normal
Stress.
Using
the
method
the
axial
load
within
3P
3P
1 internal loading is
region
FE
of
the
shaft
is
2P,
whereas
the
largest
axial
force,
3P,
occurs
within
region
EC,
Fig.
1–28b.
The
variation
of
the
region
FE
of
the
shaft
is
2P,
whereas
the
largest
axial
force,
3P,
occurs
2P
clearly
shown
onof
the
normal-force
diagram,
Fig.
1–28c.
Sinceforce,
the cross2P
region
FE
of
the
shaft
is
2P,
whereas
the
largest
axial
force,
3P,
occurs
n
ormal
Stress.
Using
the
method
of
sections,
the
axial
load
within
region
FE
the
shaft
is
2P,
whereas
the
largest
axial
3P,
occurs
hin
within regionwithin
EC,
Fig.
1–28b.
variation
of the
internal
is diagram,
clearly
shown
on
thePosition
normal-force
Fig.
1–28c. Since the crossregion
EC,The
1–28b.
variation
of loading
the
internal
loading
is to
Position
sectional
area
ofFig.
the
entire
shaft
region
subjected
ft
shown
in the
Fig.shaft
1–28a
supported
by
collar
at
C,
which
isThe
within
region
EC,
Fig.
1–28b.
The
variation
of is
theconstant,
internal
loading
is is internal
# clearly
gion
FE of
isis2P,
whereas
thethe
largest
axial
force,
3P,
occurs
within
region
EC,
Fig.
1–28b.
The
variation
ofEC
the
loading
is is subjected to
urs
shown
onthe
the
normal-force
diagram,
Fig.
1–28c.
Since
the
crosssectional
area
of
the
entire
shaft
is
constant,
region
EC
clearly
shown
on
the
normal-force
diagram,
Fig.
1–28c.
Since
the
cross(c)
maximum
average
normal
stress.Applying
Eq.Fig.
1–11,
we have
(c) Fig.
dthin
to the
shaft
and Fig.
located
on
the right
side
of
the
bearing
at
B.
clearly
shown
on
the
normal-force
diagram,
1–28c.
Since
the
crossregion
EC,
1–28b.
The
variation
of
the
internal
loading
is
Fig.
1–28
clearly
shown
on
the
normal-force
diagram,
1–28c.
Since
the
crossg is
sectional area
of the gerilmeler
entire
shaft
is dikkate
constant,
region
EC istaşınabilecek
subjected
tois 1–28
theFmaximum
average
normal
stress.Applying
we have
sectional
area
ofatthe
shaft
is1–28
constant,
region
EC
subjected
toEq. 1–11,
Fig.
P
Şafttaki
normal
alınarak
enisbüyük
kuvvet:
Fig.
ne the largest value
of P for
the
axial
forces
E entire
and
soentire
sectional
area
of
the
entire
shaft
is
constant,
region
is 3P
subjected
to EC
early
normal-force
diagram,
Fig.; 1–28c.
Since
the
crossarea
of
the
iswe
constant,
region
subjected
to
ss- shown on thethe
A
= sectional
p10.03
m2
= EC
maximumthe
average
normal
stress.Applying
Eq.2shaft
1–11,
have
P
3P
maximum
average
normal
stress.Applying
Eq.
1–11,
we
have
6
2 2
bearing
stress
onentire
themaximum
collar
does
nots
exceed
an
55110
2have
N>mm2
the
average
normal
stress.Applying
we
ctional
area
of the
region
ECallowable
is
to 1–11,
allow
;Eq.
Asubjected
= 3Pnormal
p10.03
the
maximum
average
stress.Applying
Eq.=1–11, we6 have 2
to
Pshaft is constant,
P
3P
SOLUTION
2in the shaft
s
SOLUTION
EXAMPLE
55110 2 N>m 1.15
2
allow
1s
2
=
75
MPa
and
the
average
normal
stress
or
each
possible
failure
3P
; A =
A =
p10.03
m2
b allow
e maximum
average
normalPstress.Applying
1–11,
we have
3P
; Eq.P
p10.03
m2
=
Ans.
P
=
51.8
kN
2=
6
2
2 6 failure2
;1s 2To
=sallow
p10.03
m2
=55110
To
solveAthe
problem
we
will
determine
P
for
each
possible
2
N>m
solve
the
problem
we
will
determine
P
for
each
possible
failure
s
;
A
=
p10.03
m2
=
55110
2
N>m
=
55
MPa.
t exceed
an
allowable
stress
of
allow
6
2
Ans.
P
=
51.8
kN
alue.
Why?
allow 3P s
6#
2
P
sallow # 2 t Bearing
#
#
55110
2
N>m
Stress.
As
shown
onthe
thesmallest
free-body
diagram
Fig.shaft
1–28d,
3P
55110
2Why?
N>m inThe
allow
Then m2
we will
choose
the
smallest
value.
Why?
; condition.
=
p10.03
= condition.
Then
we51.8
will
choose
value.
shown
in
Fig.
1–28a
is supported by th
Ans.
P
=
kN
6
2
P = load
51.8 of
kNAs
shown
on
the
free-body
diagram in Fig. 1–28d,
ons, s
the
axial load within
55110
2 N>m
C
the
collar
at C P
must
resist
which
acts
over
a Ans.
bearing
allow
Ans.
= Bearing
51.8
kNtheStress.
Ans.
P
=3P,
51.8
kN the
attached
to
the
shaft
and
located
on the
right si
Normal
Stress.
Using
the
method
of
sections,
the
axial
load
within
Yakadaki
normal
gerilmeler
dikkate
alınarak
taşınabilecek
en
büyük
kuvvet:
Normal
Stress.
Using
the
method
of
sections,
axial
load
within
Stress.
As
shown
on
theshown
free-body
diagram
Fig.
1–28d,
2 on the
2inresist
-3
the
collar
at
C
must
the load
of2.3P,
which3Pacts over a 3P
bearing (d)
Stress.
As
free-body
diagram
in2Fig.
1–28d,
est axial force,
3P,Bearing
occurs
BkN
60Bearing
mm
area
of
Thus,
A
=
[p10.04
m2
p10.03
m2
]
=
2.199110
m
Ans.
P
=
51.8
A
b
20
mm
Bearing
Stress.
As
shown
on
the
free-body
diagram
in
Fig.
1–28d,
3P
Bearing
Stress.
As
shown
on
the
free-body
diagram
in
Fig.
1–28d,
ns.
Determine
the
largest
value
of
P
for
the
axial
3P
2
2
-3
2
region
FE
of
the
shaft
is
2P,
whereas
the
largest
axial
force,
3P,
occurs
region
FE
of
the
shaft
is
2P,
whereas
the
largest
axial
force,
3P,
occurs
the collar
C must
resist
load
of
3P,
acts
a bearing
Pat the
P
area
ofwhich
A
[p10.04
m2
- p10.03
] = 2.199110 2 mC. Thus,
C
collar
at
Cthe
must
resist
the
load
3P,over
which
acts
over3Pm2
a bearing
of the internal loading
iscollar
b = of
mm
P
C on the collar doesCnot
the
at
C within
mustthe
resist
the
load
of
3P,
which
acts
over
ais bearing
2must
P2Fig.
earing
Stress.
As
shown
on
the
free-body
diagram
in
1–28d,
28d,
P
3P
2 80
-3
2of3P
collar
at
C
resist
the
load
3P,
which
acts
over
a
bearing
that
the
bearing
stress
-3
2
2
2
-3
2
within
region
EC,
Fig.
1–28b.
The
variation
of
the
internal
loading
3P
region
EC,
Fig.
1–28b.
The
variation
of
the
internal
loading
is
of Ab = area
[p10.04
m2
2.199110
. Thus,
(d)
2.199110
Aof= Am2
; p10.03m2
=[p10.04
-] 2=p10.03
m2=] 2-3
=m2.199110
(d)
g. 1–28c.
Since thearea
cross26 2
P
b m2
F at
22 m
-3
23P
22 m . Thus,
EAbload
-32.199110
C 2ofdiagram,
area ofthe
= normal-force
[p10.04
-Aacts
p10.03
m2
]bearing
=
22.199110
m
.2Thus,
C 2the
engcollar
C must shown
of 3P,
which
over
s
area
[p10.04
m22.199110
- p10.03
m2Fig.
]N>m
=1–28c.
mof
. # Thus,
(d)
1sb2allow =(d)75 MPa and the average
no
75110
allow
AFig.
=a1–28c.
;Since
2 m2Since
=stress
on
the
crossb =
Cthe
shown
the
normal-force
diagram,
cross# resist
# clearly
#-3 on
(b)
3P
6
2
egion
EC isclearly
subjected
to
2 P
2P
-3
2
3P
2 2 m . Thus,
sallow
75110
2 exceed
N>m an allowable stress of 1st2allow
-3
2
ea of Ab = sectional
[p10.04
-of
p10.03
m2 2.199110
]shaft
= ;2.199110
(d)
does
not
P
3P
A m2
=area
;
2
m
=
the
entire
is
constant,
region
EC
is
subjected
to
P
3P
2.199110
A
=
2
m
=
(d)
sectional
area
of
the
entire
shaft
constant,
region
EC
is
subjected
to
-3
6
2 = 55.0
P-3
2 6 kN 2
Eq. 1–11, we have A =(a)
s
2.199110
;
2; m2 =75110
26N>m
sallow
2.199110
A stress.Applying
=
2have
2 N>m
2m =
61–11,
3P
Axial P
P we
=2 55.0
kN
the maximumallow
Eq.
1–11,
we75110
saverage
the
maximum
average
normal
stress.Applying
Eq.
have
75110
2
N>m
s
-3
2 normal
allow
75110
2 N>m
allow
the
largest
load
that
can
be
applied
to the
shaft is
2.199110 2 m By
=
;
= comparison,
Force
6
2 P =3P
55.0
kN
P
P
=
55.0
kN
P
3P
s2allow
75110 2kN,
N>m Pany
By
comparison,
the largest
load
be applied toAthe shaft is
2 than
60 mm B
= 55.0
P
loadkN
larger
willkN
causethat
thecan
allowable
N>m
;
A = #
p10.03
m22 ; = since
= 655.0
A == 51.8
= P this
#that
6 p10.03
2 m2
2 is
P
comparison,
the
largest
load
can
be
applied
to
the
shaft
sBy
P
=
51.8
kN,
since
any
load
larger
than
this
will
cause
the
allowable
s
55110
2
N>m
By
comparison,
the
largest
load
that
can
be
applied
to
the
shaft
is
55110
2
N>m
allow
allow
stress
in
thethat
shaft
tolargest
be
P normal
= the
55.0By
kN
2 P shaft is
By comparison,
largest
load
can
be exceeded.
applied
to the
shaft
is
comparison,
the
load
that
can
be exceeded.
applied
to the
3P
P Ans.
= 51.8 kN,Psince
anykN,
load
larger
than
will
cause
the
allowable
normal
stress
in
the
shaft
to
be
= 51.8
since
any
loadthis
larger
than
this
will
cause
the allowableAns. F
Ans.
P
=
51.8
kN
E
P
=
51.8
kN
2P
P
=
51.8
kN,
since
any
load
larger
than
this
will
cause
the
allowable
Buna
göre
taşınabilecek
en
büyük
kuvvet
#
dir.
NOTE:
Here
we
have
not
considered
a
possible
shear
failure
of
the
ySyis
comparison,
the
largest
load
that
can
be
applied
to
the
shaft
is
P
51.8
kN, since any load larger than this will cause the allowable
in Fig. normal
1–28d, stressnormal
3P =toin
Sdiagram
T
RTERSESS S
in
the
shaft
beSexceeded.
shaft
to
be exceeded.
Position
52
CAs
H
Ain
P
Tthe
E Rstress
1
Swill
Tto
R Ethe
S exceeded.
NOTE:
Here
we
have
not
considered
a
possible
shear
failure
of
the
Bearing
Stress.
shown
on
the
free-body
diagram
in
Fig.
1–28d,
3P
Bearing
Stress.
As
shown
on
the
free-body
diagram
in
Fig.
1–28d,
3P
normal
stress
shaft
be
collar
as
in
Example
1.14.
=
51.8
kN,
since
any
load
larger
than
this
cause
the
allowable
normal stress
C in the shaft to be exceeded.
hich acts over a bearing
ble
(a)
NOTE:
Here
we
have
not
considered
a considered
possible
shear
failure
of the
(c)
asacts
in Example
1.14.
NOTE:
Here
weofhave
not
aof
shear
failure
theAxial C
-3 the
collar
attoCbe
must
resist
the
load
3P,collar
which
over
apossible
bearing
C
the
collar
at
must
resist
the
load
3P,
which
acts
over of
a bearing
ormal
stress
in2.the
shaft
exceeded.
Thus,
.199110
2m
(d)C
NOTE:
Here
we2 have
not
considered
a
possible
shear
failure
of
the
NOTE:
Here
we
have
not
considered
a
possible
shear
failure
of
the
21.14.
-3
2
2
2
-3
2
as[p10.04
in Example
Fig.
1–28
collar
as-1.14.
inp10.03
Example
area ofcollar
Thus,
A
=
m2
m2
]
=
2.199110
2
m
.
(d)
Force
area
of
Thus,
A
=
[p10.04
m2
p10.03
m2
]
=
2.199110
2
m
.
(d)
b
b as shear
collar
in Example
1.14.
OTE:
Here we have
notas
considered
a collar
possible
failure1.14.
of the
P
in Example
he
EXAMPLE
1.16
3P
3P
1 # P
2in Example
-3P 2
llar
as
1.14.
-3
2
2ON
N>m
2.199110
A =
;
2.199110
A = 2 m; =
6 supported
2 2 mby=a steel rod
6 AC
3P
sallow
The
rigid
bar
AB
shown
in
Fig.
1–29a
is
AC2
s
75110
2
N>m
The
rigid
bar
AB
shown
in
Fig.
1–29a
is
supported
by
a
steel
rod
75110
2 N>m
allow
2 AC
AB
in Fig.
1–29a
is supported
byrod
aeach
steel
rodolan
ACfailure
enthe
problem
we
willrijit
determine
P çapı
for
possible
2P alanı by
Theçelik
rigid çubuğu
bar AB shown
in Fig. 1–29a iskesit
supported
a steel
AB
çubuğu,
20mm
AC
ve
B
noktasındaki
1800
mmrod
N
inshown
Fig.
1–29a
is
supported
by
a
steel
AC
having
a
diameter
of
20
mm
and
an
aluminum
block
having
a
crosshaving
a
diameter
of
20
mm
and
an
aluminum
block
having
a
crosseter
of
20
mm
and
an
aluminum
block
having
a
crossP
=
55.0
kN
n.
Then
we
will
choose
the
smallest
value.
Why?
P
=
55.0
kN
having
a
diameter
of
20
mm
and
an
aluminum
block
having
a
C having a cross2 blok
mm andtoan
aluminum
block
olan isbir
aluminyum
tarafından
taşınmaktadır.
ACCare
veare
C noktalarında bulunan 18mmcross- Position
e0 applied
shaft
1800
mm
. The
sectional
area
18-mm-diameter
pinsatatAAand
and
mm
.2A
sectional
area
ofof1800
The
18-mm-diameter
pins
2 the
2
1800
mm
.
of
The
18-mm-diameter
pins
at
and
C
are
2
1800
mm
.
sectional
area
of
Theto18-mm-diameter
pins at A(c)
and C are
By subjected
comparison,
the
largest
that
can
be
applied
tothat
the
shaft
P and
Stress.
Using
the
method
of
sections,
the
axial
load
within
By
comparison,
the
largest
load
can
beisapplied
the shaft
is
Steel
mm
.cause
The 18-mm-diameter
pins
at
A
C
are
will
the
allowable
çapındaki
perçinler
tek
tesirlidir.
Çelik
ve
aluminyum
için
göçme
normal
gerilmeleri
subjected
single
shear.
If
the
failure
stress
forthe
the
steel
and
aluminum
toto
single
shear.
Ifload
the
failure
stress
for
steel
and
aluminum
ngle
shear.
If
the
failure
stress
for
the
steel
and
aluminum
subjected
to
single
shear.
If
the
failure
stress
for
the
steel
and
aluminum
P stress
=
51.8
kN,
any
larger
than
this
will
cause
thethan
allowable
thefailure
shaft
is
whereas
the
largest
axial
force,
3P,
occurs
Pload
= and
51.8
kN,
any
load
larger
this
will
cause the allowable
r.EIfofthe
for
steel
and
aluminum
1s
2the
=
680
2since
=
70
MPa,
is2P,
respectively,
and
the
=since
680
70
MPa,
is52
respectively,
and
the
C
H A MPa
P TMPa
E R 1and
S T1s
R1s
Eal
S2
Sfail
fail
al
fail=the
st2st
fail
80
MPa
1s
21s
=The
70
MPa,
and
respectively,
and
#
ve
#
,
ayrıca
perçinler
için2 göçme
kayma
gerilmesi
1s
2
=
680
MPa
1s
= 70 MPa,
is
and
respectively,
and the
al
fail
normal
stress
in
the
shaft
to
be
exceeded.
st
fail
egion
EC,
Fig.
1–28b.
variation
of
the
internal
loading
is
normal
stress
in
the
shaft
to
be
exceeded.
1s
2
=
70
MPa,
and
respectively,
and
the
B fail failure
Bal
tfail
900MPa,
MPa,determine
failure
shear
stressfor
foreach
eachpin
pin
determinethe
thelargest
largest al fail
==900
shear
isistfail
A stress
sible
shear
failure
of
the
B
SOLUTION
t
=
900
MPa,
ess
for
each
pin
is
determine
the
largest
t
=
900
MPa,
failure
shear
stress
for
each
pin
is
determine
the
largest
failP kuvvetini hesaplayınız.
hown
normal-force
diagram,
Fig.
Since
cross# fail
olduğuna
göre
çubuğa
uygulanabilecek
büyükfailure
tfailNOTE:
=load
900
ach
pinon
is the
determine
the1–28c.
largest
Here
we
have
not
considered
possible
shear
failure
of
the=en
NOTE:
Here
weathe
have
not
considered
possible
of the
F.S.
=2.2.shear
load
P
thatcan
can
beapplied
applied
tothe
the
bar.Apply
factor
safety
minum
PMPa,
that
be
to
bar.Apply
a afactor
ofofsafety
ofaofF.S.
num
To
solve
the
problem
we
will
determine
P for
F.S.
=
2.
be
applied
to
the
bar.Apply
a
factor
of
safety
of
F.S.
= 2.
load
P
that
can
be
applied
to
the
bar.Apply
a
factor
of
safety
of
Aluminum
l area
ofbar.Apply
thecollar
entire
is of
constant,
EC
is
subjected
0.75
m
2.Example
ed
to the
ashaft
factor
safety
ofregion
as
in Example
1.14.
collar
1.14.to
Emniyet
katsayısı
# F.S.as=in
alınacaktır.
EXAMPLE
1.16
condition.
Then
we
will
choose
the
smallest
valu
SOLUTION
1 normal
SOLUTION
mum average
stress.Applying2 m
Eq. 1–11, we have SOLUTION
UsingEqs.
Eqs.
1–9and
and
1–10,the
theallowable
allowablestresses
stressesare
are
Using
1–9
1–10,
Normal
Stress.
Using the method of section
Pand 1–10, the allowable
3P
stresses
are
Using
1–10,
stresses
are
The
rigidEqs.
bar 1–9
AB and
shown
in the
Fig.allowable
1–29a is supported
by a steel rod AC
(a)
; allowable
p10.03
m22are
=
0, the
stresses
2
region
FE
of
the
shaft
is 2P, whereas the largest
1s
2
6
21s
680
MPa
st
fail
680
MPa
st
fail
1sst2fail
55110
2 N>m
low
680 CMPa
having=a=340
diameter
and
an aluminum
680 MPa block having a crossst2fail
1s
2340
=
340MPa
MPa of 20 mm1s
1s
2stallow
==
1sst2fail
allow=
680 MPa
st=
within
region
EC,
Fig.
1–28b. The variation of
1s
2
=
=
MPa
1s
2
=
=
=
340
MPa
2
st
allow
F.S.
F.S.
2 2Ans.
st allow
=
= F.S.
mm . The
sectional
area of 1800
18-mm-diameter
pins at A and C are
P ==Steel
51.8MPa
kN
2340
P
F.S.
2
clearly
shown
on
the
normal-force
diagram, Fig.
F.S.
2
1s
2
1sal2alfailfail 7070MPa
MPa
subjected
to single shear.
stress
for the steel and aluminum
1sal
2fail on
Stress.
As
shown
free-body
diagram
in=Fig.
1–28d,
1sfailure
3P If the
70the
MPa
70 MPa
al2fail sectional
1s
2
=
=
=
35
MPa
1s
2
=
=
35
MPa
1s
2
area
of
the
entire
shaft
is
al
allow
70
MPa
al
allow
al
fail
= = resist the
= =load
MPa acts
35 MPa and the constant, reg
MPa
and= 1salC2fail == 70 MPa, =
respectively,
F.S.
F.S.
2is2 1sst2fail = 6801s
al2allow
ars
C must
of =
3P,35which
over a bearing
=alat2allow
35
MPa
F.S.
2
F.S.
2
the
maximum
average
normal stress.Applying E
A
-3 B900
2 MPa
MPa, determine the largest
failure shear stress for each pin
tfail
MPa
tfail
Ab = F.S.
[p10.04
m222-900
p10.03
m22] = 2.199110
2 m900
. Thus,
(d)t is tfail = 900 P
t
MPa
3P
900
MPa
fail
t
=
=
=
450
MPa
fail
t
=
=
=
450
MPa
2
tfail = 900 MPa
allow
allow
= 2.
of safety
of F.S. m2
Aluminum
=m450
MPa
=
tallow to
= the bar.Apply
= A = a factor
=; 450
MPa p10.03
F.S.
2 2 load P that can be applied
P= tallow
3P F.S.
= F.S. = -3
= 245020.75
MPa
s
F.S.
2
5511062 N>
allow
2.199110
;F.S.
m =
2 The 2free-body
6
2the
2m
diagram
of
the
bar
is
shown
in
Fig.
1–29b.
There
are
The
free-body
diagram
of
bar
is
shown
in
Fig.
1–29b.
There
are
SOLUTION
75110
2
N>m
low
dy
of isthe
isinshown
in Fig.There
1–29b.
There are
The free-body diagram of the bar is shown in Fig. 1–29b. There
are51.8 kN
P =
# bar unknowns.
m diagram
of the bar
shown
Fig. 1–29b.
are
three
Here
wewill
will
applythe
themoment
moment
equations
ofequilibrium
equilibrium
three
unknowns.
Here
we
apply
equations
Using
Eqs.
1–9 of
and
1–10,
the
allowable
stresses
are
(a) kN
s.e Here
we
will
apply
the
moment
equations
of
equilibrium
three
unknowns.
Here
we
will
apply
the
moment
equations
of
equilibrium
P
=
55.0
Bearing Stress. As shown on the free-body
will apply the moment
equations
equilibrium
F
orderto
toexpress
express
andF
termsofofthe
theapplied
appliedload
loadP.We
P.Wehave
have
Pof
FF
in
and
ininterms
1sstF2fail
AC
AC
BB
Çözüm:
680of
MPa
Flargest
ess F
and
ininorder
terms
of
the
applied
P.We
have
FAC=and
in is
order to express
the at
applied
load
P.Wethe
have
B of
B in terms
Fthat
must
resist
load of 3P, whic
parison,
canP.We
beload
applied
to
the shaft
FAC
and
terms
theload
applied
have
AC load
1sst2allow
=the collar
=C
340
MPa
B in the
d
+
©M
=
0;
P11.25
m2
F
12
m2
=
0
(1)
d
+
©M
=
0;
P11.25
m2
F
12
m2
=
0
(1)
2
F.S.
2
AC
B B 12 m2 = 0
AC
P11.25
m2
F
(1)
area
of
A
=
[p10.04
m2
p10.03
d
+
©M
=
0;
P11.25
m2
F
12
m2
=
0
(1) m22] = 2.1
8
kN,
since
any
load
larger
than
this
will
cause
the
allowable
AC
b
B
AC
P11.25 m2 - FAC12 m2 = 0
(1)
1s
2
70
MPa
al
fail
B
B
m2
P10.75
m2==0 0
(2)
stress in the
shaft
be
exceeded.
d +d +to
©M
==0;0;
FF
m2
(2)
P = 35
3P
B12
AA
B12
A
B- -P10.75
2
m2
-©M
P10.75
(2) m2
d + ©MA = 1s
0; al2allow = FB12 m2 =-AP10.75
0 2.199110-32 m(2)
= 2 m2
; = MPa
=
FB12 m2F-B12
P10.75
m2
= 0 m2 = 0
(2)
F.S.
6
s
We
will
each
value
thatcreates
createsthe
theallowable
allowable
1.25 meach
Here
we have
not
considered
adetermine
possible
shear
failure
the
0.75
mdetermine
We
will
now
value
ofofof
PPthat
determine
each
value
ofcreates
Pnow
that
creates
the
allowable
We will now determine
each 900
value
of allow
P that creates the allowable 75110 2 N
Feach
BFB
t
MPa
ine
value
of
P
that
the
allowable
fail
F
stress
therod,
rod,block,
block,and
andpins,
pins,
respectively.
inblock,
Example
1.14.
stress
ininthe
Brespectively.
= and pins,
= respectively.
= 450 MPa
d,
pins,
respectively.
P = 55.0 kN
stress in the rod,tallow
block,
and
pins,and
respectively.
F.S.
2
(b)
AC
.
This
requires
Rod
AC
.
This
requires
Rod
s
requires
comparison,
largest
load
AC. Thisdiagram
requires the barBy
Rod
es
The
free-body
is shown
in Fig. the
1–29b.
There
are that can be a
2
6 62 N>m
2 2] = 106.8of
1s
2Fig.
1A
340110
[p10.01m2
m2
kN
FF
1A
2 22=] ==340110
2kN
N>m2 [p10.01
] = 106.8 kN
6 = =1s
6= 51.8 2kN, since any
2 load larger than this wi
AC
allow
AC
P
1–29
AC
st22st[p10.01
allow
AC
2allow
1A
2
=
340110
2
N>m
m2
106.8
6
2
2
F
=
1s
2
1A
2
=
340110
2
N>m
[p10.01
m2
]
=
106.8
kN
AC
three unknowns.
will apply the moment equations of equilibrium
AC
stHere
allow weAC
] = 106.8 kN
#2be
0 exceeded.
C2 = 340110 2 N>m [p10.01 m2
normal
stress
in
the
shaft
to
UsingEq.
Eq.1,1, P
Using
in order to express FAC and FB in terms of the applied load P.We have
FAC
1106.8kN212
kN212m2
m2Using Eq. 1,
1106.8
NOTE:
Here
have not considered
a possib
1106.8
kN212
m2
1106.8
=171
171
P
=
P
=
1106.8
kN212 m2
d +=©M
=kN
0;
P11.25
m2 -kN212
FAC12m2
m2=
=171
0 we
(1)
B kN
=
171
kN
P
=
kN
P
=
1.25
m
1.25
m
collar
as
in
Example
1.14.
=
1.25 =m 171 kN
1.25 m
bjectedarea
to single
shear.
failure
stress
for the
steel
and
aluminum
2 at
1800
mmIf
.the
ctional
ofhaving
The
18-mm-diameter
pins
A and
C
are
of 20
mm
and
an
aluminum
block
having
a the
cross1s
1800
mm
sectional
area
pinsaand
at
Athe
C are
1s
2allow
=stress
=and
340 MPa
70 =MPa
al2steel
fail
diameter
ofsubjected
20of
mm
and
an. The
block
having
crossstfailure
Pving a diameter
to
shear.
If
for
the
1sst2aand
=1s680
MPa
1s
2single
=aluminum
7018-mm-diameter
MPa,
isMPa
and
respectively,
fail2the
alrespectively,
fail
F.S.
2=and aluminum
1sst2failarea
=single
680
2
=
70
MPa,
and
the
1s
2
=
= 35 MPa
2
al
fail
al
allow
bjected
to
shear.
If
failure
stress
for
the
steel
and
aluminum
1800
mm
.
ctional
of
The
18-mm-diameter
pins
at
A
and
C
are
subjected
to
single
shear.
If
the
failure
stress
for
the
steel
and
aluminum
1800
mm
.
sectional
area
of
The
18-mm-diameter
pins
at
A
and
C
are
B
F.S.
2
1sstMPa,
2pin
680=MPa
1s
iseach
andlargest
respectively,
and the
900 the
MPa,
failure
shear
stress
for
determine
thealMPa,
fail is=t
al2fail = 70
fail
1s
2largest
t
=
900
ilure
stress
for
each
pin
is
determine
70
MPa
fail
fail
1s
=
680
MPa
1s
2
=
70
MPa,
and
respectively,
and
the
A st2shear
bjected
to
single
shear.
If
the
failure
stress
for
the
steel
and
aluminum
fail
al2fail
1s
=failure
680Ifto
MPa
1s
=the
70 of
MPa,
istocan
and
respectively,
and 900
the MPa
subjected
single
shear.
the
stress
for
steel
aluminum
B st
al2fail
1s
22.
==of
=the
35largest
MPa
tfailand
900
shear
stress
for
pin
is
F.S.MPa,
=t=fail
2.determine
load
P that
befailapplied
thefailure
bar.Apply
aeach
factor
safety
num
allow
F.S.
=althe
ad
that=can
beB
applied
topin
the
aMPa,
factor
ofpin
safety
of
t2fail
=stress
900
lure
shear
stress
for
each
isalbar.Apply
determine
the
largest
F.S.
tallow
= the
=2
= 450 MPa
1sP
680
MPa
1s
70
MPa,
and
respectively,
and
t
=
900
MPa,
failure
shear
for
each
is
determine
the
largest
st2
fail
1s
2
=
680
MPa
1s
2
=
70
MPa,
is
and
respectively,
and
fail
st
fail
al
fail
load
P that
can be applied
the bar.Apply
Aluminum
The
rigid
AB shown in Fig.
1–29a
is supported
by atosteel
rod AC a factor
F.S.of safety2of F.S. = 2.
0.75can
m bebar
F.S.largest
= a2.factor ofthe
ad
PAluminum
that
applied
to
the
aapplied
factor
oftosafety
ofMPa,
SOLUTION
tsafety
900
MPa
tfailfor
=each
900
MPa,
lure
shear
stress
for each
pin
the
B
F.S.
= 2.
load
Pisbar.Apply
that
can
be
bar.Apply
of
faillargest
= block
900
failure
shear
stress
pin
is tdetermine
determine
OLUTION
fail the
having a 2diameter
of
20
mm
and
an
aluminum
having
a
crosst
=
=
MPa
The
free-body
is 450
shown
in Fig. 1–29b. There are
m
allow diagram of the bar =
SOLUTION
Using
Eqs.
1–9
and
1–10,
the
allowable
stresses
are
F.S.
=
2.
ad
P
that
can
be
applied
to
the
bar.Apply
a
factor
of
safety
of
2applied
F.S.
2
F.S.
=
2.
load
P
that
can
be
to
the
bar.Apply
a
factor
of
safety
of
um
sing
Eqs.
1–9
and
1–10,
the
allowable
stresses
are
OLUTION
2 m sectional area of 1800
mm . The 18-mm-diameter pins
at
A
and
C
are
SOLUTION
three
unknowns.
Here
we
will
apply
the
moment
equations of equilibrium
Using
Eqs.
1–9
and
1–10,
the
allowable
stresses
are
1s
2
680free-body
MPaand aluminum
st are
failfor
1sallowable
The
diagram
of the
bar
isinshown
inthe
Fig.applied
1–29b.load
There
are have
680
MPa
ing
1–9(a)and
1–10, Using
the
stresses
st2P
fail
subjected
to single
shear.
If
the
failure
stress
the
steel
OLUTION
Eqs.
1–9
and
1–10,
the
allowable
stresses
areF
1s
2
=
=
=
340
MPa
F
in
order
to
express
and
terms
of
P.We
a) Eqs.
st
allow
AC
B
F
1s
2
1sSOLUTION
2
=
=
=
340
MPa
AC
MPa
st fail we 680
st allow
F.S.
2
three
unknowns.
Here
will
apply
the
moment
equations
of
equilibrium
= 1–10,
680Eqs.
MPa
= allowable
70are
MPa,
is 1s1–9
and the
1sF.S.
2and
sing Eqs.
and
the
2the
680
MPa
st2fail
al2stresses
fail
stallowable
fail
1sststrespectively,
2stresses
=680
= m2
340 MPa
Using
and1s
1–10,
areMPa
fail
allow
P 1–9
dto
+=©M
= F0; ==
P11.25
Fapplied
= P.We
0
(1)
BF.S.
AC12 m2
1sstF2allow
= is t1sst21s
340
MPa
2MPa
2fail
=
340
F
in
order
express
and
in
terms
of
the
load
have
70
MPa
al=
=
900
MPa,
failure shear
stress= for
each
pin
determine
the
largest
1s
2
allow
AC
B
1s
2
fail
70
MPa
AC
al
fail
680
MPa
st 1s
fail 2
1s
2
F.S.
2
680
MPa
st
fail
F.S.
2
=
=
=
35
MPa
1s
2fail
=
==a35
MPa
Abe =
1sstalcan
22allow
==allow
340
MPa
B
al
allow
= 70
2. MPa
load P that
applied
bar.Apply
of
of340
1stoal2the
=2 F.S.
=
=
d2+safety
=F.S.
0;MPa
FB12 m2
- m2
P10.75
(2)
dfactor
+1s
©M
=©M
m2
12
= 0m2 = 0
(1)
1sF.S.
B
F.S.
70
MPa
al2failst allow
1s
=0;70AMPa P11.25
=
= F
35AC
MPa
F.S.
2
al22allow
fail
al
1.25
m
0.75
m
1sal2allow = 1s 2 = 1sal2tallow
=
35
MPa
F.S.
2
=
=
=
35
MPa
900
MPa
We will now determine
value
allowable
tF.S.
MPa
A
B MPa
70
SOLUTION
al fail 900
2= 1sfail
fail
+ ©M
FB12 m2 -each
P10.75
m2of= P0 that creates the (2)
70 MPa
FB2450
F.S.
2 MPa900
tallow
=35
A = =0;450
=
=al
# mand
# d MPa
1s 2tallow
== 1.251s
=
=fail
MPa
t35
MPa and pins, respectively.
allow
2
=
=
=
MPa
fail
stress
in
the
rod,
block,
F.S.
2
al
allow
m
Using Eqs.al0.75
1–9
1–10,
the
allowable
stresses
are
tF.S.
2 2 F.S.
900 MPa
F.S. (b)
= MPa
=
450 MPa
fail
tWe
2 900
allow
now
determine
each=value
of P that creates the allowable
failtwill
tallow
= 1s
450=bar
MPa
F.S.
2MPa
FB
t=
=RodThere
=1–29b.
450
The =
free-body
diagram
of
the
isMPa
shown
in
There
are
AC
.Fig.
This
requires
allow
680
MPa
tfailthe bar
st2is
900
MPa
The free-body diagram
of
shown
in
Fig.
1–29b.
are
F.S.
2fail
t
stress
in
the
rod,
block,
and
pins,
respectively.
900
F.S. 340 MPa
2
fail 450 MPa
1sunknowns.
== Here
tallow
=2(b)
allow Fig.
6
2
The
free-body
of
shown2 in
Fig.
1–29b.
There
are m22] = 106.8 kN
1–29
tthe
==will=
=
= 450
MPa
three
we
apply
the=diagram
moment
equations
ofisequilibrium
allow
ree unknowns.
west will
apply
equations
ofbar
equilibrium
F
=the
1sbar
2allow
1A
340110
2 N>m
[p10.01
F.S.
2Fig.
F.S.
AC
stin
AC = There
The
free-body Here
diagram
of
the
bar
is2moment
shown
in
1–29b.
There
are
F.S.
2
The
free-body
diagram
of
the
is
shown
Fig.
1–29b.
are
AC
.
This
requires
Rod
three
unknowns.
Here
we
will
apply
the
moment
equations
of
equilibrium
F
F
in
order
to
express
and
in
terms
of
the
applied
load
P.We
have
Bequations
FAC
Fapply
order
to express
in
terms
of
the
applied
have
1s
2moment
70
MPa
B
alAC
fail
ee
unknowns.
Here
weand
will
the
ofP.We
equilibrium
P
Using
Eq.
1,1–29b.
6
2
The
free-body
diagram
of
the
bar
is
shown
Fig.
1–29b.
There
are
1–29
three
unknowns.
Here
we
willload
apply
the
moment
equations
of
equilibrium
The
free-body
diagram
bar
is
shown
Fig.
There
are
F
=MPa
1sin
2allow
= the
340110
N>m
[p10.01
m22] = 106.8 kN
FAC
F
inP11.25
order
toin
express
and
terms
applied
load
P.We
1sFig.
=
=of the
=
35
AC
st=
AC2 of
B0in1A
F
alB2allow
d
+
©M
=
0;
m2
F
12
m2
(1)2 have
AC
1106.8
kN212
m2 have
AC
FACwe
order
and
in
of 12
the
load
P.We
have
+ree
©M
0; Here
P11.25
m2terms
-the
Fwe
m2
=
0
(1)
unknowns.
will
moment
equations
of
equilibrium
F.S.
2
Bapply
Fapplied
F
inForder
to
express
and
in
terms
of
the
applied
load
P.We
Bto=express
AC
three
unknowns.
Here
will
apply
the
moment
equations
of
equilibrium
AC
B
= 171
Eq.P11.25
1,
d of
+ ©M
= 0;Using
m2 - FACP
12 =m2 = 01.25 m
(1)kN
B
F
F©M
order
and
in terms
the
applied
P.We
have
B to express
t-=
900
MPa
P11.25
m2
12
m2
==
00- load
(1)
AC
B0;
FF
F
order
to
express
and
in
terms
of
the
applied
load=P.We
have
d+
©M
=
F
12
m2
P10.75
m2
=
0
(2)
fail
AC
d
+
0;
P11.25
m2
F
12
m2
0
(1)
AC
B
A
B
1106.8
kN212
m2
+©M
©MBA == 0;0; in
F
12
m2
P10.75
m2
(2)
B
AC
tBallow
=
=
= 450 MPa
B m2
Block
Bcreates
.0m2
InP-this
= 171 kN (2)
=P10.75
+ ©M
== 0;
FB=12
m2 = 0
©MB =
= 0;
0; B d + ©M
P11.25
- P10.75
FdAC
12 m2
m2
0- FACof12
(1)
F.S.
2m2
A
=12
0;
P11.25
We
will
now
determine
each
value
Pm2
that
thecase,
allowable
#
B
©M
F
m2
=
0
(2)
1.25(1)
m2
We
will
now
determine
each
value
of
P
that
creates
the
allowable
A
B
d
+
©M
=
0;
F
(2) 2
FB m
1.25 m
0.75
6
A
B12 m2 - P10.75 m2 = 0
Fdetermine
=B.1s1–29b.
A
= are
35110
N>m
[1800the
mmallowable
110-62 m2>mm2] = 63.0 kN
We
will
each
value
of P 2that
creates
free-body
diagram
the
bar
ispins,
shown
inB Fig.
There
stress in
thepins,
rod,of
block,
and
respectively.
al2allow
B
1.25
m The
In
this
case,
ress
in
and
respectively.
©M
=now
0;rod,
F
m2
P10.75
m2
=now
0Block
(2)
Bwill
F=12
We
determine
each
value
of
P
that
creates
the
allowable
A the
B
dblock,
+ ©M
0;
F
12
m2
P10.75
m2
=
0
(2)
B
We
will
now
determine
each
value
of
P
that
creates
the
allowable
A
B
AC
çubuğu
için
hesap:
mm and
veya
r=0.01
m 2
stress
thed=20
rod,equations
block,
respectively.
three unknowns.
Here
we
will
apply
theinmoment
of pins,
equilibrium
F Rod
AC
. each
This
requires
ess
in
the
rod,
block,
and
pins,
respectively.
Using
2, 6the
FB creates
=
1spins,
2allow
A
=creates
35110
2 N>m
[1800 mm2 110-62 m2>mm2] = 63.0 kN
. now
This
requires
od
(b)B determine
WeAC
will
value
of Pblock,
that
the
allowable
stress
in
theinrod,
and
respectively.
alof
B Eq.
We will
now
determine
each
value
P
that
allowable
F
F
in
order
to
express
and
terms
of
the
applied
load
P.We
have
AC
B6 Rod AC
6
2
2
) FB
.
This
requires
163.0
2
FAC
= =1s
2allow
2 2=pins,
340110
2 N>m
[p10.01
kN kN212 m2
ess
in the1s
rod,
block,
pins,
respectively.
st
AC
This
requires
dFAC
stress
inand
rod,
block,
and
respectively.
2allow
1A
2the
340110
2 N>m
[p10.01
m2
] = 2,106.8
kNm2 6] = 106.8
AC
. 1A
This
requires
AC .= Fig.
P 2=
= 168 kN
Using
Eq.
2
d + ©MstB1–29
= 0;# AC RodP11.25
m2
F
12
m2
=
0
(1)
AC
FAC
2 N>m 2 [p10.01
m2m
] = 106.8 kN
0.75
6
2 = 1sst2allow
2 1AAC62 = 340110
Using
Eq.
1,
2
AC
requires
od
1–29
163.0
kN212
m2
FAC
=. 1s
2
1A
2
=
340110
2
N>m
[p10.01
m2
]
=
106.8
kN
AC
.
This
requires
Rod
sing
Eq.
1,This
st allow
AC
FAC = 1sst2allow
1AAC2 = 340110
m2 ] = 106.8 kN
A2 N>m
or C. [p10.01
PDue
= to single shear, = 168 kN
6 Using
2Eq. 1,kN212
d += ©M
= 0;1A 21106.8
FBkN212
12 m2
- 1106.8
P10.75
m26m2
= 2m2
0]Pin
m2
2 (2)
FACEq.
1sstA2allow
=
340110
2PAC
N>m
[p10.01
=2=106.8
kN
ing
1,
AC
FPAC
=
1s
2
1A
2
=
340110
2
N>m
[p10.01
m2
]
=
106.8
kNm
171
kN
=
Using
Eq.
1,
st
allow
1106.8 kN212 m2 0.75
= 171 kN
=
6
2
2
mAcreates
FAC
=allowable
V
=single
tallowshear,
A= =171
450110
now
each
value of1.25
P Pin
that
the
1106.81.25
kN212
m2
m
or
C
.
Due
to
1106.8
kN212
m2
kN 2 N>m [p10.009 m2 ] = 114.5 kN
P
=
# P determine
sing Eq.We
1, willUsing
Eq.
1,
=
171
kN
=
1.25
m kN
Block
.1106.8
Inand
this
case,respectively.
= 171
P =kN212 m2
the
rod, B
block,
pins,
kN212
ock stress
B. Ininthis
case,
1106.8
1.25
m m2
1.25
m -6
From
Eq.
FAC
= kN
t1,
A 2= 45011062 N>m2 [p10.009 m22] = 114.5 kN
allow
6 2 =B
2 In
2== V
2
171
kN
P
=
Block
.
this
case,
171
P
=
B
deki
aluminyum
blok
için
hesap:
6
2
-6
2
2
F
1sal2requires
AB1.25
=
2 N>m
mm ] 110
2m
kN kN 12 m2
AC
This
Rod
114.5
allow
ock
. A
In
this
case,
sal2B
=. =35110
2 Block
N>m
[1800
110
2[1800
mm>mm
= 63.0
kN>mm ] = 263.0-6
mInmm
B.35110
this
case,
allow
BB
1.25
6
2
= 183
P 110
= 2 m2>mm2] = 63.0
F
=
1s
2
A
=
35110
2
N>m
[1800
mm
kNkN
From
Eq.
1,
6
2
2
6
2
2
-6
2
2
B
al
allow
B
6
2
2
-6
2
2
ock
B. F
In
this35110
case,
1s
2allow
1A
2 =case,
340110
2 N>m
[p10.01
] = kN
106.8
kN
1.25
m
s
2Eq.
[1800
mm
110
22mN>m
>mm[1800
] m2
= 63.0
B
.N>m
InAC
this
AC
stF
Using
2,
al2allow
B == Block
=
1s
2
A
=
35110
mm
110
2
m
>mm
]
=
63.0
kN
114.5
kN
12
m2
B
al
allow
B
sing
Eq.A2,
# 6
2
-6
22 By-6
comparison,
163.0
m2
6 2 110
2 2,
2 P =
2 as P reaches its smallest
= 183 value
kN (168 kN), the allowable
Using
Eq.
sal2allow
A Eq.
= 35110
2 N>m
mm
2kN212
m2>mm
] 110
= 63.0
kN
Using
1,
163.0
kN212
m2
AB =[1800
35110
2=N>m
[1800
mm
2m
] = 63.0
kN
ing Eq. 2,FBB = 1salP2allow
1.25
m
P
= 168
kN>mm
Using
Eq.
2,
163.0
kN212
m2
=
= 168
kN
normal
stress
will
first
be
developed
in
the
aluminum block. Hence,
1106.8 kN212
m2
0.75163.0
m comparison,
163.0
kN212
m m2
smallest
kN212
P = m2 as P reaches=its168
kN value (168 kN), the allowable
= By
171 kN
P
sing Eq. 2,
Using
Eq.
2,. =0.75
P
=
=
168
kN
A
or
C
Due
to
single
shear,
Pin
0.75
m
1.25 m P =normal stress will first
= 168 kN
shear,
n A or C. Due to single
P = 168 kN
Ans.
163.0
kN212
kN212 m2
0.75
m m2 163.0
m shear, be developed in the aluminum block. Hence,
#P
C.kN
Due0.75
to2=
single
Pin= A=or168
=
Block
B
.
In
this
case,
6
2
P
168
kN
2= 450110
2
or =C.V Due
toFsingle
shear,
nA
= 450110
VA0.75
=ort6allow
A
2m2
N>m
kN
C
Due
to 0.75
single
shear,
Pin
m
AC =
FAC
= tallow
A
2 .2N>m
[p10.009
] =2 [p10.009
114.5
kNm26 ] = 114.5
m
Ans.
6
2
-6
2
2 P = 168 kN2
F
A
=
35110
2
N>m
[1800
mm
110
2
m
>mm
]
=
63.0
kN
F
=
V
=
t
A
=
450110
2
N>m
[p10.009
m2
A
ve
C
deki
perçinler
için
hesap:
d=18
mm
veya
r=0.009
m ] = 114.5 kN
A=or1s
C
Due
to
single
shear,
nBF
al.2allow
B
6
2
2
AC
allow
A
or
C
.
Due
to
single
shear,
Pin
6114.5 kN
2
2
=
V
=
t
A
=
450110
2
N>m
[p10.009
m2
]
=
From
Eq.
1,
allow
FAC = V = tallow A = 450110 2 N>m [p10.009 m2 ] = 114.5 kN
omAC
Eq. 1,
6 From
114.5
12 2m2
6 m2
Eq. 1,kN
Using
Eq.
2,
114.5
kN
m2=2 [p10.009
F
=
V
=
t
A
=
450110
212
N>m
] 2=[p10.009
114.5 kNm22] = 114.5 kN
AC
allow
F
=
V
tallow
A
2 N>m
om Eq. 1,
= 183
kNkN 12 m2
= =450110
1, P
# PAC=From=Eq.
114.5
183
kN
163.0
kN212
m2
1.25
m kN
114.5
kN 12
m2
1.25
m
114.5
kN
12
m2
= 183 kN
P
=
P
=
=
168
om Eq. 1,
From
Eq.
1,
= 183
kN
P
= 114.5
By
as0.75
P reaches
its=
smallest
value
(1681.25
mthe
m
=kN),
183
kNallowable
P(168
By comparison, as
P comparison,
reaches
its
smallest
kN),
the
allowable
kN
114.5
kN 12
m2
1.25
m12 m2 value
1.25
m
By
comparison,
as
P=aluminum
reaches
itsblock.
smallest
value (168 kN), the allowable
A or
Cnormal
.first
Due
to single
shear,
Pin
will first
be
developed
in
the
Hence,
=
183
kNkN),
=stress
183 kN
P the
= value
ormal
stress
will
be
developed
aluminum
block.
Hence,
By comparison,
as# PPreaches
smallest
(168
the
allowable
Byits
comparison,
asstress
P1.25
reaches
its
smallest
value (168
kN),
the allowable
1.25
min
m
normal
will
first
be
developed
in
the
aluminum
block. Hence,
6
2
2
rmal
stress
be
developed
in
the
aluminum
block.
Hence,
FACwill
= first
V
=
t
A
=
450110
2
N>m
[p10.009
m2
]
=
114.5
kN
P
=
168
kN
Ans.
By comparison,
as
P
reaches
its
smallest
value
(168
kN),
the
allowable
normal
stress
will
first
be
developed
in
the
aluminum
block.
Hence,
allow
By comparison,
P reaches
its smallest value (168
kN), the allowable
P =as168
kN
Ans.
P = 168 kN
Ans.
rmal stress willnormal
first bestress
developed
in be
the
aluminum
block.
Hence,Ans.
will
in
the
göre
en büyük
kuvvet
P taşınabilecek
=first
168
kNdeveloped
From Eq. 1, Buna
P =aluminum
168 kN# block. Hence,
Ans.
P 114.5
= 168kN
kN12 m2
Ans.
P =
= 183
168 kN
kN
Ans.
P =
1.25 m
By comparison, as P reaches its smallest value (168 kN), the allowable
normal stress will first be developed in the aluminum block. Hence,
P = 168 kN
Ans.
#21
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