Bölüm 4. Eksenel Yük
Saint-Venant Prensibi
Bir ucundan ankastre mesnetli diğer ucundan bir P yüküne maruz bir çubuğu gözönüne
alalım. Çubuğun deformasyonu öncesi üzerine çizilen karelaj çizgilerinin deformasyon sonrası
aldığı şekli incelediğimizde çubuğun her iki ucundaki lokal deformasyondaki düzensizlikler
4.1
görülmektedir.
P
P
SAINT-VENAN
P
P
a
b
c
P
a
b
c
P
Load distorts lines
located near load
savg !
4.1
Lines located away
from the load and support
remain straight
Load distorts lines
located
P near support
savg !
A
121
SAINT-VENANT’S PRINCIPLE
#
section a–a
P
2
section b–b
(b)
P
2
P
A
se
section c–c
Fig. 4–1 (cont.)
P
A
the stress
savg !
In the same way,
distribution at the support will also even
out
and
become
uniform
over
the
cross section located the same distance
section c–c
ection b–b
section c–c
(a)
# from
away
the
support.
(c)
(b)
The fact
that stress and deformation
in this manner
is referred
Fig. 4–1
yeteri kadar
uzaklaşıldığında
gerilme ve behave
deformasyon
dağılımın
Fig. 4–1 (cont.) Bu lokal düzensizliklerden
to as Saint-Venant’s principle, since it was first noticed by the French
scientist Barré
de Saint-Venant
in 1855.
Essentially
it kesitinin
states that the
üniformlaştığı gözlenebilir. Bu uzaklaşma
mesafesinin,
en azından
yüklenen
çubuk
stress and strain produced at points in a body sufficiently removed from
en büyük boyutu kadar olması gerektiği
Gerilme
davranışı
the regionanlaşılmıştır.
of load application
willvebedeformasyonun
the same as the bu
stress
and strain
sing this idea, consider the manner in which a rectangular bar will
4
produced by any applied loadings that have the same statically equivalent
ess
atSaint-Venant
thethe
support
also even
rm distribution
elastically when
bar iswill
subjected
to a force
P applied along
its adamının
prensibi
olarak
bilinmektedir.
uygulandığı
resultant, andBu
arebilim
applied
to the “Yükün
body within
the samebölgeden
region. For
er
the
cross
section
located
the
same
distance
roidal axis, Fig. 4–1a. Here the bar is fixed connected
at
one
end,
with
example,
if
two
symmetrically
applied
forces
act
on
P>2
yeteri kadar
cisim
oluşan gerilme
ve birim deformasyon, statikçe the bar,
orce applied through
a hole uzaktaki
at its other
end.noktalarında
Due
the loading,
thedistribution
Fig.to4–1c,
the stress
at section c–c will be uniform and
eformation
in this
manner
is referred
deforms
asbehave
indicated
bybileşkeye
the once
horizontal
and
vertical
grid lines
therefore
equivalent
to
savg = P>A
as in Fig. 4–1b.doğan gerilme ve
eşdeğer
sahip
ve
aynı
bölgede
uygulanan
herhangi
bir
yüklemeden
ple, since it was first noticed by the French
wn
on the bar. Notice how the localized deformation that occurs
enant in 1855. Essentially
it states that
deformasyon
ile the
aynıuniform
olacaktır.”
ifadesi mevcuttur.
Yukarıdaki şekildeki P yüklemesini iki
ach end tends birim
to even
out and become
throughout
the
at points in a body sufficiently removed from
section
ion willof
bethe
thebar.
same
as the
stressP/2+P/2
and strainolarak yüklemek, yüklemeden yeteri kadar uzakta aynı etkileri
parçaya
ayırıp
the
material
remains
elastic equivalent
then the strains caused by this
adings that have the
same statically
rmation
are directly
related
toregion.
the stress
meydana
getirecektir.
to the body
within
the
same
For in the bar. As a result, the
ss
will applied
be distributed
uniformly
the cross-sectional
ically
forces more
on the throughout
bar,
P>2 act
bution
at section
will be
uniform
when the
sectionc–c
is taken
farther
andand
farther from the point where
in Fig.
= P>A asload
is4–1b.
applied. For example, consider a profile of the
gexternal
ation of the stress distribution acting at sections a–a, b–b, and c–c,
h of which is shown in Fig. 4–1b. By comparison, the stress tends
each a uniform value at section c–c, which is sufficiently removed
m the end since the localized deformation caused by P vanishes.
minimum distance from the bar’s end where this occurs can be
rmined using a mathematical analysis based on the theory of
ticity.
Notice how the lines on this rubber membrane
has been found that this distance should at least be equal to
the
distort after it is stretched. The localized
est dimension of the loaded cross section. Hence, section c–c should
distortions at the grips smooth out as stated by
Saint-Venant’s
principle.
ocated at a distance at least equal to the width (not the thickness)
of
bar.*
hen section c–c is so located, the theory of elasticity predicts the maximum stress to be
#42
#
Lastik membran üzerindeki çizgilerin çekildikten sonra nasıl biçim değiştirdiğine dikkat
ediniz. Tutamaklardaki
bölgesel
bozuklukları
prensibinde ifade edildiği
Notice
howbiçim
the lines
on thisSaint-Venant
rubber membrane
gibi giderek düzgünleşmektedir.
distort after it is stretched. The localized
distortions at the grips smooth out as stated by
Saint-Venant’s principle.
#43
ed loads at itsloads
endsat
and
variable
load
distributed
of the
bar
if itload
doesdistributed
not remain horizontal, or friction forces acting
concentrated
its aends
andweight
aexternal
variable
external
th. its
This
distributed
load could,
for
example,
the we wish
oncould,
the bar’s
surface. Here
ng
length.
This distributed
load
for represent
example,
represent
the to find the relative displacement d
bar of
if itthe
does
horizontal,
or friction
forces
(delta)
of oneor
end
ofacting
the
bar with
respect to the other end as caused by
ght
barnot
if itremain
does not
remain
horizontal,
friction
forces
acting
d
surface.
we wish
the
d
the bar’sHere
surface.
Heretowefind
wish
torelative
find
thedisplacement
relative
this
loading.
We willdisplacement
neglect
the localized
deformations that occur at
elta)
endofofone
theend
bar of
with
to respect
the
other
as caused
by
therespect
bar with
to
the
other
end as
causedand
by where the cross section suddenly
points
ofend
concentrated
loading
We
will neglect
theneglect
localized
that
occurthat
at occur
loading.
We will
the deformations
localized
at
changes.deformations
From
Saint-Venant’s
principle,
these effects occur within small
ncentrated
loading and
whereand
the
crossof
section
suddenly
nts of concentrated
loading
where
the
section
suddenly
regions
thecross
bar’s
length and
will therefore have only a slight effect on
4.2
Eksenel
yüklü
bir
elemanın
elastik
deformasyonu
mnges.
Saint-Venant’s
principle,
these
effects
occur
within
small
From Saint-Venant’s
principle,
these
effects
occur
within
small
the
final
result.
For
the
most
part,
the bar will deform uniformly, so the
eions
bar’s
andlength
will therefore
onlystress
a slight
effect
on effect
oflength
the bar’s
and will have
therefore
have
only
slight
on
normal
will
bea uniformly
distributed
cross section.
Değişken
kesitli
bir
çubuk
uçlarından
kuvvetlereover
ve the
L uzunluğu
boyunca bir yayılı yüke
lt.
Forresult.
the most
the bar
will
uniformly,
so
the tekil
final
For part,
the most
part,
thedeform
bar will
deform
soa the
Using
the
methoduniformly,
of sections,
differential element (or wafer) of length
will stress
be uniformly
distributeddistributed
over
cross
mal
will bemaruzdur.
uniformly
over
the
cross section.
dxthe
and
cross-sectional
area
is isolated
the bar
at the arbitrary
Çubuğun
birsection.
ucunun
diğerA(x)
ucuna
göre from
rölatif
deplasmanı
δ yı hesaplayalım.
4.2 diagram
Eof
LASTIC
Dof
EFORMATION
OF AN
XIALLY Lin
OADED
EMBER
123
method
ofmethod
sections,ofa sections,
differential
element
of length
Using the
a differential
element
(or
wafer)
length
position(or
x.wafer)
The free-body
this element
isAshown
Fig.M
4–2b.
sectional
area
A(x)
is
isolated
from
the
bar
at
the
arbitrary
and cross-sectional
area A(x)deplasman
is isolated
from
the
bar ataxial
the
arbitrary
Çubuğun
sonrası
alacağı
şekil
kesikli
ile gösterilmiştir.
The resultant
internal
force
will beçizgi
a function
of x since theÇubuğun
external sol ucundan
e free-body
diagram ofdiagram
this element
iselement
shown
in
Fig.
4–2b.
ition
x. The
free-body
of
this
is
shown
in
Fig.
4–2b.
4.2
E
LASTIC
D
EFORMATION
OF AN
A
XIALLY
L
OADED the
M
EMBER
123
distributed
loading
will
cause
it
to
vary
along
length
of
the
bar.
This
For the entire length L of the bar, we must integrate this expression to
biraxial
x be
mesafedeki
dx
elemanı
için
dδ
uzamasını
yazalım.
Oluşan
gerilmelerin
orantılılık
axial
force
will
a function
x since
the
einternal
resultant
internal
force
willload,
be aof
function
of
xexternal
since
the
external
P(x),
will
deform
the
element
into
the
shape
indicated
by
the
dashed
find d. This yields
ading
will
cause
it
to
vary
along
the
length
of
the
bar.
This
ributed loading will cause it to vary
along
thetherefore
length ofthe
thedisplacement
bar. This
outline,
and
of one end of the element with
ntire
length
Lelement
of the
bar,
we
integrate
this
expression
to
sınırından
düşük
olduğu
varsayımı
ilethe
(Hooke
kanunu
geçerlidir):
l deform
thedeform
into
themust
shape
indicated
by
the dashed
d,
P(x),
will
the
element
into
the
shape
indicated
by
dashed
dd.
respect
to
the
other
end
is
The
stress
and
strain in the element are
yields
L
herefore
the
displacement
of
one
end
of
the
element
with
line, and therefore the displacement of one P1x2
end of
dxthe element
P1x2 with
dd
dand
(4–1)
other
is dd. The
and stress
strain
in= Ethe
element
4.2
LASTIC
Din
EFORMATION
123
pect
toend
the other
end stress
is dd. The
strain
theare
element
areAXIALLY
s = OF AN
andLOADED
PM
=EMBER
0 A1x2E
L
A1x2
dx
L
P1x2
P1x2
dd
P1x2 dx dd
s =
P and
=
P the
= stress does
Provided
not exceed the proportional limit, we can apply
d s= = and
(4–1)
A1x2
dx
A1x2
dx
ire length
L of the
must
this expression to
where
! bar,
! integrate
L0 weA1x2E
Hooke’s law; i.e.,
elds does
stress
not exceed
theexceed
proportional
limit, we can
apply
vided
the stress
does not
the proportional
limit,
we can apply
d = displacement of one point on the bar relative to the other point
s = EP
i.e., law; i.e.,
oke’s
L = original length of bar
dd
P1x2
s =LEP
sdx
= EP
placement
of one
point
onaxial
the
bar
relative
the other
point P1x2 x from
P1x2
=d internal
force
at the to
section,
located
= Ea b
=
(4–1)a distance
A1x2
dx
ginal length ofP1x2
bar
P1x2
oneL0endA1x2E
dd
dd
= E a b= E a b
P1x2
ernal axial
force
the section,
distance
x from as a function
A1x2
dx located
A1x2
A1x2
= at
cross-sectional
area dx
ofathe
bar, expressed
of xdx
dd =
4
e end
A1x2E
E = modulus
of elasticity
P1x2
dx P1x2for
dxthe material
dd
= onbar,
dd
= relative
oss-sectional
areapoint
of the
expressed
as to
a function
x
lacement of one
the
bar
the otherofpoint
4
A1x2E A1x2E
x
dx
Constant
Load
and
Cross-Sectional
Area. In many cases
odulus
of elasticity
nal length
of bar for the material
bardxat
will
have
a constant
area A; and the material
will
x axial
dx the
Pthe
P2
rnal
force
section,
locatedcross-sectional
a distance x from
P(x)
P(x)
1
be
homogeneous,
so
E
is
constant.
Furthermore,
if
a
constant
external
Load
and
Cross-Sectional
Area.
In
many
cases
4.2
E
LASTIC
D
EFORMATION
OF
AN AXIALLY
L
OADED MEMBER
123
end
P2end,
P(x)then
P(x)
P(x)
dd
force is cross-sectional
appliedP2 at each
Fig.the4–3,
the internal
force P
have a constant
area
A;P(x)
and
material
will
s-sectional
area
of
the
bar,
expressed
as
a
function
of
x
dx
L
throughout
the length
of the bar
constant.
As a result,
Eq. 4–1 can
neous, so
E is constant.
Furthermore,
if is
a also
constant
external
!
4
d
dd
eulus
entire
length L of
bar,
we must integrate (a)
this expression
todd
(b)
of at
elasticity
forthe
the
material
beLeach
integrated
to
yield
pplied
end,
Fig.
4–3, then the internal
force
P
dx
dx
L
his yields
d boyunca dδ entegre edilirse:
d
Tüm
çubuk
the length
Eq. 4–1 can
Fig.
(b) 4–2
(a)of the bar is also constant. As a result,
(b)
a)
Load
and Cross-Sectional Area. In many cases
ed
to yield
PL
Fig. cross-sectional
4–2 Fig. 4–2
ave a constant
area
L
d =the material will
(4–2)
P1x2
dx A; and
AE external
eous, so E is constant.
d = Furthermore, if a constant
(4–1)
L0PLA1x2E
plied at each end, dFig.
P
= 4–3, then the internal force(4–2)
!
AE
he length of
the
bar
is
also
constant.
As
a
result,
Eq.
4–1
can
If the bar is subjected to several different axial forces along its length, or
to yieldthe cross-sectional
Sabit yük
veor
sabit
kesit
sözkonusu
olduğunda,
area
modulus
of alanı
elasticity
changes abruptly
from The vertical displacement at the top of these
displacement
ofseveral
one of
point
the
relative
to the
point
is subjected
different
axial
forces
its other
length,
or
one to
region
theon
bar
tobar
the
next,along
the
above
equation
can be applied building columns depends upon the loading
the
and to the floor attached
The vertical
displacement
at theon
top
ofroof
these
ectional
orofmodulus
of the
elasticity
changes
abruptly
from
to each
segment
of
bar where
these
quantities
remain
constant.
The applied
original area
length
bar PL
to
their
midpoint.
d
=
(4–2)
building
columns
depends upon the loading
of the bar
the next,
the
above
equation
can
bex applied
displacement
ofAE
one
end
of
the
bara with
respect
to the other
is then
found
nternal
axialtoforce
at the
section,
located
distance
from
applied on the roof and to the floor attached
!
ment
of
the
bar
where
these
quantities
remain
constant.
The
from
the
algebraic
addition
of
the
relative
displacements
of
the
ends
of
one end
to their midpoint.
nt of oneeach
end segment.
of the barFor
with
respect
to case,
the other is then found
this
general
Bir
çubuk
uzunluğu
boyunca
farklı
eksenel
yüklere maruz ise veya enkesit
alanı
cross-sectional
area
of
the
bar,
expressed
as
a
function
of
x
subjected
to
several
different
axial
forces
along
its
length,
or
gebraic addition of the relative displacements of the ends of
4
The
vertical
displacement
at
the
top
of
these
tional
area
or
modulus
of
elasticity
changes
abruptly
from
modulus
ofgeneral
elasticity
for the material
nt. For this
case,
uğruyorsa,
PLbe applied building columns depends upon the loading
f the bar to the ani
next,değişimlere
the above equation
can
d = a
(4–3)
ent
of
the
bar
where
these
quantities
remain
constant.
AE
nt Load and Cross-Sectional Area. In manyThe
casesapplied on the roof and to the floor attached
to
their
midpoint.
PL
of have
one end
of the bar
respect
to
theA;
other
then
found
will
a constant
area
and is
the
material
dcross-sectional
=with
(4–3)will
a AEdisplacements of the ends
ebraic
addition
of
the
relative
of
geneous, so E is constant. Furthermore, if a constant external
!
t.applied
For this general
case,
at each end, Fig. 4–3, then the internal force P
ut the length of the bar is also constant. As a result, Eq. 4–1 can
x
ated to yield
PL
d P= a
(4–3)
P
x
AE
PL
L
P
P
d =
(4–2)
d
AE
!
L
Fig. 4–3
d
ar is subjected
x to several different axial forces along its length, or
Fig. 4–3of elasticity changes abruptly from
-sectional area or modulus
on of the bar to the next, the above equation Pcan be applied
egment of the bar where
L these quantities remain constant. The
ment of one end of the bar with respect to thed other is then found
algebraic addition of
the
Fig.
4–3relative displacements of the ends of
ment. For this general case,
The vertical displacement at the top of these
building columns depends upon the loading
applied on the roof and to the floor attached
to their midpoint.
#44
Fig. 4–5b. They are PAB = +5 kN, PBC = -3 kN, PCD = -7 kN. This
!P
Fig. 4–5b. They are PAB = +5 kN, PBC = -3 kN, PCD = -7 kN. This
variation in axial load is shown on the axial or normal
force diagram for
variation in axial load is shown on the axial or normal force diagram for
the bar, Fig. 4–5c. Since we now know how the internal force varies
the bar, Fig. 4–5c. Since we now know how the internal force varies
throughout the bar’s length,
end A relative
to end
D the displacement of end A relative to end D
!d the displacement ofthroughout
the bar’s
length,
is determined Positive
from sign convention for P and d
is determined from
d
4–4
124
C HFig.
APTER 4
AXIAL LOAD
15 kN2L
1-3ve
kN2L
1-7 kN2L
15 kN2L
1-3 kN2LBC
1-7 kN2LCD
PL
AB
BC
PLCD
AB
İşaret
Kuralı:
Eksenel
yük
deplasmanlar
için
pozitif
yönler:
dA>D = a
=
+
+
dA>D = a
=
+
+
AE
AE
AE
AE
AE
AE
AEAE
Sign Convention. In order to apply Eq. 4–3, we must develop a
!P
sign
convention
the
force answer
and theis displacement
the other
data arefor
substituted
andaxial
a positive
calculated, it of one
If the other data are substituted and a positiveIfanswer
is calculated,
it internal
end
of
the
bar
with
respect
to
the
other
end.
To
do
so,elongates),
we will consider
means
that
end
A
will
move
away
from
end
D
(the
bar
means that end A will move away from end D (the bar elongates),
both
the
force
and
displacement
to
be
positive
if
they
cause tension
whereas
a
negative
result
would
indicate
that
end
A
moves
toward
!d
whereas a negative result would
indicate that end A moves toward
end
D elongation,
(the bar shortens).
The double
subscript
notation
is used force
to
and
respectively,
Fig.
4–4;
whereas
a
negative
and
end D (the bar shortens). The double subscript notation is used to
1dA>D2; and
indicate
this relative
displacement
however,
if the displacement
displacement
will
cause
compression
contraction,
respectively.
indicate this relative displacement 1dA>D2; however,
if the displacement
is to For
be determined
relative tothe
a fixed
then
a single
subscript
consider
bar point,
shown
in only
Fig. 4–5a.
The
internal axial
is to be determined relative to a fixed point, then
only
a example,
single
subscript
will
be
used.
For
example,
if
D
is
located
at
a
fixed
support,
then
thesegment,
forces
“P,”
are
determined
by
the
method
of
sections
for
each
will be used. For !P
example, if D is located at adisplacement
fixed support,
then
the as simply dA.
will
be
denoted
Fig. 4–5b. They are PAB = + 5 kN, PBC = - 3 kN, PCD = - 7 kN. This
displacement will be denoted as simply dA.
variation in axial load is shown on the axial or normal force diagram for
LOAD
the bar, Fig. 4–5c. Since we now know how the internal force varies
!d
throughout the bar’s length, the displacement of end A relative to end D
!
8 kN
4 kN
5 kN from
7 kN
Positive sign convention for P and d
is
determined
Sign Convention. In order to apply Eq.
4–3, we must
develop a
4
XIAL
8 kN
5 kN
4 kN
7 kN
A
B
sign convention forFig.
the4–4
internal axial force and the displacement
of one
4
C
D
LAB
LBC
LCD
end of the bar
15 kN2LAB
1 - 3 kN2LBC
1 - 7 kN2LCD
A with respect
B to the otherCend. To do so, weDwill
PL consider
d
=
=
+
+
(a)
both the force and
tension
acause
LAB displacement
LBC to be positiveA>D
Lif
CD they
AE
AE
AE
AE
and ! elongation, respectively, Fig.(a)4–4; whereas a negative force and
displacement will cause compression and contraction, respectively.
P (kN)
the other data
are substituted and a positive answer is calculated, it
5 kNshown in If
AB "bar
For example, consider Pthe
Fig. 4–5a. The internal axial
means that end A will move away from end D (the bar elongates),
5 kN
A
P (kN)
forces “P,” are
determined by the method of sections for each segment,
whereas a negative result would indicate that end A moves toward
Fig. 4–5b. They are PAB = +58 kN,
kN PBC = -3 kN, PCD = 5-7 kN. This
end D (the bar shortens). The double subscript notation is used to
variation in axial load is shown on the axial
force diagram for
PBCor
" 3normal
kN
x
indicate this relative displacement 1dA>D2; however, if the displacement
5
kN
A 54–5c. SinceBwe now know how the internal
the bar, Fig.
#3 force varies
is to be determined relative to a fixed point, then only a single subscript
PBC " 3throughout
kN
the bar’s length, the displacement of end A relative
toxend D
#7
will be used. For example,
if D is located at a fixed support, then the
PCD " 7 kN
7
kN
P and d
is determined
from
#3
displacement will be denoted as simply dA.
D
!
#7
7 kN
!
dA>D
(c)
(b)
15 kN2LAB
1-3 kN2LBC
1-7 kN2LCD
PL
= a
=
+
+Fig. 4–5
AE
AE (c)
AE
AE
8 kN
5 kN
4–5data are substituted and a positive answer is calculated, it
If the Fig.
other
A
means that end A will move away from end D (the bar elongates),
LAB
whereas a negative result would indicate that end A moves toward
end D (the bar shortens). The double subscript notation is used to
indicate this relative displacement 1dA>D2; however, if the displacement
is to be determined relative to a fixed point, then only a singlePsubscript
(kN)
PAB
will be used. For example,
if "D5 kN
is located at a fixed support, then the
5 kN
A will be denoted as simply d .
displacement
A
8 kN
5 kN
4 kN
B
7 kN
C
LBC
D
LCD
(a)
5
PBC " 3 kN
A
x
B
#3
8 kN
5 kN
PCD " 7 kN
4 kN
D B
A
LAB
(b)
C
LBC
(a)
7 kN
#7
7 kN
D
LCD
(c)
Fig. 4–5
P (kN)
" 5 kN
8 kN
5
PBC " 3 kN
x
#3
7 kN
#7
(c)
Fig. 4–5
#45
PBC " 7 kip
126
126
CHAPTER 4 AXIAL LOAD
CHAPTER 4 AXIAL LOAD
PCD " 9 kip
(b)
126
CHAPTER 4 AXIAL LOAD
EXAMPLE 4.1
EXAMPLE 4.1
! of the external loadings, the 126
ue to the application
C H Abar
PTER 4
A Xin
IAL LOAD
The A-36 steel
shown
Fig. 4–6a is made from two segments
in regions AB, BC, and CD will all be different. EXAMPLE
Thecross-sectional
A-364.1
steel barareas
shownofin A
Fig. 4–6a
is2 made
from =two2 segments
having
and2 A
in2.
AB = 1 in
BD
A
XIAL LOAD
A36ofçeliğinden
üretilmiş
çubuk
iki
farklı
kesit
bölgesine
#
ained by applyingŞekilde
the method
sections and
the
havingthecross-sectional
areas of
AABA=and
1 intheand
ABD = 2 in2.
Determine
vertical displacement
of end
displacement
force equilibrium as shown in Fig. 4–6b. This ofEXAMPLE
The
A-36
steel
bar
shown
in Fig. 4–6a is made
Determine
the vertical displacement of end A and the displacement
B relative
C.4.1
sahiptir. A ucunun düşey deplasmanını
ve Btonoktasının
C ye göre göreceli deplasmanını
n Fig. 4–6c.
having cross-sectional
areas of AAB = 1 in2
of B relative to 15
C.kip
15 kip
15 kip
15 kip
Determine
vertical
displacement
of end
Aa
The
A-36the
steel
bar
shown
in Fig. 4–6a
is ma
15 kip
15 kip
15 kip
om Athe
cover, # Est15=kip2911032 ksi.
X I A L inside
L O A D back
belirleyiniz.
of
B
relative
to
C.
having
cross-sectional
areas
of
A
=
1 in
AB
ntion, i.e., internal
forces
areshown
positive
A
Thetensile
A-36 steel
bar
inand
Fig. 4–6a is made from two segments
Determine
the
vertical
displacement
of
end
A
A
15
kip
15
kip
are negative, the vertical displacement of A
15 kip
having cross-sectional areas of AAB = 1 in2 and
ABD = 2 in2.
of
B
relative
to
C.
upport D is
Determine the vertical
the displacement
2 ft displacement of end A and
A 15
15 kip
4 kip
4 kip
kip
4 kip
4 kip 15 kip
kip]12 ft2112 in.>ft2
[+7
4 kipkip]11.5
4 kip ft21122in.>ft2
ft
of B relative
to C.
4
kip
4
kip
4
kip
4
kip
The
in2 Fig. 4–6a is made fromP two
segments
+A-36 steel
4 kipbar 43shown
kip
2
15 2kip
kip
15 kip AB " 15 kip
2 A
n22[2911032 kip>inhaving
]
12cross-sectional
in22[29110
kip>in
] of15 A
areas
and
=
1
in
A
= 2 in2.
P
B
AB
BD
4
2 ft AB " 15 kip
4 kip
4 kip
B vertical displacement of end4A
Determine the
kipand 4the
kip displacement
9 kip]11 ft21124in.>ft2
1.5 ft
8 kip
8 kip
of B relative
to C.
P " 15 kip
8 kip
8 kip
1.5 ft
2 ft
8 kip AB
8 kip
in22[2911032 kip>in2]
4 kip
4 kip
B4 kip 15 kip
8 kip
8 kip
4
15
kip
15 kip
4 kip
0127
in.
Ans.
C
PBC " 7 kip
P " 15 kip
2 ft
C
4 kip
4
1 ft
sitive, the bar elongates and so the displacement
D
between
points B and C, we
2 ft
ft
PAB " 15 1kip
D
obtain,
(a)
(a)
4 kip
B
8 kip4 kip
C 8 kip
4 kip
4 kip
4 kip
1.5 ft
kip
84 kip
1.5 ft
8 kip
ft
41kip
8 kip
PBC " 7 kip
(b)
AB
8
PBC " 7 kip
PCD " 9 kip
PCD " 9 kip
(b)
p
.5[+7
ft kip]11.5 ft2112 in.>ft2
8 kip
SOLUTION
PBC " 7 kip
DC
p
The
A-36 steel bar shown in
Fig. 4–6a
is 15made
from two
segments
=
+0.00217
in.
Ans.
P
"
kip
SOLUTION
AB
3
2
2 Due to
Internal
Force.
the application of the external loadings, the
!
12 in22[29110
2 kip>in2areas
] 0 of15 AABP=(kip)
1
ft
having
cross-sectional
and
2
in
1
in
A
=
.
BD
15
(a)
Internal
Force.
Due to AB,
the application
of will
the external
loadings, the (b)
internal
axialDforces
in regions
BC, and CD
all be different.
PBCthe
" 7displacement
kip
0
P (kip)
Determine
thethe
vertical
displacement
of end A
and
from C, since
segment
elongates.
SOLUTION
axial
forces
in regions
and of
CDsections
will alland
be different.
1.5 ft
Çözüm:
kip
8 kipby
Theseinternal
forces8are
obtained
applyingAB,
theBC,
method
the
1kip
ftB relative to C.
of
(a)
These
areforce
obtained
by applying
theForce.
method
of
sections
and
the (b) of the
15
equation
of forces
vertical
equilibrium
as
shown
in
Fig.
4–6b.
This
Internal
Due
to
the
application
0
15 kip
15 kip
15 kip
P (kip)
equation
of vertical
force equilibrium
as
shown
ininFig.
4–6b.AB,
ThisBC, and CD
SOLUTION
variation
is plotted
in Fig. 4–6c.
internal
axial
forces
regions
PBC " 7 kip
PCDis"plotted
9 kip in Fig. 4–6c.
variation
(b)
3
15
These
forces
are
obtained
by
applying
the metho
Internal
Due to2 ksi.
the application
of th
Displacement.0 From theP (kip)
inside back
cover,Force.
Est = 29110
2
1 ft
3
SOLUTION
Displacement.
From
the
inside
back
cover,
E
=
29110
2
ksi.
equation
of
vertical
force
equilibrium
asand
show
internal
axialare
forces
C
2
st in regions
Using the sign convention, i.e., internal tensile
forces
positive
and AB, BC,
Using
the
sign
convention,
i.e.,
internal
tensile
forces
are
positive
and
variation
is
plotted
in
Fig.
4–6c.
forces
are obtained
compressive
forces loadings,
are negative,
the These
vertical
displacement
of by
A applying the met
Internal Force. Due to the application
of the external
the
PCD "forces
9 kip are negative, the vertical displacement of A
P (kip)
compressive
(b)and
equation of vertical
sho
relative
the
fixed
D is
internal axial
AB,
CD
all support
be different.
Displacement.
From force
the equilibrium
inside back ascove
2will
4 kip in regions
4 kip
4 kipBC,
4 kip to
3.5 forces
to
the
fixed and
support
D isUsing
variation
is
plotted
in
Fig.
4–6c.
SOLUTION
7obtained by applying the relative
the
sign
convention,
i.e.,
internal
tensile
fo
These
forces are3.5
method
of
sections
the
[ + 15 kip]12 ft2112 in.>ft2 [ + 7 kip]11.5 ft2112 in.>ft2
PL
PAB " 15 kip
7
=2in
dA =asa
+
[24–6b.
+ 15 kip]12
ft2112
in.>ft2
[
+
7
kip]11.5
ft2112
in.>ft2
compressive
forces
are
negative,
the
vertical
equation
verticalDue
force
shown
Fig.
This
Displacement.
From
the
inside
back
co
PL
InternalofForce.
to equilibrium
the application
of
the
external
loadings,
the
3
2 kip>in2]
12+in22[2911032 kip>in2]
P (kip)
dA AE
= a 11 =in 2[29110
2
2 to
2 support
2
Di.e.,
is internal
Using
sign
tensile
variation
is plotted
in Fig.
4–6c. AB, BC, and
internal
forces
in regions
CD
will all be
different.
11 in
2[2911032 relative
kip>in
] thethe
12fixed
inconvention,
2[2911032 kip>in
]
3.5 AE
!9 axial4.5
8
kip
8
kip
7
[
9
kip]11
ft2112
in.>ft2
3
compressive
forces
are
negative,
the
These
forces
are
obtained
by
applying
the
method
of
sections
and
the
4.5 the inside back cover, E = 29110 2 ksi.
!9 From
[ + 15 kip]12 ft2112 in.>ft2 vertic
[+7
Displacement.
PL
x (ft)
+st
- 9 kip]11
ft2112
in.>ft2
=
d
=
+
2[4–6b.
3
2
relative
to
the
fixed
support
D
is
equation
of
vertical
force
equilibrium
as
shown
in
Fig.
This
A ] a
x
(ft)
2
3
2
12
in
2[29110
2
kip>in
Using the sign
convention,
i.e.,
internal
tensile
forces
are
positive
and
3.5
+
(c)
AE
11 in 2[29110 2 kip>in ]
12
2
3
7
variation is plotted
in
4–6c. the vertical displacement
12in.
in
2 kip>in2] PL
P(c)
" 7Fig.
kip
BCare
[+15 kip]12
ft2112 in.>ft2 [+
compressive
forces
negative,
of 2[29110
A
= 4.5
+ 0.0127
Ans.
!9
3 in.
= [ - 9 kip]11
dA = a
+
Fig. 4–6
ft2112 in.>ft2
0.0127
Displacement.
From
Est= = +29110
2 ksi.
relative
to the fixed
supportthe
D isinside back cover,
x (ft)is positive,
11 in22[2911032Ans.
kip>in2]
1
Since the result
the bar elongates and AE
so the
+ displacement
Fig.
4–6
2
3
2
Using the sign[+15
convention,
i.e., internal
tensile
forces
are
positive
and
12
in
2[29110
2
kip>in
]
4.5
!9
Since
the
result
is
positive,
the
bar
elongates
and
so
the
displacement
(c)
kip]12
ft2112
in.>ft2
[+7
kip]11.5
ft2112
in.>ft2
at A is upward.
PL
[-9 kip]11 ft2112 in.>ft2
are negative,
vertical
displacement of A
= forces(b)
dAcompressive
= a
+at
PCD "the
9 kip
A! is Eq.
upward.
= ++0.0127 in.
x (ft)
2
3
2Applying
3
2
4–2 between
B and C, we obtain,
!relative
AE
11 in support
2[29110 D
2 kip>in
]
12 in22[29110
2 kip>inpoints
]
to the fixed
is
Fig.
4–6
12 in22[2911032 kip>in2]
Applying(c)Eq. 4–2 between points B and C, we obtain,
SOLUTION
Since
the result is positive, the bar elongates and
[ + 7 kip]11.5 ft2112
in.>ft2
P
L
BC
BC
[+15
kip]12ft2112
ft2112in.>ft2
in.>ft2
kip]11.5
ft2112[ +in.>ft2
[-9 kip]11
PL
= + 0.0127
in.
= + 0.00217
in. Ans.
= [+7 the
dB>C
in.>ft2
PBC=LBC
A is2upward.
2 7 kip]11.5
3 at ft2112
nternal Force. dDue
to the application
of the external
loadings,
=+
+d ABC
E
12
2 kip>in
]
Fig.
4–6
A = a
2
3
2
2
3 =in 2[29110
2
= + 0.00217 in. Ans.
=
2
3
2
B>C
AE
2
3
2
11
in
2[29110
2
kip>in
]
12
in
2[29110
2
kip>in
]
kip>in
] different.ABCE
Applying
Eq.
points
and C, we
Since
the result
is between
positive, the
barBelongates
a
nternal axial forces in regions AB,12BC, and CD will
all be
in 2[29110
2 kip>in
] 4–2
Here B moves away
from C,12
since
the segment
elongates.
at
A
is
upward.
These forces are obtained by applying
the
method
of
sections
and
the
= + 0.0127
in.
Ans.
Here B moves away from C, since the segment
[ + 7 kip]11.5 ft2112 in.>ft2
[-9 kip]11 ft2112 in.>ft2
PBC Lelongates.
BC
6
=
= between
dB>C
Applying
Eq. 4–2
points B and C, w
equation
of vertical force equilibrium
as
shown3 in Fig. 24–6b. This
+
2
Since the result is positive,
the bar 2elongates
ABC E
12 in22[2911032 kip>in2]
12 in 2[29110
kip>in ] and so the displacement
variation is plotted in Fig. 4–6c.
[+7 kip]11.5 ft2112 in.>ft2
PBCaway
LBC from
at A is upward. = + 0.0127 in.
B moves
C, since the segment e
3
Ans. HeredB>C
=
=
!
Displacement.
From
the
inside
back
cover,
E
=
29110
2
ksi.
st
Applying Eq. 4–2 between points B and C, we obtain,
4–6
ABCE
12 in22[2911032 kip>in2]
Using the sign convention,
internal
tensilethe
forces
are positive
Since thei.e.,
result
is positive,
bar elongates
andand
so the displacement
Here B moves away from C, since the segment
[+7vertical
kip]11.5displacement
ft2112 in.>ft2 of A
P
compressive forces
are
negative,
BC LBC the
at
A
is
upward.
= +0.00217 in. Ans.
=
dB>C =
2
3
2
elative to the fixed support
is
Applying
points
B and C,
AD
E 4–2 between
12 in 2[29110
2 kip>in
] we obtain,
BCEq.
!
[
+
15
kip]12
ft2112
in.>ft2
[
+
7
kip]11.5
ft2112
in.>ft2
Here B moves
C,kip]11.5
since the
segment
elongates.
PL
[+7
ft2112
in.>ft2
PBCaway
LBC from
=
dA = a
+
=2 +0.00217 in. Ans.
=
=
d
2
3
2
3
2 in22[29110
2
AE
11 inB>C
2[29110
2 kip>in
] 12 in12
2 kip>in
]
ABC
E
2[2911032 kip>in
]
+
Here
B moves
away
from C, since the segment elongates.
[ - 9 kip]11
ft2112
in.>ft2
12 in22[2911032 kip>in2]
= + 0.0127 in.
Ans.
Since the result is positive, the bar elongates and so the displacement
at A is upward.
Applying Eq. 4–2 between points B and C, we obtain,
dB>C =
PBCLBC
=
[ + 7 kip]11.5 ft2112 in.>ft2
= + 0.00217 in. Ans.
#46
of 10 mm is attached to a rigid collar and passes through the tube. If a
tensile
load of 80
kN is applied to the rod, determine the displacement
EXAMPLE
4.2
EXAMPLE
4.2
4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
of the end C of the rod. Take Est = 200 GPa, Eal = 70 GPa.
The assembly
in Fig.
4–7aofconsists
of an aluminum
The assembly
shown in shown
Fig. 4–7a
consists
an aluminum
tube AB tube AB
4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
2
2
400area
mm of 400
having a cross-sectional
area
steel
rod
having a diameter
400steel
mmrod
. Ahaving
having a cross-sectional
a diameter
mmof
.A
EXAMPLE
10B mm 4.2
isto
attached
to Aa rigid
passesthe
through
the
of 10
mm of
is attached
a rigid collar
andcollar
passesand
through
tube. If
a tube. If a
tensile
80 kN is
the rod, determine
the displacement
tensile load
of 80load
kN isofapplied
toapplied
the rod,todetermine
the displacement
The assembly
shown4.2
in Fig. 4–7a consists 80
ofkN
an aluminum tube AB
EXAMPLE
endrod.
C ofTake
the rod.
Eal80=kN70 GPa.
PAB ! 80 kN
of the end
Cthe
of the
E GPa,
Est =Take
200EGPa,
= 70 GPa.
! of
st =2C200
having
a cross-sectional area
of 400 mm
a diameter
. Aalsteel rod having
of 10 mm
attached
to
a rigid
and passes
the tube. Iftube
a AB
Theisassembly
shown
in collar
Fig. 4–7a
consiststhrough
of an aluminum
400 mm
400 mm
2
80 kN
tensilehaving
load ofa80
kN600
is mm
appliedarea
to the
determine
therod
displacement
cross-sectional
of rod,
having a diameter
400 mm
. A steel
A
A
P ! 80 kN
B
B
(a)TaketoEast rigid
of the end
of the
rod.
Ealpasses
= 200
GPa,and
= 70BCGPa.
of 10Cmm
is attached
collar
through the tube.(b)
If a
kN determine the displacement
80 kNto the80
of 80 kN
rod,
4.2 Etensile
LASTIC Dload
EFORMATION
OF is
ANapplied
EMBER 80 kN
127
kN
PAB ! 80 kNPAB ! 80 kN
C 80M
CAXIALLY LOADED
4–7
mmC of the rod. Take Est = 200Fig.
of the400end
Eal = 70 GPa.
GPa,
4.2
127
4
127
ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
A
mm
600 mm 400600
mm
SOLUTION
(a)
(a)
B Force. The
Internal
B
127
4
80 kN
80 kNPBC ! 80 kNPBC ! 80 kN
(b)
(b)
80 kN
PAB ! 80 kN
C
A
4.2
free-body
diagram
of the
tube AB
and aluminyum
rod segments tüpünden müteşekkildir.
2 olan
Şekilde
görülen
birleşim
enkesiti
400
mm
–7a consists of aninaluminum
tube AB
4–7 80
Fig.
Fig. 4–7b, shows
that the rod is Fig.
subjected
tokN
a4–7
tension of 80 kN and the
80 kN
PAB ! 80 kN
C
ly
shown
Fig. 4–7a
consists
of
anaaluminum
tube
AB
A steel
rod
having
a
diameter
400
mm2. in
80 kN
600
mm
tube is subjected
to
compression
of
80
kN.
10mm
çapındaki
çelik
çubuk
rijit
bir
yakaya
bağlıdır
ve
tüpün
içinden
geçmektedir.
Çubuğa
2
P
!
80
kN
ss-sectional
area ofthrough
steel If
rod
400 mmthe
. Atube.
BC
collar and passes
a having a diameter
(a)
(b)
!
attached
to determine
a rigid
collar
and
passes
through
the tube.
If açubuğun
SOLUTION
SOLUTION
to the rod,
theçekme
displacement
Displacement.
We
will first
determine
the displacement
end C
80
kN
yükü
uygulandığında
C ucunun of
deplasmanını
600
mm
of
80
kN
is
applied
to
the
rod,
determine
the
displacement
Fig.
4–7
Psegments
!
80rod
kN segments
Est = 200Internal
GPa, EalForce.
= 70
GPa.
BC
The
diagram
of
the
tube
and
rod
with
respect
tofree-body
endThe
B.Working
in
units
of
newtons
and
meters,
we
have(b)
Internal
Force.
free-body
diagram
of
the
tube
and
(a)
C of the rod.
Take
Estshows
= 200that
GPa,
GPa. to a tension of 80 kN and the
al =is 70
in Fig.
4–7b,
theErod
subjected
# in Fig. 4–7b, shows that the rod is subjected to a tension of 80 kN and the
tube is
subjected
a compression
of 80 kN.32 of
tube is to
subjected
80 kN.m2Fig. 4–7
N]10.6
[+80110
PL to a compression
SOLUTION
400 mm
Çözüm:
dC>B =
= +0.003056 m :
=
2
9
2
A Internal
p10.005
m2
[200110
2 N>m
]andofrod
Force.
Thefirst
free-body
diagram
the
tubedisplacement
segments
Displacement.
We AE
will
determine
the of
displacement
end
C of end C
Displacement.
We
will first
determine
the
80 kN in Fig. 4–7b,
SOLUTION
shows
that
the
rod
is
subjected
to
a
tension
of
80
kN
and we
thehave
to respect
end80B.Working
in units ofPin
newtons
meters,
wemeters,
have
with
to end B.Working
units ofand
newtons
and
kN
80 kN
C with respect
AB ! 80 kN
80 kN
Pthe
80 kNrelative to
C
AB !
positive
sign
indicates
that
end
C
moves
to
right
tubeThe
is Internal
subjected
to
a
compression
of
80
kN.
Force. The free-body diagram of the tube and rod segments
3
3
end
B,
since
the
bar elongates.
2[+80110
N]10.6
[+80110
N]10.6
m2to a tension of 80 kN4and the
in
Fig.
4–7b,
shows
that
the rod80m2
is2kN
subjected
PL
PL
We =willoffirst
determine
the
displacement
of end m
C:
dDisplacement.
= tube
=kN.
+0.003056
m:
=
d
=
= kN
+0.003056
80
The
displacement
end
B
with
respect
to the
C>B P
C>B
600 mm
2
9
2
2
9
2 fixed end A is
is
subjected
to
a
compression
of
80
!
80
kN
BCAE
AE
m2
[200110
2 [200110
N>m
] 2 N>m
P
!
80 kN
p10.005
] meters, we have
BC
with respect
top10.005
end
B.Working
inm2
units
of newtons
and
(b)
(a)
!
hesaplayınız.
80 kN
4
(b)
= 0.001143
m +m0.003056
m :
= 0.00420
= 4.20 mm
m = 4.20 mm :
4
4
3
Displacement.
We will3 first[-80110
determine
the displacement
of end C
2 N]10.4
m2
PL
Fig.
4–7 The
Fig.
4–7 that
The
positive
sign
indicates
end C
moves
to moves
the right
relative
torelative to
positive
sign
indicates
that
end
C
to
the
right
d
=
=
2
N]10.6
m2
[+80110
PL
B
with respect to
end B.Working
in -6
units
newtons
and
9 meters,
2 we have
[4002mm2110
2 m2of
>mm
2 N>m
]
dend
= bar
= 2][70110
+0.003056
m:
=theAE
end B, since
the
elongates.
B,
since
bar elongates.
C>B
9
2
AE
p10.005
m2
[200110
2
N>m
]
The displacement
of
end
B
with
respect
to
the
fixed
end
A
is
The
displacement
of
end
B
with
respect
to
the
fixed
end
A
is
!
3 = 0.001143 m :
= -0.001143
2 N]10.6 m2
[+80110m
PL
d
=
= +0.003056
m:
=
C>B
rce.
The free-body
diagram
ofsegments
the
tube
and
rod
segments
2
dy
diagram
of The
the tube
and rod
positive
sign
indicates
that
end
C
movesm2
the 2right
relative to
3
392toN>m
AE
p10.005
m22[-80110
[200110
]shortens,
[-80110
N]10.4
2
N]10.4
m2
PL
Here
the
negative
sign
indicates
that
the
tube
and
so
B
PL
shows
that the
is subjected
to
a elongates.
tension
80 kN and the
is
subjected
toend
arod
tension
and
the
the
bar
dB,
= of 80
=BkN
dthe
=right
= 2 of -6
B since
22
9 2
2 9
-62][70110
2
2
moves
to
relative
to
A.
AE
AE
[400
mm
110
2
m
>mm
2
N>m
]
cted
compression
of 80 kN.
[400
mmrespect
110 2to
mthe
>mm
][70110
2 is
N>m ]
on ofto80a kN.
The The
displacement
of end
B with
fixed
end
sign
indicates
thattoend
moves
theAright
relative
Sincepositive
both displacements
are
theCright,
thetodisplacement
of C to
=
-0.001143
m
=
0.001143
m
:
=
-0.001143
m
=
0.001143
m
:
!
end
B,
since
the
bar
elongates.
ent.
We
will
first
determine
the
displacement
of
end
C
relative to the
end A is therefore
3
determine the displacement
of fixed
end C
[-80110
2respect
N]10.4to
m2the fixed end A is
PL
to
end
B.
Working
in
units
of
newtons
and
meters,
we
have
The
displacement
of
end
B
with
in units of
newtons
and
we=
have
= sign
+ 2dBmeters,
Here
the1Here
negative
that
tube
shortens,
and so2 B
the
negative
sign
indicates
that
the
tube
2d the
-6= 0.001143
2
2 m
9shortens,
dindicates
:
=
d
+
+ 0.003056
AE
C [400 B
C>B 2 m >mm ][70110
mm 110
2 N>m ] mand so B
3
moves
to
the
right
relative
to
A.
moves
to the
relative to A.
2 N]10.6
m2rightPL
[ + 80110
3
[-801103m
2 N]10.4
m2
2 N]10.6
m2 both
0PL
= -0.001143
==mto
0.001143
+
0.003056
:the
=
Since
displacements
to them
right,
themm
displacement
of C
==are
0.00420
m
4.20
::
Ans.
Since
both
displacements
are
right,
the displacement
of C
d
=
=
2
9
2
B
AE
=
+0.003056
m
:
!
2
-6
2
2
9
2
p10.005
m2
[200110
2
N>m
]
AE
2
9
2torelative
[400
mm
110
2
m
>mm
][70110
2
N>m
]
relative
the
fixed
end
A
is
therefore
to the fixed end A is therefore
2 [200110 2 N>m ]
Here the negative sign indicates that the tube shortens, and so B
==right
-0.001143
m+to=0.003056
0.001143mm :
+ 2moves
+to
e sign indicates
that
end
relative
dCCright
1:
=moves
dB relative
+dCto
dC>B
0.001143
A.
1:
2the
=the
dto
dC>B
=m
0.001143
m + 0.003056 m
B +
atthe
end
C
moves
to
the
right
relative
to
bar elongates.
Since both displacements are to the right, the displacement of C
Here
the 0.00420
negative
sign
indicates
that
the
so B
m
4.20
mm
Ans. andAns.
acement of end
B with
to the
fixed
end
A
is= :
m
4.20
mm
: tube shortens,
relative
torespect
the=fixed
end=
A0.00420
is= therefore
moves
with respect to the fixed
endtoAthe
is right relative to A.
3
+ 2 [ -Since
are to the
the displacement
of C
80110both
m2dC>B = 0.001143
d2CN]10.4
1:
= displacements
dB +
m +right,
0.003056
m
PL
= [-80110
= 32 N]10.42m2
relative
to
the
fixed
end
A
is
therefore
-6
2
2
9
2
AE
[400 mm 110 2 m >mm ][70110 2 N>m ]
Ans.
2
-6
2
2
2 = 0.00420 m = 4.20 mm :
+ 9=
m 110 2=m ->mm
][70110
] dm :
0.001143
0.001143
1m
:
22 N>m
C = dB + dC>B = 0.001143 m + 0.003056 m
43 m = 0.001143 m :
egative sign indicates that the tube shortens,
B mm :
= 0.00420and
m =so4.20
Ans.
e right
relative
to A.shortens, and so B
ates
that
the tube
h displacements are to the right, the displacement of C
A.
he fixed end A is therefore
re to the right, the displacement of C
herefore
dC = dB + dC>B = 0.001143 m + 0.003056 m
B
4
80 kN
Ans.
Ans.
#47
128
128
128
CCHHAAPPTTEERR 44 AAXXI AI AL L LLOOAADD
AXIAL LOAD
EXAMPLE 4.3
EXAMPLE
EXAMPLE 4.3
4.3
!
CHAPTER 4
90 kN
Rigid beam AB rests on the two short posts shown
200Rigid
mm beam AB rests on the two short posts shown in Fig. 4–8a. AC is
90
made
ofposts
steelshown
and has
a diameter
400
mm
90kN
kN
Rigid beam AB rests on the two
short
in Fig.
4–8a. ACof
is 20 mm, an
200
mm
AB
rijit
iki
kısa
direğe
oturmaktadır.
AChas
çelikten
yapılmış
olup
çapı
20mm
dir.
200kirişi
mm şekildeki
made
of
steel
and
a
diameter
of
20
mm,
and
BD
is
made
of
mm
aluminum
and
has
a
diameter
of
40
mm.
1
2
8
C H A P T E R400
4
A
X
I
A
L
L
O
A
D
A
made
steel
andAC
hasBisa diameter of 20 mm, and BD is made of Determin
400 short
mm
AB rests on the two
posts
shown
inofFig.
4–8a.
aluminum
and
has
a
diameter
mm.
displacement
point
FDetermine
on AB
if a the
vertical
load of 90 kN is appl
A
aluminum
a diameterofof
40
mm.
Determine
thealtında,
displacement
F mm and
BDAaluminyumdan yapılmış Bolup
çapı
40
dir. has
F noktasında
9040
kN
düşey
yükleme
B
Rigid
beam
XIAL LOAD
made of steel and has a diameter of 20 mm,ofofand
BDonisAB
madevertical
of load
point
ofofE
90
isisapplied
this
point.
Take
=kN200
GPa, Eover
= 70
GPa.
pointFFon
ABififaavertical
load
90
applied
this
point.
st kN
alover
128
C H AFPFT E R 4 A X I A L L O A D
aluminum andEXAMPLE
has
a
diameter
of
40
mm.
Determine
the
displacement
Take
E
=
200
GPa,
E
=
70
GPa.
300
mm
4.3
Take# Estst = 200 GPa, Ealal = 70 GPa.
bu noktanın deplasmanını hesaplayınız.
300
of point F on AB if a vertical load of 90 kN
is
300mm
mmapplied over this point.
90 kN
C
Rigid beam AB rests on Dthe twoSOLUTION
short posts shown in Fig. 4–8a. AC is
Take Est = 200
GPa,
E 4.3
= 70 GPa.
EXAMPLE
C 200 mm
D posts shown in Fig. 4–8a. AC is
ABalrests on the two short
AAXXI AI ALL LLOOAADD Rigid beam
C
D
made
of steel and has a diameter
of 20Force.
mm, and
BD
is made offorces acting
SOLUTION
400 mm
SOLUTION
Internal
The
compressive
(a)BD is made of
madeA of steel
and
has
a
diameter
of
20
mm,
and
aluminum
and
has
a
diameter
of
40
mm.
Determine
the
displacement
B
90 kN
Internal
compressive
forces
acting
the
top
of
post posts
are
determined
the
equilibrium
of mem
Internal
Force.
The
compressive
forces
acting
atfrom
the
topAC
ofeach
Rigid
beam Force.
AB
rests The
on
the
two short
shown
inatFig.
4–8a.
iseach
(a)
(a)
aluminum
and has
diameter
of 40 mm. Determine
the
displacement
200 mm
ofpost
point
Fdetermined
on
AB
ifhas
a vertical
load
of
kN
is
applied
over
this
point.
B
F a400
from
the
equilibrium
ofof
member
AB,
Fig.
4–8b.
These
forces
are
equal
to
the
internal
forces in each
post
are
determined
the
equilibrium
member
Fig.
4–8b.
made
ofare
steel
and
afrom
diameter
of 90
20
mm,
and
BD
isAB,
made
of
mm
90
SOLUTIONof point F on AB if a vertical load
is
applied
over
this
point.
Take
EkNforces
= 200
GPa,
Ealto
=the
70
GPa.
These
are
Fig.
4 B of 90 kN
These
are
equal
to
the
internal
forcesinineach
each
post,
Fig.4–8c.
4–8c.
aluminum
and
has
aequal
diameter
ofinternal
40
mm. forces
Determine
thepost,
displacement
200
mm stforces
A
90
90kN
kN Eal = 70 GPa.
400 mm
Take
E
=
200
GPa,
300
mm
st
Displacement.
The
displacement
of the top of ea
44
of
point
F
on
AB
if
a
vertical
load
of
90
kN
is
applied
over
this
point.
Internal Force.
The
compressive
forces
acting
at
the
top
of
each
200
mm
200AB
Fmmrests
Rigidbeam
beam
restson
on
the
twoshort
shortposts
postsshown
shownin
inFig.
Fig.4–8a.
4–8a.AC
ACisis
AB
the
two
400
mm
300 mm Rigid
400
mm
Displacement.
The
displacement
of
the
top
of
each
post
is
Displacement.
The
displacement
of
the
top
of
each
post
is
Take
E
=
200
GPa,
E
=
70
GPa.
st and
al
from
the
equilibrium
of20
member
4–8b.
C of
D of
made
of steel
steel
and
has
diameter
of
20
mm,
BDFig.
made
of
mpost are determined
made
and
has
aa diameter
mm,
andAB,
BD
isis made
of
SOLUTION
Post AC:
300 mm
!
These
forces
are
equal
to
the
internal
forces
in
each
post,
Fig.
4–8c.
DBB
aluminumand
andhas
hasaadiameter
diameterof
of40
40mm.
mm.
Determine
the
displacement
aluminum
Determine
the
displacement
SOLUTION
Post
AC:
Post
AC:
60 kN
kN
Internal Force. The30 compressive
forces acting at the top of3 each
(a)
ofCpoint
point
FkN
onAB
ABThe
ifaavertical
vertical
load
of
90kN
kN
applied
this
point.
(b) over
of
FkN
on
if
load
of
90
isisapplied
point.
[-60110 2 N]10.300 m2
PAC
AC
60Force.
kN
post
are
determined
the equilibrium
ofLmember
Internal
compressive
forces
acting
at
theover
topthis
offrom
each
60
30
D30kN
Çözüm:
SOLUTION
= AB, Fig. 4–8b.
=
3 32dN]10.300
A =
Take
E
=
200
GPa,
E
=
70
GPa.
(b)
Take
E
=
200
GPa,
E
=
70
GPa.
(b)
[
60110
m2
2
9
2
[-60110
2
N]10.300
m2
st
al
P
L
st
al
P
L
AC
AC
Displacement.
displacement
the topInternal
of
post
AAC
These
forces
areis
equal
to the internal forces
inEeach
post,
Fig. m2
4–8c.
AC
AC
post are The
determined
from theof
equilibrium
of each
member
AB,
Fig.= compressive
4–8b.
p10.010
-6[200110 2 N>m ]
-6
st at the
=
=
286110
2
m
d
Force.
The
forces
acting
top
of
each
=
=
=
-286110
2
m
d
90
kN
A
300mm
mm
A
300
2
9
(a)
60 kNin each
30
kN
Estst 4–8c.
4 These forces
are equal
to the internal forces
post,
Fig.
p10.010
m22[200110
[2001109of
N>m2]2] AB, Fig. 4–8b.
AAAC
E
AC
p10.010
m2
22N>m
200 mm
post are determined
from
the equilibrium
member
400
mm
60
kN
30
kN
60 kN
30 kN
mm
T post is
Displacement. The displacement of =the0.286
top of
each
Post DAC:
These forces are equal to the internal forces in each post, Fig. 4–8c.
D
SOLUTION90 kN
SOLUTION
= 0.286
0.286
mm
T
Displacement.
The
displacement
of
the
top
of
each
post
is
=
mm
T
4
200Force.
mm
Post
AC: at
Internal
Force. The
The
acting
at the
the top
top of
of each
each
3compressive forces
Internal
compressive
acting
Post
BD:top of each post is
400 mm
[-60110
2 N]10.300 forces
m2
PAC
LAC
Displacement.
The
displacement
of the
-6
Post
AC:
post
are
determined
from
the
equilibrium
of
member
AB,
Fig.
4–8b.
60
kN
30
kN
post are=determined from the equilibriumPost
of
member
AB, Fig.
4–8b.
Post
BD:
=
286110
2
m
dA =
BD:
9
2 in each
(b) to
[ - 6011032 N]10.300 m2
3
PAC
LACFig.
AAC
Est forces
These
forces
areequal
equal
theinternal
internal
forces
post,
Fig.4–8c.
4–8c.
p10.010
m2to2[200110
2 N>m
30 kN
These
are
the
forces
each
post,
-6 2 N]10.300 m2
[-30110
PBDLBD = - 286110
Post]din
AC:
=
=
2m
3
A
23 3 dB = 9
2
[-60110 2 N]10.300
m2kN
=
=
PACLAC
APAC
E
PAC ! 60
P
!
30
kN
N]10.300
m2 ]
[ - 30110
m2 [200110
2AN>m
PBD
BD
-6p10.010
22N]10.300
m2
BD
BD
LstLBD
-6 2[7011092 N>m2]
30 kN
m2
=
== -286110
2 m[-30110
dA! 60=kN
(c)
BD Eal = p10.020
-6
mm
d
102110
=
2
m
!
2
9
2
3
60
kN
30
kN
B
P
!
60
kN
P
!
30
kN
Displacement.
The
displacement
of
the
top
of
each
post
is
d
=
-102110
=
=
2
m
AC
BD![200110
Displacement.
The
displacement
of the
top of
each
post
is [-60110
A
B] P
PAC
!
60 E
kN
PBD
30 kN
p10.010
m2
2 N>m
N]10.300
L
st (b)
2 2[70110
9 9m2
= 0.286
mm
TAC
AAC
p10.0202m2
N>m2]2]
ACA
(c)
BD
EE
22N>m
(c)
BD
=alalT p10.020 2m2 [70110
= - 286110-62 m
dA = = 0.286
30 kN
mm
9
2 mm T
=
0.102
A
E
AC
direği:
p10.010
m2
[200110
2
N>m
]
AC
st
PostAC:
AC:
Post
= kN
0.286 mm T
0.102mm
mmTT
60
30 kN
== 0.102
Post
BD:
30kN
kN
30
Post
BD:
3
= 0.286 mm T
[ - 60110322N]10.300
N]10.300m2
m2
A diagram showing the centerline displacements
LAC
[-60110
ACL
AC
PPAC
-6
-6
PostddABD:
=
= -286110
- 286110
2m
m
A ==
=
=
2
A
diagram
showing
the
centerline
displacements
A,B,
B,
andFBy
Fon
on
2
9
2
the
beam
ism2shown in
Fig.
4–8d.
proportion o
3
3
9 2 N>m
2 ]
AAC
Estst
A diagram
centerline
displacements
atatA,
and
p10.010
m22[200110
[200110
ACE
2 N]10.300
[the
- 30110
A
p10.010
m2
2 N>m
] PBDisLshowing
2 N]10.300
m2
[-30110
P
L
BD
-6
BD
BD
the
beam
shown
in
Fig.
4–8d.
By
proportion
of
the
blue
shaded
Post
BD:
-6
triangle,
the
displacement
of
point
F
is
therefore
d=
=of-the
102110
=- 102110
=2 m
m
kN
the
is shown
in Fig. 4–8d.
blue 2shaded
Bbeam
PAC !=60 kN
PBD !330
kN
30
d30BkN
=
2 By proportion
9
2
2 N]10.300
m2
PBDLBD (c) [-30110
A
E
p10.020
m2
[70110
2
N>m
]
2
9
2
BD
al
triangle,
the
displacement
of
point
F
is
therefore
-6
ABDdE
p10.020
2 N>m triangle,
]
0.286
mm
displacement
F is therefore
= the
-102110
=TT m2 [70110
2 m of point
mm
! B al=== 0.286
D ! 30 kN
ABDEal
[-3011032 N]10.300 m2
p10.020 m22[7011092 N>m2] PBDLBD
400 mm
dB = = 0.102 mm
= - 102110-62 m
= T
PAC ! 60 kN
PBD ! 30 kN
2
9 = 0.102
2 mm + 10.184 mm2a
d
b = 0
A
E
F
p10.020
m2
[70110
2
N>m
]
Post
BD:
BD
direği:
(c)
400
mm
BD
al
= 0.102
T
Postmm
BD:
600 mm
= 0.102 mm T
d = 0.102 mm + 10.184 mm2 a400 mm b = 0.225 mm T Ans.
dFF = 0.102 mm + 10.184 mm2a 600 mmb = 0.225 mm T Ans.
A
showing
at A, B, and F on
600 mm
= 0.102 mm
T -6 the centerline displacements
N]10.300m2
m2diagram
[ - 301103322N]10.300
LBD
[-30110
BDL
PPBD
BD =
-6
the
beam
is
shown
in
Fig.
4–8d.
By
proportion
of
the blue shaded
d
=
102110
=
2
m
PBD!!
kN
Ashowing
centerline
displacements
and2F
Fmon
on
A3030diagram
the
displacements
at at
A,A,B,B,and
ddiagram
= the
600 mm
BB = A showing
kN
BD
Ealcenterline
p10.020m2
m222[70110
[70110992triangle,
2N>m
N>m22]] =the-102110
0.102 mm
BDE
A
p10.020
displacement
of
point
F
is
therefore
BD
al
400 mm
B
A
F
theshown
beam isinshown
in Fig. By
4–8d.proportion
By proportion
of
the
blue
shaded
600
mm
the beam is
Fig. 4–8d.
of
the
blue
shaded
A diagram showing
displacements
atBA, B, and F on
0.102the
mm centerline600
mm
400 mm
A
F
0.102
mm
displacement
of F
point
F is therefore
400 mm
A 4–8d. FBy proportion
the beam is shown in Fig.
of Bthe blue shaded
=the0.102
mm
triangle, thetriangle,
displacement
of Tpoint
is therefore
0.102
dF
! = 0.102 mm T
mm
triangle, the displacement of point F isd 400
therefore
0.102
mm
0.184
mm
dF = 0.102 mm 0.184
Ans.
b 0.102
= 0.225
+ 10.184
mm2daF
mm mm T
(d)
mm
F 600 mm
400
mm
0.286
mm
A
diagram
showing
the
centerline
displacements
at
A,
B,
and
F
on
0.184
mm
(d)
AdFdiagram
the centerline
displacements
at A,
B,TandAns.
F on
b
=
0.225
mm
= 0.102showing
mm + 10.184
mm2a
400 By
mmproportion
0.286
mm
(d) mm
the beam
beam isis shown
Fig. 4–8d.
4–8d.
of the
the blue
blue
shaded
400
the
Fig.
By600
proportion
of
shaded
mm
dF = 0.102
Ans.
bmm=d 0.225
mm
T 0.286
mm + shown
10.184ininmm2a
!
Ans.4–8
b = 0.225 mm T Fig.
=
0.102
mm
+
10.184
mm2a
F
triangle,the
thedisplacement
displacementof
ofpoint
point
Fisistherefore
therefore
600 Fmm
triangle,
Fig.
4–8mm
600
600 mm
400 mm
F Fig. 4–8
0.102 mm
AB çubuğu rijit olduğundan
AB boyunda eğilme yoktur.A
600 mm
0.102 mm
400
400mm
mm B
A
F
dF == 0.102
Ans.
0.225
mm
0.102mm
mm ++ 10.184
10.184
mm2 a400 mm bb == 0.225
0.102
mm
600
mm
d
Ans.
mm
TT Amm
mm2a
F
600mm
mm
0.184
0.102 mm
F
600
400 mm
0.102 mm
B
A
F
dF
0.184 mm
!
0.286 mm d
0.102 mm
0.102 mm A F
0.184 mm
A
0.286 mm
0.286 mm
(d)
600 mm 0.102 mm
600 mm
400 mm
B
F
400 mm
B
F
(d)Fig. 4–8
0.184 mm
0.184 mm
0.286Fig.
mm 4–8
0.286 mm
d
dFF
(d)
(d)
0.102 mm
0.102 mm
0.184 mm
0.286 mm
600dFmm
400 mm
(d)
dF
B
0.102 mm
B
0.102 mm
Fig. 4–8
(d)
Fig. 4–8
Fig. 4–8
Fig. 4–8
#48
EFORMATION OF AN
4.2
EXAMPLE 4.4
ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
4.2 EM
LASTIC
OF AN AXIALLY LOADED MEMBER
! XIALLY LOADED
A
EMBERDEFORMATION
129
129
129
y
member
is made
from ağırlığına
a material ve
that
a specific
weight gsahip
and bir malzemeden
BirAkonik
eleman
ɣ özgül
E has
elastisite
modülüne
modulus of elasticity E. If it is in the form of a cone having the
dimensionsBu
shown
in Fig.
4–9a,ağırlığında
determine asılı
how far
end
is displaced hesaplayınız.
üretilmiştir.
eleman
kendi
ikenitsuç
deplasmanını
r0
duehas
toy agravity
when
it is gsuspended
in the vertical
position.
y
eeight
fromga and
material that
specific
weight
and
ity
E. If the
it is in the form of a cone having the
having
r0
0
in
4–9a, determine
how farrits
end is displaced
is Fig.
displaced
SOLUTION
n it is suspended in the vertical position.
n.
Internal Force. The internal axial force varies along the member
since it is dependent on the weight W(y) of a segment of the member
4.2Hence,
ELASTIC
EFORMATIONthe
OF AN
AXIALLY LOADED MEMBER
129
below any section, Fig. 4–9b.
toDcalculate
displacement,
4.2
E
LASTIC
D
EFORMATION
OF AN
AXIALLY LOADED
Mits
EMBER
1
2
9
we
must
use
Eq.
4–1.
At
the
section
located
a
distance
y
from
free
L
The
internal
axial
force
varies
along
the
member
he member
end,
the
radius
x
of
the
cone
as
a
function
of
y
is
determined
by
nt
on
the
weight
W(y)
of
a
segment
of
the
member
the member
4XIALLY
4.2 ELASTIC DEFORMATION OF AN A
EXAMPLE
proportion;
4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
129
Fig.
4–9b. Hence,4.4
to
calculate i.e.,
the displacement,
splacement,
PLE
4.4
4.2 r ELASTIC DLEFORMATION OF AN AXIALLY LOADED MEMBER
129
1. Atits
the
section located a distanceL y from its xfree r0
rom
free
0
y
A member is made from a material that has
a specific
weight
g and
=
;
x
=
y
of
the
cone
as
a
function
of
y
is
determined
by
ermined by
EXAMPLE
4.4y
y weight
L
L
ber modulus
is made from
a material
a specific
and having
4
of EXAMPLE
elasticity
E. that
If 4.4
ithas
is in
the form
of 4a g
cone
the
r0
s ofdimensions
elasticity E.
If it in
is Fig.
in the
form
of a cone
having
theis displaced
shown
4–9a,
determine
how
far
its
end
EXAMPLE
4.4
The volume
of a cone having a base ofAradius
x and
heightfrom
y isr0 a material that has a specific weight xg and
member
is made
r0 Fig.A
rit0 is suspended
x to gravity
ons due
shown
in
4–9a,
determine
how
far
its
end
displaced
when
in Ethe
vertical
position.
y
member
is
made
from
a
material
that
has
specific
and
g EMBER
4.2
LASTIC
Dis
EFORMATION
OF aAN
AXIALLY weight
LOADED M
129
= it; is suspended
x = y in the vertical position.
modulus
of elasticity E. If it is in the form of a cone having the
2
ravity when
pr
y
L A modulus
L
1
of
elasticity
E.
If
it
is
in
the
form
of
a
cone
having
the
y
(a)
member is made from a material that2 has a specific
weight g and
0 3
in Fig. 4–9a, determine how far
pyx dimensions
= how2 far
y itsshown
V =determine
r0 its end is displaced
dimensions
shown
in
Fig.
4–9a,
end
is
displaced
modulus
of
elasticity
E.
If
it
is
in
the
form
of
a
cone
having
the
3
3L
SOLUTION
having
a base of radius x and height y is
due to gravity when it isx suspended in the vertical
tne
y is
r0 position.
x
to gravity
when
it is4–9a,
suspended
in the
vertical
PLE
4.4 dimensions
shown
in Fig.
determine
how
far itsposition.
end is displaced
ONInternal
! dueSince
y
Force.
The
internal
axial
force
varies
along
the
member
the
force
at the
section
becomes
gV, it
2 W =
topr
gravity
when
is internal
suspended
in the
vertical
1 due
(a)position.
0axial
(a)
2
3 the
l Force.
The
internal
force
varies
along
the
member
since
it
is
dependent
on
weight
W(y)
of
a
segment
of
the
member
y that has a specific weight g SOLUTION
= pyx
2
y
mber isVmade
a= material
and
2
Çözüm:
gpr
3 from
0 3
3L
SOLUTION
is dependent
onsection,
the
weight
ofHence,
a segment
ofP1y2
the member
below any
Fig.yW(y)
4–9b.
to calculate
the
displacement,
c
P(y)
+
©F
=
0;
=
y
us of elasticity E. If it is in the form of a cone havingInternal
the
Force.
The internal axial force
varies along
the member
x
3Ly2from
any
section,
Fig.
4–9b.
Hence,
tosection
calculate
the displacement,
SOLUTION
we must
use
Eq.
4–1.
AtForce.
the
located
a axial
distance
its
free the
yalong
L
r
y
Internal
The
internal
force
varies
member
internal
force
at
the
section
becomes
0
ions shown in Fig. 4–9a, determine how far its end is displaced
since
it
is
dependent
on
the
weight
W(y)
of
a
segment
of
the member
t useend,
Eq. the
4–1.radius
At
the
located
yaxial
from
L
xsection
of isthe
coneThe
asa distance
a function
of W(y)
y its
is free
determined
by
since
dependent
on
the
weight
of
a segment
of
member
Internal
internal
force
varies
along
thethe
member
2 it Force.
gravity when it is
suspended
in
the
vertical
position.
4
Displacement.
The
area
of
the
cross
section
is
also
a
function
of
below
any
section,
Fig.
4–9b.
Hence,
to
calculate
the
displacement,
gpr
e radius
x of the
cone
a section,
function
of
is determined
proportion;
i.e.,
below
any
Fig.
Hence,
displacement,
it0 yis3as
dependent
on
they4–9b.
weight
W(y) to
of calculate
abysegment
of the
member
P(y) the
P1y2 =since
P(y)
4
position
y,
Fig.
4–9b.
We
have
2
we
must
use
Eq.
4–1.
At
the
section
located
a
distance
y
from
its free
W(y)
ion; i.e.,
x Fig.At
3L must
we
the Hence,
section
located
a distance
y from its free
below
any use
section,
to
calculate
the xdisplacement,
L
r0 4–9b.
r0
xEq. 4–1.
y
end,
the
radius
x
of
the
cone
as
a
function
of
y
is
determined
by
2of y isy determined
ION
;Atthe
x cone
= yas located
the
xr0of
a function
weend,
Eq. =
4–1.
the
section
a pr
distance
from its freeby
L
r0useradius
xmust
0 2
2 proportion;
y
L
L
he
area ofofthe cross
section
is
also
a
function
of
4
i.e.,
function
A1y2
px = of2 yy is determined by
= axial
;
x
y cone
end,
x= ofLvaries
the
as the
a=function
proportion;
i.e.,
al We
Force.
force
along
member
y theLradius
L
4
b.
have The internal
W(y)y is
r
r0
i.e.,
The volume
of
cone having
of radius
and
height
x
is dependent
onproportion;
theaW(y)
weight
W(y) aofbase
a segment
ofr xthe
member
x
r0= 0 and y =y L yields = 0 ;
xheight
x = yx
0 limits
y between
2
Applying
Eq.
4–1
the
of
y
ume
of
a
cone
having
a
base
of
radius
x
and
y
is
=
;
x
=
y
pr
any section, Fig. 4–9b.
Hence,
to
calculate
the
displacement,
y
L
L
x
2
0 2
r
r
2
x
0L
pr0 3
1
(a)
A1y2 = px = 2 y
(b)
2 y
= 0y;L
= Lfree
y 2
t use Eq.
4–1. At the section
located
a distance
yfromx its
V =pr20pyx
L
L=
L1
(a)
L2 dy
LC 1gpr
>3L2of
2 ya3 D cone
dy having
The
volume
a
base
of
radius
x
and
height y is
2
3 y P1y2
0
3
3L
pyx
= a dcone
y having
Vcone
= volume
of
a base= of radius
e radius x of theThe
as a function
by x and height y isx
2= of
x y is determined
Fig. 4–9
x
3
3L
4
etween
the
limits
of
and
yields
y
=
0
y
=
L
!
yields
2
2 y is
2 height
2
y
of a force
cone having
a base of
radius
x and
tion;
i.e., W = The
0 A1y2E
L0
Since
the internal
atLthe
section
becomes
gV, volume
pr
C
1
21pr0>L 2 y D E
0
x
(b)
pr0
1
y3
V = pyx2 = (a)
y
force2at(b)
the
becomes
W =L gV, the internal
2
2 section
3
L C 1gpr
V2L=1 pyx2 =pr2 2 y3
3
3L
r0 0>3L 2 y D rdy
P1y2 dy
x
ggpr
(a)
0
0
0 33
2
2= =y
P(y)
y3 Fig. 4–9
= ; Fig.
x4–9
+ c ©Fy == 0;
P1y2
= V = y pyx
dy = 3L
gpr
2
2
20 3L 3E3L2 3
2
y P1y2
L
3L
Since
internal force at xthe sectiony becomes
= gV, the
A1y2E
P(y)
0 0;
L=
L!0Since
0
L
=
y
1pr
>L
2
y
E
C W0 = gV,2 the
D internal
!
! Wbecomes
force
at
the
section
x
3L
L a cone having
Since W
internal
atythe
section
becomesof
= area
gV,
2 force
gpr20 3 y
ume
of
a base
of the
radius
xcross
and
height
is also
2 a function
The
of the
section
is
g Displacement.
gL
x
gpr+0 c3©Fy = 0;
P1y2
=
y
y dy
= is also
Ans.
cof©F
ement.
The
area
the
a function
+4–9b.
0; section
P1y2
=gpr2 2 yof
position
y, Fig.
We
have
y =cross
3L2 P(y)
2
W(y)
3E L
6E
0 3
0
x
pr
1
3L
(a)
0 3
P(y)
0; 2 =
P1y2 =
y
! cV©F
y, Fig. 4–9b. We+have
W(y)
y
2
y
= y =pyx
2
x
3L
2
pr0 2
3
Displacement. The area yof the cross section is also a function of
3L2
2
gL2
Displacement.
The
the
is also
athe
function
=0 pxarea
= of
y cross section
NOTE: 2A1y2
As aprpartial
check
notice
how4–9b.
units
ofofthe
2 of this result,
Ans.
Ans. Displacement.
2
position
y,
Fig.
We
have
Lthe cross section
W(y
y
A1y2
=
px
=
y
The
area
of
is
also
a
function
of
position
y,
Fig.
4–9b.
We
have
force
at
the
section
becomes
W6E= gV, the internal
terms, when canceled,
W(y)
L2We havegive the displacement in units of length as
!
x
2
position
y,
Fig.
4–9b.
y
W(y)
Applying Eq. 4–1
betweengpr
the2 limits of y = 0 and y = L 2yields
expected.
xpx2 = pr0 y2
pr0 2
A1y2 =(b)
3 0 and
y
ng
4–1the
the
limits
y 0 =ythe
y of
= the
L=yields
2
ial=Eq.
check
ofbetween
this result,
notice
how
unitsA1y2
P(y)
2
2
0;of
P1y2
= of
px
=
y
yunits
L
pr
2
2
3 0 22
L
L C 1gpr2>3L
(b) x
Ly
P1y2
dy 2of A1y2
led,
give as
the displacement
in3L
units
length
as 2 2 y D 2dy
of
length
2
3 = 0px =
L
L
1gpr
>3L
2
y
dy
C
D
d =dy
=0
L
Fig. 4–9
P1y2
Eq. 4–1 between the limits
of y = 0 and y = L yields
x
2 of2 yApplying
A1y2E
Applying
between
the
0= Eq.
L0 is also
2 y D E=of0 and y = L yields
d = The area
C 1prlimits
Fig. 4–9
cement.
of L
the
cross4–1
section
a20>L
function
x
2
2
2
A1y2E Eq.
Applying
of y = 0 and y = L yields L
L0We
L0L4–1 Cbetween
(b) 2 3
1pr0>L 2the
y D limits
E
L C 1gpr2>3L
n y, Fig. 4–9b.
have
2 y D dy
P1y2 dy
3
0
L
L C 1gpr2>3L22 yW(y)
g
(b)
dy
D
P1y2
dy
0
L =
d
=
=
2
3
2
y
dy
y
L
L
g
d pr
= 2 P1y2 dy = C 1gpr0>3L 2 y D dy
L0 A1y2E
L0
>L24–9
2 y2 D E
C 1pr20Fig.
2
y dy 3E L0 2
=
0 02 A1y2E
L0
>L22 y2 D E
=
C 1pr
Fig. 4–9
y
0
3E L0 A1y2 = px d == 2 L
L
L0 A1y2E
L0
L
C 1pr20>L22 y2 D E
g
L
gL2
g
2
y dy x
=
=
Ans.
L y dy
gL
3E L0
ng Eq. 4–1
the 6E
limits of y ==g 0 and
y = L yields Ans.
= between
= 3E L0y dy
6E !
(b)
3E 0 2 3
L
L C 1gpr2L
gL2
>3L
2
y
dy
2
D
P1y2
dy
0
NOTE: As a partial check of this
result, notice how the units of the =
gL
Ans.
2
=
= result,=gL
Fig. 4–9
Ans.
Asterms,
a dpartial
check
of this
notice
of theof length as
6E
2how
2 the units
2
when
canceled,
give
the
displacement
in
units
A1y2E
6E
L0
L0 = C 1pr0>L 2 y D E
Ans.
whenexpected.
canceled, give the displacement
6E in units of length as
L
d.
NOTE: As a partial check of this result, notice how the units of the
g
#49
NOTE:
As a partial check of this result, notice how the units of the
y dy
=
whenthe
canceled,
NOTE:
As a partial check of this result, terms,
notice how
units of give
the the displacement in units of length as
3E L
0 terms, when canceled, give the displacement in units of length as
expected.
terms,
when canceled, give the displacement
in units of length as
expected.
gL2expected.
nate
Fig. 4–10a, which is subjected to the load P. In Fig. 4–10b, P is replaced
4.3its components,
Principle
of
Superposition
4.3
of
by two ofPrinciple
P =Superposition
P + P . If P causes the rod to deflect
1
2
a large amount, as shown, the moment of the load about its support,
The principle
of superposition
is often
to determinestress
the stress
The
of superposition
is often
usedofused
to
Pd, principle
will not equal
the sum of the
moments
itsdetermine
componentthe
loads,
or
displacement
at
a
point
in
a
member
when
the
member
is
subjected
orPd
displacement
in ad1member
the member is subjected
Z P1d1 + Pat
because
Z d2 Z when
d.
2d2a, point
to
a
complicated
loading.
By
subdividing
the
loading
into
components,
to a complicated loading. By subdividing the loading into components,
principle
of superposition
the resultant
the the
principle
of superposition
statesstates
that that
the resultant
stressstress
or or
4.3 Süperpozisyon
Thisprensibi
principle
will be
used
throughout
this
text whenever
we assumesumming
displacement
at
the
point
can
be
determined
by
algebraically
displacement at the point can be determined by algebraically summing
Hooke’sthe
lawstress
appliesorand
also, the bodies
that by
are considered
will be such applied
displacement
caused
load
component
thebir
orEMBER
displacement
caused
by altında
each each
load
component
applied
4.4 STATICALLY
I
NDETERMINATE
A
XIALLY
Lstress
OADED
M
137
Bu prensip genellikle
eleman
karmaşık
bir
yükleme
iken
bir
noktadaki
gerilme veya
thatseparately
theseparately
loading
will
produce
deformations
that
are
so
small
that the
to
the
member.
to
the
member.
deplasmanı hesaplamak
için
kullanılır.
Bu
prensibe
göre
bir
noktadaki
bileşke
gerilme
change
in The
position
and
of the
loading
will
be insignificant
and of veya
following
two conditions
must
be satisfied
the principle
of
The
following
twodirection
conditions
must
be satisfied
if theif principle
deplasman, elemana
yük bileşeninin
ayrı ayrı yüklenmesiyle oluşan gerilme ve
cansuperposition
beherbir
neglected.
superposition
is
to
be
applied.
is to be applied.
Axially
deplasmanın cebrik toplamı ile belirlenebilir.
Süperpozisyon prensibinin
uygulanabilmesi
için;
1. loading
The loading
be linearly
related
the stress
or displacement
1. The
mustmust
be linearly
related
thetostress
or displacement
1. Yüklemenin hesaplanacak
gerilme
veya
deplasman
iletodoğrusal
ilişkili
olması
gereklidir.
that
is
to
be
determined.
For
example,
the
equations
s = P>A
and
that is to be
ch is fixed supported at both
A determined. For example, the equations s = P>A and
2. Yükleme elemanındgeometrisini
önemli
derecede
değiştirmemelidir.
4.4
S
TATICALLY
I
NDETERMINATE
A
XIALLY
L
OADED
M
EMBER
137
d
=
PL>AE
involve
a
linear
relationship
between
P
and
s
or
d.
4.4
S
TATICALLY
I
NDETERMINATE
A
XIALLY
L
OADED
M
EMBER
137
= PL>AE involve a linear relationship between P and s or d.
g. 4–11b, equilibrium requires
4.4
STATICALLY
INDETERMINATE
AXIALLY
LOADED
MM
EMBER
137
4.4loading
STATICALLY
INDETERMINATE
AXIALLY
LOADED
EMBER
1
3 7 or or
2.
must
not significantly
change
the original
geometry
2. The
must
not significantly
change
the
original
geometry
LACThe loading
configuration
of
the
member.
If
significant
changes
do
configuration
of the member.P1If significant changes do occur,occur,
the the
Statically
Indeterminate
Axially
= 0
Statically
Indeterminate
Axially
P
direction
and
location
of
the
applied
forces
and
their
P
andC location of the applied forces and their moment
2moment
Loaded
Member Ldirection
Loaded
Member
!
"
arms
will
change.
For
example,
consider
the
slender
rod
shown
Statically
Indeterminate
Axially
Statically
Indeterminate
Axially
arms will change. For example, consider the slender rod shown in in
d1 is subjected
P which
Fig. 4–10a,
the P.
load
P. In4–10b,
Fig. 4–10b,
P is replaced
Fig. 4–10a,
which
is subjected
to thetoload
In Fig.
P is replaced
lly
indeterminate,
since the
Loaded
Member
Loaded
Member
Consider
shown
in Fig.
which
is fixed
supported
at both
d2 A the rod to deflect
onsider
thethe
barbar
shown
in Fig.
4–11a
which
is
fixed
supported
at
both
LCB
d 4–11a
A
by
two
of
its
components,
P
=
P
+
P
.
If
P
causes
by two
of
itsMcomponents,
P L=
12 . If P2 causes the
o determine
the two
reactions
4.4
STATICALLY
INDETERMINATE
AXIALLY
L4–11b,
OADED
EMBER
1OADED
3 71 +MPEMBER
4.4 Fig.
SFig.
TATICALLY
INDETERMINATE
AXIALLY
1 3rod
7 to deflect
ends.
From
the
free-body
diagram,
equilibrium
requires
f of
its its
ends.
From
the
free-body
diagram,
4–11b,
equilibrium
requires
(a)
a supported
large
amount,
as shown,
the(b)
moment
ofload
the about
load about
its support,
a
large
amount,
as
shown,
the
moment
of
the
its support,
der
the
bar
shown
in
Fig.
4–11a
which
is
fixed
at
both
#
er the bar shown in Fig. 4–11a which is fixed supported at both
AA
4
Pd,
will
not
equal
the
sum
of
the
moments
of
its
component
Pd,equilibrium
will not equal
the sum of the moments ofLits component loads,loads,
tionFrom
needed
for
solution,
it is
ends.
Fromthe
the
free-body
diagram,
Fig.
4–11b,
equilibrium
requires
nds.
free-body
diagram,
Fig.
4–11b,
requires
Fig. 4–10
AC
L
NDETERMINATE
A
XIALLY
L
OADED
M
EMBER
1
3
7
Pd
Z
P
d
+
Pd
Z
P
d
+
P
d
d1 Z# dd12 ZZ dd.2 ZACd.
2d2 , because
1 1B 1 1 2 2 ,Pbecause
displace.
Specifically,
an
Eğer yük
fazla
olur
çünkü
c ©F
minate
Axially
+©F
= 0;
+Aolursa,
F-A
-=
P# 0= 0
cbar
atically
Indeterminate
Axially
= 0;
FBF
+B F
P
r displacement is referred to
LAC
(a)
LAC
C C
aded
Member
4.4
Statikçe
belirsiz
ve
eksenel
yüklü
eleman
L
ion.
In
this
case,
a
suitable
L
F
=
0;
F
+
F
P
=
0
= 0;
FB B+ FA A- P = 0 This This
principle
willused
be used
throughout
this whenever
text whenever
we assume
principle
will be
throughout
this text
we assume
P
e which
displacement
of
one endisatofcalled
C P will
C
This
type
of
problem
statically
indeterminate,
since
the
Hooke’s
law
applies
and
also,
the
bodies
that
are
considered
be
such
Hooke’s
law
applies
and
also,
the
bodies
that
are
considered
will
be such
is
fixed
supported
both
his
type
of
problem
is
called
statically
indeterminate,
since
the
A
FA Fat
A both
shown
in Fig.
4–11a
fixed supported
L L A LCB LCB
ebar
equal
to zero,
since
thewhich
endnotissufficient
equilibrium
equation(s)
are
to
determine
the
two
reactions
that
the
loading
will
produce
deformations
that
are
so
small
that
the
that
the
loading
will
produce
deformations
that
are
so
small
that
the
m,
Fig.
4–11b,
equilibrium
requires
quilibrium
equation(s)
are
not
sufficient
to
determine
the
two
reactions
om
theoffree-body
diagram,
Fig.statically
4–11b, equilibrium
requiressince the
P Pinsignificant and
yon
condition
becomes
problem
called
indeterminate,
bar.
ype
ofbar.
problem
isiscalled
statically change
indeterminate,
since
the
in position
and direction
of theofloading
will
be
change
in position
and direction
the
loading
will
be
insignificant
and
ntype
thethe
L
LCBCB
4
brium
equation(s)
arenot
notsufficient
sufficient
todetermine
determine
two
reactions
4.4 4 STATICALLY INDETERMINATE AX
Lthe
In
order
establish
an
additional
equation
needed
for
solution,
rium
equation(s)
are
the
can
be
ACtwo
canneglected.
be
neglected.
at
In both
order
to to
establish
an
additional
needed
forreactions
solution,
it isit isLAC
A toequation
ebar.
bar.
P
= 0 to to
necessary
displace.
Specifically,
requires
ecessary
consider
thethe
barbar
displace.
Specifically,
an an
B B
FBconsider
+ FAhow
-how
Ppoints
=points
0 on on
44
order
establish
additional
equationneeded
needed
forsolution,
solution,
itis is
C itreferred
equation
that
specifies
conditions
displacement
is
to
(a)
der
totoestablish
ananadditional
equation
for
quation
that
specifies
thethe
conditions
forfor
displacement
is
referred
to
(a)
C
FA
L
sary
considerhow
howpoints
points
thebar
bardisplace.
displace.
Specifically,
B
L
as
atoconsider
compatibility
condition.
case,
a an
suitable
L
ononthe
Specifically,
an
B
sry
a tocompatibility
or or
kinematic
condition.
In In
this
case,
a suitable
FBthis
ACkinematic
Statically
Indeterminate
Axially
P
ion
that
specifies
the
conditions
for
displacement
is
referred
to
(a)
of
the
applied
loads
by
using
tically
indeterminate,
since
the
P
condition
would
require
displacement
of one
ncompatibility
that specifies
the conditions
for displacement
to
(a)
ompatibility
would
require
thethe
displacement
of one
endend
of of
problem
iscondition
called
statically
indeterminate,
since
the
LCBis referred
F
F
P
L
FA FAA A
Loaded
Member
compatibility
or
kinematic
condition.
In
this case,
a suitable
depends
onor
the
material
ent
tobar
determine
the
with
respect
to
the
toIn
be
equal
to zero,
since
Cend
ompatibility
kinematic
condition.
this
case,
a since
suitable
he
bar
with
respect
totwo
thereactions
other
end
to be
equal
to
zero,
thethe
endend CB
quation(s)
are
notthe
sufficient
toother
determine
the
two
reactions
atibility
condition
would
require
the
displacement
of
one
end
of
L
avior
occurs,
d
=
PL>AE
can
supports
fixed.
Hence,
compatibility
condition
becomes
ibility
condition
would
require
the displacement
ofbecomes
one
end of
upports
areare
fixed.
Hence,
thethe
compatibility
condition
F FA
ar
with respect
theAsolution,
end
toP be equal to zero, since Consider
the end the bar4shownFAinAFAFig.
4
4–11a which
is fixed supported at both
nwith
segment
ACtoistothe
+F
,other
andend
in
equation
needed
for
ittois
nce
the
respect
other
be
equal
to
zero,
since
establish
an additional
equation
needed
for
solution,
it isthe end
LCB
rts
fixed.
Hence,
the
compatibility
condition becomesB of its ends. FromPthe free-body diagram, Fig. 4–11b, equilibrium requires
theare
bar
displace.
Specifically,
an
4–11c,
the
above
equation
can
eactions
consider
how
points
the barddisplace.
Specifically,
ts
are fixed.
Hence,
theon
compatibility
condition
becomesan
1
PB1
P
= 0= 0
s for displacement is referreddA>B
toA>B
(a)
P
P2
P2
specifies the conditions for displacement is referred
to
(a)
!
"
ndition. In this case, a suitable
4FB!
FA FA
"
F
B
ibility
dA>B = In
0 this case, a suitable
ion, it isor kinematic condition.
FB
e the displacement of one
end of
dexpressed
0 in terms of
dby
B
+ cby
©F
FBF + FF
1 =using
A>B
(b)one
d0;
A - P = 0
This
equation
can
be
of
the
applied
condition
would
the =
displacement
end
of ! loads
1
F(c)
cally,
an
This
equation
can
besince
expressed
of! the
applied
loads
using
B in terms
A FA
A
! require
d2
d
to be equal to zero,
the end
F
F
d2
FAFP P
dzero, since
a=
relationship,
depends
on
material A A
espect
to the other end
to be equal
towhich
theonend
erred
to
load–displacement
relationship,
depends
thethe
material
(a) which
B
0load–displacement
condition
becomes
(b)
sbility
equation
can be
expressed in terms
of
the
applied
loads
by
using
Fig.
4–11
(a) (a)
F
(b)
B
ixed.
Hence,
the
compatibility
condition
becomes
behavior.
For
if için
linear-elastic
behavior
occurs,
d PL>AE
=using
PL>AE
Çözüm
düşey
denge
denklemi
yeterli
değil.
İlave denklemPgerekiyor. Süreklilik (uygunluk)
suitable
ehavior.
For
example,
if linear-elastic
behavior
occurs,
d =
cancan
equation
can
beexample,
expressed
in terms
applied
d–displacement
relationship,
whichof the
depends
onloads
the bymaterial
type
of
be
used.
Realizing
that
the
internal
force
in segment
AC
isThis
+
F
,
and
in problem Pis called statically indeterminate, since the
end
ofRealizing
e–displacement
used.
that
the
internal
force
in segment
AC
ismaterial
+F
,
and
in
A
Fig.
4–10
veya
kinematik
şart
denklemi
yazılabilir.
relationship,
which
depends
on
the
A canFig. 4–10
ior.
For example, if linear-elastic
FA FA behavior occurs, d = PL>AE
0
equilibrium
equation(s) are not sufficient to determine the two reactions
segment
the
force-F
is
-BF
4–11c,
above
equation
the
end
egment
CBCB
thethat
internal
force
,in
Fig.
4–11c,
the
B , Fig.
or.
For
example,
ifinternal
linear-elastic
behavior
occurs,
d the
=above
canincancan
dthe
=using
0 is force
ed.
Realizing
internal
segment
AC
isPL>AE
+FFequation
, and
A>B
A on
F
1LCB>L
2, so
that
Bwritten
AC
#
the
bar.
A
be
as
e
written
as
d.
Realizing
that theforce
internal force in segment AC is +FA , and in
FA
ent
CB
the internal
can
r the
reactions
become is - FB , Fig. 4–11c, the above equation
FB
In
order to establish
an additional
FB FFBB FB equation needed for solution, it is
t CB as
thethe
internal
force
is
-F
,
Fig.
4–11c,
the
above
equation
can
erms
of
applied
loads
by
using
FB
B
itten
Bu in
denklem
yük-deplasman
ilişkisi
cinsinden
ifade
edilebilir.
on candepends
be expressed
terms
ofuygulanan
the applied yüklere
loads bybağlı
using bir
necessary
to
consider
how
points(c)on(c)the
bar
displace. Specifically, an
P
which
on
the
material
ten as
FB(b) F(b)
B
P
acement
relationship,
which
depends
on
the
material
F
L
F
L
FALA
equation that specifiesFtheF conditions for displacement is referred to
AC FBLCB
B CB
AC can
behavior
occurs,
d
=
PL>AE
B
B (c)
LAC
#
= 0= 0
FA - occurs,
(b)
example,
if linear-elastic
behavior
Fig.Fig.
4–114–11
kinematic
condition. In this case, a suitable
AE
= in
P ¢segment
≤ AC is
+LFAC
, and
inLCB
AE
AEAEd = PL>AE can as a compatibility or
Brce
A
F
F
(c)
(b)
A
B
F
Bsegment AC is + F , and in
L
izing
that
the
internal
force
in
-F can
= 0
A
compatibility
condition
would
require
the displacement of one end of
by
using
Fig.
4–11c, the above
equation
FAL
L
Fig. 4–11
AC
B AE
CB
he internal force is - FAE
the
equation can the bar with respect to the other end to be equal to zero, since the end
-P 4–11c,
= above
0
B , Fig.
material
Fig.
4–11
AE
# isAE
Assuming
that
is constant,
then
=B1L
FBCB
1L>L
>L2FAC
2,supports
so that
using
FBAC
AE
can that
Assuming
AEAE
constant,
then
FAF=A F
,Bso
that
using
are fixed. Hence, the compatibility condition becomes
CB
e direction of the reactions is
FB FB
the
equilibrium
equation,
the
equations
for
the
reactions
become
,
and
in
he
equilibrium
equation,
the
equations
for
the
reactions
become
(c)
(b)
ming that AE is constant, then FA = FB1LCB>LAC2, so that using
AE
sabit
ise,
#
L
(c)
(b)
Btion
can
ngCBthat
constant,
then
so that
using
quilibrium
equation,
the
become
= 0AE
Fis
FBequations
LCB FA =forFthe
B1Lreactions
CB>LAC2, Fig.
dA>B = 0
ALAC
4–11
AE
= 0for the reactions become
ilibrium equation,
Fig. 4–11
AE the equations
AE
LCB
L
LFCB
L
AC
= ¢P ¢B ≤FB ≤andandFBF=B P=¢P ¢AC ≤ ≤
FAF=A P
LCB
LAC L L
This equation can be expressed in terms of the applied loads by using
(b)L L (c)
# P2¢, so that
olacaktır.
= AC
FB # = P ¢
≤ and
≤
FA = FB1LF
usingve
A>L
CB
L
L
a load–displacement relationship, which depends on the material
L FA = FB1LCB>LAC2,AC
L
CB
at AE
is constant,
then
ns
for the
reactions
F
and
≤ Fig.
4–11FB = P ¢ so≤that using behavior. For example, if linear-elastic behavior occurs, d = PL>AE can
A = P ¢ become
L
L
m
equation,
the
equations
for
the
reactions
become
Since
both
these
results
positive,
direction
of the
reactions
ince
both
of of
these
results
areare
positive,
thethe
direction
of the
reactions
is is
be used. Realizing
that the internal force in segment AC is +FA , and in
shown
correctly
on
the
free-body
diagram.
hown
correctly
on
the
free-body
diagram.
both of these
results are positive, the direction of the reactions
is CB the internal force is -FB , Fig. 4–11c, the above equation can
segment
LAC
F
P ¢onLresults
≤ free-body
using
nat
correctly
the
diagram.
B
oth
of=these
are positive,
theLdirection
of the reactions
is
be written
as
CB
AC
FA = P ¢ L ≤ and FB = P ¢
≤
ecorrectly
on the
L free-body diagram. L
#50
4.4
4.4
4
e, the direction of the reactions is
these results are positive, the direction of the reactions is
ram.
4.4
FALAC
FBLCB
= 0
AE
AE
4.4
# EXAMPLE 4.5
4.4
STATICALLY INDETERMINATE AXIA
139
STATICALLY INDETERMINATE AXIALLY LOADED MEMBER
4.5
STATICALLY
INDETERMINATE
AXIALLY
LOADED
MEMBER
4.4 4.4
STATICALLY
IEXAMPLE
NDETERMINATE
AXIALLY
LOADED
MEMBER
1 3 91 3 9
STATICALLY INDETERMINATE AXIALLY LOADED MEMBER
139
The steel
rod mm.
shown
in Fig.14–12a
fixed
4.4TheASsteel
TATICALLY
INDETERMINATE
AXIALLY
OADED
MEMBER
3 9 has a diameter of 10 mm. It is 139
STATICALLY INDETERMINATE
XIALLY
LOADED
MEMBER
139aLdiameter
rod
shown
in Fig. 4–12a
has
It is fixed
4.4of S10
TATICALLY INDETERMINATE AXIALLY LOADED MEMBER0.2 mm
Çapı 10 mm olan çelik çubuk A noktasından
duvara
mesnetlidir.
Yükleme
B between
P!
kNöncesi
toisthe
wall
A,ankastre
and before
it is loaded, there
is 20
a gap
of 0.2 mm
EXAMPLE
AMPLE
4.54.5to the wall at A, and before it is loaded, there
a gap
ofat0.2
mm
between
A
4.4
S
TATICALLY the
I
NDETERMINATE
A
XIALLY
L
OADED
M
EMBER
1
3
9
wall
at
and
the
rod.
Determine
the
reactions
at
A
and
B¿
4.4 Çubuğa
STATICALLY
INDETERMINATE
OADED MEMBER
1 3B¿if9 the
wall
at B¿ and the
rod.
the reactions
at A and
B¿ if the AXIALLY
ucutheile
B’
arasında
0.2
mmDetermine
boşluk
mevcuttur.
C
noktasından
20 LkN
yüklendiğinde
AB B¿
C
The
steel
rod
shown
in
Fig.
4–12a
has
a
diameter
of
10
mm.
It
is
fixed
STATICALLY
A
LOADED
Man
EMBER
39
rod
isXIALLY
subjected
to
axial
force of 1P
shown.
Neglect the
=mm20 kN as 800
he steel rod shown inrod
Fig.is4–12a
has ato4.4
diameter
of
10INDETERMINATE
mm.
fixed
0.2
subjected
an axial
force
of P It= is20
as shown.
Neglect
kN
Pthe
!
mm
P ! 20
kN20 kN0.2 mm400 mm
EXAMPLE
4.5
towall
the wall
atand
A, and
before
it is loaded,
there
is aINDETERMINATE
gap0.2
ofmm
0.2 mm
between
4.4
S
TATICALLY
A
XIALLY
Lthe
OADED
MEMBER
139
size
of
collar
at
C.
Take
=
200
GPa.
E
the
at
A,
before
it
is
loaded,
there
is
a
gap
of
between
st
size
of
the
collar
at
C.
Take
=
200
GPa.
E
A
ve
B’
mesnetlerinde
meydana
gelecek
reaksiyonları
hesaplayınız.#
of 10 mm. It is fixed
st
0.2 mm
B¿ B¿ (a)
the
wall
and
the rod.
Determine
the reactions
A and
if the A
B¿
P ! the
20 kN
4.5
EXAMPLE
4.5
eXAMPLE
wall
at
the rod.
Determine
reactions
at Aatand
the
B¿at
B¿ ifB¿
10
mm.
It
isand
fixed
B
C C
of
0.2
mm
between
0.2
mm
The
steel
rod
shown
in
Fig.
4–12a
has
a
diameter
of
10
mm.
It
is
fixed
B
P
!
20
kN
rod
isasubjected
an 10
axial
force
as shown.
Neglect
20 as
kNshown.
0.2 mm
A
LE
4.5
800 mm P ! 20 kN
B¿Neglect
ads at
has
diameter
of
mm.
is=Pfixed
is
tothe
antoaxial
force
ofIt
the the
Pof
20=kN
of
0.2subjected
mm
between
400 mm 800 mm
mm
Asteel
and
if
B¿collar
PSOLUTION
!
20
kN
to
the
wall
at
A,
and
before
it is of
loaded,
there
is fixed
a of
gap
ofmm.
0.20.2
mm
between
400
mm
Cin
A
SOLUTION
size
of
the
at
C.
Take
=
200
GPa.
E
The
rod
shown
in
Fig.
4–12a
has
a
diameter
10
mm.
It
is
B¿
B
The
steel
rod
shown
Fig.
4–12a
has
a
diameter
10
It
is
fixed
st
4.5
eatof
collar
C.
Take
= 200between
GPa.
Est mm
B¿
aded,
there
athe
gap
ofwall
0.2
0.2 mm
20 kN(a) A0.2 mm
Athe
and
the
B¿ ifisat
P ! 20 kN
shown.
Neglect
C the
the
and
rod.
Determine
theisEquilibrium.
reactions
at mm
A
and
B¿mm
B¿Pif!the
(a)
Ais0.2
Bthe
rod
shown
inA,
Fig.
hasitatat
ais
diameter
of800
10
Itof
fixed
to the
wall
at
and
before
loaded,
there
aon
gap
mm
between
C
B¿
to4–12a
the
wall
A,
and
before
itismm
ismm.
loaded,
there
a gap
of 0.2
between
Equilibrium.
As
shown
free-body
diagram,
Fig.
4–12b,
we
400
As
shown
on
Fig.
4–12b,
we
0.2
mmthe free-body diagram,
B
P
!
20
kN
A
hown.
Neglect
the
A
B¿
mine
the
reactions
at
A
and
ifthe
the
800
mm
B¿
isassume
subjected
an
axial
force
of
asatshown.
Neglect
the
Pand
=reactions
20
kN
C
atshown
A,
and
before
it the
isrod
loaded,
there
isB¿
ato
gap
of
0.2
mm
between
800
mm
wall
at
and
rod.
Determine
reactions
at
A
if
the
B¿
B¿
400
mm
dthe
in
Fig.
4–12a
has
a
diameter
of
10
mm.
It
is
fixed
the
wall
at
and
the
rod.
Determine
the
A
and
if
the
B¿
B¿
will
that
force
P
is
large
enough
to
cause
the
rod’s
end
B
to
(a)
B
will
assume
that
force
P
is
large
enough
to
cause
the
rod’s
end
B
to
C
0.2 mm
C
A P ! 20 kN
SOLUTION
B¿ 400Bmm
B
of
as
Neglect
the
P
=
20rod.
kN
OLUTION
size
ofthere
the
collar
at
C.
Take
=B¿
GPa.
trod
and
the
Determine
the
reactions
atThe
AE
and
if=the
B¿and
is
subjected
an
axial
force
of
asstof
shown.
Neglect
the
P
=
20
kN
A,
before
it isto
loaded,
iswall
ato
gap
of
0.2
mm
between
800
mm
rod
isshown.
subjected
an
axial
force
as
shown.
Neglect
the
P200
20
kN
contact
the
at(a)
problem
is
statically
indeterminate
since
B¿.
C
800
mm
contact
the wall at400
The problem
is statically
since
B¿.mm
800
mm
(a)indeterminate
400
mm
A
B¿B
Equilibrium.
As
shown
on
free-body
diagram,
Fig.
4–12b,
we
400 mmP ! 20 kN
# 200
jected
to
axial
force
of
Neglect
the
Pcollar
=two
kN
size
ofthe
thean
collar
at
C.
=the
200
GPa.
E
and
rod.
Determine
the
reactions
atshown.
Adiagram,
and
if Fig.
the
B¿
=quilibrium.
200
GPa.
800 mm
As
shown
on
the
free-body
4–12b,
weare
there
are
unknowns
and
only
one
equation
of 400
equilibrium.
size
ofTake
the
at
C.as
Take
=
GPa.
E
st 20
C
st B¿
there
two
unknowns
and
only
one
equation
of
equilibrium.
mm
B
(a)
F
will
assume
that
force
P
is
large
enough
to
cause
the
rod’s
end
B
to
(a)
FB FA4
A
ected
at C.
Take
EstP
to an
axial
force
of=isP200
Neglect
the end B to (a)
= GPa.
20 enough
kN as shown.
llcollar
assume
that
force
large
to cause
the rod’s
800(a)
mm
SOLUTION
am,
Fig. 4–12b,
we
3
Çözüm:
+B¿.
400
mm
contact
the wall
at
The
problem
is
statically
indeterminate
since
3
(1)
:
©F
=
0;
F
F
+
20110
2
N
=
0
+
P
!
20
kN
(b)
x
A
B
ollar
atthe
C.
Take at
= The
200 GPa.
EstB¿.
ntact
wall
problem is statically indeterminate :
since
(1)
©Fx = 0; (a) P !-F
m,
Fig.
4–12b,
we
A - FB + 20110 2 N = 0
20 kN
the
rod’s
end
to Equilibrium.
SOLUTION
there
are
twoB
unknowns
and onlyAs
oneshown
equation
SOLUTION
on of
theequilibrium.
free-body diagram,
Fig. 4–12b,
we
F
4
ere
are
two
unknowns
and
only
one
equation
of
equilibrium.
F
A
B
P
kuvvetinin
B
noktasını
B’
ile
buluşturacağını
varsayarak
hesap
yapalım.
N rod’s end Bsince
he
to shown
ndeterminate
4
FB
will
assume
force
Pon
large
enough
to
the
B no
to
Amove
Equilibrium.
As
on the
diagram,
Fig.
4–12b,
Pthat
!free-body
20
kN force
Compatibility.
The
P causes
point
Bcause
toFwe
to4–12b,
with
B¿,end
Equilibrium.
As
shown
the
free-body
diagram,
Fig.rod’s
we
3 is
+ ©F = since
Compatibility.
The
force
P
causes
point
B
to
move
to B¿, with no
(1)
:
0; diagram,
-free-body
FFig.
-P F
+
20110
2
N
=
0
(b)
determinate
xshown
A
B
3
equilibrium.
e
free-body
4–12b,
we
um.
As
on
the
diagram,
Fig.
4–12b,
we
! force
20
kN
contact
the
wall
at B¿.
The
istostatically
force
large
cause
thethe
rod’s
end
B the
toindeterminate
further
displacement.
Therefore
condition
forsince
the
(1)
:will
©Fassume
FAP -is F
20110
N
= 0problem
assume
that
P2 to
is
large
enough
cause
rod’s
end B
to(b)
FP ! 20 kN
x = 0; that will
B + enough
4
FBcompatibility
A
further
displacement.
Therefore
the
compatibility
condition
for the
quilibrium.
me
that
force
Pcause
iscontact
large
enough
the
rod’s
end
to
there
are
twoto
unknowns
only
one
equation
of equilibrium.since
contact
thetowall
at
The
problem
isBstatically
indeterminate
since
B¿.free-body
enough
the
rod’s
end
toand
m.
As shown
on
diagram,
Fig.
4–12b,
we
rod
isthe
wall
atcause
The
problem
isFBBstatically
indeterminate
B¿.
Fthe
4 with
AThe
P
!
20
kN
P
!
20
kN
F
Compatibility.
force
P
causes
point
B
to
move
to
no
B¿,
4
FB
(1)
F
F
A
(b) indeterminate since rod is
A
A
he
wall
at two
is
statically
B¿.
# problem
there
are
unknowns
and
only
one
equation
of equilibrium.
that
force
P isThe
large
enough
to
cause
the
rod’s
end
Btotoequation
P ! 20 kN
ompatibility.
The
force
Ptwo
causes
point
B
toonly
move
noequilibrium.
B¿, with
there
are
unknowns
and
one
of
FA
blem
is
statically
indeterminate
since
3 20
(1)
+
further
displacement.
Therefore
the
compatibility
condition
for
the
F
4
(b)
F
F
P
!
kN
A
4
FB
B
A
two unknowns
only
one
of equilibrium.
(1)
:
©F
=equation
0;
-FA -dB>A
F
20110
2mN = 0
=+ 0.0002
(b)
B
at
problem
is xstatically
indeterminate
since
B¿. Theand
FB
rther
displacement.
Therefore
the compatibility
condition
Ffor the P ! 20 kN
FB FB 4m
3
ywall
one
equilibrium.
dB>A = 0.0002
+rod
3 A
+of©F
move
toisequation
with
no
B¿,
(1)
:
©F
=
0;
F
F
+
20110
2
N
=
0
(1)
:
=
0;
F
F
+
20110
2
N
=
0
(b)
x
A
B
(b)
FFAB
x equation
A FA B
unknowns and only
4(hiperstatik) (c)
do 0;
is
3 of equilibrium.
# - Fone
….FAof
(1) the
Statikçe
belirsiz..
=
- Fno
+ 20110
2 N The
= 0 force
This
displacement
can be
expressed
FA(1)
ove
to B¿, with
(b)unknown
y condition
for
the
A
B
FA move
Compatibility.
P causes
pointinB terms
to
to B¿,
withFAno FB 4
3
d
=
0.0002
m
This
displacement
can
be
expressed
in terms
B>A the load–displacement
FB
FB of the unknown
reactions
relationship,
Eq.
applied
condition
for
the
(1)
+ 20110
2AN
=
0+force
(b)
The
P3using
BPto
move
tocompatibility
no condition
B¿,
(1)
0;Compatibility.
-F
- Compatibility.
F
20110
2causes
= F0point
displacement.
Therefore
the
for
The
force
causes
point
Bwith
to move
to
with
nothe
B¿,4–2,
dB>A
=N0.0002
m
(b)
Bfurther
Fig.
4–12Eq. 4–2, applied
F
F
F
B
B
A
A
using
the load–displacement
relationship,
bility.
The force Ptocauses
pointAC
B and
tocompatibility
move
to B¿,
withWorking
no reactions
segments
CB,inFig.
4–12c.
inthe
units
of newtons
and (c)
further
Therefore
the
condition
for
rod
is
further
displacement.
Therefore
the
compatibility
condition
for
the
Thisdisplacement.
displacement
can
be
expressed
terms
of
the
unknown
FA
FA
to segments AC and CB, Fig.
FA 4–12c. Working Fin
splacement.
Therefore
the
compatibility
condition
(c)
meters,
we
have
A units of newtons and
This
displacement
can
be
expressed
inno
terms
of for
the
FB¿,
Fthe
rod
isThe
ity.
force
causes
point
B towith
move
to
no
B¿, with
B
Bunknown
rod
isload–displacement
reactions
using
the
relationship,
Eq.
4–2, applied
uses
point
B Pto
move
to
Fwe
#
FAFAFig. 4–12
FA
meters,
A have
actions
using
the
load–displacement
relationship,
Eq.
4–2,
applied
d
=
0.0002
m
F
F
B 4–12c. Working
BFnewtons
Therefore
compatibility
conditioninB>A
for
theof
LAC
FBLCB FA
to
AC andthe
CB,
Fig.
units
and
FA
FB
FB
(c)
A
Fig. 4–12
the
compatibility
condition
for
slacement.
of segments
the
unknown
dB>Athe
=mind0.0002
= m
dB>A =Working
0.0002
=m
0.0002
segments
AChave
and CB, Fig. 4–12c.
units
of
newtons
and
B>A
F
F
meters,
we
F
L
F
L
F
F
(c)
B
B
B
B
AE
AE
A AC
B CB
= 0.0002 m
ofEq.
the4–2,
unknown
p,
appliedd
FA
FB
can
be expressed
in termskadar
ofFA the
unknown
dB>A
= 0.0002 m = FB
- (c)
Çubuk
aradaki
boşluk
miktarı
uzayabilecektir.
eters, we have # B>AThis displacement
Fig.
4–12
F
F
(c)
(c)
AE
AE
, Eq.
applied
F
L
F
L
A
A
F
10.4
m2
nits
of4–2,
newtons
and
This
displacement
can
be
expressed
in
terms
of
the
unknown
A
AC
B
CB
This
displacement
can
be
expressed
in
terms
of
the
unknown
A
using
the
load–displacement
Eq. 4–2,(c)applied F
dB>A
=expressed
0.0002
m inmFig.
= 0.0002
=
splacement
can
bedreactions
of F
the
unknownrelationship,
FB
0.0002
m =
Fig. 4–12
B>A
B
Fterms
L4–12
A
AC
BL
CB
ts
of0.0002
newtons
and
2 Fig.
9için
2 applied
reactions
using
the
load–displacement
relationship,
Eq.
4–2,
Uzamaları
AC
ve
CB
bölgeleri
ayrı
ayrı
yazıp
cebrik
toplamını
hesaplayalım.
AC bölgesi
reactions
using
the
load–displacement
relationship,
Eq. of
4–2,
applied
AE
AE
FAand
10.4
m2
to
segments
AC
and
CB,
4–12c.
Working
in
units
newtons
=
m
p10.005
m2
[200110
2
N>m
]
d
=
0.0002
m
=
Fig.
4–12
using the load–displacement
relationship, Eq. 4–2, applied
B>A
Fig.
4–12
F
F
B
B
(c)Fig. 4–12
0.0002
m
=
AE
AE
to
segments
AC
and
CB,
Fig.
4–12c.
Working
in
units
of
newtons
and
lacement
can
be
expressed
in
terms
of
the
unknown
to
segments
AC
and
CB,
Fig.
4–12c.
Working
in
units
of
newtons
and
meters,
we have
9
2
FA10.4
m2 in units of newtons and
nts
AC and CB, Fig.
4–12c.
Working
BLCB
p10.005
m22[200110
2 N>mçalışacaktır.
]
çekme
etkisinde
uzamaya
çalışırken
basınç
etkisinde
kısalmaya
FB(c)
10.8
m2
0.0002
m have
= terms
meters,
ing
the we
load–displacement
relationship,
Eq.
4–2,
applied CB bölgesi
meters,
we
have unknown
F
10.4
m2
2
9
2
A
xpressed
in
of
the
L
e
have
CB
Fig.
4–12
AE
F
L
F
L
p10.005
m2
[200110
2
N>m
]
A
AC
B
CB
0002
m =CB, Fig. 4–12c.
2
9
AC and
Working
of newtons
9 in units
d2F
0.0002
mand
=Ap10.005
m2
[200110
2 N>m2]
FB10.8 m2
FBL
B>A
F
LAE
F
p10.005 m22[200110
2 N>m
]FALL=
ement relationship,
Eq.
applied
AC
CB
E
AC
BLAE
CB
FA4–2,
L
AC
B
CB
d
=
0.0002
m
=
have
d
=
0.0002
m
=
Fig.
4–12
B>A
F
10.8
m2
B>A
B
d
=
0.0002
m
=
B>A
AE AE
AE
p10.005 m22[20011092 N>m2]
12c. Working
in# units
of newtons
and
or
- AE
AE
AE
FAF
10.4
m2
10.8
2 m2
9
2
B
F
L
F
L
m2 [200110 2 N>m ]
A
AC
CB10.4 m2
0.0002
m m2
= -- B Fp10.005
F=A10.4
A 2 2
dB>A = F0.0002
m
m2
9 m22] =or
2 3141.59 N # m
(2)
FAE
m2
FB910.8
m2
[200110
2 N>m
A10.4
0.0002
m = A10.4
p10.005
m2
[200110
22 N>m=15707950
]
0.0002
m
= =#p10.005
AE
Burada
#
N
2
9
2
0.8
m2
2
9
=
or
p10.005
m2
2
N>m
]
2
9 [200110
2
p10.005
m2
[200110
2
N>m
]
p10.005
m2
[200110
2
N>m
]
LAAC
Solving
AF
BLCB Eqs. 1 and 2 yields
10.4
8 m2 F
2 m2F
(2)
FB10.8 m2 FA10.4 m2 - FB10.8 m2 = 3141.59 N # m
200110
] -F 10.4 m2 - F 10.8 m2 = 3141.59 N # m
m
= 92 N>m
(2)m2
A
B
F
10.8
m2
F
10.8
2
9
2
9 AE
2
B kN
#
….
(2)
B
Ans.
F
=
16.0
F
=
4.05
kN
AE
F
10.8
m2
2
9
2
A
B
p10.005
m2
[200110
2
N>m
]
00110 2 N>m ]F 10.4 m2 - F 10.8 m2
# m p10.005
m2 [200110
2 N>m
]
- B= 3141.592N Solving
Eqs. 91 and
22 yields
- B
2(2)
Solving Eqs. 1Aand 2 yields
2
9 [200110
2 92 N>m
p10.005
m2
]2[200110
p10.005
m2
2 N>m
] wall at
p10.005
m2
[200110
2
N>m
]
Since
the
answer
for
is
positive,
indeed
end
B
contacts
the
F
ve (2)= denklemlerinin
F 10.8 m2B ortak çözümünden,
Ans.
FA = 16.0 kN
FB = 4.05 kN
or
lving
1 and (2)
2(1)
yields
Ans.
16.0 kNBassumed.
FB = 4.05 kN
99or
N # mEqs.
A originally
B¿F-as
2
or
2
9
2
]
p10.005
[200110
2 N>m
]
#N>m
Ans.
kN m2FF
= 4.05
Since
theatN
answer
(2)# FAfor= F16.0
N2Since
m the
# m for FB is(2)positive, indeed end B contacts the wall at
B
10.4
m2end
-kN
F
10.8
m2 =the
3141.59
answer
BBcontacts
wall
Aindeed
B is positive,
#
#
(2)
F
10.4
m2
F
10.8
m2
=
3141.59
N
m
#
(2)
F
10.4
m2
F
10.8
m2
=
3141.59
N
m
A
B
A
B
as
originally
assumed.
B¿
(2)
F
10.4
m2
F
10.8
m2
=
3141.59
N
m
A
B
as originally
B¿the
If Findeed
wereend
a negative
nce
answerAns.
for
B contactsquantity,
the wall the
at problem would be
FNOTE:
B is positive,
Fassumed.
10.8
m2Eqs.
1 Band 2 yields
kN
BSolving
Solving
Eqs.
1assumed.
and
2statically
yields
Solving
Eqs. determinate,
1 and 2 yieldsso that FB = 0 and FA = 20 kN.
as1originally
2 yields
Nqs.
Fand
m2Ans.
- FB210.8 m2 =9 3141.592NF#Am= 16.0 kN(2) FB = 4.05 kN
A10.4
contacts
theIfwall
at
p10.005
m2
[200110
N>m
] B= =16.0
NOTE:
aAns.
negative quantity, the problem would be
NOTE:
were
a
negative
the
problem
would
F
4.05
Fquantity,
kNkN
F
kNbeIf FB wereAns.
Ans.
FA = B16.0FkN
F2BkN
= 4.05
A = 16.0
AFkN
B = 4.05Ans.
ontacts
the
wall
at
.OTE:
1statically
and 2If
yields
statically
determinate,
so
that
determinate,
so
that
and
F
F
=
0
=
20
kN.
Since
the answer
for FBthe
isApositive,
indeed
end
B contacts the wall at FB = 0 and FA = 20 kN.
B
negative
quantity,
problem
would
becontacts
FB were
answer
for
is end
positive,
indeed
endwall
B
the wall at
F
Since the
foraFthe
positive,
indeed
Bthe
contacts
the
at
answer
foranswer
positive,
end
B
wall
at
FB isSince
Bcontacts
B isindeed
as
originally
B¿
Ans.
FA = 16.0 kN
FBFB= =4.05
kN FA = 20 kN.
atically
determinate,
sooriginally
that
and
0 assumed.
assumed.
B¿ as
assumed.
B¿ as originally
inally
assumed.
problem would be
(2) the wall at
10.8 m2
3141.59
m end B contacts
nswer
for F=B is
positive,N
indeed
0 kN. would
roblem
be NOTE: If FB were a negative quantity, the problem would be
ally
assumed.
NOTE:
If FB were
negative
quantity,
the problem
would be
NOTE:
If FaB negative
were a quantity,
negative
quantity,
the problem
be
If
the aproblem
would
bewould
kN.FB were
statically determinate,
so that FB = 0 and FA = 20 kN.
statically
determinate,
so that
FB20=kN.
0 and FA = 20 kN.
statically determinate,
FA =
FBF=A 0=and
determinate,
so that
FBso=that
0 and
20
kN.
Ans. would be
NFB were
FB a=negative
4.05 kNquantity, the problem
terminate, so that FB = 0 and FA = 20 kN.
#
ve, indeed end B contacts the wall at
e quantity, the problem would be
B = 0 and FA = 20 kN.
#51
A
P ! 9 kip
EXAMPLE
4.6
The aluminum post shown in Fig. 4–13a is reinforced with a brass
core. If this assembly supports an axial compressive load of P = 9 kip,
applied to the The
rigidaluminum
cap, determine
the average
normal
in with a brass
post shown
in Fig. 4–13a
is stress
reinforced
P ! 9 kip
3
the
aluminum
and
the
brass.
Take
and
E
=
10110
2
ksi
core.
If
this
assembly
supports
an
axial
compressive
load
of P = 9 kip,
al
140
C H A P T E R 4 A X I A2 Lin.L O A D
1 in.
3
140
C H A P T E R 4 A X IE
AL LOAD
=
15110
2
ksi.
applied
to
the
rigid
cap,
determine
the
average
normal
stress in
br
1.5 ft
the aluminum and the brass. Take Eal = 1011032 ksi and
EXAMPLE
4.6
Ebr = 1511032 ksi.
CHAPTER 4 AXIAL LOAD
SOLUTION
AD 140
1.5 ft
2 in.
1 in.
# EXAMPLE 4.6
Equilibrium.
The free-body
diagram
of the with
post aisbrass
shown in
The aluminum
post shown
in Fig. 4–13a
is reinforced
P ! 9 kip
aluminyum
prinç
çekirdek
ileresultant
güçlendirilmiştir.
Bu
aksama
rijit
bir başlık
The
aluminum
post
shown
in
Fig.
4–13a
is
reinforced
with a brass
Pbrass
!core.
9 kolon,
kip IfFig.
in Fig.2 in.
4–13a is Şekildeki
reinforced
with
a
4–13b.
Here
the
axial
force
at
the
base
is
represented
by
SOLUTION
this assembly supports an axial compressive load of P = 9 kip,
1 in.
(a)
EXAMPLE
4.6
vasıtasıyla
P=9
kip
eksenel
basınç
yükü
uygulandığında
aluminyum
ve
prinç
(brass)
core.
If
this
assembly
supports
an
axial
compressive
load
of
9 kip, in
rtsThe
an axial
compressive
load
of
P
=
9
kip,
the
unknown
components
carried
by
the
aluminum,
and
brass,
F
,
2
in.
1
in.
applied
to the rigid
cap,
determineThe
the free-body
average normal
stress
al of in
aluminum post shown in Fig. 4–13a
is reinforced
withEquilibrium.
a brass
diagram
the postPis= shown
3
applied
to
the
rigid
cap,
determine
the
average
normal
stress
in
e core.
aluminum
post
shown
in
Fig.
4–13a
is
reinforced
with
a
brass
malzemelerdeki
ortalama
normal
gerilmeleri
hesaplayınız.
The
problem
is
statically
indeterminate.
Why?
.
F
determine
the
average
normal
stress
in
aluminum
brass.
Take
andbase is represented
Eal = axial
10110force
2 ksiat the
1 4 0an axial the
C
HAPT
ER
A X I Aof
Land
LPOFig.
A=Dthe
br 4 load
If this
assemblypost
supports
compressive
94–13b.
kip,
Here
the4–13a
resultant
by
4 The
The
aluminum
post
shown
inand
Fig.
is reinforced
with
a brass
P !shown
9 kip
aluminum
incompressive
Fig.
4–13a
is
reinforced
with
a brass
3
3
3 of P
the
aluminum
the
brass.
Take
and
E
=
10110
2
ksi
e.applied
Ifbrass.
this assembly
supports
an
axial
load
=
9
kip,
Vertical
force
equilibrium
requires
Take
and
E
=
10110
2
ksi
(a)
=
15110
2
ksi.
E
al
al
to
the
rigid
cap,
determine
the
average
normal
stress
in
br
the
components
carried load
by the
aluminum,
#
ve#axial compressive
alınacaktır.
1.5
ft
core. If this assembly
supports
an
axial compressive
of P
= 9 kip, Fal , and brass,
core. If 2this
supports
an
load
of Punknown
= 932kip,
in. assembly
1 in.
3br = 15110
ksi.
Estress
plied
toaluminum
the rigid
cap,
determine
the Take
average
normal
in
the applied
and
the
brass.
and
= 10110
2F
ksi
1.5
ftEcal
The
problem
is
statically
indeterminate.
Why?
.
(1)
+
©F
=
0;
-9
kip
+
F
+
F
=
0
applied
to
the
rigid
cap,
determine
the
average
normal
stress
in
to
the
rigid
cap,
determine
the
average
normal
stress
in
Çözüm
br
y
al
br
3
3140
4 the Cbrass.
2 ksi and
EXAMPLE
H
A P T E R Take
4 AXIE
A L4.6
L=
O A10110
D
= 15110
2 and
ksi.
Ealuminum
al
br the
force
equilibrium
the
aluminum
and
brass.
Take requires
E = 1011032 ksi and
SOLUTION
aluminum
and the brass.
Take
and
E = 101103Vertical
2 ksi the
al
al
3
= 15110
2 ksi.
3
The rigid cap at the top of the post causes both the
2 ksi.
ECompatibility.
SOLUTION
Ebr = 1511032 ksi. P ! 9 kip1.5 ft
br = 15110
Equilibrium.
The
free-body
the
post+ F
is
shown
in is reinforced with(1)
The
aluminum
post
shown
in
a brass
P
!
9
kip
+ c ©F
=diagram
0; the of
-9amount.
kip
+ F4–13a
y displace
al Fig.
br = 0 ….(1)
aluminum and # brass
to
same
Therefore,
SOLUTION
Fig.
4–13b.
Here
the
resultant
axial
force
at
the
base
is
represented
by
Equilibrium.
The
free-body
diagram
of
the
post
is
shown
in
ody
diagram of
the
post
is
shown
in
core.
If
this
assembly
supports
an
axial
compressive
load
of
P
=
9 kip,
1 in.
EXAMPLE
4.62 in.
(a)
Statikçe
belirsiz..
(hiperstatik)
d
=
d
the
unknown
components
carried
by
the
aluminum,
and
brass,
F
,
LUTION
nt
axial
force
at
the
base
is
represented
by
Fig.
4–13b.
Here
the
resultant
axial
force
at
the
base
is
represented
by
Compatibility.
The
rigid
cap
at
the
top
of
post
causes
both
the
al
al
br
Equilibrium.
The
free-body
diagram
of
the
post
is
shown
in
applied
to
the
rigid
cap,
determine
the
average
normal
stress
in
SOLUTION
SOLUTION
P ! 9 kip
(a) brass,
3
The
problem
is
statically
indeterminate.
Why?
.
F
carried
by
the
aluminum,
and
F
,
the
unknown
components
carried
by
the
aluminum,
and
brass,
F
,
br
aluminum
and
brass
to
displace
the
same
amount.
Therefore,
Fig.
4–13b.
Here
the
resultant
axial
force
at
the
base
is
represented
by
the
aluminum
and
the
brass.
Take
and
E
=
10110
2
ksi
al
al
The
aluminum
post
shown
in
Fig.
4–13a
is
reinforced
with
a
brass
P
!
9
kip
uilibrium.
The free-body
diagramdiagram
of theEquilibrium.
postthe
is post
shown
The
free-body
diagram
of the post is shown in al
Using
the
load–displacement
Equilibrium.
The free-body
of
is in
shown
in relationships,
3
Vertical
force
equilibrium
y. 4–13b.
indeterminate.
Why?
The
problem
is
statically
indeterminate.
Why?
. this
Fbr
the
unknown
components
carried
by
the
aluminum,
and
brass,
FHere
,Ifrepresented
=by15110
2 ksi.
Erequires
core.
assembly
supports
an
axial
compressive
load ofby
P = 9 kip,
al
the
resultant
axial
force
at
the
base
is
represented
by
br
2
in.
1
in.
Fig.
4–13b.
the
resultant
axial
force
at
the base
Fig.Here
4–13b.
Here
the
resultant
axial
force
at
the
base
is
1.5
ft
4
dal =isdrepresented
br
F
L
F
L
m unknown
requires
al
br
Vertical
force
equilibrium
requires
problem
is
statically
indeterminate.
Why?
Fbr . The
(a)
applied
to
the
rigid
cap,
determine
the
average
normal
c ©F
components
carried by
the +aluminum,
(1) brass, stress in
0;Fal , andcomponents
9 kipbrass,
+ carried
Fal=+ Fby
the
unknown
the0 aluminum, Fal , and
the unknown
components
carried
by
the
aluminum,
Fbrass,
y =
br =
al ,-and
Vertical
force
equilibrium
requires
A
E
A
E
Using
the
load–displacement
relationships,
the + caluminum
the
Take
Eal= =0 1011032 ksi and (1)
al and
alindeterminate.
br brbrass.Why?
problem
isystatically
Fbr . The
is statically
indeterminate.
Why?
. The+ Fproblem
problem
is statically
indeterminate.
Why?
. The
(1)
kip
F
+
F
=
0
©F
=
0;
-9
kip
+
F
+
F
br
SOLUTION
al
br
al
br
3cap at the top of the post causes both the
4
Compatibility.Ebr
The
rigid
= 15110
2(1)
ksi. requires A
force
equilibrium
Vertical
Vertical
force
P ! 9equilibrium
kip-requires
Eal FbrLdiagram of the post is shown in
+ c ©Fforce
9 kip + requires
Fal +1.5Fftbr = Vertical
0
FalalL
y = 0;equilibrium
Equilibrium.
The
aluminum
and
brass
to
displace
the
same
amount.
Therefore,
FalThe
= Frigid
bfree-body
aat
cap at the top of the post
causes
both
the
br a cap
=
Compatibility.
theb top of the post causes both the
F
A
E
c
P
!
9
kip
br
(1)base is represented by
+
©F
=
0;
9
kip
+
F
+
F
=
0 Abraxial
c
Fig.
4–13b.
Here
the
resultant
at the
bralE
br
(1)
+
©F
=
0;
9
kip
+
F
+
F
=
0
A
Eamount.
(1)
©F
=
0;
-9
kip
+
F
+
F
=
0
y
al
br
y
al
br
al
br forceTherefore,
y
al
br
Compatibility.
The Therefore,
rigid cap at the top of
causes
both the
lace
the same amount.
aluminum
anddal
brass
displace the same
= dto
(a)the postSOLUTION
br components carried by the aluminum, F , and brass,
the
unknown
al
3
aluminum and brass to
Therefore, The rigid
Faldisplace the same amount.
A
p[12
in.2
- 11top
in.2
2alksi E
Compatibility.
cap
at2 the
of2d] the=10110
post
causes
both
the
al
captop
at Using
thethe
top
ofload–displacement
the
postboth
causes
both
the
dalCompatibility.
= dbr The rigidThe
The
problem
is
statically
indeterminate.
Why?
.
F
mpatibility.
at the
of
post
causes
the
d
the
relationships,
Equilibrium.
The
free-body
diagram
of
the
post
is
shown
in
Pcap
! 9rigid
kip
br
al
br
F
=
F
a
b
a
b
Falbrass
= Fbr
R
alR B Therefore,
br 3
(b)4 to displace
2equilibrium
aluminum
and
toBdisplace
the
same
amount.
Fbr
aluminum
brass
amount.
Therefore,
A2brbase
damount.
brsame
Vertical
force
requires
minum
and brassand
to displace
the dsame
Therefore,
al =the
p11
in.2
ksi Eisbrrepresented
Fig.
4–13b.
Here
the
resultant
axial
force15110
at
the
Rijit
başlık
aluminyum
ve
prinç
malzemelerde
aynı
kısalmaya
sebep by
nt relationships, Uygunluk denklemi:
F
L
F
L
Using
the
load–displacement
relationships,
al
br
(a)
the
unknown
components
carried
by
the
aluminum,
and brass,
F
,
=
d
=
d
al
d
=
d
Using the load–displacement
br
br
(1)
+ c ©F
0; 2F
- 92 kip
+in.2
Fal2]+ F10110
dalrelationships,
= dbr Falal
br =
(2)032 ksi
p[12
- 11
Abry E=Fbral
al =
br in.2
alEal
L
FbrL
F
L
F
L
olacaktır.
#
problem
is
statically
indeterminate.
Why?
Fbr . TheA
al
br
F
=
F
B
R
B
R
= Using4 the load–displacement
al
br
Using the load–displacement relationships,=
2
relationships,
Falrelationships,
L
F
L
br(b)
p11
in.2
1511032 ksi
Solving Eqs.
1 and 2force
simultaneously
Vertical
equilibrium
requires
Abr
Ebr
Aal Ayields
Ealal rigid
ing
load–displacement
Compatibility.
A
al the
=
alEThe
brEcap
br at the top of the post causes both the
P ! 9 kip
Fal F= L
Fbr a Fb L
a
b
AalEalFalL AbrEbr
aluminum
brass
to displace the same amount. Therefore,
FbrL
Fbr
Aand
E
br brkip
Aal
E
Fbr+
= al3F
(1)
+ c ©Fy = 0; Falal = 6=kip
-9
+kipF
= 0E
FbrL=
A=br 2F
(2)
al FalL
br F
=
F
a
b
a
b
AalEal
AbrEbrFal = Fbr aal al bdabr=al db
=
al
br s ! 0.955 ksi
A
E
A
E
A
E
al
al
al
al
br
br
2
2
3 A
Abrbr FE
al
br
F
al brAalEal F A=
E
Ebr causes both the
b a Since
b the resultsSolving
p[12
in.2
-1 indeed
11
in.2
]theatstress
10110top
2 ksibr
are positive,
will
be
compressive.
al br Fbr
brbra
Eqs.
and
2 cap
simultaneously
yields
The
rigid
Abr A Ebr EFalCompatibility.
ARalB the
Eal3andof
= Fbr Bnormal
Rthe post
! 9 kip
sal !P0.637
ksi
al Thealaverage
2 load–displacement
stress
in
the
aluminum
brass
isTherefore,
therefore
Using
the
relationships,
2
2(b)
3
Fin.2
= displace
Fbr a
b15110
a same
b2 ksi
brass
the
amount.
2
Eal b a
baluminum and p11
al to
in.2 - 11 in.2 ] 10110
2 ksi FalFA
al=al Fbr a
p[12
11 in.2F
]br =10110
Fbralin.2
= 2E
6-br
kip
3 kip32 ksi
FF
A
br
2
2
3
A
E
=
F
a
b
a
b
br
br
al
br - 11 in.2 ]
R
B
R
p[12
in.2
10110
2
ksi
6
kip
F
=
F
B
R
B
R
F
L
F
L
br
Abr Ebr R B
br (2) 3
2
ksi
dp11
dalbr
p11 in.22 Fal = F
15110
ksi0.955(b)
al ==
B s3br2 !
R Since
Ans.
sal =theFal
0.637
al = 2F
br
in.2
15110
2be
ksicompressive.
=the
br
22are
22 indeed
3ksi stress
2
3
results
positive,
will
Fal
2
2
3
p[12
p11in.2
in.2 - 11sin.2
15110
2 ksi2 ksi
p[12 in.2
in.2 -- 11
11 in.2
in.2 ]] Aal10110
Eal 2Aksi
p[12
] ksi
10110
brEbr
0.637
2
al !Eqs.
Using
the
load–displacement
relationships,
F2alsimultaneously
= F
Solving
and
R in
B= the
R and brass is therefore
The
normal
aluminum
Falp[12
= Fbr
B 2 - 11 in.2(2)
R32B1ksi
Rbr Baverageyields
Fal = 2Fbr
in.2
] 2 10110
(b)
2 stress
3
(2)
F
2F
3
al 15110
br 2 ksi A
p11 in.2
in.2
15110
2 ksi
Eal
(2)
Fal2p11
= 2F
Fal = Fbr B
RB
R
al
br
3
kip
3
F
L
F
L
F
=
F
a
b
a
b
F
=
6
kip
F
=
3
kip
6
kip
al 0.955 ksibr
al
br
p11 in.2
15110
2 ksi Solving
neously yields
salbr Eqs.
Ans.
= 1 and F
=br
2s
simultaneously
yields
Fbr
A= 0.637
Ebrksi
Ans.
=2F=br
(2)
Solving
Eqs.ksi1 and
2 simultaneouslyFyields
al al=
sbr ! 0.955
2
2 br
p11(2)
in.22A
al = 2Fbr
E
A
E
#
….(2)
p[12
in.2
11
in.2
]
al
al
br
br
Since
the
results
are
positive,
indeed
the
stress
will
be
compressive.
6 kip
Fbr = 3 kip
(c)
(2)
Fal ksi
= 2Fbr
Falyields
= 6 kipp[12 F
2= 3 kip
FalSolving
salF2
!simultaneously
0.637
br therefore
1 andthese
2 simultaneously
in.2
-are11shown
in.22] in1011032 ksi
FbrThe
=
3average
kip Eqs.
Solving Eqs. 1 and
yields
normal
stress
in the aluminum
and
brass
is
NOTE:
Using
results,
the stress
distributions
al = 6 kip
A
E
al
al
s
!
0.955
ksi
FFalal==F
Ba
RB
R
br
e,
indeed
stress
will
compressive.
b astress
b will
3Fbrkip
(b)Fig. 4–13c. Since the results are positive,
2
ving
Eqs.the
1 and
2 simultaneously
yields
the
Fig.be
4–13
p11
1511032 ksi Ans.
6 kip
=br 3 A
kip
Fbr6 kip
results(1)
are
the
stress
be
Ebrin.2
Falisindeed
=
Fortak
=ksi3çözümünden,
kipcompressive.# Fal6 =kip
sbr = Fbr indeed
=
0.955
ksi be compressive.
br
br will
vepositive,
(2)
denklemlerinin
sal ! 0.637
ssSince
in thethe
aluminum
and
brass
therefore
2
sal The
= average
=
0.637
normal stress
inin.2
theksi
aluminum Ans.
and brass is therefore
p11
sbr ! 0.955
ksi
The average
normal
and brass
is therefore
Falare
=stress
6positive,
kipFin the
Fbraluminum
= 3the
kip
p[12positive,
in.22 - indeed
11 in.22the
] stress
Since
are
will be
(2)
= 2Fbr
2
2 Fcompressive.
3
Since the results
indeed
stressthe
willresults
be compressive.
(c)
al 10110
al
6 kip
p[12
in.2
11
in.2
]
2 ksi
sal ! 60.637
ksi
6
kip
NOTE:
Using
these
results,
the
stress
distributions
are
shown
in
kip
The
average
normal
stress
in
the
aluminum
and
brass
is
therefore
Ans.
=
0.637
ksi
The average
normal
stress
in stress
the aluminum
and brass is therefore
FSolving
B
R B =yields
R
al = Fs
brEqs.
ce
indeed
will
compressive.
Ans.
0.637
2 results
2 s
(b) the
3 ksi
al = 1 and
2 simultaneously
Ans.
= be
0.637
ksi
kip
2 2
in.2the
- 11 in.2are
] positive,
al =
p11
2 ksi
2
Fig.34–13c.
4–132]and brass iss therefore
p[12
in.2in.2
- 11 in.22] 15110
p[12
- Fig.
11
in.2
Ans.
=
=
0.955
ksi
6
kip
The average normal stress
in in.2
the aluminum
br
6 kip
2
=al0.637
ksi
= 6 kip
Fbr =Ans.
3 kip (2)
al =in.2 Ans. 2
sal =
= 0.637 ksi sp11
Fal2]=F2F
2
2
br
ip
p[12
in.2 - 113 in.2
(c)
p[12
in.2
11
in.2
]
6
kip
kip
s
!
0.955
ksi
3
kip
br
Ans.
= 0.955 ksi
NOTE:
thethe
stress
are
shown
Ans.Ans.
sal s
= br =
= 0.637Using
ksi these results,
Since
are positive,
indeed
theinstress will be compressive.
Ans.
= distributions
= 0.955
ksi
br results
2 2 = 0.955
2 ksi
n.22
2ssimultaneously
2 yields
p[12p11
in.2in.2
- 11 in.2Fig.
] 4–13c. sal !Solving
0.637 ksi Eqs. 1 and
p11
in.2
Fig.# 4–13
3 kipaverage normal stress in the aluminum and brass is therefore
The
3 kip
(c)
sbr =
Ans.
0.955
ksi F = 3 kip
s = are shown
Ans. 2F=
= in
0.955 ksi
ts, the stress distributions
=
6 kip
2
br
NOTE:
Using
theseal results,
the stress
distributions
are shown in
p11
in.2
NOTE: Using these3 br
results,
the
stress
distributions
are
shown
in
6
kip
p11
in.2
kip ksi
! 0.955
Ans.
s
=
=
0.637 ksi
al
Fig.
4–13c.
Fig. 4–13c. sbr =sbr (c)
Ans.
= Fig.
0.955
ksi
4–13
2
results
areinpositive,
indeed
the stress
will
compressive.
NOTE: Since
Usingthe
these
results,
the stress
distributions
are
shown
p[12
in.22 11be
in.2
] in
NOTE: Usingp11
these
are
shown
in.22results, the
sal !stress
0.637 ksidistributions
Fig. 4–13c. The average normal stress in the aluminum and brass is therefore
Fig. 4–13
Fig. 4–13c.
OTE: Using these results, the stress distributions are shown in
3 kip
s6brkip
Ans.
=
0.955ksi
ksi
Ans.
sal =
==0.637
. 4–13c.
2 2
p11
in.2
p[12 in.22 - 11
in.2
]
#
(c)
Fig. 4–13
(c)
Fig. 4–13
NOTE: Using these results, the stress distributions are shown in
3 kip
Fig.s4–13c.
Ans.
= 0.955 ksi
br =
p11 in.22
NOTE: Using these results, the stress distributions are shown in
Fig. 4–13c.
#52
4.4
# EXAMPLE 4.7
AXIALLY LOADED MEMBER
141
steelINDETERMINATE
bars
shown
in
Fig.
4–14a
are
pinelemana
connectedmafsallarla
4.4
Sçeliğinden
TATICALLY
AXIALLY
LOADED
MEMBER
1to
4 1a
ÜçThe
adetthree
A36A-36
imal
edilmiş
çubuk
rijit
bir
STATICALLY INDETERMINATErigid
AXIALLY
LOADEDIfMthe
EMBER
member.
applied load141
on the member is 15 kN, determine
LY INDETERMINATE
141
STATICALLY INDETERMINATE AXIALLY LOADED MEMBER
B
bağlanmıştır. 15 kN
D
F
the force
developed
each bar.oluşacak
Bars AB kuvvetleri
and EF each
have a cross-AB ve EF çubuk alanları 50
yükleme
altında
herbirinçubukta
hesaplayınız.
sectional area of 50 mm2, and bar CD has a cross-sectional area of
141
0.5 m
ected to a
4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER
141
bars shown in Fig. 4–14a are pin connected to a
determine
B
D
F
napplied
connected
loadto
onathe member is 15 kN, determine
ve
a4.7
crossB
D
F
E15
E MEMBER
kN,
determine
SOLUTION
4.4 STATICALLY AINDETERMINATECAXIALLY LOADED
in
bar. Bars AB
EF each
al each
areaEXAMPLE
of
Band
D have a crossF
4.7
2
ach
have
a
crossbar shown
CD has
cross-sectional
area
mm steel
, andbars
0.5of
m diagram
Equilibrium.
Thepin
free-body
A-36
inaFig.
4–14a are
connected
to a of the rigid member is shown
sectional area of in Fig. 4–14b. This problem is statically indeterminate since there
0.5 m
are
threeload
A-36onsteel
bars shown
inkN,
Fig.determine
4–14a are pin connected to a
er. If theThe
applied
the member
is 15
0.4 m
0.5 m
B
D
F
EXAMPLE
4.7
0.2 m 0.2 m
three
unknowns
and
only
twomember
equations.
member.
If the
applied
load
onhave
the
15 kN, determine
evelopedrigid
in each
bar.
Bars
AB
and EF
each
aavailable
cross- is equilibrium
4.4 STATICALLY
INDETERMINATE AXIALLY LOADED
B
D
F
C
E
2 A
4
force
developed
in each
bar. Bars AB area
and EF
rea of 50the
bar CD has
a cross-sectional
of each have a cross- E
mm
, and
A bars shownCin Fig. 4–14a
are pin connected to a
2The three A-36 steel
r is shown
sectional area
of
and
bar
CD
has
a
cross-sectional
area
of
50
mm
,
0.5
m
C
+Ac ©Fy = 0; rigid
FEAIf+the
FCapplied
+ FE -load
15 kN
= 0member is 15(1)
member.
on the
kN, determine
is shown
B
D
F
eree-body
there are
0.5 m
30diagram
mm2. of the rigid member
EXAMPLE
4.7
0.4
m
the
force
developed
in
each
bar.
Bars
AB
and
EF
each
have a cross- 15 kN
member
is
shown
roblem is statically0.2indeterminate
since there are
m 0.2 m
ons.
2
0.4
mm2
- FA10.4
d + ©MC = 0; sectional
m2 of
+ 50
15 mm
kN10.2
m2bar
+ FCD
= 0 (2)
E10.4
area
has
a cross-sectional
area of
since
there
are equilibrium
0.2 ,mand
two
available
equations.
A0.2 msteel bars
C shown in Fig.
E
The
A-36
4–14a are pin connected to a
4 three
Neonly
0.4
2 m
(a)
0.
0.2 m 0.230
m mm .
equations.
A on the4member
C
E determine
member. If the applied load
is 15 kN,
SOLUTION
B
D
m. The(1)
free-body diagram of the rigid member is rigid
shown
4
the rigid
force
developed
in each bar.
Compatibility.
The
applied
load
will cause
theishorizontal
line Bars
ACEAB and EF each have a crossEquilibrium.
The
free-body
diagram
of
the
member
shown
(1)
FAproblem
+ FC + isFstatically
15
kN
=
0
b. This
indeterminate
since
there
are
E
2
15 kN
m CD has a cross-sectional
sectional
of 50line
mm
, and0.4bar
FA
FCarea of
FE
shown
in
Fig. is
4–14c
to move
to the area
inclined
0 0 and
0.2 m
0.2are
mA¿C¿E¿. The
inonly
Fig.(1)
4–14b.
This problem
statically
indeterminate
since
there
owns
two
available
equilibrium
equations.
A
C
E
(2)
=
2
15
kN
SOLUTION
0.4 m
# 30 Emm
. be related by similar triangles.
displacements
of
points
A,
C,
and
can
0.2
m
0.2
m
(2)
10.4 m2 three
+ 15 kN10.2
m2 and
+ F(a)
10.4
m2
=
0
unknowns
only
two
available
equilibrium
equations.
15
kN
4
E
Equilibrium.
Thethat
free-body
of the rigid member
is shown
Thus the compatibility
equation
relatesdiagram
these
displacements
is
0.4 m2 = 0 (2) Çözüm
C
4
(a)
in
Fig.
4–14b.
This
problem
is
statically
indeterminate
since
there
are
(1)
0;
F
+
F
+
F
15
kN
=
0
A
C
E
l line ACE
0.4 m
(a)
A
C
0.2 m 0.2 m
c ©Fwill
SOLUTION
three
and
two
equations.
(1)
+load
0;
+ unknowns
FC +ACE
15only
kN =
0 available15equilibrium
y = cause
e applied
the horizontal
line
FA
FF
FEFE kN
The
C¿E¿.
C A
dA=-0Equilibrium.
dE (2) dFAC - dEThe Ffree-body
line
ACE
+ 15
kN10.2 line
m2 +A¿C¿E¿.
FE10.4 m2
FE
diagram
of15the
c0;triangles.
to -F
move
tom2the
inclined
The
A10.4
kN rigid member is shown
C
rizontal
0.4 m
=
F
F
F
0.2 m 0.2 m
(2)
-F
d+
©M
=
0;
10.4
m2
+
15
kN10.2
m2
+
F
10.4
m2
=
0
ne
The
A¿C¿E¿.
0.8
m
0.4
m
A
C
E
in
Fig.
4–14b.
This
problem
is
statically
indeterminate
nts
A,
C,
and
E
can
be
related
by
similar
triangles.
C
A
E
(a)
ments is
C + c ©Fy = 0;
(1) since there are
FA + FC + FE - 15 kN = 0
0.2 m 0.2 m
similar
triangles.
three unknowns and
equations.
(a)
ty
equation
that relates these displacements is
C only two available equilibrium
15 kN
15 kN
1
1
lity.
The applied
load will cause theChorizontal line ACE
placements
is
dC-F
= A10.4
dA +
(2)
d+ ©M
0; cause
m2 d+E15FkN10.2
m2 +
C =
(b)
FC FE10.4 m2FE= 0
Compatibility.
The inclined
applied
load
will
horizontal
line ACE
Fig. 4–14c
to move to the
line
The
A¿C¿E¿.
A
2the
2
c
(1) (a)
+
©F
=
0;
F
+
F
+
F
15
kN
=
0
0.4
m
y
A
C
E
F
F
FE
shown
in
Fig.
4–14c
to
move
to
the
inclined
line
The
A¿C¿E¿.
nts of dpoints
Edcan
be m
related by similar triangles.
A
C
A - dA,
E C,dand
C -0.2
E
m 0.2
0.4 m
=
15 kN
displacements
of
points
A,
C,
and
E
can
be
related
by
similar
triangles.
mpatibility
that
these
displacements
is applied
0.2 Eq.
m
0.2
m we
Cthe horizontal line ACE 0.4 m
0.8 mequation
0.4
m relates
Using
the load–displacement
4–2,
have
0.4
m
Compatibility.
The
load
will
cause
0.4 m relationship,
(2)
F
d
+
©M
=
0;
10.4
m2
+
15
kN10.2
m2
+
F
10.4
m2
=
0
C
A
E
15
kN
Thus the # compatibility
equation
that
relates
these
displacements
is
0.2 m 0.2 m
CA
#
EFC
CFA
F
shown in Fig. 4–14c
to move 15
to kNthe inclined line A¿C¿E¿.
The
dE
1
1
(b)
dC = ddA Bilinmeyen
+- d dE d -iç dkuvvet
displacements
of
points
A,
C,
and
E
can
be
related
by
similar
triangles.
15
kN
FCL sayısı1 3, yazılan
FAL denge denklemi
1 (b)FELsayısı 2, statikçe
E¿
dA ! belirsiz
dE
sistem.
E
2 A 2E = C
applied
load
= d c -Compatibility.
d + cThe
d displacements
0.4will
m cause the
C ¿ ACE C dE
Thus
the
compatibility
relates
is horizontal line
2-(b)
2equation that
2 these
d
d
d
A
E
C
E
0.2
m
0.2
m
2
2
dCThe
150shown
mm 2E
150 mm
2Est to the inclinedA¿ line0.4dCA¿C¿E¿.
0.8 m
0.4130
m mm 2Est =
FA
inst Fig. 4–14c
to move
dA m! dE
0.4 m denklemi:
0.4 m
Uygunluk
0.2 m 0.2 m
0.8 m E 0.4 displacements
m
acement relationship,
A Eq. 4–2, we
C have
of
points
A,
C,
and
E
can
be
related
by
similar
triangles.
(c)
0.4 m
0.4kN
m
15
1
1
dE
0.3FA
0.3F
e have
A+dcompatibility
- dE(b) Ethat(3)
0.4FmC = Thus
the
relates these
is
dC = dA + d0.4
A - dEE C dC equation
Em
15 kN displacements
1
1
Fig. 4–14
dE
0.4 m
E¿
=
2 A 2
dA ! dE
E
C
d
=
d
+
d
0.2 m 0.2 m
d 1
A dE
E
0.8 m
0.4 m
C ¿ LC
(b)
F
L
F
1
E
¿
d
d
!
d
A
E
2
2
E
A
E
Est = c
A¿d + dcC ! dE
d
dE
C¿
1–3d2Csimultaneously
dA
FEL 2 150 mm22E
E¿yields A¿
dA Solving
! dE
2 Eqs.
d1m
Eoad–displacement
150Eq.
mm4–2,
2EC¿
dC - dE
15 kN
d
!
d
C dA - dE
relationship,
we
have
st
st
st
1
C
E
d
d
0.4
0.4
dA
(c)
E
=m
d
=
d
+
d
d
mm22E(3)
C
A
E
A¿
d
!
d
C
relationship,
Eq. 4–2, we haveA2 (c) 2 0.8
0.2 m
stUsing the load–displacement
(b)
E0.4mm
C m
0.4
E
0.4 m
dA C
(3) FA = 9.52# kNdE
FC = 0.3FA# + 0.3FEFig. 4–14
Ans.
A
E
C
(c)
FCL
FAL
FEL
1
1
dE1
d ! dEFig. 4–14
1E¿
= (3)c
d +Fig.
c4–14the
d C =1 3.46AkN
C¿
d
=
d
+
ddE
Ans.
F
C
A
Using
load–displacement
relationship,
Eq.
4–2,
we
have
F
L
F
L
F
L
2
2
2
1
E¿
d
!
d
0.4 m
0.4 m
C 2E
A 2E
E
A
2 150 mm
2 c 150 mm
dCE2
30 mm 2Est
2 E
A¿ d
d
!
d
st
st
=
d
+
c
C
E
dE
C¿
d
ultaneously yields
A
A
E
C
2
2
2
2 150 mm 2Est FE =2 2.02
dC
130 mm 2Est
150 kN
mm 2Est
A¿
Ans.
dE
(c)
dA dC ! dE
(3)the
FC = 0.3FA + 0.3FE
Ans.
FCLUsing
FAL
FEL Eq. 4–2,
1 load–displacement
1relationship,
! dE
(c) wedAhave
0.4Em
= c
d +4–14c
d
Fig.
C¿ A
(3)
FC = 0.3FAns.
FA = 9.52 kN
A + 0.3F
2 E
2
2
2
2
d
130
mm
2E
150
mm
2E
150
mm
2E
A¿
st
st
st
Ans. Ans. #
Fig. 4–14
dA dC ! dE dE C
Ans.
FC = 3.46 kN
s. 1–3 simultaneously
yields
FCL
FAL
FEL
1
1
dA !
(c)dE
Ans.
= cE
d + c (3)
d
Yazılan
3 denklemi ortak
FC = 0.3F
Solving
1–3 simultaneously
yieldsçözecek olursak,
Ans.Eqs.
2 A + 0.3F
2
2
2
2
130 mm 2Est
150 mm 2Est
150 mm 2Est
A¿
FE = 2.02 kN
Ans.
Fig. 4–14
dA dC
Ans. FA = 9.52 kN
Ans.
(3)
FC = 0.3FA + 0.3FE
Solving
Eqs.
1–3
simultaneously yields Ans.
FA =
9.52
kN
Fi
Ans.
FC = 3.46 kN
Ans.
FC = 3.46 kN
Ans. Eqs.F1–3=simultaneously
Solving
yields
Ans.
9.52 kN
# FE = 2.02 kN
A
FE = 2.02 kN
Ans.
Ans.
FC = 3.46 kN
Ans.
FA = 9.52 kN
FE = 2.02 kN
Ans.
Ans.
FC = 3.46 kN
4.4 STATICALLY INDETERMINATE
AXIALLY LOADED MEMBER
2 ve CD
2
2
mm
. çubuk alanı 30 mm dir.
30 mm
FE = 2.02 kN
Ans.
#53
PTER 4
AXIAL LOAD
142 142 142C H A PCTHEAR PC4THE AR A
L XLI O
4X I AA
A LA D
LOAD
4
HAPTER
AXIAL LOAD
EXAMPLE
4.84.8 4.8
# EXAMPLE
EXAMPLE
142
E 4.8
CHAPTER 4
AXIAL LOAD
The
bolt
in4–15a
Fig.
4–15a
is 2014-T6
made
of aluminum
2014-T6
aluminum
and is
Şekilde aluminyum
alaşımdan
yapılmış
bir
bulon,
magnezyum
alaşımdan
The bolt
shown
in shown
Fig.
is made
of
alloy
andalloy
is
The bolt
shown
in 4–15a
Fig.
is
made
of 2014-T6
aluminum
alloy
and is
tightened
it compresses
a cylindrical
tube
of Am
1004-T61
tightened
so itso
compresses
a cylindrical
tubetube
made
of made
Am
1004-T61
tightened
it so
compresses
a cylindrical
made
of Am
1004-T61
1 in tir.
The bolt shown
in Fig.
4–15a
is made
of
2014-T6
aluminum
alloy
and
is 1 1 of
yapılmış
bir
tüpün
içine
yerleştirilmiştir.
Tüpün
yarıçapı
0.5
magnesium
alloy.The
tube
has
andış
outer
it is assumed
in.,
EXAMPLE 4.8
magnesium
alloy.The
tubetube
has an
radius
of radius
is assumed
magnesium
alloy.The
hasouter
an
outer
radius
of 2and
and
itand
is assumed
in.,2it
2 in.,
tightened so
it compresses a cylindrical
tube
made
of Am
1004-T61
that
both
the
inner
radius
oftube
the
tube
andradius
the
radius
ofbolt
the1are
bolt1are 1 in.
3 in.
that that
bothboth
thebulon
inner
radius
of
and
the
ofoldukları
the
are
the inner
radius
oftube
the
andradius
the
ofbolt
the
3 in. 3 in. Tüpün
1 the
4 in. 4 in. 4
iç yarıçapı
ile
yarıçapı
0.25
in
alınarak
aynı
magnesium
alloy.The
tube
has
an
outer
radius
of
and
it
is
assumed
in.,
The
shown
in
Fig.
4–15a
is
made
2014-T6
alloy
and
isto beto be
The
washers
atand
the
top
andof
bottom
of aluminum
the
are
considered
2 and
1
1
The bolt
washers
at
the
top
bottom
of
the
are tube
considered
to be
The
washers
at
the
top
bottom
of tube
are considered
1
1the tube
in.
in. 1 in. 2 in. 1that
in. 1both
in. 4the
inner radius
of
the
tube
and
the
radius
of
the
bolt
are
in.
tightened
so
it
compresses
a
cylindrical
tube
made
of
Am
1004-T61
2
4
3 in.
2
4
rigid
and
have
a
negligible
thickness.
Initially
the
nut
is
hand
tightened
4
rigidBulon
and have
a negligible
thickness.
Initially
the10.025
nut
hand
tightened
rigid
and
have
a negligible
thickness.
Initially
the isnut
hand
tightened
somunu
başlangıç
konumundan
inisyukarı
doğru
142
C H AThe
PTER
4 varsayılmıştır.
A X Iat
A L the
L O Atop
Dmagnesium
washers
and
bottom
of
the
are
considered
to
be
alloy.The
tube
has
an
outer
radius
of
and
it
is
assumed
in.,
1
snugly;
then,
using
a
wrench,
the
nut
is
further
tightened
one-half
2 tightened
snugly;
then,then,
usingusing
a wrench,
the nut
further
tightened
one-half
turn.
in.
snugly;
a wrench,
the isnut
is further
one-half
turn.turn.
1
4
rigid
a negligible
thickness.
Initially
the
nut
is
hand
tightened
that
both
the
inner
radius
of
the
tube
and
the
radius
of
the
bolt
are
in.
sıkıştırıldığında
bulonda
meydana
gelecek
gerilmeyi
hesaplayınız.
3 in.and have
If
the
bolt
has
20
threads
per
inch,
determine
the
stress
in
the
bolt.
4
If theIfbolt
has 20
per inch,
determine
the stress
in the
the bolt
hasthreads
20 threads
per inch,
determine
the stress
inbolt.
the bolt.
the nutatisthe
further
tightened
one-half
turn.
The
washers
top
and
bottom
of
the
tube
are
considered
to
be
1
1snugly; then, using a wrench,
in.
in.
C H A P T E R EXAMPLE
42 A X I A L L O A4If
D 4.8
the bolt has 20 threads
per
inch,
determine
the thickness.
stress in the
bolt. the nut is hand tightened
rigid
and
have
a negligible
Initially
SOLUTION
SOLUTION
SOLUTION
(a) (a) (a)
snugly; then, using a wrench, the nut is further tightened one-half turn.
Equilibrium.
The
free-body
diagram
of
aaluminum
of
the bolt
and
Equilibrium.
free-body
diagram
of
section
ofsection
the
bolt
and
the
Çözüm:SOLUTION
Equilibrium.
diagram
of
a section
the
bolt
and
The
bolt
shown
inThe
Fig.free-body
4–15a
is made
of a2014-T6
alloy
and the
is the
If
the
bolt
has
20The
threads
per
inch,
determine
the
stress
inofthe
bolt.
a)A D 4
4
Fig.
4–15b,
is considered
in
to
relate
the
force
in bolt
the bolt
tube,
Fig.tube,
4–15b,
considered
order
to order
relate
the
force
in the
4
tube,
Fig.so4–15b,
is considered
in order
to relate
theofforce
in bolt
the
tightened
itiscompresses
aincylindrical
tube
made
Am
1004-T61
PLE 4.8
Equilibrium. The free-body
diagram
of
atube,
section
bolt
and
therequires
that
the
Fthe
. Equilibrium
toFthat
in
the
tube,
Equilibrium
requires
Fb magnesium
Fttube
. tube,
in
thein
Equilibrium
requires
Ft .of
alloy.The
has
outer
radius
of 12 in., and it is assumed
b to
tan
b toFthat
SOLUTION
tube,
Fig.
4–15b,
is
considered
in
order
to
relate
the
force
in
the
bolt
F
F(a)
F
tThe bolt shown in Fig.
t
t
4–15a
isc the
made
of 2014-T6
aluminum
alloy
and
is of the bolt are 14 in.
both
inner
radius ofF
the
tube
and
the
radius
3 in.
c.that
c
Magnezyum
tüpte
basınç,
aluminyum
bulonda
çekme
etkisi
oluşacaktır.
(1) (1) (1)
+
©F
=
0;
F
=
0
+
©F
=
0;
F
F
=
0of the bolt and the
shown in Fig. 4–15a is made
2014-T6
aluminum
alloy
and
is
+
©F
=
0;
F
F
=
0
Equilibrium.
The
free-body
diagram
of
a
section
y
b
t
y
b
t
that
in
the
tube,
Equilibrium
requires
Fbofto
F
y
b
t
t The washers
a cylindrical
tubeand
made
of Am
1004-T61
at
the
top
bottom
of
the
tube
are
considered
to be
1
1 tightened so it compresses
d so 4it compresses
1004-T61
in. aFbcylindrical
tube,
Fig.has
4–15b,
is considered
1in order to relate the force in the bolt
F4 bin. tube made of Am
Fb
2
magnesium
alloy.The
tube
an
outer
radius
of
and
it
is
assumed
in.,
rigid
and
have
a
negligible
thickness.
Initially
the
nut
is
hand
tightened
c
1
+ ©F
FCompatibility.
Ftalloy
=When
0 and
Compatibility.
the
nut
is
on
the
bolt,
thebolt,
tubetube
2
When
thetightened
nuttightened
is (1)
tightened
on bolt,
the
the tube
Compatibility.
When
the
nut
is
on the
the
y =made
bin-the
um alloy.The
hasin
anFig.
outer
radius
of0;2 in.,
and
itbistoassumed
The bolttube
shown
2014-T6
aluminum
that
tube,
Equilibrium
F
Fthe
1
# 4–15a
t . is
thatisboth
theofinner
radius
of1shorten
the
tube
and
radius
ofbolt
therequires
bolt
are
in., Fig.
snugly;
then,
using
a
wrench,
the
nut
is
further
tightened
one-half
turn.
3 in.
will
shorten
and
the
bolt
will
elongate
Fig.
4–15c.
Since
the
d
,
d
,
4
will
shorten
and
the
will
elongate
Fig.
4–15c.
Since
d
,
d
,
will
and
the
bolt
will
elongate
4–15c.
Since
the the
d
,
d
F
t
b
t
b
t
b
tcompresses
h the inner
radiussoofitthe
tube
and
the
radius ofWhen
the
bolt
are
in.
tightened
a cylindrical
tube
made
1004-T61
4=of
1 1 1 1(1)
1 1
Compatibility.
the
nut
isAm
tightened
on
the
bolt,
the
tube
The washers at
the
bottom
ofthreads
theone-half
tube
considered
tothe
be
+ ctop
©F
0;
Fare
Fdetermine
=it 0advances
1
If
the
bolt
has
20
per
inch,
stress
in
the
bolt.
1and
nut
undergoes
one-half
turn,
it
advances
a
distance
of
(
)(
in.)
=
y
b
t
nut
undergoes
one-half
turn,
it
advances
a
distance
of
(
)(
in.)
nut
undergoes
turn,
a
distance
of
(
)(
in.) =
in.
2
20
Statikçe
belirsiz.
2 20 2 20 =
magnesium
alloy.The
tube
hastube
an outer
radius of 2 in.,
and it is assumed
hers at
the
top and
bottomwill
ofrigid
the
considered
tothickness.
be
4
and have
a negligible
Initially
the
nut
is hand
tightened
shorten
the
bolt
will
elongate
4–15c.
Since
theof these
dare
d1bolt.
, Fig.
t , and
bthe
F
along
the
bolt.
Thus,
the
compatibility
displacements
0.025
in.
along
the
Thus,
the
compatibility
of
these
displacements
0.025
in.
along
bolt.
Thus,
the
compatibility
of
these
displacements
0.025
in.
b
both thethickness.
inner radius
of thethe
tube
and
the
radius
of the bolt
are 4 in.
1 1
d havethat
a negligible
Initially
nutone-half
is hand
tightened
Compatibility.
the tightened
nut of
is (tightened
on the bolt, the tube
snugly;
then,
using
a wrench,
nutWhen
is further
one-half
nut
undergoes
turn,
it the
advances
a distance
requires
2 )(
20 in.) =turn.
SOLUTION
requires
requires
The washers
the
topis and
bottom
of theone-half
tube
are
considered
to
be
(a)
hen, using
a wrench,atthe
nut
further
tightened
turn.
will
shorten
and the of
bolt
willdisplacements
elongate
dt , determine
db , Fig. 4–15c. Since the
If in.
thealong
bolt has
threads
per
the
stress
in the bolt.
the20
bolt.
Thus,
theinch,
compatibility
these
0.025
1 and the
and have
negligible
thickness.
Initially
is hand
tightened
Equilibrium.
The free-body
diagram
cbolt.
lt has rigid
20 threads
per ainch,
determine
the stress
in nut
thethe
c 21+
dt =itd0.025
1+
2 nut
in. -of
1+
= d0.025
in.
-section
c 2 one-half
undergoes
turn,
adabdistance
ofthe
(12)(bolt
in.db - dbof
t advances
requiresthe
t = 0.025
20 in.) =
snugly;
then,
using
a
wrench,
nut
is
further
tightened
one-half
turn.
4
tube,
Fig.
4–15b,
is considered
in order to relate
the displacements
force in the bolt
along
the bolt.
Thus,
the compatibility
of these
0.025
in.the
SOLUTION
Taking
moduli
of
elasticity
fromfrom
the
table
on table
the
Taking
the
moduli
of
elasticity
the table
on inside
the
backback
(a) If the bolt has 20 threads
per
inch, determine
the
stress
in
the
bolt.
Taking
the
moduli
of
elasticity
from
the
on inside
theback
inside
c
to
that
in
the
tube,
Equilibrium
requires
F
F
.
d
1+
2
=
0.025
in.
d
b
t
t
b
requires
ON
cover,
and
applying
Eq.
yields
cover,
and
applying
Eq.
4–2,
Equilibrium. The
free-body
diagram
of a4–2,
section
of the
bolt and the
Ft
cover,
and applying
Eq. yields
4–2,
yields
(1)
©Fthe
=in0;order
Fbforce
-in.
Ft in
0b bolt
Taking
of
elasticity
the to
table
the
inside
ium. SOLUTION
The free-body diagram
of athe
section
of the
c+2cand
y from
tube,
Fig.moduli
4–15b,
is1bolt
considered
relate
the
dton
+
0.025
-=back
dthe
F=t13
Fin.2
in.2
13
in.2
t13 F
t
cover,
and
applying
Eq.
4–2,
yields
g. 4–15b, is considered in
relate
force
in Fthe
bolt
F orderFto
that the
in the
requires
= = =
b to diagram
t . Equilibrium
2the the
2 tightened
Equilibrium. Theb free-body
of tube,
aTaking
section
of the
bolt
and
2
the
moduli
of
elasticity
the2][6.48110
table
on32 the
back
Compatibility.
When
is
tube
2from
2 32 ksi]
3 inside
p[10.5
in.2
- 10.25
in.2
][6.48110
p[10.5
in.2
- nut
10.25
in.2
ksi]
Ft the tube, Ft . Uygunluk
denklemi:
at in
Equilibrium
requires
p[10.5
in.2
10.25
in.2
][6.48110
2 bolt,
ksi] the
F
in.2-dinF
(b) is(b)
tube, Fig. 4–15b,
considered
order to cover,
relate
the
force
the
t13
(b)+ c ©Fin
(1)
F
= bolt
04–2,
applying
Eq.
willand
shorten
the yields
bolt=willF elongate
Fig.
4–15c.
Since
the
d
,
y = 0;
b
t , tand
b
in.2
b13Fin.2
b13F
in.2
1 1
2
3
theFtube,
requires
Fb to that inBulondaki
Ft .=Equilibrium
b13
(1)in.220.025
=F0;
0 vep[10.5
in.2
- 10.25
][6.48110
nut
undergoes
one-half
turn,
it
advances
a
distance
of
(
)(
in.)
=
in. 0.025
-2in.ksi]
b - Ftuzama
tüpteki
kısalma
miktarlarının
toplamı,
somunun
yukarı
doğru
ilerleme
0.025
b
2 20
F
13the
in.2
-in.2
2
3
2
3
)
Compatibility. When
the nut is the
tightened
ontin.
bolt,
the
tube
2 2 ksi]
3
p10.25
[10.6110
p10.25
in.2
[10.6110
2 =ksi]
p10.25
in.2
[10.6110
2
ksi]
bolt.
Thus,
the
compatibility
of
these
displacements
(1)
+ c ©Fy = 0;
Fb - Ft = 0 0.025 in. along
Fb13
in.2 2 d , Fig. 4–15c.
willolacaktır.
shorten
and
the
elongate
Since32the
dt , bolt,
p[10.5
in.2 - b10.25 in.22][6.48110
ksi]
tibility. When the
nut is tightened
on the
the bolt
tubewill
miktarına
eşit
0.025
in. requires
(2) (2) (2)
0.78595F
=-25
-1251.4414F
(b)
11.4414F
2
30.78595F
t = 25
t
b
0.78595F
=
- b=
1.4414F
[10.6110
ksi]
b
nut undergoes
one-half
turn,
itin.2
advances
a2distance
)(20
in.)
rten dCompatibility.
elongate
4–15c.
Since
thebolt,
db , isFig.
t , and the bolt will
Fofb13(t2in.2
When
the
nut
tightened
onp10.25
the
the tube
1 c 2Eqs. 1 and
dt these
1+
= 0.025
in.
d
0.025
in.
- 2of
2and
simultaneously,
we
along
the
bolt.
Thus,
compatibility
displacements
0.025
Solving
Eqs.
1Since
simultaneously,
we -get
ergoeswill
one-half
itδadvances
ain.
distance
ofSolving
(d12)(
in.)
=the
δtüp
0.025
inelongate
veya,
shortenturn,
the=
bolt
will
4–15c.
the
dt ,+and
2 get
bulon
Solving
Eqs.
1 2and
simultaneously,
we3b2get
b ,20Fig.
(2)
0.78595F
1.4414F
t = 25 1 bp10.25
in.2
[10.6110
ksi]
1
requires
alongnut
the undergoes
bolt. Thus, the
compatibility
these displacements
one-half
turn, it of
advances
a distance
of moduli
(2)(20 in.)
=F
11.22
kip
kip kip
Taking the
of=Felasticity
from
the
table
on the inside back
b =FF
b t=F
t==F11.22
=
11.22
b
t
Solving
Eqs.
1 and 2 simultaneously,
we get 0.78595Ft = 25 - 1.4414Fb
(2)
Thus,
compatibility
of these
displacements
0.025 in. along the bolt.
Final
cover,
and
applying
Eq.
4–2,
yields
d
1+ cFinal
2the Final
=
0.025
in.
d
t in the
The The
stresses
andb tube
are therefore
stresses
inbolt
theinbolt
and tube
are therefore
position
position
The
stresses
the
bolt
and
tube
are
therefore
requires
Fb =Eqs.
Ft 1=and
11.22
kip
Solving
2 simultaneously,
we
get
dt = 0.025 in.
- db position
F
in.2inside
Fb the
kip
Taking
the moduli of elasticity from
table11.22
ont13the
back
Fb 11.22
kip
Final
F
11.22
kip 57.2
c
=
db stresses
Ans.Ans.
sb therefore
=sb = = F
= 257.2
ksi ksi
1position
+ 2
in.bolt
- dand
ddbt = 0.025
=2 b b==Ft =
The
in applying
the
tube
are
b
2 = kip
3 ksi
2 11.22
cover,
and
Eq.
4–2,
yields
d
Ans.
s
=
=
57.2
0.025
in.
A
b
0.025
in.
b
A
10.25
in.2
][6.48110
2
ksi]
in.2
the moduli of elasticity dfrom
the table on0.025
thein.inside back p[10.5
p10.25
in.2
b in.2
2
d
bp10.25
t
t
A
Final dt
p10.25
in.2
(b)
b
F
11.22
kip
Taking
the
moduli
of
elasticity
from
the
table
on
the
inside
back
b
The
stresses
in
the
bolt
and
tube
are
therefore
nd applying
Eq. 4–2, yields position
Ft13=in.2
Fb13
in.2
db
Ans.
sb =
=
57.2
Ft ksi
11.2211.22
kip
Ft
kip
Eq. 4–2,
yields
F-t3 kip =
11.22
kip 3= 19.1
0.025applying
in.
0.025
Ab
ksi ksi
p10.25
in.2s2t =sF
=
= 19.1
Initial
dt cover, and
2
2b= = in.
Initial
11.22
t
2
2
2
Ft13 in.2
p[10.5 in.2 - 10.25
in.2
][6.48110
2=p10.25
ksi]2in.2
s=A
= p[10.5
ksi
A
in.2
[10.6110
Initial
p[10.5
in.2
10.25
in.2
] 22ksi]
t
10.25
in.2
] 2 = 19.1Ans.
t
t
d
s
=
=
57.2
ksi
2
position
position
b
(b)
b
A
F
13
in.2
2
p[10.5
in.2
10.25
in.2
]
=
t
t
0.025 3in.position
A
F
11.22
kip
p10.25
in.2
2
2
d
b
t
in.2
t
= are F
stresses
less
the
reported
yield
stress
for
each
material,
b13than
p[10.5Initial
in.2 - 10.25 in.2
2tksi]
stresses
are
less
than
thet reported
yield
stress
for each
material,
(c) ][6.48110 s
(2)
0.78595F
=the
25reported
- 1.4414F
=2 0.025
=These
= 19.1
ksi
3These
b stress for each material,
in.
- These
2 stresses are2 less than
yield
p[10.5 in.22 (c)
- 10.25
2 ksi]
(c) in.2 ][6.48110
At 1sp[10.5
2 in.2
3 22
in.2
10.25
]
and
1s
(see
the
inside
back
cover),
and and
2
=
60
ksi
2
=
ksi
F
11.22
kip
and
1s
(see
the
inside
back
cover),
1s
2
=
60
ksi
2
=
22
ksi
position
Y al Y alp10.25 in.2
Y mgY mg
t [10.6110
2
ksi]
Fb13 in.2
and 1sY2mg
inside
1sEqs.
2tal =1=and
60 =
ksi
= we
22 ksi
=the
19.1
ksi back cover), and
Solving
2 analysis
simultaneously,
get (see
Ys
Initial
Fb13
in.2
Fig. 4–15
2 is valid.
therefore
this
“elastic”
is each
valid.
Fig.
4–15 3 are
therefore
thisA
“elastic”
analysis
stresses
less
than
the reported
yield
stress
for
material,
(c) 0.025 in. p[10.5
in.2
- 10.25
in.22(2)
]
t - “elastic”
0.025These
in. 2position
Fig. 4–15
therefore
this
analysis
is
valid.
0.78595F
=
25
1.4414F
p10.25
in.2
[10.6110
2
ksi]
t
b
2
3
#
#and
Fbback
= Ftcover),
= 11.22
kip for each material,
1s
(see
the
inside
and
1sY2al p10.25
= 60 ksi
=ksi]
22 ksiare
in.2
[10.6110
Y2mg2stresses
These
less
than
the
reported
yield
stress
(c)
Final
Fig. 4–15 0.78595Ft = therefore
this
“elastic”
analysis
is
valid.
Solving
Eqs.
1
and
2
simultaneously,
we
get
(2)ksi in
25
- 1.4414F
b
the1sbolt
therefore
and
(see
the inside back cover), and
1sYThe
2al b=stresses
60
2mgand
= tube
22 ksiare
0.78595F
1.4414F
Y(2)
positiont = 25 (1) ve (2) denklemlerinin
ortak
çözümünden,
Fig.
4–15
F
=
F
=
11.22
kip
therefore bthis “elastic”
analysis
is valid.
t
Eqs. 1Solving
and 2 simultaneously,
we
get
F
11.22
kip
b
Eqs. 1 and 2 simultaneously, we get
Final
db
Ans.
=
= 57.2 ksi
b =therefore
The
stresses
in the bolt and tube s
are
0.025
in.
A
position Fb = Ft =dt11.22
p10.25 in.22
kip
b
F
=
F
=
11.22
kip
b
t
#
Fb
11.22 kip
Ft ksi
11.22 kip Ans.
sses inThe
thedstresses
bolt
and
intube
the are
bolttherefore
and tube are s
therefore
=
b
b =
st2 == 57.2
=
= 19.1 ksi
Initial
0.025 in.
Ab
p10.25 in.2
2
dt
A
p[10.5
in.2
- 10.25 in.22]
Fb
11.22 kip
Fb
11.22
kip
t
position
== 57.2 ksi 2 = 57.2Fksi
Ans. 11.22 kipAns.
sb =
= sb =
t These stresses
are less than the= reported
Ab 2 (c)
Ab Initial
p10.25
in.2
p10.25
in.2
s
=
=
19.1 ksi yield stress for each material,
#
t
2
2
A
and
1s
1s
2
=
60
ksi
2
=
22
ksi (see the inside back cover), and
p[10.5
in.2
10.25
in.2
]
t
Y al
Y mg
F
Ft position
11.22
kip4–15 11.22 kip
t
Fig.
therefore
this
“elastic”
analysis
is
valid.
s
=
=
=
19.1
ksi
These stresses
are less
t
st =(c) =
= 19.1
ksi than the reported yield stress for each material,
At 2 - p[10.5
in.222] - 10.25 in.22]
At
p[10.5 in.2
10.25
1sY2alin.2
= 60 ksi and 1sY2mg = 22 ksi (see the inside back cover), and
stresses
arereported
less than
the reported
yield
for each
material,
ressesThese
are
than the
yield
stress
eachstress
material,
Fig.less
4–15
therefore
thisfor
“elastic”
analysis
is valid.
and
1s
(see
the
inside
back
cover),
and
1s
2
=
60
ksi
2
=
22
ksi
Y
al
Y
mg
60 ksi and 1sY2mg = 22 ksi (see the inside back cover), and
#54
therefore
this
“elastic”
analysis
is
valid.
e this “elastic” analysis is valid.
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