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6
!
!
!
!
!
!
!
!
!
!
!
4.6
!
! !
!
!
!
caused
onlyload
bydonly
the
external
loads acting
on
theexternal
member
plusofacting
(vectorially)
displacement
at this atthe
displacement
relationship,
such
as
=and
PL>AE.
support
caused
support
by redundant
the external
caused
only
loadsby
acting
the
on
member
loads
plus
on
(vectorially)
the the
member
the
plus
displacement
• Express
the
external
displacements
inthe
terms
the
loadings
by
using
a(vectorially)
load–
•support
Once
established,
the
equation
can
then
be
solved
for
the
magnitude
of the
redundant
force. thisdisplace
4compatibility
•
Once
established,
the
compatibility
equation
can
then
be
solved
for
the
magnitude
of
the
redundant
force.
Compatibility.
support
caused
only
by
the
redundant
reaction
acting
on
the
member.
support
caused only
support
by
the
redundant
caused
only
reaction
by the
acting
redundant
on the
reaction
member.
acting
on the member.
displacement
relationship,
such
as
d
=
PL>AE.
4
• Once
established,
the
compatibility
equation
can
then
be
solved
for
the
magnitude
of
the
redundant
force.
4
Equilibrium.
• •Equilibrium.
Express
the
load
and
redundant
in
the of
loadings
bydo
using
ausing
load–
• Express
external
• Express
load
and
the redundant
external
load
displacements
and equation
redundant
terms
displacements
in
terms
byforce.
of
the
loadings
a load– by us
Choose
oneexternal
ofthe
the
supports
as
redundant
and
the
ofof
compatibility.
To
this,
the
known
• Once
established,
the
compatibility
equation
candisplacements
thenwrite
be
solved
forterms
thein
magnitude
ofthe
theloadings
redundant
Equilibrium.
displacement
relationship,
such
assuch
dsupport,
= as
PL>AE.
Equilibrium.
4
displacement
relationship,
displacement
relationship,
d =which
PL>AE.
as dzero,
= PL>AE.
displacement
at the
redundant
is such
usually
is equated to the displacement at the
• Draw
a free-body diagram and
write
theaappropriate
equations
of equilibrium
for
theappropriate
member using equations
the
•write
Draw
free-body
diagram
and
write
the
of
•• Draw
free-body
diagram
and
the
appropriate
equations
of plus
equilibrium
for
the
member
using
thethisforce.
support
caused
only
by•the
the
external
loads
acting
on
the
member
(vectorially)
the
displacement
at
Once
the
compatibility
equation
can
then
beequilibrium
solved
for
the
magnitude
the
redundant
force.
• aestablished,
Once
established,
Once
compatibility
established,
equation
the compatibility
can
then
beequation
solved
for
canreactions.
the
then
magnitude
beof
solved
of
for
the
the
redundant
magnitude
of the redu
• Draw Equilibrium.
a free-body
diagram
and
write
appropriate
equations
of
for
the
member
using
the
calculated
result
for
thethe
redundant.
Solve
these
equations
for
any
other
4.4
S
TATICALLY
I
NDETERMINATE
A
XIA
calculated
result on
for
redundant.
result for
redundant.
Solvereaction
these equations
for the
any
other
reactions.Solve these equations for any
4
support
only
the redundant
member.
4calculated
4 theby
calculated
result
for thecaused
redundant.
thesethe
equations
for acting
any
other the
reactions.
• Draw
a free-body
diagram Solve
and write
appropriate
equations
of equilibrium for the member using the
Equilibrium.
Equilibrium.
Equilibrium.
• Express
thefor
external
load and
redundant
displacements
terms
of the loadings by using a load–
Example
4.5’in
süperpozisyon
prensibi
ile equations
çözülmesi:
calculated
result
the redundant.
Solve
these
for any in
other
reactions.
displacement
relationship,
such
as
d
=
PL>AE.
• Draw
free-body
diagram
and write
the
appropriate
equations
of
for
the for
member
using the
• aDraw
a free-body
•diagram
Draw
aand
free-body
write
the
diagram
appropriate
and
write
equations
theequilibrium
appropriate
of equilibrium
equations
the
of equilibrium
member
using
for the
the memb
EXAMPLE
4.5
EXAMPLE
4.9
•
Once
established,
the
compatibility
equation
can
then
be
solved
for
the
magnitude
of
the
redundant
force.
calculated
result
for
redundant.
Solve
these
equations
for
any
other
reactions.
calculated
result
for
calculated
the
redundant.
result
Solve
for
the
these
redundant.
equations
Solve
for
these
any
other
equations
reactions.
for
any
other
reactions.
!
EXAMPLE
4.9
EXAMPLE
4.9
4
The steel4.9
rod shown in Fig. 4–12a has a diameter of 10 mm. It is fixed
EXAMPLE
The A-36 steel
rodankastre
shown in mesnetlidir.
Fig. 4–17a hasYükleme
a diameter
of 10 B
mm. It is
Çapı
10 mm
duvara
öncesi
Equilibrium.
EXAMPLE
4.9olan çelik çubuk A noktasından
toshown
the
wall
at
A,
and
before
itdiameter
is loaded,
is a gap
0.2
mm
Thefixed
A-36
steel
rod
shown
inhas
Fig.
hasloaded
athere
diameter
mm.
It between
is
to
the
wall
at4–17a
A, and
before
it is
there
isisof
a10gap
between
The A-36 steel
rod
in
Fig.
a4–17a
of
10 mm.
Itof
A
the
wall
at
and
the
rod.
Determine
the
reactions
at
A
and
if the in Fi
B¿
B¿
fixed
to
the
wall
at
A,
and
before
it
is
loaded
there
is
a
gap
between
•
Draw
a
free-body
diagram
and
write
the
appropriate
equations
of
equilibrium
for
the
member
using
the
The
A-36
steel
rod
shown
the
wall
at
and
the
rod
of
0.2
mm.
Determine
the
reactions
at
A
B¿
fixed
to
the
wall
at
A,
and
before
it
is
loaded
there
is
a
gap
between
ucu
ile
B’
arasında
0.2
mm
boşluk
mevcuttur.
Çubuğa
C
noktasından
20
kN
yüklendiğinde
A
Thethe
A-36
steel
rod
shown
in an
Fig.
4–17a
hasDetermine
diameter
ofas
10shown.
mm. ItNeglect
is A the
EXAMPLE
EXAMPLE
4.9
EXAMPLE
4.9 4.9
rod
isB¿
subjected
to
axial
force
ofa C.
Preactions
= 20 E
kN
wall
at
and
the mm.
rod
of
0.2other
mm.
the
at
P ! 20 kN result0.2
calculated
for
the
Solve
these
equations
for
any
reactions.
and
Neglect
the
size
of
the
collar
at
Take
B¿.
=reactions
200
GPa.
the redundant.
wall fixed
at B¿ to
and
the
rod
ofA,0.2
Determine
the
at
A
mm
st
fixed
to
the
wall
at A, and bef
the
wall
at
and
before
it
is
loaded
there
is
a
gap
between
C
P !mesnetlerinde
20 kN
size
ofofthe
collar
C.
Take
200
GPa.
EstEat=C.
and B¿.
Neglect
the
sizeat
ofat
the
collar
Take
Est = 200 GPa.
0.2 mmmeydana
P ! 20 kNve B’
reaksiyonları
hesaplayınız.!
andB¿B¿. gelecek
Neglect
the
size
the
collar
C.
Take
=
200
GPa.
mmC
st
A 0.2
A-36
steel
rod
shown
The
A-36
in Determine
Fig.
steel
4–17a
rod
shown
has
areactions
diameter
inof
Fig.
4–17a
ofand
10
mm.
a diameter
Itrod
is of
of
A-36
shown
Fig.
4–17a
has
athe
diameter
mm.
Ithas
is
theThe
wall
atThe
androd
the
rod
ofin
0.2
mm.
the
at
A
B¿steel
SOLUTION
wall
at10B¿
the
C
800 mm B¿
A
P ! 20400
kNmm B¿
fixed
to
the
wall
at
fixed
A,
and
to
before
the
wall
it
is
at
loaded
A,
and
there
before
is
it
a
is
gap
loaded
between
there
is
a
g
A
fixed
to
the
wall
at
A,
and
before
it
is
loaded
there
is
a
gap
between
andSOLUTION
Neglect
the
size
of
the
collar
at
C.
Take
B¿.
E
=
200
GPa.
0.2 mm
st
SOLUTION
PCompatibility.
! 20 kN
Here
theB¿.
support
at the
B¿ as
Neglect
size
ofthe
the
0.2
mm we will considerand
800
C
(a) mm
800mm
mm
the
wall
at
and
the
the
rod
wall
of
at
0.2
mm.
and
the
Determine
rod
of
0.2
the
mm.
reactions
Determine
at
A
rea
B¿
B¿
the
wall
at
and
the
rod
of
0.2
mm.
Determine
the
reactions
at
A
SOLUTION
B¿
400
C
Compatibility.
Here
we
will
consider
the
support
at
as
B¿
400 mm A EXAMPLE
redundant.
Using
principlethe
of superposition,
Fig.as
4–17b, we have
Here we
willthe
consider
support at B¿
4.9 B¿ Compatibility.
SOLUTION
(a)20 kN
(a) P ! 20 kN
P800
!
and
the
and
sizeB¿
of
the
Neglect
collar
the
at
C.
size
of
the
C. Takewe
B¿. Neglect
B¿.
Estcollar
= we
200athave
GPa.
Est = 200
mm P ! 20
0.2
mm
and B¿.the
Neglect
the
size
ofshown
the
collar
atFig.
C.free-body
Take
EstTake
= have
200
GPa.
AkN
mm0.2 mm 0.2
redundant.
Using
the
principle
of
superposition,
Fig.
4–17b,
Equilibrium.
As
on
the
diagram,
Fig.
4–12b,
SOLUTION
redundant.
Using
principle
of
superposition,
4–17b,
we
+
C
C
C kN
400 mm P ! 20
Compatibility.
Here
we
will
consider
the
support
at
as
B¿
(1)
1
:
2
0.0002
m
=
d
d
0.2
mm
800
mm
A (a)
A
B¿ SOLUTION
B¿
A
B¿+
A-36
steel
rod
shown
in
4–17a Phas a Bdiameter
of 10
mm.end
It isB to
will
assume
that
force
P Fig.
ismlarge
to cause the
rod’s
SOLUTION
SOLUTION
+The
:
2mm
0.0002
= dPenough
- dCompatibility.
P 0.2
! 20
kN
Using0.0002
the principle
of- superposition,
we have(1)
! 1400
0.2mm
mm800 mm
Here
we w
(1)
1:
2 redundant.
m
=
d
d
P ! 20 kN
B Fig. 4–17b,
800
mm
mm
P
B
800
The
deflections
and
are
determined
from
Eq.
4–2.
d
d
fixed
to
the
wall
at
A,
and
before
it
is
loaded
there
is
a
gap
between
contact
the
wall
at
The
problem
is
statically
indeterminate
since
B¿.
(a)
P
B
400
mm
400
mm
Compatibility.
Here
Compatibility.
we
will
consider
Here
we
the
will
support
consider
at
the
as
support
B¿
400 mm
Compatibility.
Here
we
will
consider
the
support
at
as
B¿
+The
(a)
(a)deflections
redundant.
Using
the
principle
(a)
(1) at
1:
2the
0.0002
m
=0.2donly
-of
dDetermine
and
are
determined
from
Eq.
4–2.
dtwo
dB the
P ! 20 kN
at
the
rod
of
the
reactions
A have
0.2 mm InitialThe deflections
and
are
determined
from
Eq.
4–2.
dPwall
dUsing
there
are
unknowns
and
one
of equilibrium.
Pmm.
B equation
Pand
Çözüm:
B B¿
redundant.
Using
redundant.
principle
Using
superposition,
the
principle
Fig.
of
4–17b,
superposition,
we
Fig.
4–1
3 superposition,
redundant.
the
principle
of
Fig.
4–17b,
we
have
d
F
P
2 N]10.4
m2Take E = 200 GPa.
A
position
PLAC the size[20110
PInitial
! 20 kN
anddB¿.=Neglect
of the collar
at C.
0.2
mm
Initial
st 0.5093(10 - 3) m
+
=
=
3
The
deflections
and
are
determined
from
Eq.
4–2.
d
d
+
+
3
3
P
C
+
d
P
Bmm
+!
dP kN
2P
kN
P
[20110
22AN]10.4
(1) 0.000
1::
2[20110
:m2
0.0002
m9220110
=N>m
P ! PL
20 Uygunluk
kN
position
PL
2 N]10.4
0.2
©F
-F
- =FBdm2
2] 2
NdB0.0002
= 0 m- 3 = dP(1)- dB(1)
P !position
20 kNP ! 20
0.2
mm
1AC
:
2 20
0.0002
d1BdP:
AE
P ! 20 kN
AC
p10.005
[200110
x = 0;
0.2
mm 0.2 mm
P+denklemi:
! 1m2
! 2m
AP ! 20 kN Initial
B¿
SOLUTION
dP= =
)m
= 2
=- 30.5093(10
dP =
)m
= 90.5093(10
4.6
THERMAL S
2
2
9
2
3
AE
800 mm dP
p10.005
m2
2m2
N>m
] Eq.
p10.005
[200110
2 N>m
] m2
FinalAE deflections
[20110
N]10.4
The
The
deflections
determined
and
from
are
Eq.
determined
4–2.
from
Eq.
4–2.
dPB and
d[200110
dPThe
ddeflections
position
PL
F
11.20
P ! 20400
kNmm
are
determined
from
dAB
BBare
B4–2.
FBLm2
ACdeflections
and
d
P and
- 3 at d
dB positionThe
Compatibility.
Here
we
will
consider
the
support
as
B¿
9
P
B are d
B =to 76.3944110
move) m
to B¿, 2F
with
d = dCompatibility.
= 0.5093(10
= =
= The2 force 2P9 causes92 point
(a)Final
BInitial
B no
Initial
2
Final F LPredundant.
AE
Initial
F
11.20
m2
F
11.20
m2
p10.005
m2
[200110
23superposition,
N>m
FBLAB
AE
Using
the
principle
4–17b,
have the
p10.005
m2 of
[200110
2]compatibility
N>m
B
3
ddB position
dP B AB further
dPB Initial
3[20110
-we
9 m2
-Fig.
9
Therefore
the
2 AC
N]10.4
m2=][20110
N]10.4
PL
PL
position
P ! 20dBkN
Pd P =
2 N]10.4
m2
dB ==position
2FB-for
= 2AC [20110
76.3944110
2FB 2condition
=
76.3944110
position position
PLACdisplacement.
P ! 20 kN
3
3
B
B ! 20FkN
9
2
9 d2dP= 2
-3
d
=
) m2 [20110
=
=
=
0.5093(10
= 0.5093(
AE
AE
Final
+
position
P
P
p10.005
m2
[200110
2
N>m
]
p10.005
m2
[200110
2
N>m
]
PL
d!
=2BL
) m9 (1)
=AE
= 0.5093(10
Thermal
Stress
Fp10.005
m2
P1Substituting
kN
rod
is
2=
2
2
F
P 20
AC
B11.20
:
0.0002
m
P ! 20 kN
AB into
2
9 dP - p10.005
29d
Eq.
1,
we
get
dB position F
0.2F mm
AE
B
9
m2
[200110
2
N>m
]
m2
[200110
2
N>m
]
AE
F
[200110 2 N>m ]= 76.3944110
B
B
dB =
2FB=
= p10.005 m2
A
dP =
2
9
2
(b)
AE
p10.005
m2
[200110
-23 N>m ]from Eq. 4–2.
- AE
9
Substituting
into
Eq.
1, we
get
Substituting
into
Eq.
1,
we
get
p10.005
m22
Final
Final
The
deflections
and
are
determined
d
d
0.0002
m
=
0.5093(10
)
m
76.3944110
2F
F
11.20
m2
F
11.20
m2
d
=
0.0002
m
Final
P
B
F
L
F
L
A
change
in
temperature
can
cause
a
body
to
change
its
dimensions.
B
!
BB AB
B
F
F
11.20
m2
B>A
B
AB
d
d
F
L
B
9
B
20
kN
B
B
F
B
AB
dB position
-9
3.39 kN position
dB = =position
dB body
=-33 2will-expand,
2FB= 76.394
= the
(b)A
(b)
9= 9 = whereas
dBtemperature
=m into
76.3944110
9 =2 76.3944110
InitialGenerally,
9B
if0.0002
the
increases,
if 2F
Substituting
Eq.
1,m
we
=AE
0.5093(10
)=mp10.005
- 276.3944110
2F
9=- 4.05
2B2 N>m-2]m2
AE
AE
0.0002
=B- 3get
0.5093(10
)N
m
76.3944110
2F
m2
[200110
p10.005
[200110
2Ans.
N>m2]
3 Final
4.05110
2
kN
F
B
d
FB
4.6
T
HERMAL
S
TRESS
1
5
1
p10.005
m2
[200110
2
N>m
]
F
L
P
20 kNFA
[20110
2
N]10.4
m2
FA
20
kN
position
PL
P ! 20 kN 3.39 kN
Borthe
AB
This displacement
candBOrdinarily
be expressed
terms of
FB decreases,
ACit willFcontract.
3.39temperature
kNFB
the
this in
expansion
- 3 unknown
B
(b)
-33 position
dFP 0.0002
== 4.05110
)m
==320.5093(10
=- 92F
0.5093(10
d
=
=
B
m
)
m
76.3944110
Ans.
N
=
4.05
kN
2
9
2
Ans.
=
4.05110
2
N
=
4.05
kN
F
B
From
the
free-body
Fig.
reactions
load–displacement
relationship,
Eq. 4–2, applied
(c)
Substituting
into
Eq.
Substituting
1,m2
we[200110
get diagram,
into
Eq.
1,] we
getB AE
AE using
B
contraction! Substituting
is Equilibrium.
linearly
related
top10.005
the
temperature
increase
or4–17c,
decrease
2 N>m
20 kN
FA
into
Eq.
1,the
we
get
p10.005 m2
3.39 kN
3 material
(b)
(b)
to
segments
AC
and
CB,
Fig.
4–12c.
Working
in
units
of
that
occurs.
If
this
is
the
case,
and
the
is
homogeneous
and
F
3
--Ans.
39newtons and
-9
+
B
(b) Fig. 4–17 Equilibrium.
From
the
free-body
diagram,
Fig.
4–17c,
=
4.05110
2
N
=
4.05
kN
F
(c)
Final Equilibrium.
From
the
free-body
diagram,
Fig.
4–17c,
3
9
(c)
:
©F
Ans.
=
0;
F
+
20
kN
4.05
kN
=
0
F
=
16.0
kN
B
0.0002
m
=
0.5093(10
0.0002
)
m
m
=
76.3944110
0.5093(10
)
2F
m
76.3944110
2F
F
11.20
m2
x F
A
B
0.0002
m A=experiment
0.5093(10
) m the
- 76.3944110
2Fof
BLAB
dB position
Thermal Stress
-B9
B a
20
kN
20
kN
F
F
isotropic,
it
has
been
found
from
that
displacement
meters,
we
have
A
A
3.39
kN
3.39
kN
20
kN
d
=
2F
=
=
76.3944110
FA
Substituting
into
Eq.
1,
we
get
B
B
3.39
kN
+
2
9
2
+ ©F
3 = Fig.
: ©Fx Equilibrium.
=having
0;:
-aFxlength
20
- +4.05
=m2
04.05
FkN
kN
From
the
free-body
AE
[200110
2 0kN
N>m
3diagram,
Fig. 4–17 (c) Fig. 4–17
A =+ 0;
-F
20
kN
=16.0
F=A] 4.05110
= Ans.
16.0 3kN
member
LkN
can
calculated
using
formula
Ans.
4.05110
2the
N
=
4.05
kN
2 N =Ans.
4.05 kN
FkN
F4–17c,
B =Ans.
=p10.005
4.05110
2N
=A 4.05
FABbe
FB!
FABLAC
FBLCB
Yatay
kuvvet
dengesi:
(b)
in temperature can cause
a body
to change
its dimensions.
+
d
=
0.0002
m
=
0.0002
m
= 4–17c,
0.5093
: ©F
0; -Finto
20From
kN
-the
4.05
kN = 0From
FA151
= 16.0
B>A
4.6
Sfree-body
TRESS
Fig.the
4–17
A +
Substituting
Eq.
we
get
Equilibrium.
diagram,
free-body
Fig. kN
4–17c,Ans.
diagram, Fig.
(c)
(c)x20=Equilibrium.
AEthe
AE
Equilibrium.
From
theT1,HERMAL
free-body
diagram,
Fig.
4–17c,
(c)
FAwhereas
if the temperature increases,
body
will expand,
ifkN
3.39
kN
3
+ ©F 0.0002
++ ©F
rature decreases, it will contract.(b)
Ordinarily
or :
dT m
= F
a A¢TL
:
204.05
kN
=- kN
-0;
4.05
+=20
0 kN
F-A(4–4)
-92F
=4.05
16.0
kN
kN
= B0 Ans.
=
0.5093(10
)m
76.3944110
+ Fig.
F
10.4
=F4.0511
F
Fig. 4–17this expansion
x = 0;
x m2
A F
A = 16.0
B
A:
©F4–17
Ans.
+ -20
kN
=--F
0kN
20 kN Fig. 4–17
FA
x = 0; m -F
A = 16.0 kN
3.39or
kNdecrease0.0002
= A
n is linearly related
to the
temperature increase
2
9
2
3
p10.005
m2 [200110
Ans.
2 N 2=N>m
4.05 ]kN
FB = 4.05110
s. IfThermal
this is the case,
and the material is homogeneous and
! Stress
Equilibrium. From
the free-b
(c)
Most traffic
bridge
t has been found from experiment
of a
From the free-body diagram, Fig. 4–17c,
(c) that the displacementEquilibrium.
expansion
joints
to
F
10.8
m2
+ ©F B = 0; -thermal
nge ina temperature
can
cause a body
change
its dimensions.
4.6 be
Isıcalculated
Gerilmeleri
aving
length L can
usingtothe
formula
- :
FA2 +movement
20 kN o
Fig.
4–17
x
+
2
9
where
: ©Fxif = 0; - FA + 20 kN - 4.05 kNp10.005
Ans.
= 0 Fm2
16.0 kN2 avoid
N>m
] thermal str
any
A =[200110
ally, if the temperature increases,Fig.
the4–17
body will expand, whereas
mperature decreases, it will contract. Ordinarily this expansion or
or
ction is linearly related
the temperature increase or (4–4)
decrease
dT = to
a ¢TL
a homogeneous
= a property of
the material,Freferred
to as the linear coefficient #
ccurs. If this is the case, and the material is
and
(2)
A10.4 m2 - FB10.8 m2 = 3141.59 N m
of
thermal
expansion.
The
units
measure
strain per degree
of
4
pic, it has been found from experiment that the displacement of a
Most traffic
bridges
are designed with
Solving
1 and
2 yields
temperature.
They Eqs.
are 1>°F
(Fahrenheit)
in the FPS system,
er having a length L can be calculated using the formula
expansion joints to accommodate the
and
1>°C
(Celsius)
or
1>K
(Kelvin)
in
the
system.
Ans.
F
=
16.0
kN
FB =Typical
4.05 kN
A the deckSI
α: Doğrusal ısı genleşme katsayısı!
thermal movement of
and
thus
values are given
on any
thethermal
inside stress.
back cover
avoid
Since the answer for FB is positive, indeed end B contacts the wall at
δT: Eleman
değişim
¢T =cebrik
the algebraic
change in temperature of the member
dT = auzunluğundaki
¢TL
(4–4)
B¿ as originally assumed.
L
=
the
original
length
of the member
ΔT: Elemandaki
ısı değişimi
4
operty of the material,
referred to ascebrik
the linear
coefficient
dT = the algebraic change
in the
length
of the
Most
traffic
bridges
are member
designed with
NOTE:
If
were
a
negative
quantity,
the
problem would be
F
hermal expansion. The units measure strain per degree of
B
expansion
joints to accommodate the
statically
determinate,
so
that
and
F
F
=
0
=
20
kN.
perature. They are 1>°F (Fahrenheit) in the FPS system,
thermal movement of the Bdeck and thusA
avoid
any
thermal
stress.
1>°C (Celsius) orStatikçe
1>K (Kelvin)
in
the
SI
system.
Typical
belirli (izostatik) sistemlerde ısı değişimi elemanlarda gerilme oluşturmaz. Aksine
ues are given on the inside back cover
change in length of a statically determinate member can easily be
sistemlerde
ısı The
değişimi
ısı gerilmelerine sebep olur.
algebraic change hiperstatik
in temperature
of the member
calculated using Eq. 4–4, since the member is free to expand or contract
a property
of the
material,
referred to as the
linear
when
it coefficient
undergoes a temperature change. However, in a statically
original
length
of the
member
of thermal expansion. The units measure strain
per
degree
of
indeterminate
member,
these thermal displacements will be constrained Long extensions of
algebraic
change
in are
the 1>°F
length
of the member
temperature.
They
(Fahrenheit)
inby
thethe
FPSsupports,
system, thereby producing thermal stresses that must be carry fluids are subj
climate that will ca
and 1>°C (Celsius) or 1>K (Kelvin) in the SI
system. Typical
considered
in design. Determining these thermal stresses is possible and contract. Expans
values are given on the inside back cover using the methods outlined in the previous sections. The following one shown, are used
stress in the material.
the algebraic change in temperature of theexamples
member illustrate some applications.
nge in length of a statically determinate member can easily be
the original
length
the
memberis free to expand or contract
using
Eq. 4–4,
sinceof
the
member
ndergoes
a temperature
in a statically
the algebraic
change in thechange.
length ofHowever,
the member
PAGE !57
ate member, these thermal displacements will be constrained Long extensions of ducts and pipes that
pports, thereby producing thermal stresses that must be carry fluids are subjected to variations in
climate that will cause them to expand
in design. Determining these thermal stresses is possible and contract. Expansion joints, such as the
!
The A
betwee
R 4
AXIAL LOAD
0.5 in.
raised
Applying the thermal
and load–displacement
we
the thermal andrelationships,
load–displacement
relationshi
develo
152
C H A P T E R 4 A X I A L Applying
LOAD
have
have
152
CHAPTER 4 AXIAL LOAD
A
10 The A-36 steel bar shown in Fig. 4–18a is constrained to just fit
FL
FL
2 between
C H A Ptwo
T E R fixed
4 A Xsupports
I A L L O A Dwhen T = 60°F. If the temperature is
0 = a¢TL SOLUT
1
0 = a¢TL EXAMPLE 4.10
AE
AE
raised to T2 = The
determine
theshown
average
normal
thermal
stress
120°F,
A-36
steel
bar
in
Fig.
4–18a
is
constrained
to
just
fit
F
F
Equilib
EXAMPLE 4.10 A36 çeliğinden yapılmış bir çubuğun hareketi sıcaklık T =60 Fo iken,
2 ft
152
C
H A P TIf
E Rthe
4 temperature
A Xthe
I A L L O A Dis
1
developed
in the
bar. in two
between
fixed
whenThus,
Tto
60°F.
from
the
data
on
inside
back
cover,
0.5 in.
he A-36
steel bar
shown
Fig.(b)
4–18asupports
is constrained
fit
1 =just
Thus,
from
the
data
on
the
inside
back
cover,
The
A-36
steel
bar
shown
in
Fig.
4–18a
is
const
Fig.
4–
(b)
XAMPLE
4.10
raised
determine
theThe
average
normal
thermal
stress
=O
120°F,
TL 2L0.5
etween
when
If the
temperature
is
TA1in.
60°F.
A-36
steel
bar
shown
in
Fig.
4–18a
is
constrained
to
just
fit
o
between
two
fixed
supports
when
If
th
T
=
60°F.
opposi
1 50.5
2 in. two fixed
C H A Piki
Tsupports
E Rankastre
4 Ato
X
I Amesnet
D= arasında
1
engellenmiştir.
Sıcaklık T2=120
ye çıkartıldığında çubukta
F = a¢TAE
F =F a¢TAE
C H Ato
P T ET
R 4= 120°F,
A X I A L developed
Ldetermine
OAD
inthe
theaverage
bar.
between
two
fixed
supports
when
If
the
temperature
is
T
=
60°F.
ised
normal
thermal
stress
0.5
in.
raised
to
determine
the
average
norm
=
120°F,
T
d
d
1
2
T
2
T
0.5 in.
SOLUTION
The A-36 steel
bar shown in
Fig. -6
4–18a is constrained-6to just 2fitB
3 thermal
EXAMPLE
4.10
0.5 in.
raised= to
determine
the average
normal
120°F,
Thesaplayınız.
eveloped in the bar.
2>°F]1120°F
- 60°F210.5
in.2
[29110
2 kip>in2] stress
[6.60110
2>°F]1120°F
in.22[29110+32ckip
=Ifdeveloped
[6.60110
in the
bar.
2 = T
©F
gelecek olan
ortalama
ısı bar
gerilmesini
E=!
4
between
two
fixed
supports
when
the
temperature
is - 60°F210.5
1 = 60°F.α=!
Equilibrium.meydana
The free-body
diagram
of the
is
shown
in
developed
in
the
bar.
EXAMPLE
4.10
d
d
0.5
in.
F=equal
determine
the average =
normal
thermal
F load,
Ais
2.871The
kip A-36stress
Fig.
4–18b.C152
Since
there
isHXno
the
at A
C
A P Texternal
E R raised
4 A
X Ito
AL T
L 2O A=force
D 120°F,
2.871
kip
SOLUTION
0.5
in. but
steel (a)
bar shown in Fig. 4–18a
is
HAPTER 4
A
IAL LOAD
LE152
4.10
The
A is,developed in the bar.
opposite
to
the
force
at
B;
that
OLUTION
Equilibrium.
The
free-body
diagram
of
the
bar
is
shown
in
between
two
fixed
supports
when
T
=
60°
SOLUTION
0.5 in.
1 determ
Since
also
represents
internal
force within
the b
The A-36 steel Since
bar shown
Fig. 4–18a
isinternal
constrained
tothe
just
fit theaxial
F also in
represents
the F
axial force
within
bar, the
2 ft
SOLUTION
0.5 in. A The free-body
Fig. 4–18b.
Since
there
isthe
no
load,
the
force
Aaverage
is equal
but
0.5
The
A-36
steel
barexternal
shown
in supports
Fig.
4–18a
isat in.
constrained
toIfisraised
just
fit to TThe
determine
the
averag
120°F,
quilibrium.
diagram
of
bar
isfixed
shown
in
normal
compressive
stress
is
thus
2 =
between
two
when
the
temperature
is
T
=
60°F.
Equilibrium.
free-body
diagram
of
the
b
average
normal
compressive
stress
thus
F
1
+ c ©Fy = 0;
FAforce
=two
Fat
=B;Fthat
ft
EXAMPLE
4.10
Bfixed
opposite
to the
is,A= is120°F,
Equilibrium.
The
diagram
of inthe
bar
shown load,
in theComp
EXAMPLE
4.10
between
when determine
If free-body
the
temperature
isSince
T1 = 260°F.
g. 4–18b.
Since there
is0.5noin.external
load,
the
force
at
equal
but
developed
the
bar.
raised
tosupports
the
average
normal
thermal
stress
T
Fig.
4–18b.
there
is noisexternal
force
SOLUTION
2
2 ft
4–18b.
Since there
is no
external
load,
force
is equal
but
that oc
0.5 in.
pposite to the
force at B; that raised
is,
to T2developed
determine
the average
normal
thermal
stress
= 120°F, in
the Fig.
bar.
opposite
thethe
force
at at
B; A
that
is,
kip
F
The
free-body
diagram
ofF the
bar kip
is to
shown
in 2.871
The problem
since
this
force
cannot
be
A steel
in. Equilibrium.
The
bar
shown
inis2.871
Fig.
4–18a
is
constrained
to =just
fitksi
0.5 in.
The
A-36
steel
bar
in
Fig.
4–18a
constrained
to
just
fit
+2isftcstatically
©Fy developed
= 0.5
0;indeterminate
FA =
FBopposite
=A-36
Fshown
to the
force
at
B;
that
is,
to pus
s
=
=
11.5
in
the
bar.
s = the =force at A2is=equal
Ans.
11.5 ksi
Since there
isB notwo
external
load,
A from equilibrium. Fig. 4–18b.
10.5
in.22
between
fixed
supports
when
If the
temperature
is
Tthe
= temperature
60°F.Abut
A
10.5
in.2
1
between
two
fixed
supports
when
If
is
T
=
60°F.
c ©Fdetermined
SOLUTION
=
0;
F
=
F
=
F
1
y
A B B opposite to the force at B; that is,
+ c ©Fy = 0;
FA = FB = F condit
4
A
0.5
in.
raised
tosince
determine
the
average
normal
thermal
stress
120°F,
T=2 =
c
0.5
in.
+
©F
0;
F
=
F
=
F
raised
to
determine
the
average
normal
thermal
stress
=
120°F,
T
The
problem
is
statically
indeterminate
this
force
cannot
be
SOLUTION
y
A
B
2
4
Equilibrium. The free-body diagram of
Compatibility.
Since dA>B = 0, the thermal displacement dT at A
2
ft
developed
in
the
bar.
determined
from
developed
in(c)
the
bar.
B is Fig.
(a)
1that
+ cthi
2c
The that
problem
statically
indeterminate
since
thisforce
force
be
NOTE: ofof
From
the
magnitude
of F,
it
should
apparent
(c) equilibrium.
Equilibrium.
The
free-body
the
bar
isbeshown
in
NOTE:
theFdiagram
magnitude
F,The
itFig.
should
apparent
that
problem
is statically
indeterminate
since
occurs,
4–18c,
isSOLUTION
counteracted
by the
Fcannot
that
isFrom
required
4–18b.
Since
there
ischanges
no be
external
load,
th
FA =
= F
2 ft
(a) + c ©Fy = 0;
B statically
The
problem
is
indeterminate
since
this
force
cannot
be
etermined
from
equilibrium.
in
temperature
can
cause
large
reaction
forces
in
st
Fig.The
4–18b.
Since
there
is no external
the
force
at in
A from
isto
equal
but in atstatically
A to its original
determined
equilibrium.
to push theAbar dF back
position.
The
compatibility
in
temperature
cause
large
reaction
forces
Equilibrium.
free-body
diagram
of can
the load,
bar
shown
opposite
the
force
B; that is,
Fig.the
4–18
Compatibility.
thermal
displacement
at A
dT is
2 ft
Fig. 4–18Since dA>B = 0,
determined
from
equilibrium.
App
indeterminate
members.
opposite
toisthe
force
at B;
that
is, force at A this
condition(a)
at A becomes Fig. 4–18b.The
indeterminate
members.
Fload,
Since
there
external
the
is equal
isno
statically
forcebut
cannot be
SOLUTION
occurs,
isproblem
counteracted
by
required
SOLUTION
ompatibility. Sincethat
the F4–18c,
thermal
displacement
atBindeterminate
Aforce F that is since
dA>B
= 0, Fig.
dT the
Compatibility.
Since dA>B = 0, the thermal
disp
have
opposite
to
the
force
at
B;
that
is,
c
determined
from
equilibrium.
+
©F
0; displacement
= FB = F
Compatibility.
Since
the
thermal
A
d
=
0,
dTFat
tois push
the bar by
back
to
its
original
position.
The
compatibility
dF the
y =shown
A
B
A>B
at occurs,
Fig.
4–18c,
counteracted
force
F
that
is
required
4
Equilibrium.
The
free-body
diagram
of
the
bar
is
shown
in
Equilibrium.
The
free-body
diagram
of
the
bar
is
in
that
occurs,
Fig.
4–18c,
is
counteracted
by
the
force
c
c
dA>B = 02 ft
1+ 2
=+ dT©F
- yd=
FA 4–18c,
= FB =is F
! occurs, Fig.
F 0;
F
that
by
the
forcebut
is required
condition
A becomes
push
to2 ftits atoriginal
position.
The
compatibility
Fig.
4–18b.
Since
there
isload,
nocounteracted
external
load,
force
atFAthat
is
but position. T
B the bar dF back
Fig. 4–18b.
Since
there
is(a)
no0,external
the
atthe
Athe
equal
toforce
push
bar
to equal
its original
dF back
Compatibility.
Since
the
thermal
displacement
d
=
dis
c
A>B
T at A
+
©F
=
0;
F
=
F
=
F
to
push
the
bar
back
to
its
original
position.
The
compatibility
d
y
A
B
The
problem
is
statically
indeterminate sin
F
ondition at A becomes
opposite
to
the
force
at
B;
that
is,
opposite
to
the
force
at
B;
that
is,
(a)
condition
at
A
becomes
Çözüm:
Applying the thermal and load–displacement
relationships,
we by the force
that occurs,
Fig. 4–18c,
is counteracted
thatforce
is required
The problem
is statically
indeterminate
sinceFthis
cannot beF
determined from equilibrium.
dA>B = 0 = condition
1+ c2
dT - dF at A becomes
have
Thus, f
to
push
the
bar
back
to
its
original
position.
The
compatibility
d
determined
from
equilibrium.
(a)
F
B
B
(b)
statically+indeterminate
since this force cannot=beF
dA>B = The
+ c2
0 = problem
dT - d+Fcis©F
0;! c ©Fy = 0; F ! FA = FB = FF!A1 +=cF
dA>B = 0 = dT - dF
2B
yA=becomes
4
condition
at
F
Düşey
kuvvet
dengesi:
c
dA>Bwe= Compatibility.
1+ 2
0 = dT - dF
Since d
= 0, the therma
determined
from FL
equilibrium.
Applying
the thermal
and load–displacement
F
Compatibility.
Since dA>B = relationships,
displacement
0, the thermal
dT at A d A>B
0 = a¢TL
T
that
occurs,
Fig.
4–18c,
is
counteracted
by
the
F the thermal
Applying
relationships,
we
(a)
(a)haveand load–displacement
AE
The
problem
indeterminate
this be
forceand
cannot
be
occurs,
Fig. 4–18c,
counteracted
by the
force
F force
that
iscannot
required
problem
statically
since
Applying
the
thermal
load–displacement
dis
1 + cthat
2 The
= indeterminate
0is=statically
ddisplacement
A>Bthermal
T - dF and load–displacement
Compatibility.
Since
A since
=is Applying
0,
dthis
the
thermal
we
T at
Uygunluk
denklemi:
! dA>B
! the
to
push
the bar drelationships,
ave
F back to its original pos
to
push
the
bar
back
to
its
original
position.
The
compatibility
d
determined
from
equilibrium.
determined
from
equilibrium.
have
F
FL by the force F that is required
Thus, from the data on the
back
thatinside
occurs,
Fig.cover,
4–18c,
is
counteracted
d
have
condition at A becomes F
0 =ata¢TL
condition
becomes
F
F
Applying
theAthermal
and load–displacement
relationships, we
FL
AE
push
the
bar
to
its
original
compatibility
dF back
FL
F = a¢TAE 0 = to
Compatibility.
Since
thermal
displacement
dA>B position.
= Since
0, thedThe
dT at A 0 =da¢TL
Compatibility.
the
thermal
=
0,
a¢TL
FL displacement
A>B
T at A have
Sinc
AE
condition
at
A
becomes
0
=
a¢TL
!
c
AE
that
occurs,
Fig.
4–18c,
is
counteracted
by
the
force
F
that
is
required
that
occurs,
Fig.
4–18c,
is
counteracted
by
the
force
F
that
is
required
d
1+
2
=
0
= daverag
-6 from the data on
2
3
2 = 0 = d - d
Thus,
cover,
c 2 inside
A>B
T - d
1 + the
AE
)
T
F
2>°F]1120°FF - 60°F210.5
in.2back
[29110
2 kip>in
]
= [6.60110
F dA>B
FLtoposition.
to itsdForiginal
The position.
compatibility
dF back
to push
the bar
back
itsThus,
original
The compatibility
hus, from the data on the inside back cover,to push the bar
fromcover,
the data on the inside back cover,
0=
=d a¢TL
- on the inside
F =1+a¢TAE
c 2 (b)
(b)
Thus,
from
the
data
back
d
=
0
d
=
2.871
kip
A>B
T
F
condition
at
A
becomes
condition
A becomes
AE
Applying thewe
thermal and load–displace
dT
Applying the
thermalat and
load–displacement
relationships,
F = a¢TAEF
-6
2
3
F = a¢TAE
2>°F]1120°F
- within
60°F210.5
in.2
[29110
2 kip>in2]
= [6.60110
have
have
F
=
a¢TAE
Since F also
represents
the
internal
axial
force
the
bar,
the
dT back cover,
Thus,
from
the dataand
on the
-6
2 inside
d the
(b)
Applying
we
c thermal
1+in.2
22[2911031+
= 0-6= drelationships,
- d=
c 2load–displacement
2>°F]1120°F
- stress
60°F210.5
2 kip>in
] dA>B
dA>B
-6 2
2
T
F 0 = dT - dF
dF= [6.60110
average
normal
compressive
isT thus
2
- 60°F210.5
in.2
= [6.60110
=
2.871
kip
FL
2>°F]1120°F
- 60°F210.5
in.22>°F]1120°F
[2911032 kip>in
]
=
[6.60110
FL[2
! have
F = a¢TAE
0
=
a¢TL
0
=
a¢TL
= 2.871 kip dT
dF load–displacement
AE the
dF Applying
= 2.871 kip we
AE
theApplying
thermal
and
relationships,
also represents
the internal
axial
force
within
(c)
the
thermal
and
= 2.871
kip the bar,
NOTE
F Since F F
F FL
2.871 kip = [6.60110-62>°F]1120°F
- 60°F210.5
in.22load–displacement
[2911032 kip>in2] relationships, we
have
Since F also represents
the
internal
axial
force
within
the
bar,
the
average
normal
compressive
stress
is
thus
0
=
a¢TL
s =
Ans. back cover, Since F also represents the internal axial force
=
= 11.5the
ksihave
in tem
data on the inside
w
Thus, axial
from force
the data
on the inside
also
within
bar, theback cover,
A
AErepresents the internal
(b) dF
10.5 Thus,
in.22 =from
Fig.
4–18
verage normal
compressive
2.871 kip Since F (b)
F
! stress is thus
indete
average
normal
compressive
stress
is
thus
FL
average
normal
compressive
stress
is
thus
FL
F =
0 = a¢TL - 0 = a¢TL - F = a¢TAE
Thus, fromSince
the data
ona¢TAE
the
inside
cover,
2.871
kip back
dT
(b)
FFalso
represents
internal
within theAE
bar, the
AE
s
=
Ans.
= apparent
= the
11.5
ksi dT axial force
NOTE: From the
magnitude
of
F,
it
should
be
that
changes
-6
2
3
2
F F
2
2.871Fkip
2>°F]1120°F
60°F210.5
in.2
[29110
2=kip>in
] -62>°F]1120°F
=
[6.60110
A
average
normal
compressive
stress
is
thus
10.5
in.2
2.871 kip- 60°F210.5
[6.60110
F
=2 a¢TAE
2.871
kip
s = can =cause F large
Ans.
= 11.5
ksi
F
in temperature
reaction
forces
in
statically
Thus,
from
the
data
on
the
inside
back
cover,
s
=
dT
Thus, from the data on s
the= inside
Ans. = 11.5 ksi
= back cover,
= 11.5 ksi =
(b) AdF 10.5(b)
in.2
2
A
indeterminate members.
-6= 2.871 kip
2
A 32 kip>in
10.5 in.22
dF 2[29110
10.5 in.2
=
2.871
kip
2>°F]1120°F
60°F210.5
in.2
]
= [6.60110F
2.871 kipthat changes
NOTE:
From the magnitude=ofa¢TAE
F, it should
apparent
FF
=be
a¢TAE
dT
s =
=the internal
= axial
11.5 ksi
Since F also represents
force within theAns.
bar, the
d
d
T
2in statically 2 Since3 F also 2represents the internal axial f
F
OTE: From the magnitude
of F, it (c)
should
because
apparent
thatreaction
changes
-6 A
in temperature
forces
= can
2.871
kip
10.5
in.2
-6-is
2magnitude
3 that
(c)
2>°F]1120°F
60°F210.5
in.2
[29110
2
kip>in
]
= large
[6.60110
NOTE:
From
the
ofchanges
F,2]it should be appa
NOTE:
From
the
magnitude
of
F,
it
should
be
average normal
compressive
stress
thus
4–18
[29110
2 kip>in
= [6.60110 2>°F]1120°F - 60°F210.5 in.2apparent
temperature can indeterminate
cause large members.
reaction forces in statically
average
normal
compressive
stress
is thus for
in
temperature
can
cause
large
reaction
forces
in
statically
in
temperature
can
cause
large
reaction
dF Since F also
thekip
internal
axial force within the bar, the
dF represents
= 2.871
determinate members.
Fig.
4–18
Fig.
4–18
=
2.871
kip
(c)
indeterminate
NOTE: compressive
From the magnitude
it should
that changes
indeterminate
members.
average normal
stress is thus
2.871
kip be apparent
Fof F, members.
s
=
=
=
11.5
ksi
Since F alsocan
represents
the
internal
axial
force
within
the
bar,Ans.
the
F the2.871 kip
in temperature
cause
large
reaction
forces
in
statically
2 the internal axial force within the bar,
Since F also
A represents
Fig. 4–18
10.5 in.2
s =
=
= 1
average normal
compressive
stress
is
thus
indeterminate
members.
normal
A
10.5 in.22
2.871
kip compressive stress is thus
Faverage
s =
Ans.
=
= 11.5 ksi
2
(c)
10.5 in.2F
NOTE: FromAthe magnitude
of2.871
F, it should
apparent that changes
kipF be
2.871
kip in From
= (c) = large
= 11.5
ksi
NOTE:
the magnitude
of F, it should b
in temperature cans cause
forces
statically
2 =
=reaction
Ans.
= 11.5
ksiAns.
A
10.5sin.2
Fig. 4–18
2
A
10.5in
in.2temperature can cause large reactio
indeterminate
members.
(c)
NOTE: From the magnitude of F, itFig.
should
4–18be apparent that changes
indeterminate members.
in
temperature
can
cause
large
reaction
forces in statically
(c)
NOTE: From the magnitude of F, it should be apparent that changes
Fig. 4–18
(c)
indeterminate members. NOTE: From the magnitude of F, it should be apparent that changes
in temperature can cause large reaction forces in statically
in temperature can cause large reaction forces in statically
Fig. 4–18
Fig. 4–18 indeterminate members.
indeterminate members.
0.5 in.
1+ c 2
1+ c d2 A>B = 0 = dT - dF dA>B = 0 = dT - dF
PAGE !58
4.6
THERMAL STRESS
4.6
153
THERMAL STRESS
153
AMPLE 4.11
EXAMPLE 4.11
!
300 mm
300 mm 150 kN/m
e rigid beam shown
in Fig. 4–19a is fixed to the top of the three
mm
The rigid
beam
in Fig.ve4–19a
is fixed
to the
top
three
sts made of A-36
steel
and
2014-T6
aluminum.
The
posts
each
have
4.6of Tthe
HERMAL
STRESS
153 300 mm 150 kN/m
Şekildeki
rijit
kirişshown
A36
çeliği
aluminyumdan
yapılmış
direklerin
üstüne300
mesnetlidir.
4.6
THERMAL STRESS
4.6 153
THERMAL S
made
A-36
and 2014-T6
ength of 250 mmposts
when
no of
load
is steel
applied
to the aluminum.
beam, andThe
theposts each have
4.6
T
HERMAL
S
TRESS
153
o
length
of 250Cmm
when
no load
is applied
the beam,
thedir. Rijit kirişe 150 kN/m
T1Determine
=20
başlangıç
uzunlukları
250and
mm
mperature is T1Sıcaklık
the force
bytoeach
=a 20°C.
4.6
THERMAL
STRESS de dikmelerin
153 supported
60 mm
4.6 THERMAL
temperature is T1 = 20°C. Determine the force supported by each
AMPLE
4.11
mm STRESS
4.6 T250
HERMAL
153
60 mm
st if the bar is subjected
to a uniform
distributed
load
of4.11
150
kN>m
o ye çıkarılması halinde
EXAMPLE
4.11
250
mm
4.6
T
HERMAL
S
TRESS
1
5
3
üniform
yayılı
yük
uygulanması
ve
sıcalığın
T
=80
C
dikme
EXAMPLE
40
mm
40
mm
2
post if the bar is subjected to a uniform distributed load of 150 kN>m
40 mm
40 mm
d the temperatureand
is raised
to T2 = 80°C.
the temperature
is raised to T = 80°C.
300 mm
300 mm 150 kN/m o
e rigid beam shown in Fig. 4–19a is fixed to the top 2of the three
kuvvetlerini
Est=200
GPa,
Ealis=73.1
ve αal=23µ/C
300 mm
st=12µ/C
The rigidbulunuz.
beam shown
in Fig.
4–19a
fixed GPa,
to theαtop
of the othree
EXAMPLE
4.11
Theposts
rigid each
beam
shown4.6in Fig.
4–19a
is fixed
to the1 5top
of the three 150 kN/m300 mm 300
Steel
Steel
Aluminum
sts made of A-36 steel
and
2014-T6 aluminum.
The
have
THERMAL
STRESS
OLUTION
EXAMPLE
4.11
Steel 3Aluminum
Steel
SOLUTION
posts
made
of
A-36
steel
and
2014-T6
aluminum.
The
posts
each
have
EXAMPLE
4.11
posts
made
of
A-36
steel
and
2014-T6
aluminum.
The
posts
each have
ength
of 250ismm
when
no
isofapplied
to the beam, 300
andmm
the 300 mm 150 kN/m
Fig. 4–19a
fixed
tomm
theload
top
the
three
300
300
mm
300 mm
3
quilibrium.
diagram
of
the
beam
is
shown
in
Fig.4–19b.
the three The free-body
The
rigid
beam
shown
in
Fig.
4–19a
is
fixed
to
the
top
of
the
three
Equilibrium.
Thethe
free-body
diagram
ofis
the
beam
is
shown
inload
Fig.4–19b.
a length
of 250
mm
when
noFig.
load
applied
to
the
the
150shown
kN/m
(a)to the beam,
asupported
length
of
250
when
nobeam,
isthe
applied
and the
(a)mm
300
300 mm 150 kN/m
mperature
is Taluminum.
Determine
force
by
each
rigid
beam
in
4–19a
ismm
fixed
toand
the
top
ofand
three
and
2014-T6
The
posts
each
have
1 = 20°C. The
60
mm
posts
made
of
A-36
steel
2014-T6
aluminum.
The
posts
each
have
300
mm
300
mm
each
have
TheMoment
rigid
beam
shown
in
4–19a
is fixed
to
the
top
ofDetermine
the
three
oment
equilibrium
about
the
beam’s
requires
the
forces
the
equilibrium
about
the
beam’s
center
requires
the
forces
in the
temperature
isof
Determine
the
supported
by
each
TA-36
= Fig.
20°C.
150 kN/m
temperature
is
thehave
force250
supported
T
= in
20°C.
mm
1center
1 force
PLE
4.11
60 mm by each
st if
the
bar is
is subjected
to
athe
uniform
distributed
load
150
kN>m
posts
made
steel
and of
2014-T6
aluminum.
The
posts
each
n
no
load
applied
to
beam,
and
the
apost
length
ofbar
250is
mm
no
load
is
applied
to the
beam,
and
the 250 mm60 mm
40of
mm
40
mm
posts
made
of
A-36
steel
andon
2014-T6
aluminum.
Thewhen
posts
each
have
m,
and
the
steel
posts
to
be
equal.
Summing
forces
on
the
free-body
diagram,
el
posts
to
be
equal.
Summing
forces
the
free-body
diagram,
post
if
the
bar
is
subjected
to
a
uniform
distributed
load
150
kN>m
if
the
subjected
to
a
uniform
distributed
load
of
150
kN>m
d
the
temperature
is
raised
to
=
80°C.
T
a
length
of
250
mm
when
no
load
is
applied
to
the
beam,
and
the
90
kN
40
mm
40
mm
kN
2
40 mm
40
Determine
force
supported
by iseach
temperature
Determine
the force90supported
by each
Tto
20°C.
ainlength
of
250
mm when
no
load
isTtemperature
applied
beam,
the80°C.
1 =the
we
have
d have
by
eachshownthe
60 mm
mm
the
temperature
raised
tothe
=60is
80°C.
300 mm
eC.
gid
beam
Fig.
4–19a
is fixed
top
ofthe
and
raised
toand
T300
2 three
60and
mm
2 =mm
temperature
istoTthe
Determine
theisforce
supported
by
each
20°C.
150
kN/m load 60
4
1 =mm
250
mm
mm
4
Steel
Steel
post
if
the
bar
is
subjected
to
a
uniform
distributed
of
150
kN>m
Aluminum
d
to
a
uniform
distributed
load
of
150
kN>m
250
is0;T1 = 20°C.
Determine
the
force
by each (1)
3
OLUTION
kN>m
c2014-T6
40 mm
3 st
made
of =
A-36 steeltemperature
and
The
posts
each
have
mm
250 mm
+ 2F
©F
+ =
Fal0to
- 90110
2 Ndistributed
=supported
0 (1)
40 mm
mm load of 150
y+=if
post
the
is 2F
subjected
uniform
kN>m 60Steel
c150
Steel
Aluminum 250
mm
40
mm
©Fto
Faluminum.
-bar
90110
2N
Aluminum
mm Steel
the atemperature
is 40
raised
to kN>m
T2 = 80°C.
y T 0;= 80°C.
st
SOLUTION
40 mm
40 mm
post40
ifSOLUTION
the
baris
isal
subjected
to
aand
uniform
distributed
load
of 150
2mm
hsed
of 250
when
no
load
applied
to
the
beam,
and
the
uilibrium.
The
free-body
diagram
of
the
beam
is
shown
in
Fig.4–19b.
40 mm
40 mm
(a) the
and the temperature
is raised
to T2 =and
80°C.
Compatibility.
Due
to
load,
geometry,
material
symmetry,
and
the
temperature
is
raised
to
=
80°C.
T
Equilibrium.
The
free-body
diagram
of in
the
beam
is
shown
in Fig.4–19b.
Steel
2theby
Alumin
Equilibrium.
The
free-body
diagram
of the beam is shown in Fig.4–19b.
ature is
Determine
thecenter
force
supported
each
T1 = 20°C.
(a)
SOLUTION
oment
equilibrium
about
the
beam’s
requires
forces
the
ompatibility.
Due
toAluminum
load,
geometry,
and
material
symmetry,
the
(a)
Steel
Aluminum
60 mmSteel
Steel
top
of
eachequilibrium
post
isSteel
displaced
by
an
equal
amount.
Hence,
Steel
Steel
Aluminum
!
SOLUTION
Moment
about
the
beam’s
center
requires
the
forces
in
the
250
mm Aluminum
Moment
equilibrium
about
the
beam’s
center
requires
the
forces
in
the
the
bar
is
subjected
to
a
uniform
distributed
load
of
150
kN>m
Steel
Steel
el
posts
to
be
equal.
Summing
forces
on
the
free-body
diagram,
p of each post isSOLUTION
displaced by an equal amount.
Hence, The free-body
Equilibrium.
diagram
of
the
beam
is
shown
in
Fig.4–19b.
40 mm
40 mm90 kN
(a)
toinbe
equal.
Summing
forces
the
free-body
diagram,
ody
diagram ofisthe
beam
is
shown
Fig.4–19b.
(2) on requires
1+steel
T2Equilibrium.
=equilibrium
dalto be
steeldposts
equal.
Summing
forces
the free-body
diagram,
eFig.4–19b.
temperature
raised
to
= 80°C.
T2posts
stdiagram
The
free-body
ofon
the
beam
isthe
shown
in
Fig.4–19b.
have
(a)
(a)
Moment
about
beam’s
center
the forces
inkN
the Fst
90
Fst
Fal4 (a)
90
Çözüm:
Equilibrium.
The
free-body
diagram
of
the
beam
is
shown
in
Fig.4–19b.
(2)
T
2
d
=
d
(a)
stforces
al
we
have
the
ut
theinbeam’s
center
have
3 of
Moment
equilibrium
about
the
beam’s
requires
the
forces
inon
thethe
The
final
position
the
top
of
each
post
iscenter
equal
toSumming
its
displacement
F
Fal free-body
Fstdiagram,
steel
posts
to be
equal.
forces
crces
st
Steel
Steel
Aluminum
(1)
©F
2Frequires
Fal -the
90110
2 Nin
=the
0we
4
Moment
about
the
beam’s
center
requires
the
forces
in
the
TION
y = 0;
st +equilibrium
3
The
final position
of
the
top
post
isKuvvet
equal
todengesi:
displacement
diagram,
(b)
to diagram,
beSumming
equal.
Summing
on
the
csteel
Summing
forcessteel
on
the
free-body
caused
by
the
temperature
increase,
plus
its
causeddiagram,
by32 N = 0
+
©F
=of
0;each
2F
+chave
Fforces
90110
2displacement
Nfree-body
=2F
0 +free-body
we
(1)
©F
=its
0;
Fdiagram,
y posts
st+
aly al - 90110(1)
posts
to
be
equal.
onforces
the
90
kN
90 kN
90st kN
rium.
The
free-body
diagram
of
the
beam
is
shown
in
Fig.4–19b.
ompatibility.
Due
to
load,
geometry,
and
material
symmetry,
the
90
kN
(b)
(a)
we
have
used by the temperature
increase,
plus
its
displacement
caused
by
3
the
internal
axial
compressive
force,
Fig.
4–19c.
Thus,
for
the
steel
and
c
(1)
+
©F
=
0;
2F
+
F
90110
2
N
=
0
we have
4
y
st
al
4
Compatibility.
Due
to
load,
geometry,
and
material
symmetry,
the
nt
equilibrium
about
the
beam’s
center
requires
the
forces
in
the
4
! Hence,
Compatibility.
Due
to load, geometry, and material
symmetry, the
each post
is displaced
by anpost,
equal
amount.
4
3
we
have
ep of
internal
4–19c.
for
steel
3 aluminum
(dal)T
(1)
+=0c0;
©Fforce,
=on
0;Fig.
2FThus,
+90110
Fal
90110
2 Nand
= 0
3- the
y
st
(1)
+
F
- axial
90110compressive
2c N
=
(
d
)
(1)
+Summing
©F
2F
+
F
2
N
=
0
osts
toal(1)
be
equal.
forces
the
free-body
diagram,
top
equal
amount.
st T
top
of an
each
post
isDue
displaced
by geometry,
an equal
Hence,symmetry,
y of each post is displaced
st
al by
Compatibility.
toHence,
load,
and material
the
Uygunluk
denklemi:
T2
dst = dal
90 kN amount.
uminum
post, we have
1+ T2
dst = -1dst2T + 1dst(2)
2F
(d )
Fstby an
Fal amount.
F
e
Initial
Position
Compatibility.
Due
to
load,
geometry,
and
material
symmetry,
the
stHence, al T
(
d
)
top
of
each
post
is
displaced
equal
metry,
the
st
T
(2)
1+
T2
d
=
d
(2) (dal)F
dst =
Due
tois load,
geometry,
thedal
st and
al material symmetry,
4
load,
geometry,
and
symmetry,
the
The
final
positionCompatibility.
of thematerial
top
of=
each
equal
toT2
its ! by
displacement
! 21+
Fst
Fal
Fst Fst
3 post
Fal
T
2
d
1d
2
+
1d
top
of
each
post
is
displaced
an
equal
amount.
Hence,
st
st
T
st
F
1+
T2
d
=
-1d
2
+
1d
2
(1)
=
0;
2F
+
F
90110
2
N
=
0
al
al
T
al
F
1
+
T
2
d
=
d
y by an equal top
st of each
al
post
displaced
by an
equal
amount.
Final Position
st Position
al is equal
(b)
(dst)(Fdal)F (2)
The
final isposition
the
top
offinal
each
post
isHence,
equal
itsInitial
displacement
ed
amount.
Hence,
The
position
of thetotop
of
each
post
used
by the temperature
increase,
plus
its of
displacement
caused
by
dst !to
dalits displacement
Fst
F
Sıcaklık
artışı
dikmelerde
1d+ force,
T 2Eq. 1d
dthe
= temperature
dand
2Fig.
gives
(b) by
st
al
position
the top
of each
post
is(2)
equal to its
displacement
(b)
increase,
plus
its of
displacement
caused
by
Tinternal
2 d(2)axial
24–19c.
+ 1d
2symmetry,
(2)
TApplying
2caused
dThe
=final
dby
eatibility.
compressive
Thus,
for
the
Due
to1 +load,
geometry,
and
the
increase,
plus
its
displacement
caused
al =by-the
altemperature
Tmaterial
alcaused
F
st
al steel
F
F
F
Final
Position
st
al
st
(2)
=
d
(dstst)F
Fal
Fst uzama deplasmanı
Fal
st
al Fst
d
st
st !plus
(b
by
temperature
increase,
its Fdisplacement
caused Fby
The
final
position
the
top
each
is
equal
toforce,
itsFddisplacement
FThus,
the
internal
axial
compressive
Fig.
for
the
2top
+caused
1deach
2force,
=ofFthe
-1d
2post
+ to
1d
each postpost,
is displaced
an
equal
amount.
uminum
we
have
the
axial
Fig.aland
4–19c.
st
al
ststeel
stHence,
T of
stinternal
F post
al4–19c.
Tcompressive
al2F
(Thus,
dal)T for the steel and
Theby
final
position
of-1d
the
of
is
equal
its
displacement
placement
pplying
Eq.
2
gives
(
d
)
st
T
oluşturacaktır.
top of each post is equal
to(b)
itspost,
displacement
(b)
the
internal
axial
compressive
force,
Fig.
4–19c.
Thus,
for
the
steel
and
caused
by
the
temperature
increase,
plus
its
displacement
caused
by
aluminum
we
have
aluminum
we haveon the
(b)
(dal)T
caused
the
temperature
pluspost,
its
displacement
caused
Using
4–4
the material
properties
insideby
back
(2)
d=stEqs.
= d4–2
caused
by
T2
d2stby
2Tand
+ 1d
2and
st
F increase,
(dst)T
(dst)T
(b)
(dal)F
-its
1dstdisplacement
+ -1d
1dstalst2aşağı
=
1d
+ 1dalkabul
2force,
Initial
Position
Fst Thus,
Fst
aluminum
post,Fig.
we 4–19c.
have
Deplasmanlar
doğru
pozitif
edildiğinden
re increase, plusthe
caused
Tthe
F
al2by
T
F4–19c.
internal
axial
compressive
forFalthe
steel and
internal
axial
compressive
force,
Fig.
Thus,
for
the
steel
and
cover,
we
get
(c)
(dst)T
1+
T2
d
=
-1d
2
+
1d
2
steelposition
and of the top of each post is equal to its
1+
T2
dst = -1dst2T + 1dst2FInitial Position
stdisplacement
st T
st F
final
(dPosition
Initial
T2 Eqs.
dand
=post,
-1d
2the
+ steel
1dwe
2Fhave
al)F
aluminum
sive
force,
Fig.
4–19c.
Thus,
for
and
al
al
Tpost,
alproperties
sıcaklık
deformasyonunun
negatif
işaret
1önünde
+ T 2 on
ddst !=d - 1d
2
+
1d
2
(
d
)
sing
4–2
and
4–4
the
material
the
inside
back
Final
Position
st
T
st
F
aluminum
we
have
al
T
(
d
)
(dst()dTal)T
Initial Position
by the temperature increase,
plus its displacement
caused2 by+ 1d 2
al (b)
(dal)T
Fst 10.250
m2 st F
1+ Dikmelerdeki
T2
d1+
=
T2 -1d
2F (c) (dst)T Fig.
al d
al1d
T 2 +ve
al F
st2)T
pplying
Eq.
2 gives
FinaldPosition
deplasmanlar
sıcaklık
yayılı
1 +(dT-6
= +2 dalst = -1d
ver, we
getcompressive
(dal)alT22T ++ 1d
(dst)4–19
2>°C]180°C
m2
stst2+
TF (1d
(dst)F
dst ! dPosition
F
1vardır.
+ T-[12110
2 force,
dst- =20°C210.250
1d
ernal
axial
Fig.
4–19c. Thus,
for
the
and
!ddalal)F
1+ 1d
T 2stst2steel
1dal2alPosition
2F Initial
al
dstst
)T Fdal =
T
2 - 1d
al 9T Initial
(dal)F stdst(!
(dst
p10.020 m2
[200110
2 N>m
]
dal
Applying
Eq.
2 gives
(+
dal)1d
Initialyük
Position
Eq.
2 gives
FApplying
etkisinden
kaynaklanan
iki
farklı
deplasmanın
toplamı
olarak
yazılabilir.
+1 1d
2
=
-1d
2
2
dum
- 1d
1d
post,
we
stst22T
st
F
al
T
al
F
st =
st2have
T +-1d
F
(
d
)
+ T2
dInitial
= +
-Eq.
1d
22Tgives
+ m2
1dal2F
Applying
2
F
10.250
al T
al
al
(
d
)
Position
st
(
d
)
1
+
T
2
d
=
1d
2
1d
Final Position
al m2
F
(dstFinal
)F Position
al
al T
al F
-6
F
(dstd)alF
-1d
1d
= -1d
1deşit
2 1d
al10.250
!Fig.
dal al4–19
2stFT = -1d
1d
2Fdst !
Bu
deplasman
birbirine
olduğundan,
[12110
2>°C]180°C
- Applying
20°C210.250
m2
+2iki
st
T +
st2Finside
al2-1d
T +st2
al+
ing Eqs. 4–2
and d4–4
the
properties
on
the
back
-6 1d
Position
al22Tdst++
dst dand
=
1dst(2material
d )+
2 m2
2 st
st2
F
2TTN>m
+F1d
Eq.
2Final
gives
=alF -[23110
2>°C]180°C
-p10.020
20°C210.250
+1d9st
st !
st22F = - 1d
al T (dal2)F1dal2F
dal = - 1dal2T + Applying
1d
Initial
Position
Eq.Tst2 Fgives
m2
[200110
2
]
al2
9
Position
(p10.030
d ) on m2
ver, we get
(c)
[73.1110
2 N>m ]
Using Eqs. 4–2 and 4–4 and the material properties
theFinal
inside
back
300 mm
dst !4–2
dal and 4–4st and
F
UsingEqs.
Eqs.
material properties
on
4–2
and
the
properties
onthe
theinside
insideback
back
- 1d
1d
=
- 1d
2 Fand
+ 1dthe
dal = - 1dal2T + 1d
stst22TF +
F
al
al2Fmaterial
al2stF2T +
- 1d
1dUsing
= we
- st1d2get
1d4–4
Final Position
al2F
T al+
al3d2T
(dst)F
!
10.250
m2
cover,
we
get
(c)
!
dal
cover,
st
F
10.250
m2
-6
cover,
we
get
=
1.216F
165.9110
2
F
(3)
st
st
al
-6
ng-Eq.
22>°C]180°C
gives2>°C]180°C
Fig. 4–19
Using
Eqs.
4–2+and
andm2
4–4+
and the material
properties
on the
inside
back
=[12110
[23110
-4–2
20°C210.250
Using
Eqs.
and m2
4–4
the
material
properties
on
the9 inside
back
- 20°C210.250
2 m2
+ 1d
2
9 2[73.1110
2
st2F = - 1dal2T + 1dal2F
FFstst10.250
m2
p10.030
m2
2FN>m
]
nside
back
st10.250
p10.020
m2
[200110
2
N>m
]
10.250
m24–19
cover,
we
get
(c)
-6
-6
we
get
(c) Fig.
- 1dst2cover,
+
1d
2
=
1d
2
+
1d
2
-6 2>°C]180°C
T
st be
F consistent,
al T all numerical
al -F 20°C210.250
-[12110
2>°C]180°C
m2 been
+
- 20°C210.250
20°C210.250
m2
To
data
has
expressed
--[12110
[12110
2>°C]180°C
m2
+
2 in terms
9 +
2
2
99
2
d the material properties
on(c)
the
inside back
2
3 F 10.250 m2 p10.020 m2F[200110
2
N>m
]
p10.020
m2
[200110
2
N>m
p10.020 m2 [200110 2 N>m]2]
m2
= 1.216F
- 165.9110
2 alCelsius.
Fnewtons,
(3)
Eqs. 4–2 and
the
properties
on the
inside
back Solving
meters,
and
degrees
Eqs.
and
3
st110.250
F
10.250
m2
stmaterial
al
-6 4–4 andof
st
-6
Fig.
4–19
(c) 2
=50-[23110
2>°C]180°C
- -6
m2
+
Fig. 4–19
-20°C210.250
[12110 2>°C]180°C
- 20°C210.250
- [12110
2>°C]180°C
- 20°C210.250
m22[73.1110
+ m292+N>m
2(c)
F
m2
we m2
get
FFalal10.250
simultaneously
al910.250
2]
2 92 N>m2]
10.250m2
m2
p10.030 m2
p10.020
m2
[200110
Fig. 4–19
-6yields
-6p10.020
m2
[200110
2
N>m
]
-6
=
-[23110
2>°C]180°C
20°C210.250
m2
+
=
-[23110
2>°C]180°C
20°C210.250
m2
+
=
[23110
2>°C]180°C
20°C210.250
m2
+
9 consistent,
2
To
be
all
numerical
data
has
been
expressed
in
terms
F
10.250
m2
2
9
2
2
99
22
2
st
00110
2
N>m
]
F
10.250
m2
p10.030
m2
[73.1110
2
N>m
]
3
F
=
-16.4
kN
F
=
123
kN
Ans.
2 2N>m
]]
Fig. F
4–19
p10.030m2
m2[73.1110
[73.1110
N>m
st
st
al
Fm2
10.250 m2 p10.030
-6
= 1.216F
165.9110
2
Fstand
(3)
al4–19
10.250
Fig.
20°C210.250
m2- +
al2 al
-6
10
2>°C]180°C
20°C210.250
m2
+
newtons,
meters,
degrees
Celsius.
Solving
Eqs.
1
and
3
-6
9
= -m2
[23110
2>°C]180°C
20°C210.250
2 29 m2 + 2 m2 3+
= The
[23110
2>°C]180°C
p10.020
[200110
2F20°C210.250
N>m
]
2
9
2
3
3
negative
value
for-F
indicates
that
this
force
acts
opposite
to
10.250 m2
p10.020
m2
[200110
2
N>m
]
2
9
2
st = 1.216Fal - 165.9110
(3) ] 2 N>m
1.216F
-[73.1110
165.9110
Fst2 =
(3)
p10.030
m2
1.216F
22 ]
(3)
st
p10.030
m2
[73.1110
2 165.9110
N>m
al al
multaneously
yields
To
numerical
data
has
been
in the
terms
shown in
Fig.10.250
4–19b.
In expressed
other
words,
steel posts are in tension
2 be consistent,
9
2allthat
F
m2
F
10.250
m2
[73.1110
2
N>m
]
3
al
al
-6
=kN
1.216F
- 3165.9110
2
(3)
newtons,
meters,m2
and
degrees
Celsius.
Eqs.
1 aland
32 all
=Solving
- consistent,
165.9110
(3)terms
st
the
aluminum
post
is=1.216F
inF
compression.
-be
16.4
kN
F
123
Ans.
23110
2>°C]180°C
-F
20°C210.250
m22 F
+
To
be
numerical
data
has been
been expressed
expressedininterms
terms
20°C210.250
+=
To
consistent,
all
numerical
data
expressed
in
st al
al
To
all
data
has
st
2be2 consistent,
9has been
2 numerical
9 m2
!
p10.030
[73.1110
2
N>m
]
m2
[73.1110
2
N>m
]
multaneously
yields of p10.030
of
newtons,
meters,
and
degrees
Celsius.
Solving
Eqs.
1
and
(3)
newtons,
meters,
and
degrees
Celsius.
Solving
Eqs.
1
and
3
ofnumerical
newtons,
meters,
andto degrees
Celsius. Solving Eqs. 1 and 3 3
e negative valueTo
forbeFconsistent,
indicates
this
force
opposite
stTo
be consistent,
all
has
been
expressed
allthat
numerical
data acts
hasdata
been
expressed
in terms in terms
3
simultaneously
yields
3
(3)Ans.
simultaneously
yields
F=st 1.216F
=simultaneously
-16.4
kN165.9110
Fyields
= 2123(3)
kN
st! 4–19b.
almeters,
= 1.216Fin
- F165.9110
2 alother
In
words,
steel
posts
are
in tension
of
newtons,
and degrees
Celsius.
Solving
Eqs. 31 and 3
newtons,
meters,
and the
degrees
Celsius.
Solving
Eqs. 1 and
datinshown
terms al Fig.of
F
=
16.4 kN
kN F
FAns.
Ans.
st kN
al = 123 kN
negative
valuesimultaneously
for
Fstissimultaneously
indicates
that this
force
acts kN
opposite
to123
F
=
-16.4
F
=
yields
F
=
-16.4
Ans.
dee1the
aluminum
post
in
compression.
st
al
yields
st
al = 123 kN
consistent,
and 3 all numerical data has been expressed in terms
umerical
has been
expressed
in terms
The
negative
value
for
F
indicates
that this
this force acts opposite to
at shown
indata
Fig.and
4–19b.
In other
words,
the
steel
posts
are
in tension
The
negative
value
indicates
force
acts
opposite
negative
value
for
FststkN
indicates
wtons,
meters,
degrees
Celsius.
Eqs.
116.4
3123
FFstThe
Fthis
123
Ans.force acts opposite to
st =
FstSolving
= for
- 16.4
kN
Fand
=that
kN=
Ans.that to
al
al kN
that
shown
in
Fig.
4–19b.
In
other
words,
the
steel posts are in tension
the
aluminum
post
is
in
compression.
ddneously
degrees
Celsius.
Solving
Eqs.
1
and
3
thatThe
shown
in Fig. 4–19b.
In
other
words,
the4–19b.
steel
posts
are
inopposite
tension to
yields The negative
that
shown
in
Fig.
Inopposite
other
posts are in tension
negative
for
that
this
actswords,
value forvalue
Fst indicates
that
this
force
acts
to the steel
st indicates
andFthe
aluminum
post
isforce
in compression.
Ans.
and
the
aluminum
post
is
in
compression.
and
the
aluminum
post
is
in
compression.
Fig.kN
otherthe
words,
steel
are in tension
Fst that
= -shown
16.4that
kNinshown
Fal 4–19b.
=in123
Ans.
Fig.
In4–19b.
other In
words,
steel the
posts
areposts
in tension
pposite
to
16.4
kN
F
=
123
kN
Ans.
and
the
aluminum
post
is
in
compression.
and
the
aluminum
post
is
in
compression.
al
gative value for F indicates that this force acts opposite to
in tension
st
own
in Fig. 4–19b.
other
words,
steel posts
that In
this
force
actsthe
opposite
to are in tension
st indicates
eInaluminum
post is
in steel
compression.
other words,
the
posts are in tension
in compression.
PAGE !59
(c)
FigF
A D 154
4 AXIAL
R
CHAPTER 4
L O A D 154
AXIAL LOAD
CHAPTER 4
154
154
AXIAL LOAD
CHAPTER 4 AXIAL LOAD
P LT ELRO4
4C154
APXTI EARL 4
L O ACADH154
HA
XAI A
AD AXIAL LOAD
EXAMPLE
4.12
C H A P T E R EXAMPLE
4
A X I A L L O A D 4.12
54
C
APTER 4
CH
HAPTER 4
A
AD
A XX II AA LL LL O
OAD
EXAMPLE
EXAMPLE 4.12
4.12
2
2
A 2014-T6 aluminum
tube having
area
of 600
is a cross-sectional
2
EXAMPLE
4.12a cross-sectional
A 2014-T6
aluminum
tubemm
having
area
of 600 mm2 is
EXAMPLE
4.12
2having a
12aluminum
PLE
4.12tube! having
T6
a cross-sectional
area
of 600amm
is
A
tu
600
mm2aluminum
A
2014-T6
aluminum
tube
cross-sectional
area of
is
600
mm
A
2014-T6
aluminum
tube
having
cross-sectional
area
of
is
A 2014-T6
2014-T6
aluminum
tu
used as a sleeve
for an A-36 steel boltused
having
a sleeve
cross-sectional
area
of bolt having a cross-sectional area of
as
a
for
an
A-36
steel
EXAMPLE
4.12
sleeve for an2A-36used
steelas
bolt
having
a
cross-sectional
area
of
used
as
a
sleeve
for
an
A
used
as
a
sleeve
for
an
A-36
steel
bolt
having
a
cross-sectional
area
of
a
sleeve
for
an
A-36
steel
bolt
having
a
cross-sectional
area
of
used
as
a
sleeve
for
an
A
2
2
2
2
2
400 mm , Fig. 4–20a.2When the
Talanı
= 2014-T6
15°C,
temperature
isFig.
thehaving
nut
2 is tüp,
2 of 600
1nut
22 is
600
mm
enkesit
bulunan
bir
aluminyum
enkesit
alanı
400
mm
400
mm
, tube
T
=
15°C,
4–20a.
When
the
temperature
the
nut
600
mm
A
tube
having
a
cross-sectional
area
of
mm
A
2014-T6
aluminum
tube
a
cross-sectional
area
is
2 aluminum
1
, Fig. 4–20a. When400
T
=
15°C,
the
temperature
is
the
600
mm
A
2014-T6
aluminum
tube
having
a
cross-sectional
area
of
is
600
mm
A
2014-T6
aluminum
having
a
cross-sectional
area
of
is
400
mm
,
Fig.
4–20a.
W
1
400
mm
,
T
=
15°C,
Fig.
4–20a.
When
the
temperature
is
the
nut
mm
,
T
=
15°C,
Fig.
4–20a.
When
the
temperature
is
the
nut
400
mm
,
Fig.
4–20a.
W
1
1
2
holds the assembly
in a such
snug that
position
suchasthe
that
the
axial
force
insteel
the
o de,
holds
assembly
in
athe
snug
position
such
that
the
axial
inT
the
used
a sleeve
for
an
having
aposition
cross-sectional
area
ofaxial
used
as
sleeve
for
an
steel
bolt
a cross-sectional
area
of
600
mm
2014-T6
aluminum
tube
having
abulonun
cross-sectional
area
ofhaving
isthe
e assembly in a snug
position
axial
force
in the
olan
çeliğinden
yapılmış
biraxial
üzerine
geçirilmiştir.
C
used
as
a assembly
sleeve
for
an
A-36
steel
bolt
having
aaA-36
cross-sectional
of
used as
aA
sleeve
for
an
A-36
steel
bolt
having
abolt
cross-sectional
area
offorce
1=15
holds
the
in
holds
assembly
inA-36
aarea
snug
such
that
force
in assembly
the
holds
the
inthe
aA36
snug
position
such
that
the
force
in
the
holds
the
assembly
in aa
2
2 =
T
80°C,
bolt
is
negligible.
If
the
temperature
increases
to
2
2
2
400
mm
,
T
=
15°C,
Fig.
4–20a.
When
the
temperature
is
the
nut
400
mm
,
T
=
15°C,
Fig.
4–20a.
When
the
temperature
is
the
nut
T
=
80°C,
bolt
is
negligible.
If
the
temperature
increases
to
used
as
a
sleeve
for
an
A-36
steel
bolt
having
a
cross-sectional
area
of
150
mm
T
=
80°C,
negligible.
If
the
temperature
increases
to
1
1
400
mm
,
T
=
15°C,
Fig.
4–20a.
When
the
temperature
is
the
nut
2
400
mm
,
T
=
15°C,
Fig.
4–20a.
When
the
temperature
is
the
nut
2
bolt
is
negligible.
150
mm
somun,
aksamda
bir
iç
kuvvet
oluşturmayacak
derecede
sıkıştırılmıştır.
1
T2 =is 80°C,
bolt is negligible.
If 1 the temperature 150
increases
tobolt
mm
bolt
isin negligible.
Ifsleeve.
the150
150 mm
negligible. If
If
mm
2 temperature increases to T2 = 80°C,
force
the bolt and
holds
the
assembly
in
a
snug
position
such
that
the
axial
force
in
the
400
mm
,
T
=
15°C,
Fig.
4–20a.
When
the
temperature
is
the
nut
holds
the
assembly
in
a
snug
position
such
that
the
axial
force
in
the
determine
the
force
in
the
bolt
and
sleeve.
ne thedetermine
force in thethe
bolt
and
sleeve.
o
1
holds
the
assembly
in
a
snug
position
such
that
the
axial
force
in
the
holds
the
assembly
in
a
snug
position
such
that
the
axial
force
in
the
determine
determine
the bulonda
force in the
andmeydana
sleeve.
determine the force
in the bolt
and C
sleeve.
Sıcaklık
T2=80
ye
çıkarılırsa
vebolt
tüpte
gelecek kuvveti
determine the
the force
force in
in tt
80°C,too T2 = 80°C,
is
negligible.
If negligible.
the
temperature
to T
150
mm is holds
themm
assembly
intemperature
a snug
position
that
the
is
temperature
increases
150
2in=the
TIf2 =the
80°C,
bolt is negligible.
If bolt
the
temperature
increases
to such
T2axial
= 80°C,
bolt
negligible.
If
thebolt
increases
toincreases
150 mm
150 mm
oforce
hesaplayınız.
E
=200
GPa,
E
=73.1
GPa,
α
ve
α
stthe
albolt
st=12µ/C
al=23µ/C
determine
force
in
the
and
sleeve.
T = 80°C,
bolt is negligible.determine
If the temperature
increases
150 mm
the force in the
bolt and to
sleeve.
2
the force inthe
the
bolt in
and
force
thesleeve.
bolt and sleeve.
ON SOLUTION determinedetermine
SOLUTION
SOLUTION
SOLUTION
SOLUTION
SOLUTION
determine
the
force
in
the
bolt
and
sleeve.
free-body
of a top
segment of the
um. Equilibrium.
The free-bodyThe
diagram
of adiagram
top Equilibrium.
segment
of the
The
free-body
diagram
of
a
top
segment
of
the
Equilibrium.
Equilibrium.
The The
free-body
diagram
of a top segment
of the diagram of a top segment
Equilibrium.
The fre
fre
The free-body
of the The
FEquilibrium.
assembly
is shown
in Fig.
4–20b.
forces
b andSOLUTION
s are produced
Fb andSOLUTION
Fs are Fproduced
y is shown
in Fig.
4–20b.
The
forces
SOLUTION
SOLUTION
Fb and
Fsforces
assembly
is shown
inFFig.
4–20b.
Thein
forces
are produced
assembly
is
shown
in
Fi
F
assembly
is
shown
in
Fig.
4–20b.
The
forces
and
are
produced
assembly
is
shown
in
F
F
F
assembly
is
shown
Fig.
4–20b.
The
and
are
produced
(a)
b
s
(a)
b
s
since
sleeve
has a higher
coefficient
of thermal
than thediagram of a top (a)segment of
(a)
SOLUTION
sleeve
has the
a higher
coefficient
of thermal
expansion
thanexpansion
the
Equilibrium.
The
free-body
the
Equilibrium.
The
free-body
diagram
ofthermal
athan
topthe
segment
of the
since
the
sleeve
has
a
higher
coefficient
of
thermal
expansion
Equilibrium.
The
free-body
diagram
of
a
top
segment
of
the
Equilibrium.
The
free-body
diagram
of
a
top
segment
of
the
since
the
sleeve
has
a
hig
since
the
sleeve
has
a
higher
coefficient
of
thermal
expansion
than
the
hi
since
the
sleeve
has
a
higher
coefficient
of
expansion
than
the
4therefore
4
bolt, and
therefore
sleeve
willwhen
expand
more
when
the
temperature
4
4
the sleeve
will the
expand
more
the
temperature
Fbwhen
F
assembly
is
shown
in
Fig.
4–20b.
The
forces
and
Equilibrium.
The
free-body
diagram
of
as are
top
segment
ofproduced
s are
Fbthe
assembly
shown
in
Fig.
4–20b.
The
forces
and
produced
F
Fs F
assembly
isassembly
shown
inis sleeve
Fig.
4–20b.
The
forces
and
are
produced
F
shown
in therefore
Fig.
4–20b.
The
forces
and
produced
bolt,
and
the
sleeve
will
expand
more
the
temperature
bolt,
and therefore the s
s are
and therefore
will
expand
more
when
the
temperature
b is
bthe
bolt,
and
therefore
sleeve
will
expand
more
whenFthe
temperature
a)
(a)
Itbolt,
is(a)
required
that (a)the
Çözüm:
sed. Itisisincreased.
required
that
since
the
sleeve
has
a
higher
coefficient
of
thermal
expansion
than
the
F
F
assembly
is
shown
in
Fig.
4–20b.
The
forces
and
are
produced
since
the
sleeve
has
a
higher
coefficient
of
thermal
expansion
than
the It is require
b
s
since
the
sleeve
has
a
higher
coefficient
of
thermal
expansion
than
the
since
the
sleeve
has
a
higher
coefficient
of
thermal
expansion
than
the
is
increased.
is
increased.
It
is
required
that
(a)
is
increased.
It
is
required
that
require
is
increased.
It
is
required
that
4
bolt,
and
therefore
the
sleeve
will
expand
more
when
the
temperature
since
the
sleeve
has
a
higher
coefficient
of
thermal
expansion
than
the
bolt,
and
therefore
the
sleeve
will
expand
more
when
the
temperature
bolt, and therefore
sleeve will
expandwill
more
whenmore
the temperature
bolt, andthe
therefore
the sleeve
expand
when the temperature
Fbolt,
F
increased.
It is
is
required
that
= 0; + c ©Fy = 0; is increased.
F = Fb increased.
(1)
s = is
b required
and
therefore
the
sleeve will
expand
more
when the temperature
It (1)
is required
that
It
+ c ©Fyy = (1)
0;
+ c ©Fs y = is
0; It is required
= that
Fb+increased.
+ cisthat
©Fy F=s 0;
(1)
c
©F
=
0; Fs = Fb (1) Fs = Fb
y
is
increased.
It
is
required
that
Düşey kuvvet dengesi: !
(1)
y =
+ cThe
©Fytemperature
= +0;c ©Fy =increase
Fs0;=the
F
0;+ c ©Fcauses
Fsy ==and
F0;b boltFs = Fb (1) Fs = Fb(1)
+ bcsleeve
©F
(1)
ibility.Compatibility.
The temperature
increase
causes
the
sleeve
and Bulonun
bolt
Uygunluk
denklemi:
deplasmanı
= Tüpün deplasmanı (1)
c
+
©F
=
0;
F
=
F
y
s
b
Compatibility.
The te
Compatibility.
Thethe
temperature
causes
the sleeve
and
bolt
Compatibility.
The
temperature
increase
causes the
sleeve causes
and bolt
Compatibility.
The
temperature
increase
the sleeve and bolt
2T and
1db2However,
expand
However, increase
the
redundant
forces
1db2T1d, sFig.
d 1ds2to
4–20c.
redundant
forces
T , Fig. 4–20c.
T and
F
1d
2
1dbb
to
expand
and
s
1dsthe
2Tand
2to
to
expand
and
Fig.
4–20c.
However,
redundant
forces
Fshorten
F1d
1ds2Consequently,
1db2Tthe
, uzamasına
and
Fig.
4–20c.
However,
the redundant
forces
T ,expand
s theBulondaki
çekme
içsleeve.
kuvveti
bulonun
olacağından
(+)
1d
1d
2T , Fig.
to
expand
and
4–20c.
However,
the
redundant forcesss TT
s bConsequently,
Fb and
Fsbolt
elongate
bolt
shorten
the
the
T
Fss elongate
the
and
sleeve.
the
s2T
bsebep
Compatibility.
The
temperature
increase
causes
the
sleeve
and
bolt
F
F
and
elongate
the
b
Compatibility.
The
temperature
increase
causes
the
sleeve
and
bolt
Compatibility.
The
temperature
increase
causes
the
sleeve
and
bolt
Fab final
Fs elongate
and reaches
theposition,
bolt
and
shorten
theand
sleeve.
Consequently,
Compatibility.
Thethe
temperature
increase
causes
the
sleeve
bb and bolt
ss the
Fbnot
F
and
elongate
bolt
andasshorten
thethe
sleeve.
Consequently,
the
F
Fsame
bolt
and the
shorten
the
sleeve.
Consequently,
end of reaches
the assembly
a final
which
is
not
the
he assembly
which
is
the
as
s same
b the
s, elongate
Fs position,
1d
2
1d
2
to
expand
and
Fig.
4–20c.
However,
redundant
forces
Compatibility.
The
temperature
increase
causes
the
sleeve
and
bolt
s
T
b
T
Fs
Fs
F
end
of
the
assembly
rea
1d
2
1d
2
,
to
expand
and
Fig.
4–20c.
However,
the
redundant
forces
1d
2
1d
2
,
to
expand
and
Fig.
4–20c.
However,
the
redundant
forces
deplasmana,Tüpteki
basınç
iç
kuvveti
tüpün
kısalmasına
sebep
olacağından
(-)
end
of
the assembly
acondition
final
position,
which
as
1dassembly
2aTnot
1dreaches
2T , Fig.
expand
andthe
However,
theasredundant
forcesas
s reaches
s Tcondition
bend
s T
T of
T toend
the bassembly
reaches
final
position,
which
not
the same
sis
bsame
position.
Hence,
the
compatibility
becomes
of
the
a 4–20c.
finalisConsequently,
position,
which
is not the same
its initial
position.
Hence,
the
compatibility
becomes
Fits
Fbolt
Fss2shorten
and
elongate
the
bolt
and
shorten
theand
sleeve.
1d
1dthe
,FFig.
to
expand
and
4–20c.
However,
theshorten
redundant
forces the
sF initial
bthe
T
b2
Tsleeve.
its
initial
position. Henc
F
and
elongate
the
and
Consequently,
the
F
F
and
elongate
the
bolt
and
shorten
the
sleeve.
Consequently,
the
position.
Hence,
compatibility
condition
becomes
F
and
elongate
the
bolt
the
sleeve.
Consequently,
the
b
s
b
s
its
initial
position.
Hence,
the
compatibility
condition
becomes
b
s
its
initial
position.
Hence,
the
compatibility
condition
becomes
Fb
Fb
Fdeplasmana
end
of
the
assembly
reaches
a
final
position,
which
is
not
the
same
as
F
sebep
olur.
Sıcaklık
artışı
heriki
malzemede
de
uzamaya
sebep
olur
ve
F
F
and
elongate
the
bolt
and
shorten
the
sleeve.
Consequently,
the
b end
b
b
s
end of the
assembly
reaches
a
final
position,
which
is
not
the
same
as
of the assembly reaches
a final
position,
which
is not
the same
as which is not the same as
end
of the
assembly
reaches
a final
position,
= b1d
-end
2of
1 + Td2 = 1db2T + 1d
db2=F 1d
2Ts2+T 1d
2Fsits
=F initial
1ds2assembly
1ds2Freaches
position.
Hence,
compatibility
condition
b1d
T the
athe
final
position,
which
is not becomes
the
same
as becomes
its
initial
position.
Hence,
the
compatibility
condition
becomes
its
initial
position.
Hence,
the
compatibility
condition
becomes
F
(b)
(b) F
its
initial
position.
Hence,
the
compatibility
condition
b
1+
d = 1
1
+
T
2
d
=
1d
2
+
1d
2
=
1d
2
1d
2
1 + T2
T2
Fb
b T
b F
T 1d 2 s +
F 1d 2d == 1d
(b) (+) deplasmandır.
(b)
b
F1+
=
1dbs22TT +- 1d
1dbecomes
1+dsT2
b T2position. Hence,
bcompatibility
T
b F
bs2F = 1ds2T - 1ds2F
its
initial
the
condition
Fb
g Eqs.Applying
4–2 and 4–4,
the mechanical
properties
from
Eqs.and
4–2using
and
and using1+
the
mechanical
properties
(b) 4–4,
d1d
= 21d
2Tfrom
1db2F = 21ds2T - 1ds2F
b1d
(b)
(b)
1+ T2 we Eqs.
d (b)
= 4–4,
1dbT2
2and
+=
1d1d
= +1d
2+Fproperties
1 + T4–2
2 and
T d
b2F
s2T
s1d
b2the
T
b F =
s2Td-=1d
1+
T2
1ds bFfrom
2T + 1db2F = 1ds2T - 1ds2F Applying Eqs. 4–2 and 44
Applying
using
mechanical
on the
cover,
have
theinside
tableback
on the
inside back
we
have
! cover,
Applying
Eqs.
4–2
4–4,
and usingproperties
the mechanical
Applying
Eqs.
4–2
and
4–4,
and
using
mechanical
from properties from
(b)
1+ T2
= 1db2T + 1db2F
=and
1dthe
s2T - 1ds2F
the table on the inside
back cover, we dhave
the table on the inside b
the
table
on
the
inside
back
cover,
we
have
the
table
on
the
inside
back
cover,
we
have
Applying
Eqs.
4–2
and
4–4,
and
using the properties
mechanicalfrom
properties from
-6 15°C210.150
[12110-62>°C]180°C
m2
+
Applying
Eqs.
4–2
and
4–4,
and
using
the
mechanical
properties
from
Applying
Eqs.
4–2
and
4–4,
and
using
the
mechanical
[12110 2>°C]180°C -6- 15°C210.150 m2
+
Applying
Eqs. 4–2 and 4–4, and using the mechanical properties from
-6
the
inside
back
cover,
we mechanical
have
Applying
Eqs.on
4–2
and
4–4,
and
using
properties from
[12110
-the15°C210.150
m2
+ the
[12110-6
2
-6
theFtable
on
the
inside
back
cover,
we
have
the
table
on 2>°C]180°C
the table
inside
back
cover,
have
-6we
the
table
on
the inside
back
cover, we
[12110
2>°C]180°C
-have
15°C210.150 m2 +
10.150
m2
[12110
2>°C]180°C
15°C210.150
m2
+
b
Fbthe
10.150
m2
table onFthe
inside
back
cover,
we
have
10.150
m22 -62>°C]180°C - 15°C210.150 m2 +
2
9 -6 2 2
-6
9 [12110
1400 mm22110-6
m2>mm
2[200110
2 N>m
] b-62>°C]180°C
m2
Fb+10.150
m2 + Fb10.150
-6 m2
1400
mm22110
m2>mm
2[200110
2 N>m
] - 915°C210.150
[12110
2>°C]180°C
15°C210.150
m2[12110
[12110
2>°C]180°C
- 15°C210.150 m2 +
2
-6
2
2 -6
2
1400 mm 2110 m [12110
>mm 2[200110
2
N>m
]
mm
1400 mm
-6
2>°C]180°C
15°C210.150
m2
+
2
-6
2
2
2
-6 1400
2 mm
9 >mm2 2[20011092 N>m2]
-6 15°C210.150 m2
= [23110 2>°C]180°C
2110 m
b10.150
2110
>mm2m2
2[200110
2 N>mm2
]
Fb10.150
m2
= [231102>°C]180°C -6- 15°C210.150
m21400
Fbmm
10.150
m2Fm
F
10.150
Initial
-6
(db)T
b
= m2
[23110 2>°C]180°C
- 15°C210.150
2 m2
-6
2 m2
2
9
2
= [23110 -62
Initial
-6
b10.150
Initial
1400
mm
2110
mposition
2[200110
2 -N>m
]
2
-6
22
2-6 -6
9F2=
2 >mm
2
9 2>°C]180°C
2
Fs10.150
(db)T
(dbb)TT 2
[23110
15°C210.150
m2
2
-6
2
2
9
=
[23110
2>°C]180°C
15°C210.150
m2
1400
mm
2110
m
>mm
2[200110
2
N>m
]
F
10.150
m2
1400
mm
2110
m
>mm
2[200110
2
N>m
]
(db)F position
position
1400 mm 2110
]
(db)T
(ds)9T m >mm
- Final
(dbs)T 9
-6
2 2[200110
d
d 2 N>m
m22-6
2 F 10.150
Initial
- -6 m2>mm
((ddb))F22110
mm
m2>mm-22[200110
2 N>m
]
(dbb)FF
(ds2)2T2[73.1110
1600 mm
-6 -6
2 2 N>m
2 d] -6 s 1400
9[23110
22110
=
2>°C]180°C
15°C210.150
m2
s T Initial
Final
F
10.150
m2
Final
position
d
1600
mm
2110
m
>mm
2[73.1110
2
N>m
]
(
d
)
F
10.150
m2
s
-6
= [23110
2>°C]180°C
position
[23110
2>°C]180°C
m2
((ddbb))TF =
s 2 2>°C]180°C
2
2-b F15°C210.150
2Final - 15°C210.150
9 [23110
=m2
- 15°C210.150 m2
Initial
-6
(db)T
position
position
1600
mmFinal
2110-6
m
>mm
2[73.1110
2 N>m
]
Tposition
1600 m
m
(d(sd)bT! )gives
Eq.
1
and
solving
=
[23110
2>°C]180°C
15°C210.150
m2
(
d
)
2
-6
2
2
9
2
d
b
T
2
-6F1600
2 mm
2 m2
9 >mm 22[73.1110 2 N>m ]
(d )
Using
Eq.
1 and
position
position
s10.150
2110 m
(db)T (db)F position
d
d solving
1600
mm
2110
m
>mm
2[73.1110
2
N>m
]
F
10.150
m2
(ds)Tgives
)F s T
F
10.150
m2
(
d
)
(
d
)
(
d
)
Final
s
s
s
F
d
b
F
b
F
Fs10.150
m2
Using Eq.
and
solvi
(ds)T
Using Eq. 1 and solvin
(db)FAns.
(ds)F position
2 F 10.150
-6
2 m2 2
9
2
d 1Final
- solving
-gives
Fs =Final
(dF
Final
s)Fb = 20.3 (kN
1600
2110
2 N>m
]
gives
2 kN-6
2Using
91
2>mm
2and
-6 mm
2 Eq.
2-sandmsolving
9 2[73.1110
2
Using
Eq.
12>mm
solving
gives
Fb Final
=mm
20.3
Fs db=)F1600
Ans.
!
position
position
2110
m
2[73.1110
2
N>m
]
1600
mm
2110
m
>mm
2[73.1110
2
N>m
]
2
-6
2
2
9
2
position
(c)
2110 92mN>m
>mm
2 N>m ]
(ds)F
-6
2
2 (c)
20.3
kN 22110gives
Fs = F
Ans.
b =11600
position
m21600
>mmmm
2[73.1110
] 2[73.1110
Using
Eq.
and mm
solving
(ds)F elastic material
s)Since
F
linear
behavior
was1(c)
assumed
in gives
this
=
F
=
20.3
kN
F
Ans.
Using
Eq.
1
and
gives
Using
Eq.
and
solving
(ds)solving
(c)
s
b
F
=
F
=
20.3
kN
F
Ans.
s
b
NOTE: Since
in1this
Usinggives
Eq.
and solving gives
(ds)F linear elastic material behavior was assumed
Using
Eq.
1
and
solving
Fig.
the4–20
average normalNOTE:
stresses(c)
should
be
checked
to
make
sure
Fig.
4–20
Since linear elastic
material behavior
was
assumed
in this
=kN
Fb = 20.3
kN
Fssure
Ans. NOTE: Since linear e
(c)
(c) average normal stresses
analysis, the
should
kN
Fs =beFchecked
Ans.
= to
Fbmake
= Since
20.3
Ans.
Fig.the
4–20
b =F20.3
sNOTE:
linear
elastic
behavior
was assumed
inAns.
thisaverage no
(c)
Fig. 4–20
do not exceed the analysis,
proportional
limits for
material.
=material
Fwas
= assumed
20.3
kN Ans.
F
NOTE:
Since
linear
elastic
material
behavior
in
this
the
average
normal
stresses
should
be
checked
to
make
sure
s
b
analysis,
the
(c)
=
F
=
20.3
kN
F
that they do not exceed
the proportional limits for theanalysis,
material.
s
b
Fig.
4–20
the
average
normal
stresses
should
be
checked
to
make
sure
NOTE:
Since
linear
elastic
material
behavior
was
assumed
in
this
analysis,
the
average
normal
stresses
should
be
checked
to
make
sure
that
they
do
not
exceed
the
proportional
limits
for
the
material.
that
they
do
not exceed
Fig. 4–20 Fig. 4–20 NOTE: Since
linear
elastic
material
assumed
this
NOTE:
Since
linear
elasticbehavior
materialwas
behavior
wasinassumed
in this
Fig.
4–20
NOTE:
Since
linear
elastic
material
behavior
was
assumed
in
this
that
they
do
not
exceed
the
proportional
limits
for
the
material.
analysis,
the
average
normal
stresses
should
be
checked
to
make
sure
Fig. 4–20
NOTE:
Since
elastic
material
was
assumed
in this
that
they
dolinear
not
exceed
proportional
for
thesure
material.
analysis, the
average
normal
stresses
should
bethe
checked
tobehavior
makelimits
sure
analysis,
the
average
normal
stresses
should
be
checked
to
make
analysis,
the
average
normal
stresses
should
besure
checked to make sure
that they
do not
exceed
proportional
limits
for the
analysis,
the
average
normal
stresses
should
be
checked
to material.
make
that they do
notthey
exceed
theexceed
proportional
limits the
for
the
material.
that
do not
the proportional
limits
for the
material.
that
they
do
not
exceed
the
proportional
limits
for
the material.
that they do not exceed the proportional limits for the material.
PAGE !60
ing. The chapter begins with a discussion of
r and moment diagrams for a beam or shaft.
d torque diagrams, the shear and moment
means for determining the largest shear and
d they specify where these maximums occur.
nt at a section is determined, the bending
Eğilme
ted. First we will 6.
consider
members that are
cross section, and are made of homogeneous
6.1 Kayma ve Moment Diyagramları
erward we will discuss special cases involving
Boyuna
eksenlerine
dik doğrultuda yük taşıyan ince uzun elemanlara kiriş denmektedir. Basit
d members made
of composite
materials.
be given to curved members, stress
mesnetli kiriş, konsol kiriş, çıkmalı kiriş gibi mesnetlenme türüne göre isimler verilmiştir.
ending, and residual stresses.
256
C H A P T E R 6 B E N D I N G256
CHAPTER 6 BENDING
Moment Diagrams
er and support loadings that are applied
ngitudinal axis are called beams. In general,
bars having a constant cross-sectional area.
s to how they are supported. For example, a
pinned at one end and roller supported at the
red beam is fixed at one end and free at the
ng beam has one or both of its ends freely
rts. Beams are considered among the most
elements. They are used to support the floor
a bridge, or the wing of an aircraft. Also, the
boom of a crane, even many of the bones of
P
w0
A
A
D
B
x1
6
C
x2
x3
6
Fig. 6–2
w(x)
Positive external distributed load
V
V
Positive internal shear
M M
Positive internal moment
Beam sign convention
Fig. 6–3
Because of the applied loadings,Because
beams develop
an internal
shearbeams dev
of the applied
loadings,
Simplyand
supported
beammoment that, force
force
bending
in general,
vary from
point that,
to point
and bending
moment
in general, va
along the axis of the beam. In along
order the
to properly
design
a beam
it to pro
axis of the
beam.
In order
therefore becomes necessary to therefore
determinebecomes
the maximum
shear
necessary
to and
determine th
moment in the beam. One way moment
to do thisinisthe
to beam.
expressOne
V and
wayMtoasdo this is
functions of their arbitrary position
x along
the beam’s
axis.
These x along
functions
of their
arbitrary
position
shearCantilevered
and moment
functions
can
then
be
plotted
and
represented
by be plot
shear
and
moment
functions
can
then
beam
graphs called shear and moment graphs
diagrams.
Theshear
maximum
values of
V
called
and moment
diagrams.
Th
and M can then be obtained from
these
graphs.
since the
shear
and
M can
thenAlso,
be obtained
from
these graph
and moment diagrams provide detailed
information
about
the variation
and moment
diagrams
provide
detailed informa
of the shear and moment along the
axis,
they
are often
used
of beam’s
the shear
and
moment
along
theby
beam’s axis
engineers
to decide
reinforcement
materials
within
engineers
to decide
where to
placethe
reinforcem
Overhanging
beam where to place
beam or how to proportion thebeam
size of
at variousthe
points
or the
howbeam
to proportion
size of the b
Fig.length.
6–1
along its
along its length.
In order to formulate V and M in terms
of xtowe
must choose
theMorigin
In order
formulate
V and
in terms of x w
Kirişlerin
tasarımı
için for
iç kuvvetlerin
(moment
veis kesme
and
the positive
direction
x. Although
the choice
arbitrary,
most
and the positive
direction
for x. Although
the
often
the origin
is
located
at
the
left
end
of
the
beam
and
the
positive
often
the
origin
is
located
at
the
left
end
of
the
kuvveti)
maksimum
değerlerine
gerek
vardır.
İç
kuvvetler
255
P
w0
direction
is
to
the
right.
direction
is
to
the
right.
kiriş enkesit değişimine, mesnetlenme ve yüklenme
In general, the internal shear and
moment
functions
of shear
x will and
be moment
general, the
internal
durumlarına
göre kiriş boyuncaIndeğişiklik
discontinuous, or their slope willdiscontinuous,
be discontinuous,
at points
or their
slopewhere
will bea discontin
B
D iç kuvvetlerin kiriş boyunca
göstermektedir.
Bu nedenle
distributed
load changes
or where
concentrated
forces or
distributed
load changes
or couple
where concentr
C
x1değişimini gösteren diyagramların çizilmesi
moments are applied. Because of moments
this, the shear
and moment
functions
are
applied.
Because
of this, the shea
x2
gerekmektedir.
must
be determined
for
each
region
of
the
beam
between
any
two of the b
must
be
determined
for
each
region
x3
x1, x2, and
discontinuities of loading. For example,
coordinates
discontinuities
of loading.
For x
example,
coordi
3 will
Fig.
6–2 to describe the variation
have to be
used
of used
V and
throughout
the
have to be
to M
describe
the variation
of V
length of the beam in Fig. 6–2. These
be valid
only coordin
length coordinates
of the beamwill
in Fig.
6–2. These
x1, from
within the regions from A to B for
B to Cfrom
for xA
within
the regions
to Bfrom
for x1, from B
2, and
C toİşaret
D for xKuralı:
C to D for x3 .
3.
w(x)
Beam Sign Convention. Beam
Before Sign
presenting
a method for
Convention.
Before pre
determining
the
shear
and
moment
as
functions
of
x
and
later
plotting
determining the shear and
moment
as function
Positive external distributed load
these functions
diagrams),
is firstand
necessary
these
functionsit(shear
momenttodiagrams)
V (shear and moment
V
establish a sign convention so asestablish
to definea “positive”
and “negative”
sign convention
so as to define “p
values for V and M. Although values
the choice
of
a
sign
conventionthe
is choice o
for V and M. Although
Positive here
internal
shear
arbitrary,
we
will use the one
often
used
in
engineering
practice
arbitrary, here we will use the one often used
M Fig. 6–3. The positive directions are as follows: the
M in
and shown
and shown in Fig. 6–3. The positive directio
distributed load acts upward ondistributed
the beam; load
the internal
shearonforce
acts upward
the beam; t
causes
a
clockwise
rotation
of
the
beam
segment
on
which
acts;
causes a clockwise rotationit of
theand
beam segme
Positive internal moment
the Beam
internal
moment causes compression
in themoment
top fibers
of thecompression
segment in the t
the internal
causes
sign convention
such that it bends the segment so such
that that
it holds
water.the
Loadings
are it holds w
it bends
segmentthat
so that
opposite Fig.
to these
6–3 are considered negative.
opposite to these are considered negative.
PAGE !61
Shear and Moment Fun
segment of the beam is sh
this segment, wx, is repre
258
C H A P T E R 6 B E N D I2N5G8
(a)
CHAPTER 6 BENDING
segment is isolated as a fre
centroid of the area comp
wx
x>2 from the right end. A
x
$ EXAMPLE 6.1 ŞekildekiEXAMPLE
üniform yayılı6.1
yüklü basit mesnetli kirişin
moment ve kesme
yields
2
L
M
Draw the
and shown
moment
Draw the shear and moment
for shear
the beam
in diagrams
Fig. 6–4a. for the beam
A diagrams
+ c ©Fy = 0;
kuvveti diyagramlarını çiziniz.
x
w
258
CHAPTER 6
BENDING
SOLUTION
Support Reactions.
w
wL
2
V
SOLUTION
(b)
Support
Reactions.
The
reactions are
The support
reactions
are shown
in support
Fig. 6–4c.
wL
Shear
Moment
Functions.
A free-body
Shear and Moment Functions.
A and
free-body
diagram
of the
left
d+ ©M
= 0;
-a
L
2
segment of the beam is shown in Fig. 6–4b. The di
segment of the
6 beam is shown in Fig. 6–4b. The distributed loading on
wx,
this
segment,
is
represented
by
its
resultant
this segment,
(a) wx, is represented by its resultant force only after the
(a)
segment
is isolated
as acts
a free-body
This fo
is isolated
as afor
free-body
diagram.
This
throughdiagram.
the
AMPLE 6.1 Çözüm:
Draw the shearsegment
and moment
diagrams
the beam
shown
in
Fig.force
6–4a.
centroid
of
the
area
comprising
the
distributed
lo
EXAMPLE
6.1
EXAMPLE 6.1
centroidwxof the area comprising the distributed loading, a distance of
wx
x>2
from
the
right
end.
Applying
the
two
equa
Draw
the
shear
and
moment
diagrams
for
the
beam
shown
in
Fig.
6–4a.
PLE 6.1
x
Öncelikle mesnet xreaksiyonlarını
hesaplayalım.
x>2 from the right end. Applying the two wequations of
equilibrium
w
Shear
and Moment Dia
SOLUTION
yields
2and
Draw
thethe
shear
moment
diagrams
forfor
the
beam
shown
inin
Fig.
6–4a.
Draw
shear
and
moment
diagrams
the
beam
shown
Fig.
6–4a.
yields
2
shown in Fig. 6–4c are obt
M
Support
Reactions.
The
reactions
Draw
the shear
diagrams
forsupport
the beam
shown inare
Fig.shown
6–4a. in Fig. 6–4c.
M and moment
wL
w
A
wL + Lc ©Fy = 0; wL
SOLUTION
zero
shear
can-beVfound
- wx
= 0 fr
A
wL
c ©Fy x= 0;
+
wx
V
=
0
w w
2
SOLUTION
xSupportShear
SOLUTION
2
2
2
and
Moment
Functions.
A
free-body
diagram
of
the
left
V
Reactions. The support reactions are shown in Fig. 6–4c.
V
V
L w
wLReactions.
SOLUTION
V
segment ofSupport
the
beam
isReactions.
shown in Fig.
6–4b.
The reactions
distributed
loading
onininFig.
The
support
are
6–4c.
Support
The
support
reactions
areshown
shown
Fig.
6–4c. L
wL
L the left
2
V = wa - xb
(b)
Shear
and
Moment
Functions.
A
free-body
diagram
of
2Support
wx,
this
segment,
is
represented
by
its
resultant
force
only
after
the
wL
Reactions.
The
support
reactions
are
shown
in
Fig.
6–4c.
V = wa - xb
(1)
(b)
2
L
(a)
2 acts through
Shear
and
Moment
Functions.
A
ofofthe
segmentsegment
of the beam
is shown
in
Fig.
6–4b.
distributed
loading
ondiagram
Shear
and
Moment
Functions.
Afree-body
free-body
diagram
theleft
left
2The
is
isolated
as
a
free-body
diagram.
This
force
the
Kesme kuvvet
fonksiyonlarıni
elde
edebilmek
için
kirişi
mesnetler
arası
Lve
L moment
x birloading
wL
x
wx,ofFunctions.
this
segment,
is
represented
by
its resultant
force
after
the
segment
the
beam
isthe
shown
in
Fig.
6–4b.
The
distributed
onon
Shear
and
Moment
A
free-body
diagram
ofonly
the
left
segment
of
the
beam
is
shown
6–4b.
The
distributed
loading
centroid
the
areaofcomprising
distributed
loading,
wL Lin Fig.
(a) wx
d+
©M x= a0;distance
-ofa
b x + 1wx2 a b + M
L
wL
segment
is
isolated
as
a
free-body
diagram.
This
force
acts
through
the
wx,
this
segment,
is
represented
by
its
resultant
force
only
after
the
segment
of
the
beam
is
shown
in
Fig.
6–4b.
The
distributed
loading
on
d+
©M
=
0;
a
bx
+
1wx2a
b
+
M
=
0
wx,
this
segment,
is
represented
by
its
resultant
force
only
after
the
2
2
2
"
x>2
from
the
right
end.
Applying
the
two
equations
of
equilibrium
noktadan
keselim.
(x=0) olarak sol mesneti
alalım.
x
(a)(a) Referans
2
2loading,
2 force
6thenoktası
NOTE:
From
the mome
centroid
of
area
comprising
the
distributed
a
distance
of
segment
is
isolated
as
a
free-body
diagram.
This
acts
through
the
wx,
this
segment,
is
represented
by
its
resultant
force
only
after
the
segment
is
isolated
as
a
free-body
diagram.
This
force
acts
through
the
6
yields
2
wx
(a)
2
M
w
wL
point
on
the
beam
wher
x>2 is
from
the right
end.
Applying
the
two
equations
of
equilibrium
ofdiagram.
thethe
area
comprising
the
distributed
a adistance
of
segment
isolated
as a centroid
free-body
This
force
acts
through
the loading,
centroid
of
area
comprising
the
distributed
loading,
distance
M
x
M
max !
wx wx
w
M = of1Lx - x22
wL
2
A2
8
Eq. 6–2 (see
- equations
x2
from
thethe
right
end.
Applying
the
two
equations
ofofequilibrium
centroid
of the
the
distributed
of
2 (2)Sec. 6.2) th
from
right
the1Lx
two
equilibrium
+ carea
©Fy comprising
= x>2
0;x>2
- end.
wxloading,
-Applying
VM
= a0=distance
x yields
x
wx
xM
2
2 equations of equilibrium
have
yields
x>2
the
right
end.
Applying
the
two
2 from
yields
V
2
wL
x
A
Shear and Moment
Diagrams. The shear an
- wx - V =wL
0
MM
x
y = 0;
w
yields+ c ©F
2 x wL
wL
wLL
2
Shear
and
Moment
Diagrams.
Theinshear
and are
moment
diagrams
A
A
2 MV (b)
V
=
wa
xb
(1)
shown
Fig.
6–4c
obtained
by plotting Eqs.
c ©F
+ c+©F
=
0;
wx
V
=
0
=
0;
wx
V
=
0
y y
Mmax1
2
wL
x x
2 obtained
shown
in
Fig.
6–4c
are
by
plotting
Eqs.
1
and
2.
The
point
of
2
2
wL
A
(c)
zero
shear
can
be
found
from
Eq.
1:
L
V yV= 0;
+ c ©F
wx
V
=
0
L
wL
x2
=wL
wcan
a be
- found
x b wL
(1)
shear
from Eq. 1:
2zero V
(b)
wL2
L
wL wLwL
x
2
V
LL
2
L
d+
©M
=
0;
a
bx
+
1wx2a
b
+
M
=
0
2
2
Fig.
6–4
V
=
wa
xb
V
2
2
V
=
wa
xb
(b)(b)
wL
V = (1)
w(1)
a - xb = 0
L
2
2
2L
2 - (1)
V
2
2
wL
x
V
=
wa
xb
V
=
wa
xb
=
0
(b)
wL- a
d + ©M = 0;
b x + 21wx2 a b + M = 0
2
xx
2
2 wwL
wL
wL
2
2 1wx2ab b+ +
d+ ©M = =0; 0; M =
- a- a1Lx
bxbx
MM= =0 0 (2) x = L
-x+x+1wx2a
2L
2
wLd+ ©M
x
2
22
2
2
L
2
d+
©M
=
0;
a
bx
+
1wx2a
b
+
M
=
0
x =
x
6 6
w
L
2M =
2
21Lx - x22 " wL 2
(2)
wL
2 w wNOTE: From the moment diagram, this value
2
2 w
Shear "and Moment Diagrams.
The shear
and moment
diagrams
MM= = diagram,
1Lx
(2)
1Lx- -x2this
x2 22 value of x represents
(2)the
2
Diyagramları çizelim: shown in 2Fig.
NOTE:
From
the
moment
M 6–4c
w
wL2 by plotting 2
point
on
the
beam
where the maximum
mom
are
obtained
Eqs.
1
and
2.
The
point of
2
M
!
w
max - x 2
M
=
1Lx
(2)
2
ShearwLand
Moment
Diagrams.
and moment
diagrams
M
8 shear
point
on from
theThe
beam
where
theEq.
maximum
moment
occurs,
sinceVby= dM>dx =
6–2
(see
Sec.
6.2)
the
slope
zero
shear
can
be
found
Eq.
1:
2
M
!
max
wLmoment
L
wL
Kesme
kuvveti ve
x’e
sınır
değerleri
Fonksiyonların
shown
in wFig.fonksiyonlarını
6–4c Shear
are
obtained
by
plotting
Eqs.
1 vererek
and
2.The
The
point
of
8w
Vçizelim.
=shear
dM>dx
=moment
0. From
Eq.and
6–2
(see
Sec.
6.2)
the
slope
Eq. 2, we
Moment
Diagrams.
The
shear
and
moment
diagrams
Shear
and
Moment
Diagrams.
and
diagrams
have
2
2
w
L
zero
shear
can be Diagrams.
found
from
Eq.
1:6–4c
V
shown
in
Fig.
6–4c
are
obtained
by
plotting
Eqs.
1
and
2.
The
point
ofof
have
Shear
and
Moment
The
shear
and
moment
diagrams
shown
in
Fig.
are
obtained
by
plotting
Eqs.
1
and
2.
The
point
wL
L
x
V
wa - xb = 0
maksimum
ve sıfır noktalarını grafik üzerinde
not= edelim.
L
w
L
L 2
2
2 shown in Fig. 6–4c arex obtained
zero
shear
can
be
found
from
Eq.
1:
by
plotting
Eqs.
1
and
2.
The
point
of
zero
shear
can
be
found
from
Eq.
1:
M
=
L
a
b
a
b
2
B
wL
L
L
wL
wL
L
wL
max
2
wL
L
w
L
L
V
2
2
2
V =
a - x sıfır
b =Molduğu
0
from
Eq.
1:w(c)
2 2 wL
2
=
La
b
a
b
Kesme
kuvvetinin
x
mesafesi:
B
R
2 L
max
2zero shear can be2 found
L
2
LL
2 = =0 0 2
V V
(c)
x =
x
V V= =wawa2 - -xbxb
2
L
L
2
Fig.
6–4
2
2
wL2
wL
V = x
wa= L- xb = 0
2 wLwL
"
=
2
x
Fig.
6–4
2
2
wL of x represents the
8
L
2 2
NOTE: From the moment
diagram, this
value
2
LL
wL
x x= = =moment
x xon the beam where the maximum
2M
"2
8
wL
point
occurs,
since
by
L
Mmax !L L 2
NOTE: From
moment
this value 2of2x represents the
wL
wLthe
xSec.
= 6.2)diagram,
x 2 82
V = dM>dx = 0. From Eq. 2, we
Eq.
6–2
(see
the
slope
"
"
L
2
2
M
point on the2 beam
where the
maximum
moment
occurs,
since
byofof
2 NOTE:
From
the
moment
diagram,
this
value
x xrepresents
Mmax ! wL " wL
NOTE:
From
the
moment
diagram,
this
value
representsthe
the
have
2
8
V
=
dM>dx
=
0.
Eq.
(see
Sec.
6.2)
the
slope
From
Eq.
2,
we
2 6–2
2
MM 2
wL
point
on
the
beam
where
the
maximum
moment
occurs,
since
by
wL
point
on
the
beam
where
the
maximum
moment
occurs,
since
by
NOTE:
From
the
moment
diagram,
this
value
of
x
represents
the
x
Momentin maksimum
değeri: L 2
Mmax
! !
Mmax
L2
have
wmoment
Lslope
8 8 the beam where
V
=
dM>dx
=
0.
Eq.
6–2
(see
Sec.
6.2)
the
From
Eq.
2,
we
V
=
dM>dx
=
0.
Eq.
6–2
(see
Sec.
6.2)
the
slope
From
Eq.
2,
we
wL
point
on
the
maximum
occurs,
since
by
Mmax = B La b - a b R
Mmax ! 2
x
2
8
2Eq. 2, we
(c)
have
= 2L
0. From
Eq. 6–2 (see Sec. 6.2) have
the
slope V
L
w = dM>dx
L2
M
=
L
a
b
a
b
B
R
max
2
x x
have
2
2 2
2 ww LL
L L
(c)
L L2 2
Fig. 6–4
wL
x2 2
M
=
La
b
a
M
=
La
b
a b bR R
B
B
2
L
w
L =
Lmaxmax 2 2
22
22
(c) (c)
2b - a
8
M
=
La
b
B
R
Fig.
6–4
max
2
2 = wL2
2
(c)
2 2
8
Fig.
6–4
Fig. 6–4
wL
wL
=
=
2
Fig. 6–4
wL
88
=
8
CHAPTER 6
BENDING
258
258 6.1 C HCAHP AT EP RT E6R 6LB EBNEDNI NDGI N G
EXAMPLE
CHAPTER 6 BENDING
PAGE !62
6.1
6.1
SHEAR AND MOMENT DIAGRA
259
SHEAR AND MOMENT DIAGRAMS
EXAMPLE 6.2
6.1
Draw the shear and moment diagrams for the beam shown in Fig. 6–5a.
! EXAMPLE 6.2 Şekildeki üçgen yayılı yüklü konsol kirişin kesme kuvveti ve moment
w0 L
2
Draw the shearçiziniz.
and moment diagrams for the beam shown
w0 in Fig. 6–5a.
diyagramlarını
EXAMPLE
6.2
w0 L
EXAMPLE 6.2
w0 L
w0
6.1
6.1
SHEAR AND MOMENT DIAGRAMS
Çözüm:
w0 L2
3
L
(a)
2
L
3
Mesnet
r and moment diagrams
forreaksiyonları:
the beam shown in Fig.
6–5a.
SOLUTION
wn in Fig. 6–5a.
L
Support Reactions.
w0 L
w0
2 force and
w0
L Fig. 6–5a.
ownw0in
6.1
SHEAR AND MOMENT D
2 w0
Draw the shear and moment diagrams for the beam shown in Fig. 6–5a.
DrawSthe
and
moment
diagrams for the
wshear
6.1
HEAR
AND M
OMENT
DIAGRAMS
259beam shown in Fig. 6–5a.
0L
EXAMPLE
6.2
2
2
2
w0 L
259
w0 L
L
L
w0 L
w0
3
w0
w
20
Draw the shear and moment
for the beam shown2in Fig. 6–5a.
w0diagrams
3
259 (a)
(b)
w0 L
SHEAR AND MOMENT DIAGRAMS
6.2
2
w0
6.1 SHEAR AND M
SHEAR AND MOMENT DIA
w0 L
(b)
w0
L
2
w0 L
2
2
w0 L
2
2
w0 L22
2
w0
L
L
3
0 L is replaced
The distributed wload
3 3 by its resultant
(b)
3
(a)
the reactions
have been determined as shown
in Fig. 6–5b.
(b)
(a)
2
1 w0 x
x
2 L
SOLUTION
w0
w0 L2
w0 L
L
L
w0 L
2
3
Support Reactions. TheShear
distributed
load is replaced
by its resultant
x
2
1 w
3
w0 L
and Moment
A free-body
diagram(b)of
a 0 beam
x
2
(a) asFunctions.
w0
w0 x
L
force
and
the
reactions
have
been
determined
shown
in
Fig.
6–5b.
2
2
segment of length
x is shown in Fig. 6–5c. Note wthat
the intensity wof" L
SOLUTION
0L
SOLUTION
the
triangular
load
at
the
section
is
found
by
proportion,
that is,
2
1x
Shear
and Moment
Functions. A Support
free-bodyReactions.
diagram of The
a beam
2
distributed
load is replaced
by itswresultant
2
0L
w0 L2
L
L
Support
Reactions.
The
distributed
is
replaced
by
its
resultant
or
With
the
load
intensity
known,
the M
w>x
=
w
>L
w
=
w
x>L.
1 3w
0 6–5c. Note that
0
3
segment of length
x is shown SOLUTION
in Fig.
the intensity
of
3 6–5b.
force
and
the
reactions
have
been
determined
as
shown
in
Fig.
x 2 L
3
force and
thedistributed
reactions have
been
determinedfrom
as shown
in Fig.
6–5b.
ofis the
loading
isthat
determined
the area
under
w0 L
a)
the triangular load(b)at theresultant
section
found
by proportion,
is,load is2 replaced
w
L
Support
Reactions.
The distributed
by its1resultant
0
V
x
w
L
2 1(c)
the
diagram.
Thus,
0
Shear
and
Moment
Functions.
A free-body
diagram
of a beam
b)
With and
the
load
intensity
known,
thefree-body
w>x = w0>L or w = w0x>L.Shear
3a beam
2
force
the
reactions
have
been
determined
asdiagram
shown in
6–5b.
2
6
and
Moment
Functions.
A
of Fig.
3
Kesme
kuvveti
moment loading
fonksiyonlarını
elde
edelim.
Referans
noktası
olarak
ankastre
x that the intensity of
w0 L
segment
ofw
length
xarea
is
shown
in
Fig.
6–5c.
Note
resultant
of thevedistributed
is determined
from
the
under
segment
of
length
x
is
shown
in
Fig.
6–5c.
Note
that
the
intensity
of
L
w
x
1
0
0
b)
Shear
and
Moment Functions.
A
free-body
diagram
of(c)a beam that is,2
the triangular
load
by proportion,
+ c ©F
= 0;
- section
-found
V = is
0byfound
¢at the
≤ xissection
the diagram. Thus,
ytriangular
loadyayılı
at the
proportion,
that is,
mesneti
seçelim, eğer boş ucuthe
seçersek
trapez
yük
ile
uğraşmak
zorunda
kalırız.
2
2
L
w0 L2
2
segment
of
length
x
is
shown
in
Fig.
6–5c.
Note
that
the
intensity
of w0 Lthe
orx>L.
With
the
load intensity
known,
w = With
w0x>L.
0>L
or ww
the
load
intensity
known,
the
w>xw= xww>x
>L =
= 1w
3
tions. The distributed load is replaced bywitsLresultant
0
0
0x
1the triangular
3
load
atwdistributed
the
sectionloading
is found
by proportion,
that
is, under
0 resultant
of0 the
is determined
from
the area
xw
x
2
c ©Fy = 0;as shown in0 Fig.
resultant
of
the
distributed
loading
determined
from
the
area
under
x
V
=
0 w02is
¢
≤
w
L
x
L
reactions
have been +determined
6–5b.
2
2
by its resultant
0
1 w0 x
or w =Thus,
w>xL= w
w0x>L.
(1) the w0 L3
V
= w "With
1L - the
x 2 load intensity known,
0>L
2
2
the
diagram.
x
w
L
0
the
diagram.
2L L is determined from the area under 2 w0
x
w0 x
wn in Fig. 6–5b.
2 L
resultant
of2Thus,
the distributed loading
w " of a beam
doment
by its resultant
Functions. wA
L
0 L free-body
1 wdiagram
0x
w0diagram. Thus,
wL
wx
the
x
w2 L w0L
1 0wM0x-(1)11 ¢ w00xw≤0 x
L -1 V = 0
w0 xV = of 1L2+-c ©F
L the intensity
wn in
Fig.
6–5b.in Fig.26–5c. Note
ngth
xofis
x22y =w0;
2 that
0L 0
ram
ashown
beam
c2L
+
©F
=
0;
x
V =
0 xb + M = 0
¢
≤
w " M d+ ©M
=
0;
1x2
+
¢
≤2xa
y
2
2
L
w
L
L
0 found by proportion,
w0 L2
2
2
L
load
at
the
section
is
that
is,
w
L
w
x
he intensity of
3 0 1 x 2 V1
L
3
0 2
2
2
c
+
©F
=
0;
x
V
=
0
w
L
¢
≤
V
3
gram
a 0beam
y0
or
w of
= that
w
x>L.
3
ortion,
is, With2 the load intensity
2 L
w0 2 6 2
M2
1 x V wknown,
wthe
3 1 w0x x 12
0L
0L
w
w0 L
w0 L
w
the
intensity
of
0 V
w
L
=
1L
x
2
2
2
0
0
d+
©M
=
0;
1x2
+
xa
xb
+
M
=
0
¢
≤
ety
distributed
loading
is
determined
from
the
area
under
2
3
2
3
known, the
3
0L
(1) 2 w(1)
V1-2L
=
1L
-2L
6
wx0 L2 x 2
(2)
M
=
+
3L
x
3
2
2
L
3
2
3
w0 2
portion,
x
2w L
6L(c)V =2L
1x V
hus,
the area that
underis, w L2
0
(1)
1L - 3x22 V
0
ity known, the
3
2L
2
6
(c)
w0L
L2 Eqs.
w0L6–1
w0 results
1 1wof
1
2 applying
3
0xSec. 6.2,
0 Land
3 can be
2 checked
3w0Lby
x
w
w
1w
w0under
L
1 w0x
0
0x+ 6–2
d
+
©M
=
0;
1x2
x
a
x
b
+
M
=
0
¢
≤
(2)
M =These
1-2L
+
3L
x
x
2
m the area
2
w0 L2
d+
xb + M
- ¢
≤ x - V = 0 (c)
30L1x2 + 221¢ L
3 = 0 w0 L2
6L
that
is,©M = 0;
w30L2- w
w0x≤ x2 a 3 1L
2
2
2 L
V
3
M
d + ©M = 0;
1x2 + ¢w x ≤ x a x b + M = 0 w3 L2 V
dV 36–2ww0 2
0
0 Sec.
0L
2 6.2,
3
x
These results can be checked by applying Eqs.w6–1
and
of
w
w
L
OK
=
=
10
2x2
=
0
3
2
3
w0 M = 3 0 1 - 2L
w30(2)
L V
w0 2
+ x3L
w0
dx
3 x - x 2
2
that
w0 L
(2)
M = 2L
1 - 2L 6L+ 3L2x L
-M
2
(1)
V = Diyagramları
1L is,- x22 çizelim:
w
2
w
L
0
0
2L w L
2 dM w0w
(2) 2 2 x
=6L 12- 2L3 2+ 3L2wx0 - x2 32 2
w
xM
dV
0
0
(1)
w0 0 10 w
=
=
2x2
=
V
=
=
10
+6L
3L -OK
3x
=
1LEqs.
- x
2 andOK
0L
These
results
can
be
checked
by2applying
6–1
6–2
ofwSec.
6.2,
2
!
2
dx
2L
L
These
results
can
be
checked
by
applying
Eqs.
6–1
and
6–2
of
Sec.
6.2,
dx
6L
2L
Bu
sonuçları
bölüm
6.2’deki
bilgilerle
kontrol
edelim:
3
w0L
w0L w L 1 w0x
1
that
is,
These
can be checked by applying Eqs. 6–1 and 6–2 of Sec. 6.2,
(d)
(1)
1x20 + ¢
+ Mthat
= 0is, wresults
≤xa xb w
2
M
wDiagrams.
0
0
1
M
2
2
2 The
and
Moment
graphs
1 and
3
2
2 V
L = dM
3 = Shear
2
w0 w0 Lof2 Eqs.
w0x2 are
dVOK
is,2 0-L
10 that
+ 3L
3x
2
=
1L
x
2
w
w
x
dV
x b + M = 0 w L2
V
0
0
OKM
= ! 10=3 -- 2x2
dx
6L
2L
shown in Fig.3 6–5d.
0
OK
w = dVw= = wdx
w0x= - L
3
010 - 2x2
Fig
2L
w0 3 3V
(d)
dx
2L
L
w
L
OK
w
=
=
10
2x2
=
1
0
2
3
dx
2L
L
(2)
- 2L
+
3L
x
x
2
xb +M
M== 0 1 w
2
Shear
and
Moment
Diagrams.
The
graphs
of
Eqs.
1
and
2
are
w
w
2
0
0
2
2
6L 0wL0 L
3
dMV =wdM
w0 L2
0
0 22 2=
=
- w3x
1L
-OK
x22 w OK
2
210 + 3L
2
(2)
w
3shown
V = dM
= wdx
x2Fig.
22 6–5
0L 2 !
010 + 3L
01L 2-2L
2 V in Fig. 6–5d.
2- 3x 22 =
6L
3
V = dx =6L 10 + 3L - 3xx 2 =2L 1L - x 2
OK !! w3 0 L
an be checked by applying
Eqs. 6–1 and 6–2 of Sec. 6.2,
w0 L
dx
6L
2L
3
(2)
Shear
and Moment
Diagrams.
The graphs 1ofand
Eqs. 1 and 2 are
x
2
6–2 of Sec. 6.2,
Shear
Moment
Diagrams.
M
Shear and
and
Moment
Diagrams. The
Thegraphs
graphsofofEqs.
Eqs. 1 and2 2are
are
shown
in
Fig.
6–5d.
w0
w0x
dV
x
shown
ininFig.
6–5d.
shown
Fig.
6–5d.
M
OK
w
=
=
10
2x2
=
x
d 6–2 of Sec.
dx6.2, 2L
L
x
OK
w0
w
M
dM
0
=
OKx ! w0 L2
=
10 + 3L2 - 3x22 =
1L2 - x22
dx2
6LOK w L2
2L
2
3
OK ! 0
- x2
(d)
3
oment Diagrams. The graphs of(d)
Eqs. 1 and 2 are
2
OK ! w0 L2
L
x22 2 are
–5d.
qs.
1 and
qs. 1 and 2 are
3
Fig. 6–5
(d)6–5
Fig.
Fig. 6–5
PAGE !63
BENDING
EXAMPLE
EXAMPLE 6.3
6.3
EXAMPLE
6.3
EXAMPLE
6.3
2 kip/ft
6 kip/ft
SOLUTION
EXAMPLE
6.3
Draw
the
shear
and
moment
diagrams
the
beam
shown
in Fig.
6–6a.
Draw the shear and momentDraw
diagrams
forfor
the
beam
shown
indistributed
Fig.
6–6a.
the Draw
shear
and
moment
diagrams
for
theload
bea
Support
Reactions.
The
the
shear
and
moment
diagrams
f
260
CHAPTER 6 BENDING
260
CHAPTER 6 BENDING
6 kip/ ft
6 kip/ft
6 kip/ ft
and
rectangular
component
loadings
and
6 kip/ ft
260
CHAPTER 6 BENDING
18 ft
2 kip/ ft
replaced by their resultant forces. Th
2 kip/ft
2 kip/ ft
SOLUTION
2 SOLUTION
kip/
ft
SOLUTION
SOLUTION
6 ki
(a)
Draw
and moment
for the
beam shown
Fig. 6–6a.
determined as shown on the beam’s free-bo
B E N Dthe
I N Gshear
EXAMPLE
6.3
Draw
the sheardiagrams
and moment
diagrams
for
theinbeam
shown
inReactions.
Fig. 6–6a. The
Support
The
distributed
load
is divided
into
triangular
Support
Reactions.
The into
distributed
load is d
Support
Reactions.
distributed
load
is Reactions.
divided
triangular
EXAMPLE
6.3
Support
The
distributed
p/ ft
2loadings
kip/ ft
36
kip 36 kip
6 kip/ ft EXAMPLE
6.3 Şekildeki trapez yayılı
and
rectangular
component
and
these
loadings
then
and
rectangular
component
loadings
and
thes
and
rectangular
component
loadings
and
these
loadings
areare
then
and
rectangular
component
loadings
!
yüklü
basit
mesnetli
kirişin
kesme
kuvveti
vemoment
Shear
and
Moment
Functions.
AforfreeDraw
the
shear
and
diagrams
the
18
ft
4resultant
kip/ft
18
ft
18
ft
18
ft
replaced
by
their
forces.
The
reactions
have
been
SOLUTION SOLUTION
replaced
by
their
resultant
forces.
The
ret
Draw
the
shear
and
moment
diagrams
for
the
beam
shown
in
Fig.
6–6a.
replaced
by
their
resultant
forces.
The
reactions
have
been
replaced
by
their
resultant
forces
segment
is
in
Fig.
6–6c.
As
above,
moment
diyagramlarını
çiziniz.
(a)
6the
kip/beam’s
ft for thefree-body
Draw the
shear
moment
diagrams
beam
shown
in Fig.
6–6a.
(a) and as
he shear and moment diagrams for the beam(a)
shown in Fig. 6–6a.
2on
kip/ft
(a)
determined
shown
diagram,
Fig.
6–6b.
determined
as
shown
on
the
beam’s
free-body
d
determined
as
shown
on
the
beam’s
free-body
diagram,
Fig.
6–6b.
determined
as
shown
on
the
beam’s
fre
6
kip/ft
Support Reactions.
The
distributed
load
is
divided
into
triangular
by rectangular and triangular dis
Support Reactions. The distributed
load
divided into triangular
C H A P T E R 6 B E N D Ireplaced
NG
6 kip/
ft is
2 kip/
ft 2 6 0
SOLUTION
36
kip
36
kip
36
kip
36
kip
and rectangular
component
loadings
and
these
loadings
are
then
intensity
of
the triangular
at
the
sectio
18 ft load
36
kip
kip
and
rectangular
component
loadings
and
these
loadings
then
36 kip 36 kip
2 kip/ft
Draw
the shear and
moment
diagrams for
the
beam
shown
inare
Fig.36
6–6a.
and
Moment
Functions.
A
free-body
diagram
of
left
SOLUTION
Shear and
Moment
Functions.
Athe
free-body
2 kip/ ft
Shear
and Moment
Functions.
A
9Shear
ft
and
Moment
Functions.
free-body
diagram
of
the
left
4 SOLUTION
kip/
ft Shear
4 kip/
ft
ION
4 kip/
ft A
replaced by
their
resultant
The forces.
reactions
have
been
4 kip/ft
Support
Reactions.
The
distributed
load
The
resultant
force
and
the
location
of
eac
replaced
by theirforces.
resultant
The
reactions
have
been
(a)
6 kip/ft
segment
is
shown
in
Fig.
6–6c.
As
above,
the
trapezoidal
loading
is
segment
is
shown
in
Fig.
6–6c.
As
above,
the
t
12
ft
segment
is
shown
in
Fig.
6–6c.
As
abo
Support
Reactions.
The
distributed
load
is
divided
into
triangular
segment
is shown
in
Fig.
As
above,
the
trapezoidal
loading
is equat
2 Support
kip/ ft6–6b.
2 kip/
ft 6–6c.
shown
on the
beam’s
free-body
diagram,
Fig.
and
rectangular
component
loadings
and
as shown
on
the beam’s
free-body
diagram,
Fig.
6–6b.
Reactions.
The
distributed
isrectangular
divided
intoand
triangular
also
shown.
Applying
the
equilibrium
2 kip/
ftload
rtdetermined
Reactions.asdetermined
The
distributed
load is
divided
into
triangular
2 kip/ft
EXAMPLE
6.3
replaced
by
rectangular
and
triangular
distributions.
Note
that
the
replaced
by
triangular
distribu
18
ft
replaced
by
rectangular
and
triangula
and
rectangular
component
loadings
and
these
loadings
are
then
SOLUTION
ft
replaced by18component
rectangular
and triangular
distributions.
Note
that
the The
36 kip resultant
36
kipthen
replaced
bytriangular
their
forces.
and
these
loadings
are
loadings are30and
then
kip rectangular
42 kip loadings
pctangular component loadings and18 these
intensity
of
the
triangular
load
at
the
section
is
found
by
ft
intensity
of
the
load
atproportion.
the
section
intensity
offound
the
triangular
load
at1theiss
replaced
byleft
their
resultant
forces.
The
reactions
have
been
(a)
18
ftThe
Shear
and
Moment
Functions.
A
free-body
diagram
of
the
intensity
of
the
triangular
load
at
the
section
is
by
proportion.
Shear
and
Moment
Functions.
A
free-body
diagram
of
the
left
Support
Reactions.
distributed
load
is
divided
into
triangular
9
ft
9
ft
kip/
ftk
determined
as
shown
on
the
beam’s
free-bod
replaced
by
their
resultant
forces.
The
reactions
have
been
c ©Fy of
d by 4their
(b)
kip/ ft resultant forces.9 ftThe
=resultant
0; 30
14di
+
kip
- 12
kip>ft2x
-4each
ft
Draw
the
shear
and
moment
diagra
(a) reactions have been The 9resultant
force
and the
location
each
distributed
loading
are
The
resultant
force
and
the
location
of
The
force
and
the
location
o
Çözüm:
determined
as
shown
on
the
beam’s
free-body
diagram,
Fig.
6–6b.
(a)
segment
is
shown
in
Fig.
6–6c.
As
above,
the
trapezoidal
loading
is
segment
is
shown
in
Fig.
6–6c.
As
above,
the
trapezoidal
loading
is
2
The
resultant
force
and
the
location
of
each
distributed
loading
are
12
ft
12
ft
and
rectangular
component
loadings
and
these
loadings
are
then
12
ft
determined
as
shown
on
the
beam’s
free-body
diagram,
Fig.
6–6b.
ined as 2shown
36
kip 36 kip
kip/ ft on the beam’s free-body
12 ft diagram, Fig. 6–6b.
6also
kip/
ftshown. Applying
also
shown.
Applying
the
equilibrium
equations,
we
have
2
kip/
ft
also
shown.
Applying
the
equilibrium
equations,
the
equilibrium
e
t replaced by rectangular
18
ft
1shown.
x Applying
replaced
rectangular
and
triangular
thethe
18
ft Note
distributions.
Note
that
the
Shear equations,
and Moment
Functions. A free-b
equilibrium
we have
36 triangular
kip 36 kip
18that
ft
replacedbyand
by
their
forces. distributions.
The also
reactions
been
4
x have
4 kip/
ft
18resultant
ft
kip
x2
30 kip 36 kip 36
42kip
kip
Shear
and
Moment
Functions.
A
free-body
diagram
of
the
left
30
42
kip
2x
2
kip/
ft
2
18
30
kip
42
kip
intensity
of
the
triangular
load
at
the
section
is
found
by
proportion.
Shear
and Moment
Functions. A
free-body
diagram
ofV
segment
is
in Fig.
6–6c.
SOLUTION
intensity
of the
triangular
at theon
section
by
proportion.
4iskip/ft
and
Moment
Functions.
A
free-body
diagram
of
left
x the
1 shown
= ¢left
30As-1above,
2x - th
30 kipload
42
kip
determined
as shown
the
beam’s
free-body
diagram,
6–6b.
1
4found
kip/
ftthe
2-kip/
ftabove,
c ©Fy4Fig.
(b)
1-0;
cabove,
=xin(b)
0;
14yrectangular
kip>ft2
ax kip
+(b)
30
kip
12
kip>ft2x
-is
V 14
= kip>ft
0 dist
=+the
+
©F
30
kip
12
kip>ft2x
is
Fig.
6–6c.
As
loading
ctrapezoidal
= 0;- 30
©F
-b-xis
12
kip>ft2x
-9
y the
kip/ft
The
resultant
force
and
the
location
ofsegment
each
distributed
loading
are
segment
isshown
shown
in
Fig.
6–6c.
As
trapezoidal
loading
replaced
by
and
triangular
The
resultant
force
and
the
location
of
each
distributed
loading
are
nt
is
shown
in
Fig.
6–6c.
As
above,
the
trapezoidal
loading
is
c
(b)
9
ft
2
kip/ft
=
0;
14
kip>ft2a
+
©F
30
kip
12
kip>ft2x
bx
V
=
0
2
18
ft
2
Support
Reactions.
The
distribu
18
y
2
kip
2 kip/
ft replaced by rectangular
6 A
and
distributions.
Note
thethe
2 of 12
18 that
ft that
also
shown.
Applying
equations,
replaced
by
rectangular
and
triangular
distributions.
Note
Shear
andequilibrium
Moment
Functions.
free-body
of
the
lefttriangular
intensity
triangular
load
at the section
d also
by rectangular
and
triangular
distributions.
Note
that
the1 wediagram
shown.
Applying
the
equations,
we
have
1 the
xequilibrium
xhave
d+ ©M
=and
0;the2ft
rectangular
component
4 kip/ft
12 kip/ft
x
2loadi
4
x
4
x
9
ft
4 loading
x load
intensity
of18
the
triangular
thethe
section
is is
found
byby
proportion.
x force
x each
x found
2 x1 in is
intensity
of
load
section
proportion.
2 xthe
segment
isthe
shown
Fig.
6–6c.byAs
above,
isft atVat
18
kip triangular
ty of42the
load at
section
proportion.
and
location
of
2 trapezoidal
2 the
x 2 triangular
18
ft
by
fo
= ¢The
30
-resultant
2x replaced
-2found
kip
≤ 18
V
= their
30are
ki
x
9 ft9 ft 2x 2 24 18118x x
¢the
1xThe resultant
=2x¢1
30(1)
- ≤2x
Mand
2 kip/ft
12 ftx force
xV- resultant
x force
the
location
of
each
distributed
loading
x
(1)
9
9kip>ft
x
c
V
=
30
2x
kip
¢
≤
The
resultant
and
the
location
of
each
distributed
loading
are
(a)
sultant
force
and
the
location
of
each
distributed
loading
are
=
0;
14
kip>ft2
a
+
©F
30
kip
12
kip>ft2x
b
x
V
=
0
replaced
by
rectangular
and
triangular
distributions.
Note
that
the
also
shown.
Applying
the
equilibrium
equati
30
kip
42
kip
4x kip>ft2a
kip/ ft
4 018Vftkip/ ft4
-30
kip1x2
+
12
kip>ft2xa
b
+
14
14
+ c ©Fy = 0; 30 kip y- 1212kip>ft2x
b
x
V
=
ft
kip/
ft
determined
as
shown
on
the
beam’
12 ft
18kip/ft
18
9 have
2 also
18
ft
18 equilibrium equations, we
2
2
shown.
Applying
the
24 we
18
ft6 also
3 Applying
6
6 at the
the42equilibrium
equations, we
have
own. Applying theintensity
equilibrium
equations,
have
of the
triangular
load
section
is dfound
30
kip xshown.
kip
1818
ft ft
(b)
218kip/
ft
kip/by
ft= proportion.
x2©M
+
0;
d
+
©M
=
0;
2
kip/
ft
6
36
kip
36
kip
1
d
+
©M
=
0;
2
3030
kipkip
42242
kip
3
2 kip/ft
kip
The
resultant force and
the
location
of xeach
distributed
Functions.
©M
= 0; loading are
0; x
©Ffty =Shear
30xand
kip -Moment
12 kip>ft2x
- 2 14xki
2d+
2 (b)
x¢M
1+41ckip/
x
1
(1)
M
x
1
x
1
x
x
V
=
30
2x
kip
≤
2
M 12
30
kip
x
1
(1)
c=
(b)
M
=
30x
x
¢
V
=
30
2x
kip
=
0;
14
kip>ft2a
+
©F
30
kip
12
kip>ft2x
bx
V
=
0
¢
≤
c
(b)
x
y
=
0;
14
kip>ft2
a
+
©F
30
kip
kip>ft2x
b
x
V
=
0
x
also
shown.
Applying
the
equilibrium
equations,
we
have
1
x
=
0;
14
kip>ft2
a
30
kip
12
kip>ft2x
b
x
V
0
is 12
shown
As
x
y- 30 kip1x2
1218kip>ft2x
abkip>ft2x
14M
kip>ft2
a
9
a302bkip1x2
+
14+segment
kip>ft2
xxba +in
b Fig.
y ft
M18 ft
a+
b 6–6c.
+= 014
p/
271k
4ftax+
V x (c) V+ 12 kip>ft2xx221
kip/2ft 30 kip1x2
2
x 3 V 9
2 b 3+ 2M2 =and
18
ftxrectangular
3 1 kip1x2
2
2replaced
x 2 18 ft
-30
kip>ft2a
bxa
0 2trian
4
x3 + 12 kip>ft2xa b2 + 14
18
by
42 kip
x
V
1
x
2
x
x 2 x x18 x
1 x
2 Equation
2x2 2 2 may
18be
ft checked
3= ¢by
0; 14 4x xx23xx
x
V
30
2x
≤
6
kip/ft
3
noting
that
d + ©M = 0; d ++©M
3
c ©F=y2x=
x
x - V = 0
intensity
of
xx - 12 kip>ft2x - 14
kip
bx
2 kip>ft2a
2
4 the triangular
kip/
ft 2 load
x at29 td
21818
x
2 x2 0;
2x 30
(1)
4 29 ft kip/
2≤ kip
VftV= =¢ 30
kip
2(1)
2x-- ≤xw
#
¢ 30-= -2x
V = ¢ 30 - 2x ≤2x kip x
18
2 (1) 30xkip 2 18 ft
3 dV>dx
30 kip
(2)
30 2x
kip
M
=
30x
x
kip
¢
≤
M
30x
kip
ft
¢
≤
1
x
18
M
=
30x
x
¢
Also,
This
equa
=
=
2
x.
2
x 27resultant
force 9and the27locatio
9 29 The
2x
kip/ft
4 491 kip/ft
# ft2 kip/
30kip>ft2xa
kip1x2
+x12
kip>ft2x
a(c)ftb + x614bkip>ft2
a
b(c)
x a b 12+ft(c)
M =M
0 =6¢ 30xdKontroller:
30 kip
ft when x(2)= 18 ft,
- ©M
kip
1818
- 30 kip1x26 6+ -12
b +
14 kip/
kip>ft2a
=also
0;
and
x+
=x 0,-w
=≤ -shown.
2 kip>ft,
2 ft x a 3xb2 +18M
ft = 0 23 kip/ ft
(c) 2
Applying
the equilibriu
27
2
2
18
18 ft (1)
2 kip/ft
2 kip/ ftV = ¢ 30 -6 2x
d+
©M
=
0;
d
+
©M
=
0;
= 0;
6 kip/be
ft checked
Equation
2 may
that
dM>d
kip/ft3 ≤ kip
Equation
by noting
thatbedM>dx
that is,1by
Eq.
1. t
= by
V,xnoting
6 kip/
ft kip
Equation
2checked
may be
checked
noting
M2 may
30 kip
42
M
9
x
x
3
2that
kip/ft
2noting
-Shear
30
kip1x2
+
12
kip>ft2x
ax. -bsince
+-Eq.
14
kip>ft2
2 equation
x
2
6
kip/ft
and
Moment
Diagrams.
Equati
Equation
2
may
be
checked
by
that
is,
1.
dM>dx
=
V,
M
x
x
1
x
x
#
M
x
1
x
x
V
Also,
This
w
=
dV>dx
=
2
x
1
x
x
Also,
This
equation
checks,
when
w
=
dV>dx
=
2
x.
(2)
Also,
This
w
=
dV>dx
=
2
x.
2
kip/
ft
- x - -30
kip
≤ kip1x2
(b)
2 kip/
ft
kip/ftft(2)
9 kip
30
12
2 =
29 kip>ft2x
9kip>ft2a
3+ +
Vb 0;
x -M
x2a =- ¢ 30x
1212
kip>ft2xa
b b+ + 14
bxa
M
=
0 -0of
¢14x30x
kip>ft2x
14
kip>ft2
a+ c ©F
b yx a=
b+the
+
M
p1x2
+ 12 kip>ft2x
bM+=
kip>ft2
b≤xkip
a b# ft+ M
0x2kip1x2
27-=30
t
Fig.
6–6d.
point
maximu
d+a ©M
Also,
This
equation
since
=2a42-2
-2x292x.
= in
0,
w
-18
2ftkip>ft,
x Fig.
=when
18
ft,xw= =1
when
6–6a.
x x= w
0, =
w dV>dx
= - 2 kip>ft,
xx==
ft,
w
=and
6when
kip>ft,
and
when
=
0,
w
= 3checks,
18
2kipand
18
ftx3Since
32-kip>ft,
2 2=kip/ft
20;3 3 V V
18 27
ft
3
30 kip
x
x x
x x
2 x = 20, w = 1-2 kip>ft,
x3 1
x
(Eq.
6–2),
then,
from
Eq.
dM>dx
=
V
=
0
and
when
Fig.
6–6a.
x
=
18
ft,
w
=
-6
kip>ft,
2
4
x
6 kip/ ft
V(kip)
Equation22 may
be
noting
that
that
is,
Eq.
1.
dM>dx
=
V,
x
1
x
x
3
3
30
kip
3checked by
M
=
30x
x
¢
≤
2
2
2and
x b2 +Moment
2 noting
Equation 2 may-30
be checked
that
Eq.bxa
1. (c)
= kip>ft2a
V, that is,Shear
2
2 by
xShear
x and
x
18 = 0 Diagrams.
Shear
Moment
Diagrams.
Equations
1 and 2(2)
are
plotted
and
Moment
Diagrams.
kip1x2
b + 14
M
2 2 30
2dM>dx
= ¢ 3027E
#
#
#
kip
(2)VEquations
kip
(2) 18
kip
M
=
30x
x
kip
ft
MAlso,
=30¢30
30x
x2+2 -12(c)kip>ft2xa
kip
ft
¢
≤
M
=
30x
x
kip
ft
≤
¢
≤
30
x
This
equation
checks,
since
when
w =- dV>dx
=
2
x.
2
2
ft
3
9
(c)
(c)
4
kip/
Fig.27
Since
point
of0plotted
maximum
Also, w = dV>dx = - 2 - 9 x. 27
This
equation checks, since
27Equations
inwhen
Fig.
6–6d.
Since
theftin
point
of6–6d.
moment
occurs
when
Shear
and
Moment
Diagrams.
1the
and
2Vare
inmaximum
Fig.
6–6d.
Since
the
point
of
mam
=
= 30
- 2x
18 ft
kip
kip
kip
42 kip Equation 2 may be checked
6–6a.
x = 0, w =30 2 kip>ft, and when x =3042
18
ft, 36
w 30= kip
- 6 kip>ft, Fig. 42
6kip
kip/
by
noting
that
d
(Eq.
6–2),
then,
from
Eq.
1,
dM>dx
=
V
=
0
when x = 18 ft, w = - 6V(kip)
Fig.
6–6a.
x = 0, w = - 2 kip>ft, andV(kip)
kip>ft,
(Eq.
6–2),
then,
from
Eq.
1,
dM>dx
=
V
=
0
(Eq.
6–2),
then,
from
dM>dx
=
V
=
0
in
Fig.
6–6d.
Since
the
point
of
maximum
moment
occurs
when
2 kip/ ft x(ft)
x 1. V(kip)
d + ©M
=that
0;
21. 1.
630x
kip/
Equation
2
may
be
checked
by
noting
that
that
is,
Eq.
dM>dx
=
V,
on 2 may be checked by noting
is,Eq.
=
V, ftthat
30 kip that dM>dx
#
Equation
2
may
be
checked
by
noting
that
is,
Eq.
dM>dx
=
V,
(2)
M = 6 ¢kip/ft
- 42
x2kip
kip
ft
≤
Also,
This
equat
w
=
dV>dx
=
2
x.
9.735 ft = V = 0 (Eq. 6–2), then,
2 kip/ ft dM>dx
Choosing
root, 9 6 kip/ ft
from Eq.the
1, positive
2 and
Moment
Diagrams.
Equations
1w and
2 are =plotted
30
x12
27are
30
30
V(kip)
x2since
x 18- ft,
Thisx
equation
checks,
since
when
dV>dx
-29 x.29 x.This
This
equation
checks, 1since
when
w Shear
= dV>dx
-Shear
22 kip/ft
equation
when
wplotted
= =dV>dx
= x- 2- 2- M
2 kip/
and= Moment
Equations
andAlso,
2Also,
and
when
== 0,
w--=checks,
2-kip>ft,
x
=
9 x.ftDiagrams.
V
=
0
=
30
2x
V
=
0
=
30
V
=
0
30
2x
30
kip1x2
+
12
kip>ft2x
a
b
+
in Fig.
6–6d.
the
of maximum
moment
when
2 18
kip/
x = 9.735
Vand
30 xpoint
x62kip>ft,
when
Fig.that
win=642
-kip
2 kip>ft,
=Since
18be
ft, checked
w
=point
- 6 by
kip>ft,
when
Fig.
6–6a.
x=
=
=-=2-V,
2occurs
kip>ft,
=18
#42
9 Fig.
and
when
6–6a.
x 6–6a.
0, 0,
w w=when
kip>ft,
x x= x(ft)
ft,ft,
wftw
= =-6-kip>ft,
2 ft 92
Fig.
6–6d. and
Since
the
of
maximum
moment
occurs
3 is,
kip/ft
x(ft)
Equation
2
may
noting
that
Eq.
1.
dM>dx
x(ft)
V = 0 Shear
=Thus,
30 -from
2x from Eq.M(kip"ft)
1,
dM>dx = V = 0 (Eq. 6–2), then,
x ft Mxmax ! 163 kip"ft
and
Moment
Diagrams.
Equatio
9.735 ftChoosing
Eq.
2, root,
9.735
9.735
Choosing
the
positive
9the
then,
from
1,2 x. This equation
dM>dx = V = Also,
0 (Eq.w6–2),
Choosing
positive
root,
the
positive
checks,
since
when root,
= dV>dx
=ft -Eq.
21 -and
x(ft)
9 2 are plotted
2
2 Diagrams.
2and
and Moment Diagrams.
Equations
Shear
and
Moment
Diagrams.
Equations
1and
and
2 the
are
plotted
Shear
Moment
Equations
1
2
are
plotted
in
Fig.
6–6d.
Since
point
of
maximum
x
30
kip
M
=
30x
¢
9.735
ft V
Diyagram
x when
= 9.735xft=
and
Fig.
6–6a.
= 0, wof=maximum
-çizimi:
2 kip>ft,
= 2x
18
ft,
wChoosing
= 6–6d.
- 6 kip>ft,
30
42
kip of maximum
the positive
root,
9.735
x = 9.735
ft
= when
0 =x2occurs
30x - Fig.
(c)
#42
6–6d. Since thexpoint
moment
when
inkip
Since
the
point
moment
occurs
#42
#42
in
Fig.
6–6d.
Since
the
point
of
maximum
moment
occurs
when
(Eq.
6–2),
then,
from
Eq.
1
dM>dx
=
V
=
0
30
kip
42
kip
9
V
=
0
=
30
2x
V(kip)
M(kip"ft) M(kip"ft) M
30 kip
42 kip! 163
M(kip"ft)
163M
kip"ft
!then,
163from
kip"ft
kip"ft
M
Thus,
from
2, M
(Eq. 6–2),
then, from
Eq. 1,
= V = 0x(ft)
max2, = 3019.7352 - 19.735
maxthen,
Thus,
Eq.
(Eq.
6–2),
from
Eq.
1,Eq.
dM>dx=Thus,
=V V=from
=0max
0!
x =Eq.
9.735
ft from
max
Eq.
2,
9
(Eq.
6–2),
1,
dM>dx
V(kip)
#42
x(ft)
Shear
and
Moment
Diagrams.
Equations
1
and
2
are
plotted
V(kip)
30
6 kip/30
ft kip
Equation 2 may be checked
by not
Choosing M(kip"ft)
the positive root,
42 kip
163 maximum
kip"ft
Mmax ! of
=3 kip
0 =# ft
30 -2 2x 2 x2
Thus,
from(d)
Eq. 2, when
Choosing the positive
= V163
x2 point
in
6–6d. Since the
moment
occurs
3030Fig.root,
x
Also,
w
=
dV>dx
=
2
V(kip)
19.7352
2
kip/
ft
42#42
kip
x = 9.735 ft
V = 0 = 30 - 2x 9 x.2 Th
V = 0=x(ft)
=3030- -2x2x
- -- 19.7352
2 = 3019.7352
M0,
-and
19.7352
M
= 3019.7352
-19
Eq. 1, 9.735 ft Fig. 6–6 VM=max0 =
dM>dx = V =x 0= (Eq.
3019.7352
96–2),
maxw9
9.735
ft then, from
max
when
x
=
=
2
kip>ft,
x
9
3
x(ft)
x(ft)
x(ft)
! 163 kip"ft
30
Choosing
the positive
root,
x(ft)
27
19.7352
Thus, from Eq. 2,
x(ft)
2
2
9.735
#
(d)
#
ngThus,
the positive
root,
Choosing
the
positive
root,
(d)
#
from Eq.
2,
=
163
kip
ft
9.735
ft ft
M
=
3019.7352
19.7352
(d)
x
= x163
kip
= 163 kip ft
Choosing the positive root,
max
=Diagrams.
9.735ftft
Shear and27Moment
V = 0 = 30 - x(ft)
2x 3
#42
19.7352
x(ft)
x = 9.735 ft
x
=
9.735
ft
Fig.
6–6
9
Fig.
6–6
#
M(kip"ft)
Fig.
6–6
(d)
= 9.735
ftft from
#42
3 2
=x 163
kipThus,
Mmax ! 163 kip"ft
x(ft)
inEq.
Fig.
#42
Mmax = 3019.7352
19.7352
- 30 kip
19.7352
9.735
ft2, 6–6d. Since the point of
M(kip"ft)
42 kip
2 kip"ft
!
163
M
27
M(kip"ft)
rom Eq. 2,x(ft)
max
Thus,
from
Eq.
2,
Choosing
the positive
root,
Mmax = 3019.7352
19.7352
! 6–6
163 kip"ft
M-max
dM>dx = V = 0 (Eq. 6–2), then, fr
Thus, from Eq.
2,
Fig.
V(kip)
= 163 kip # ft 27
x
=
9.735
ft
30
= 163 kip # ft
#42
19.73523
19.7352
M3 3 = 3019.7352 - #42
19.7352
2
2 19.7352 max
M(kip"ft)
= 0 = 3
x(ft) - 19.7352
Mmax = Thus,
3019.7352
19.7352
M
=
3019.7352
Mmax ! 163Vkip"ft
2
ax ! 163 kip"ft
from-Eq.
2,
max
M
=
3019.7352
19.7352
max
27 x(ft)
x(ft)
27
(d)
= 163 kip # ft
27
# ft x(ft) Choosing
(d)
= 163 kip # ft
9.735
ft = 163 kip
3
the
positive root,
#
(d)
=
163
kip
ft
19.7352
Fig. 6–6
M
= 3019.7352 - 19.73522 Fig.
6–6
max
x = 9
Fig. 6–6
#42
x(ft)
27
x(ft)
M(kip"ft)
Mmax ! 163 kip"ft
Thus, from Eq. 2,
= 163 kip # ft
(d)
x(ft)
(d)
Fig. 6–6
PAGE !64
Fig. 6–6M
max
= 3019.7352
= 163 kip # ft
6.1
261
6.1
SHEAR AND MOMENT DIAGRAMS
261
SHEAR AND MOMENT DIAGRAMS
EXAMPLE 6.4
EXAMPLE 6.4
6.1 SHEAR AND MOMENT DIAGRAMS
261
Draw the shear and moment diagrams for the beam shown in Fig. 6–7a.
nt diagrams for theDraw
beamthe
shown
in and
Fig. 6–7a.
shear
moment diagrams for the beam shown in Fig. 6–7a.
15 kN
6.1 SHEAR AND MOMENT DIAGRAMS
261
EXAMPLE
6.4
5 kN/ m
!
Ölçü
ve
yükleme
durumu
şekilde
15 kN
6.1 SHEAR
AND MOMENT DIAGRAMS
261
15AND
kN MOMENT DIAGRAMS
80 kN!m
6.1 SHEAR
261
5 kN/m
6.1 SHEAR
AND
M
OMENT
DIAGRAMSkirişin
2 6 1kuvveti ve5 kN/
m
80 kN!m
verilen
basit
mesnetli
kesme
moment
80
kN!m
Draw the shear80and
moment
for the
beamDshown
6.1diagrams
SHEAR AND
MOMENT
IAGRAMSin Fig. 6–7a.
261
6.1
SHEAR
AND
OMENT
DIAGRAMS
2 16 1
kN!m
6.1
SHEAR
AND
MM
OMENT
DIAGRAMS
26
C
M
diyagramlarını çiziniz.
A
C
C
EXAMPLE 6.4 A
15 kN
V
5 kN/ m
ms for the
in
Fig. 6–7a.
B x1
EXAMPLE
6.4
.4 B beam shownEXAMPLE
6.1
80
kN!m
6.4
Draw
the
shear
and
moment
diagrams
for
the
beam
shown
in Fig. 6–7a.
m
m Çözüm:
5.75 kN
5m
5m
ment
diagrams5for
the beam
shown in Fig. 6.4
6–7a.
EXAMPLE
M
6.1
261
SHEAR
B AND MOMENT DIAGRAMS
x1
V5 m
580
m kN!m
SHEAR AND MOMENT DIAGRAMS
M
(a)
5.75 kN
261
5
Nwn in Fig. 6–7a.
C
(a)
(b)
Draw
the
shear
and
moment
for
the
beam
shown
Fig.
6–7a.
(a)diagrams
(b)
nd moment
diagrams
the beam
shown
inmoment
Fig.
6–7a.
5 kN/ m for
Draw
the
shear
for
the
beam
shown
inin
Fig.
6–7a.
A80and
kN!m
15
kN diagrams
x1
V
AB
ve
BC
bölgelerinden
kesimler
yapalım:
B
SOLUTION
EXAMPLE
15 kN
5
kN/
m
Draw the6.4
shear and moment diagrams
for
the
beam
shown
in
Fig.
6–7a.
M
80 kN!m
5
kN/m
15
kN
80
kN!m
15
kN
5(x2 " 5)
15 kN 515
5(x2supports
" 5)
5 m 80 kN!m
m kN Support
m
5.75 kN
Reactions. The
reactions
have been
80 kN!m
M 15atkNthe
C SOLUTION
5 kN/
5 kN/m
5 kN/
mm
kN!m
Draw the shear
and moment
diagrams
the beam
in Fig. 6–7a.
80 for
kN!m
8080
kN!m
M shown
15 kN
(a)
80
kN!m
(b)
x
M
determined
and
are
shown
on
the
free-body
diagram
of
the beam,
V
80
kN!m
80
kN!m
e reactions at the
supports
have beenThe1 80
C supports5 kN/
Support
Reactions.
reactions
at the
kN!m
m been
80 kN!m 80 kN!m
M have
MM
CA
x
V
80
kN!m
Fig.
6–7d.
1
n on5 the
diagram ofand
the
beam,
B on the
15 kN
determined
are
shown
free-body
diagram
C the
m free-body
M beam,
M
kN
M
Cof
CV5.75 A
x1
V 5 kN/ m
B
x1
15 kNV 5(x2 " 5)
SOLUTION
A
80 kN!m
x
Fig.
6–7d.
6.1
SHEAR
AND MOMENT Dx
IAGRAMS
261
V
x
V
80
kN!m
5
m
5
m
B
C
1 1
5.75 kN
1
(b)
B
B
Shear
and
Moment
Functions.
Since
there
is
a
discontinuity
of
M
5m
5 5.75
m Support
kN5 m B at the supports
x1
V
Reactions.A The
reactions
V have been
5m
kN
V
(a)55.75
(b)
80
kN!m
5 m 5.75 kN
52m
x
"
5
x
"
5
5.75
kN
x
"
5
x
"
5
m
5
m
C
5.75
kN
2
5
m
5
m
distributed
load
and
also
a
concentrated
load
at
the
beam’s
center,
2
2
ctions. (a)
Since there
is aand
discontinuity
Shear
Moment
Functions.
Since
there is a discontinuity
of
A are of
(b)
determined
shown on515
the
free-body
of the beam,
(b) and
5(x2 " 5) 25diagram
x1 kN(b)(b) V 2
2
m regions
2
5.75
Bm kN
(a)(a)
(a)
ofcenter,
x must be considered
in order
to describe
theMshear and
(b)attwo
a concentrated
load
at
the
beam’s
center,
distributed
load
and
also
a
concentrated
load
the
beam’s
x
x
Fig. 6–7d.6.4
2 2 " 5)
15 kN 5(x
EXAMPLE
SOLUTION
(a) 2
(b)
ons
at
the
supports
have
been
5
m
5
m
moment
functions
for
the
entire
beam.
5.75kN
kN
nsidered in order two
to describe
80and
kN!m
regionsthe
of xshear
must
be
describe
5.75 kN in order to 15
kN 5(x2the
" 5)shear and 5.75
15
kN
5(xconsidered
5m
V5)
2 " 5)
15
kNkN5(x5(x
5)
SOLUTION
25"
Support
Reactions.
Thethe
reactions
at the
supports
have
been
x15
(a)
2"
free-body
ofSOLUTION
the
beam,
(b)
15
kN
5(x
2 " 5 x2 "
80of
kN!m
2 " 5)
entire
beam.diagram
Shear
and
Moment
Functions.
Since
there
is
a
discontinuity
moment
functions
for
entire
beam.
M
(c)
(c)
Draw
the
shear
and
moment
diagrams
for
the
beam
shown
in
Fig.
6–7a.
Fig.
6–7b:
0
…
x
6
5
m,
15
kN
5(x
SOLUTION
2 " 5)
1
Thehave
reactions
at 80
the
supports
have on
been
2
2
rts
been
determined
and are
shown
free-body
diagram
the
beam,
80reactions
kN!m
kN!m
Reactions.
The
reactions
atthe
the
supports
have
been 8080
Support
Reactions.
The
at
have
been
M
distributed
load
and have
alsothe
abeen
concentrated
load
atofsupports
the
beam’s
center,
kN!m
kN!m
ions. The reactions
atSupport
the supports
x2
80
kN!m
15
kN
5(x
"
5)
Fig.the
6–7b:
0 SOLUTION
… determined
xdiagram
6
5 m, of
own
free-body
beam,
Support
Reactions.
reactions
at
the
supports
have
been
15free-body
kN
m
of on
thethe
beam,
2
15 kN
Fig.
6–7d.
1determined
80 k
80 kN!m
and
are
shown
onthe
the
free-body
diagram
ofthe
the
beam,
+ cdiagram
©F
=the
0;
5.75
515
mkNThe
are
shown
on
ofshear
M
V
6beam,
M
6
two
regions
of xand
must
be
considered
order
and
are
shown
free-body
diagram
of
the
beam,
MM
5.75
kN kN - V = 0
5ykN/m
x2in
"the
5 x2 free-body
" 5to describe
5 kN/ m
Mbeam,
Since
thereon
is the
a discontinuity
of
determined
and 80
are
shown
on
diagram
of
the
5
m
Support
Reactions.
The
reactions
at
the
supports
have
been
5
kN/
m
V
Fig.
6–7d.
M
Fig.
6–7d.
!m kNbeam.
80 kN!m
kN!m
8080kN!m
cmoment
(1)
5.75 kN - VShear
= 0+ and
for thekN
entire
©FyMoment
= functions
0;80 kN!m
5.75
- V
2 = 02
V = 5.75
x2 "kN
5 (c)
x2 " 5
Since
of
Fig.
trated load at
the beam’s
center,
determined
andFunctions.
are shown on
free-body
diagram of the
x2 there
5 m6–7d.
5 m is a discontinuity
V the
V beam,
52Mm
5m
V
M
V
2
x
"
5
x
"
5
x
"
5
x
"
5
A
(1)
5
m
(1)
2 a concentrated
V
"2 "
5= 5x0
"
5 5
V = 5.75
kN
V = 5.75
kN
dbeam’s
+2 ©M
kN
2"
2x
2x
discontinuity
of
distributed
load
also
load
at
the
center,
unctions.
Since
there
a and
discontinuity
C0;
C
1 + xM
5xm
in order
to describe
shear
Shear
Moment
Functions.
Since
there
a"
discontinuity
of m - 5.75
Shear
Since
there
is isxa22=
discontinuity
x2 kN
Fig.
6–7b:2ofFunctions.
0the
…
xis
5and
V
Fig.
6–7d.
5 x2 " 5 - 80 of
5.75
15 kN
1 6 and
ment
Functions.
Since
there
ism,
aMoment
discontinuity
of
x2 " 25 2x2 " 5
C
2kN 2
6
2
2
2
2
Shear
and
Moment
Functions.
Since
there
is
a
discontinuity
of
5 kN/
m (2)
A also
AkN
A beam’s
two
of=
x 0;
must
be
in
order
toBxdescribe
shear
and
beam’s
center,
2 beam’s
2 5.75
so
at
distributed
and
a aconcentrated
load
at
the
center,
and
also
concentrated
load
the
beam’s
center,
ma -concentrated
5.75
x1regions
+
=distributed
0the
m.
51V
mB
xload
d load
+M
©M
-considered
80center,
kN
m
5.75
kN
+ xM
=the
0at
V 2
M
= 15.75x
+ 802 kN
(c)
xx2 x"
2 load
2 m
1
and
also
a kN
concentrated
load
the
beam’s
center,
5 x "2 5
B5.75
80 kNx!m
c ©F
xa2 discontinuity
+functions
= at
0;Moment
- V 1=there
0 25 mis
distributed
load
and
also5kN
a concentrated
load
at the
beam’s
center,
m
5m
52 m
5.75
Shear
and
Functions.
Since
of
y regions
x2 2
moment
for
the
entire
beam.
e
the
shear
and
considered
in
order
to
describe
the
shear
and
two
regions
of
x
must
be
considered
in
order
to
describe
the
shear
and
two
of
x
must
be
considered
in
order
to
describe
the
shear
and
(c)
5.75
kN
5.75 kN+ 802 kN m5 m 6 x … (2)
5.75
kNkN
5.75
2
2
(2)
=
15.75x
m
M
= a15.75x
Fig.
6–7c:
10(1)
m, 5.75
must
be considered
ordertwo
to describe
the
shear
and
1 + 802 kNin
1 55.75
kN
2
regions
of
xalso
must
be
considered
in
order
to
describe
the
shear
and
distributed
load
and
concentrated
load
at
the
beam’s
center,
5
m
m
15
kN
5.75
kN
V
kN
x
V
= 5.75
kN (c)
5.75
kN
34.25 6kN
2
5.75 kN
moment
functions
for
the
entire
beam.
moment
functions
for
the
entire
beam.
he
entire beam.
(c)
(c)(c) 34.25 kNC
m (c) the
Fig. of
6–7b:
0 …beam.
x51 two
m,
for
the entire
beam.5tokN/
15
kN
xfunctions
must
be considered
in order
describe
shear
and
c:s for the entire
Fig.
6–7c:
m6 65regions
xmoment
m,
(c)
5.75
kN
(a)
(b)
2 … 10
+ c ©F
- 15 kN - 5 kN>m1x
- 5m m2 -6V = 0
V (kN)
VA(kN)
kN!-m80 kN
B
- V = 0
d +0©M
0;65805m,
m - beam.
5.75 kN x1 + M
= 0y = 0; 5.75 kN
52 kN/
moment
for
the
entire
Fig.
6–7b:
Fig.
6–7b:
0… …x=1functions
x16
15(c)
kNkN5 m
15 kN m,
:g. 6–7b:
15
5m
15 kN 15 kN
80 kN!m 6
65.75
6
6
+ c ©F
=
0;
5.75
kN
V
=
0
Fig.
6–7b:
0
…
x
6
5
m,
c
15
kN
y
15
kN
5
kN>m1x
5
m2
V
=
0
1
+
©F
=
0;
5.75
kN
15
kN
5
kN>m1x
5
m2
V
=
0
5 kN/
m(3)
5 kN/
m
5 kN/
m
2
(2)
y(1)
2
5 kN/m
.75 kN
6 115.7515 6
M =
15.75x
m
=
5x5(x
C
1 + 802 kN
222kN
5 kN/
m
5 kN/m5.75V
kNkN
" 5)
Fig.
6–7b:
5y=m,
80
kN
!
m
c1©F
80 =
kN00!m…+ x
80
kN
!
m
c6
kN
34.25
kN
0;
5.75
kN
V
=
0
15
+
©F
=
0;
5.75
kN
V
=
0
80
kN
!
m
(1)
y
5.75 kN5.75
-SOLUTION
V
V
=
5.75
kN
6
5.75
5.75
80
kN
!
m
c
C
80
kN
!
m
A
+
©F
=
0;
5.75
kN
V
=
0
kN
V
=
0
5
kN/
m
B
y (3)
(3)- 80 kN m - 5.75 kN x + 15 kN1x - x(m)
5=kN
x1 + -M5x=220kN5 m 6 x2 … 10
115.75
6–7c:
= 115.75
5x22 supports
kN d + ©M
m, Fig.V
= x(m)
0;been
5 m2
VkN
(kN)
Reactions.
reactions
at=--=5+
the
have
2
2
80
!m B
(1)80
VV
5.75
kN
(1)
m
m
A (1)
+=c 0;
©F
5.75
kN
V
5.75
kN!m
CC
y = 0;
(1)m5 dVSupport
+ ©M
- 80 kNThe
5.75
x
M=
=0kN
0kN
CkN
V
=
5.75
kN
(1)
V
=
5.75
1
(1)
C
=
5.75
kN
C
(2)
802 kN
C
5
m
5
m
determined
and
are
shown
on
the
free-body
diagram
of
the
beam,
c
A
-+ 5.75
kN xm
+
15
kN1x
5
m2
d
+
©M
=
0;
m
5.75
kN
x
+
15
kN1x
5
m2
80
kN
+
©F
=
0;
5.75
kN
15
kN
5
kN>m1x
5
m2
V
=
0
"9.25
"9.25
A
B
(1)
2 A
2©M
25.75
2
y©M=B=
2
M
V
=
5.75
kN
d
+
0;
80
kN
m
5.75
kN
x
+
M
=
0
B
5.75
kN
34.25
kN
x
5
m
d
+
0;
80
kN
m
kN
x
+
M
=
0
A
C
1 1x + M
A
2
(2)
M
=
15.75x
+
802
kN
m
B
B
A
kN-80
m kN
- 5.75
kN
x
+
M
=
0
d
+
©M
=
0;
80
kN
m
5.75
kN
=
0
1
B1
1 kN 5xm
5m
m Fig.
- 5.75
5¢m5 m
m = 0
+ 5 kN>m1x
≤ 5+m5 M
6–7d.
1 + M = 0 5m
kN 5.75
2 - 5 m2
A
5xmkN
5m
B 2 34.25 kN 5 m
d + ©M = 0; V (kN)
- 80 kNV
m
-15.75x
5.75
+
M
=m
0 5 m 5.75
m
MM
kN
(3)(2)
(2)(2)
5 6mkN
55+x802
12
==M115.75
kN
(2)x2 5+-m802
=
15.75x
802
kNkN
m
2 1--1+5x
2m
Fig.
6–7c:
x
…
10
m,
=
15.75x
+
802
m
x(m)
(2)
5
m
5
m
M
=
15.75x
m
2
1
5.75
kN
34.25
kNkN
(2)
5
m
M
V
M ¢= 115.75x
+
kN
5.755.75
kNkN
34.25
V (kN)
5 m2
+V M
+ 5 kN>m1x
- 15.75x
5kN
m2 ¢ + 802 kN
≤ 802
≤ +m M = 034.25 kN
5.751kN
kN
25 234.25
34.25
kN
x2 " 5 x2 " 5
(2)
5.75
kN>m1x
=m
0=6x0m
M
= 6–7c:
5.751kN
34.25
2 - Shear
2
Fig.
6–7c:
5©M
m
6m
10
m,
Moment
Functions.
Since
there
discontinuity
ofM kN
25 m2 d-and
2 + is15akN1x
Fig.
5
…
10
m,
2x2…
(4)
Fig.
6–7c:
5
6
x
…
10
m,
=
1
2.5x
+
15.75x
+
92.52
kN
m
V
(kN)
5.75
kN
34.25
kN
+
=
0;
m
5.75
kN
x
5
m2
80
kN
"9.25 2
V2 (kN)
c ©F
2 = 0
2
V (kN)
= 0; x5.75
kN2m,
- Fig.
15 kN
- 5 kN>m1x22 - 5 m2 - V
M (kN!m)
M (kN!m)
V (kN)
Fig. 6–7c: +distributed
m,–7c:
y
"34.25
"34.25center,
5.75
6–7c:2V (kN) V (kN)
5 m 6load
10 also
a concentrated
loadkN>m1x
at the beam’s
2 …and
x2
-2 5x215.75x
2 kN 2 + 92.52
V (kN)
c (3)
+kN
©F
=M
0;(4)
5.75
kN
- -15
kN
-x-25 5
m2
V
=
0
(4)
c+©F
=
1
2.5x
+
15.75x
+
92.52
kN
m
5x
m
x(m)
+
0;
5.75
kN
15
kN
5
kN>m1x
5
m2
V
=
0
y©F
2
c
2
5.75
2- 2-15+
VkN
= 0- 5 two
=
0;
5.75
kN
15
kN
5
kN>m1x
5
m2
V
=
0
y =
2
108.75
108.75
These
results
can
be
checked
in
part
by
noting
that
w = dV>dx
5
m
y
2
(3)
2
regions
of
x
must
be
considered
in
order
to
describe
the
shear
and
kN>m1x
5
m2
V
=
0
V
=
115.75
5x
2
kN
80
75 kN - 15 kN - 5+kN>m1x
- 5+5.75
m2
-kN
V -= 15
0 5 m2
c2©Fy =2 0;
x(m)
-2 ¢5 kN>m1x2 ≤- +5 M
m2 =- 0V = 0 5.75 kN
5 kN>m1x
5.75
2V-kN
5.75
N x2in+ part
15
kN1x
5
m2
5.75
(3)
and
Also,
when
Eqs.
1
and
2x(m)
give
V
=
dM>dx.
x
=
0,
"9.25
5.75
=
115.75
5x
2
kN
2
(3)
1
These
results
can
be
checked
in
part
by
noting
that
w
=
dV>dx
moment
functions
for
the
entire
beam.
2
ked
by
noting
that
w
=
dV>dx
V
=
115.75
5x
2
kN
2
(3)
(3) d + ©M = 0; - 80 kN m(3)
V
=
115.75
5x
2
kN
(c)
2
5.75
2
5.75
x(m)
80
80
x(m)
5.75
x(m)
- 5.75
x2 + 15
kN1x
m25.75 kN and
(3)kN
V = 115.75
- 5x22-kN
"9.25m; when x = 10 m, Eqs. 3 and 4
(3) Mx(m)
V == 115.75
2=kN
2 -V5 =
V
= 115.75
-0,
5xEqs.
2and
= M
80(kN!m)
kN
22 kN
x(m)
"34.25
x(m)
2
and d5x
when
1kN1x
and -2 5give
dM>dx.
x1 =kN
when
1V
give
0, Eqs.
2 5.75
1
x2 - 5xm
+d5©M
=2=0;
mm
x
+x92.52
15+
m2
-Also,
80
kN
"9.25
(4)
2
2
+
©M
0;
5.75
kN
x
+
15
kN1x
5
m2
80
kN
M
=
1
2.5x
+
15.75x
+
kN
m
Fig.
6–7b:
0
…
x
6
m,
d
+
©M
=
0;
m
5.75
kN
15
kN1x
5
m2
80
kN
"9.25
"9.25
15
kN
2
2
2
2
2
2
1
-80
5
m2
"9.25
108.75
give
V
=
34.25
kN
and
M
=
0.
These
values
check
with
+
M
=
0
¢m
≤
6 the
x
5
m
and
when
x
Eqs.
3
and
4
V
=
5.75
kN
M
=
80
kN
m;
=
10
m,
when
x
Eqs.
3
and
4
kNkN
m;
=
10
m,
d
+
©M
=
0;
m
5.75
kN
x
+
15
kN1x
5
m2
80
kN
"9.25
- 25.75
x22 kN
+ 15
- 5 m2
m -kN5.75
x2kN1x
+ 152 kN1x
2
"9.25 2
"9.25
2 2
5 kN/ m
2 - 5 m2
+ 5 kN>m1x
- 5 m2 ¢
+-xM
=m
0m reactions
≤xvalues
5
m
2 and
support
shown
on
the
free-body
diagram,
Fig.
6–7d.
2
5
x
5
80
kN
!
m
c ©F
M
(kN!m)
give
V
=
34.25
kN
M
=
0.
These
check
with
the
These
results
can
be
checked
in
part
by
noting
that
w
=
dV>dx
=
0;
5.75
kN
V
=
0
M = 0. These+values
check
with
the
2
2
y
x(m)
80
+ +5 +
kN>m1x
525
m2
=M
x
m ≤ ≤+ ≤+M+
5 kN>m1x
5¢2m2
0 (kN!m)
2 2- 5
kN>m1x
m2
Mx(m)
=0 =
0M
¢- ¢5"34.25
xm
5(4)
m
xkN
-support
5and
2 2 - V
2 diagram,
=kN>m1x
02 free-body
2Eqs.
+5 92.52
m
+
5shown
kN>m1x
-5.75
5free-body
m2
Mand
=(1)
0 2 give
¢1 = 0,
≤2 +1Fig.
reactions
on
the
diagram,
6–7d.
Also,
when
=
dM>dx.
x
n
the
Fig.
6–7d.
"34.25
2=
2
108.75
5.75x
5
m2
+
M
=
0
¢
≤
1x
m2
+
M
=
0
¢
≤
V
kN
2
2
2
MM
(kN!m)
C "34.25
and
Equations
1 through
4 "34.25
are plotted
(4) Moment Diagrams.
M (kN!m)
= and
1 - 2.5x
15.75x
+ 92.52
m m, Eqs.
(kN!m)
(d)
(d)2 kN Shear
2 (kN!m)
108.75
M
"34.25
x2 =
4 (4)M (kN!m)
V
=25.75MkN
M 2=
=+
kN
m;
2 22 when
2(kN!m)
"34.25
"34.25
A(4)
M2+
M
180
2.5x
+
15.75x
+ +210
92.52
kN
m3mand
rt
by
noting
that
w
=
dV>dx
M
(kN!m)
M
=
1
2.5x
+
15.75x
+
92.52
kN
m
B
(4)
2
2
d
+
©M
=
0;
80
kN
m
5.75
kN
x
+
M
=
0
2
in
Fig.
6–7d.
M
=
1
2.5x
15.75x
92.52
kN
108.75
"34.25
108.75
1
80
2
2
and
Equations
1w
through
4"34.25
are
(4)
rams.
Equations
1give
through
4-be
are
plotted
M
=Diagrams.
1 -in2.5x
+ 0.
15.75x
+values
92.52
kN
m
m= 1-2.5x
2TheseShear
2 that
V
= Moment
kN
M2 =by
These
withplotted
the
2 (4)
5m
m 108.75
108.75
can
checked
part
noting
=check
dV>dx
(4)
Fig.5108.75
6–7
x(m)
M12.5x
92.52
kN
m
Fig.
6–7
(4) and
15.75x
+results
92.52
kN
m34.25
80
Eqs.
and
2These
give
= 0,
2 1+ 215.75x
2 +
108.75
2 +
108.75
(2)
in
Fig.
6–7d.
M
=
15.75x
+
802
kN
m
results
can
be
checked
in
part
by
noting
that
w
=
dV>dx
These
results
can
be
checked
in
part
by
noting
that
w
=
dV>dx
1
support
reactions
shown
on
the
free-body
diagram,
Fig.
6–7d.
These
results
can
be
checked
in
part
by
noting
that
w
=
dV>dx
8080 80
and
Also,
when
Eqs.
1
and
2
give
V
=
dM>dx.
x
=
0,
These
results
can
be
checked
in
part
by
noting
that
w
=
dV>dx
1
5.75
kN
34.25
kN
w
=
dV>dx
when
x2 = 10 in
Eqs.by
3 and
4 =that
m,80part
80
(d)
necked
be checked
noting
w
=kN
dV>dx
Also,
when
Eqs.
2give
V
dM>dx.
x=1x10
=xm,
and
Also,
when
14and
2give
give
V=Fig.
dM>dx.
=
0,Eqs.
in partV5by
that
wdM>dx.
dV>dx
10,
and
Also,
when
1 and
2 give
V
0,Eqs.
6–7c:
m noting
6and
x2and
…
10
m,
80x
and
when
3 1and
=
5.75
kN
M
==dM>dx.
80
m;
Also,
Eqs.
1Eqs.
2and
Vthe
=
x21 =
0,
1 =
80when
nd
2
give
V (kN)
These
values
check
with
x(m)
and
when
x
Eqs.
3
and
4
V
=
5.75
kN
M
=
80
kN
m;
=
10
m,
and
when
x
Eqs.
3
and
4
V
=
5.75
kN
M
=
80
kN
m;
=
10
m,
Also, when
1and
give
x.
x1V
0,V
Shear
and
Moment
Equations
through
2with
o,
when
Eqs.
1Eqs.
2and
give
= 0,V
and
3plotted
and
4
= and
5.75
kN
M
80 kN
m; when
m,
x2check
3are
and
4
5.75
kN
M
m; when
=2x1
10
m,10Eqs.
== =34.25
kN
and
M2=Diagrams.
=80
0.=kN
These
values
the4Eqs.
2 =
Fig. 6–7
x(m)
qs.
3=and
4x1give
e-body
diagram,
6–7d.
cFig.
+m;
©F
=xV
0;
5.75
15
kN
-and
5 kN>m1x
- 5values
m2
- check
V =check
0with
give
V
=334.25
-kN
34.25
kN
=
0.
values
with
the
VV
=shown
-kN
kN
and
M
=M=These
0.
These
the
y give
2 These
d
when
Eqs.
3
and
4
MkN
80
kN
=
10
m,
in
Fig.
6–7d.
x(m)
x(m)
2
give
=
34.25
and
M
=
0.
values
check
with
the
give
=
34.25
kN
and
M
0.
These
values
check
with
the
when
x
Eqs.
and
4
80
m;
=
10
m,
(d)
support
reactions
on
the
free-body
diagram,
Fig.
6–7d.
x(m)
2
x(m)
kkN
with
the
x(m)
support
reactions
shown
on
the
free-body
diagram,
Fig.
6–7d.
support
reactions
shown
on
the
free-body
diagram,
Fig.
6–7d.
5.75
(d)
and
M
=
0.
These
values
check
with
the
support
reactions
shown
on
the
free-body
diagram,
Fig.
6–7d.
support
reactions
shown
on
the
free-body
diagram,
Fig.
6–7d.
(3)
x(m)
nd
M
=
0.
These
values
check
with
the
V
=
115.75
5x
2
kN
Equations
through 4 are plotted
2
x(m)
x(m)
g. 6–7d. on1 the
(d) (d)(d)(d)
Fig. 6–7 1 through 4 are plotted
free-body
diagram,
Fig. 6–7d. Equations
Shear
and Moment
Diagrams.
(d)
ns shown
on the free-body
diagram,
Fig.
6–7d.
Shear
and
Moment
Diagrams.
Equations
1
through
4
are
plotted
Fig.
6–7
Shear
and
Moment
Diagrams.
Equations
1
through
4
are
plotted
(d)
Shear
Moment
Equations
through
4 are4 plotted
©M
= Shear
0; and
m -Diagrams.
5.75
kN x2 +
15Equations
kN1x
51 m2
- 80
kNMoment
and
Diagrams.
through
are plotted "9.25
(d) 21 Fig. 6–7
ind +
Fig.
6–7d.
6–7
Fig.Fig.
6–7
gh 4 are plotted
Fig.
6–7
in Fig. 6–7d.
#
#
#
#
#
#
#
#
#
#
#
#
#
# #
## #
#
#
#
#
## #
#
#
#
#
#
#
#
#
#
#
## #
#
#
#
#
#
#
#
#
#
#
# # # #
# # # #
#
inin
Fig.
6–7d.
in Fig.
6–7d.
ment
Diagrams.
Equations
1 6–7d.
through
4 are plotted
Fig.
6–7plotted
Fig.
agrams.
Equations
1 through
4 are
+ 5 kN>m1x2 - 5 m2 ¢
Fig. 6–7
x2 - 5 m Fig. 6–7
≤ + M = 0
2
M = 1 - 2.5x22 + 15.75x2 + 92.52 kN # m
M (kN!m)
(4)
These results can be checked in part by noting that w = dV>dx
and V = dM>dx. Also, when x1 = 0, Eqs. 1 and 2 give
V = 5.75 kN and M = 80 kN # m; when x2 = 10 m, Eqs. 3 and 4
give V = - 34.25 kN and M = 0. These values check with the
support reactions shown on the free-body diagram, Fig. 6–7d.
"34.25
108.75
80
Shear and Moment Diagrams. Equations 1 through 4 are plotted
in Fig. 6–7d.
PAGE !65
x(m)
(d)
Fig. 6–7
right side,
where
0 6 and
k 6moment,
1 [for example,
if w(x)
is uniform,
both thethe
internal
resultant
shear
acting on
the right
face of k = 2 ].
Applying
of equilibrium
to the
have
the segment,
mustthe
beequations
changed by
a small amount
in segment,
order to we
keep
the
segment in equilibrium. The distributed load has been replaced by a
resultant force w1x2 ¢x that acts at a fractional distance k1¢x2 from
the right side, where 0 6 k 6 1 [for example, if w(x) is uniform, k = 12 ].
Applying the equations of equilibrium to the segment,
wex have
w(x)"
w(x) yöntemle elde edilmesi
6.2. Kesme kuvveti ve moment diyagramlarını grafik
Yayılı yük bölgeleri:
w(x)"x
F
w(x)
k(" x)
w(x)
6.2
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS
V
+ c ©Fy = 0;
F
6.2
M
w(x)
V + w1x2 ¢x - 1V + ¢V2 = 0
¢V = w1x2 ¢x
M0
GRAPHICAL METHOD FOR CONSTRUCTING
SHEAR AND MOMENT DIAGRAMS
"
x
V
x
M
x2 ¢x - 1V + ¢V2 = 0
¢V = w1x2 ¢x
x
M0
d+ ©MO = 0;
"x
263
k("x)
O
M ! "M
V ! "V
"x
M ! "M
-V ¢x - M - w1x2 ¢x[k1¢x2] + 1M
+ ¢M2
= 0
Free-body diagram
¢M = V ¢x + w1x2 k1¢x22
O
of segment " x
V !M"V
(b)
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND
OMENT DIAGRAMS
263
6.2 GRAPHICAL
M
ETHOD FOR CONSTRUCTING
S
HEAR AND MOMENT DIAGRAMS
263
"x
Fig. 6–8
6.2G
GRAPHICAL
M
ETHOD
FOR
CONSTRUCTING
SHEAR
AND
MOMENT
DIAGRAMS
M - w1x2 ¢x[k1¢x2] + 1M + ¢M2
= 0by ¢x and taking6.2
RAPHICAL
M:
ETHOD
FORabove
Cdiagram
ONSTRUCTING
SHEAR
AND M
OMENT
DIAGRAMS
¢x
0,
Dividing
the
limit
as
the
two
equations
Free-body
c
(a) 6.2
263 2
+ ©Fy = 0;
V + w1x2 ¢x - 1V + ¢V2 = 0
of
segment"x
become
2
V ¢x + w1x2 k1¢x2
+ c ©Fy = 0;(a)
V + =w1x2
¢V
w1x2¢x
¢x- 1V + ¢V2 = 0
(b)
c ©F=y 0;
= 0;
V w1x2
+ w1x2
¢x
- 1V
+ ¢V2
+ c+©F
V¢V
+
¢x¢x
- 1V
+ ¢V2
= 0= 0
y
= w1x2
Fig. 6–8¢V = w1x2 ¢x
¢V = w1x2 ¢x
he limit as ¢x : 0, the above two equations
dV
d + ©MO = 0; - V ¢x - M - w1x2 ¢x[k1¢x2] + 1M += ¢M2
w1x2 = 0
dx
2
d+ ©MO = 0;¢M-V
- M
- w1x2
¢x[k1¢x2] + 1M + ¢M2 = 0
= ¢x
V ¢x
+ w1x2
k1¢x2
+ ©M
= 0;- V-¢x
V ¢x
- M
- w1x2
¢x[k1¢x2]
+ 1M
+ ¢M2
d +d©M
- M
- w1x2
¢x[k1¢x2]
+ 1M
+ ¢M2
= 0 = 0 (6–1)
O 0;
O =
slope
of 2 distributed
¢M = V ¢x + w1x2
k1¢x2
2
2
= ¢x
V ¢x
+ w1x2
k1¢x2
¢M¢M
= V
+ w1x2
shear
diagram
load
intensity
= k1¢x2
dV
by ¢x and taking the limit as ¢x : 0, the above two equations
= Dividing
w1x2
at each point
at each point
dx
become Dividing by ¢x and taking the limit as ¢x : 0, the above two equations
¢x
:giderken
0, the
Dividing
taking
the
limit
as ¢x
above
equations
¢x
: 0,
Dividing
by by
andand
taking
theböler
limit
as ¢x
the
above
twotwo
equations
Denge
değerine
ve
limit
alırsak;
(6–1)
becomedenklemlerini
ope of
distributed
become
become
agram = load intensity
dV
dM
point
at each point
= w1x2
= V
dx
dx
dV
dVdV
= w1x2
= w1x2
= w1x2
dx
(6–1)
(6–2)
slope of
distributed
dx dx slope of
shear
w ! w(x
6
shear diagram slope
intensity
= load
(6–1)
dM
diagram
= at each
of moment
distributed
(6–1) (a)
(6–1)
Bu iki denklem bize kesme kuvveti
ve
diyagramlarını
hızlı bir
biçimde
elde etme
slope
of distributed
distributed
slope
ofmoment
= V
at each point
at each
point
dx
at
each
point
point
shear diagram
load
intensity
=diagram
shear
intensity
shear
diagram
intensity
= load
= load
imkanı sağlar.
at each at
point
at
each at
point
A
at each
point
at each
point
each
point
each
point
(6–2)
C
D
slope of
shear
w ! w(x) wB
6
nt diagram = at each
w = negative incr
These dM
two(a)equations provide a convenient means for quickly
V
slope = negative
in
V and moment diagrams for a beam. Equation 6–1
0
6
obtainingdx
the =shear
each point
point
"wC
dMthe slope
states that at a point
of the shear diagram equals the intensity
dMdM
"wD
= V
A
VA
=example,
V= V consider theBbeam in Fig. 6–9a.
dx
of the distributed
loading.
For
(6–2)
w
B
slope of
sheardx Cdx D
(b) w ! w(x)
The distributed loading is negative
and increases from zero to wB .
w = negative increasing
provide a convenient meansmoment
for quickly
diagram = V at each
(a)
(6–2)
(6–2)
Therefore, the shear
willslope
curve
has
a negative
slope,
wB
(6–2)
=anegative
increasing
slope0diagram
ofslope
shear
w ! w(x)w w
of
!Bw(x)
slope
ofbeshear
shear that
w ! w(x)
oment diagrams for a beam. Equation
V = positive decreasing
at each6–1
point
point
"w
C
-w
.
w
=
0,
-w
,
-w
,
increasing
from
zero
to
Specific
slopes
and
B at each at each
C
D
momentmoment
diagram
slope = positive decreasing
=diagram
ope of the shear diagram equals the intensity
diagram
= = at each A
moment
(a)
(a) (a)
"w
D
-wB are shown inVAFig. 6–9b.
A
M
B
"
For example, consider the beam in Fig. 6–9a.at each at
point
point
each
point
point
atEq.
each
point
point
D
x the slope Cof the
In a similar(b)manner,
6–2
states
that
at a point
negative and increases from zero to wB .
VD
0
A
negative increasing
Aw = A
moment
diagram
is equal to the
shear.
that the shear
diagram
in
These two equations
provide
a convenient
means
forNotice
quickly
B
B
V
"VB V 0
m will be a curve that has a negative slope,
C negative
D C increasing
DC D
slope =
C
obtaining the shear and
diagrams
for
a= positive
beam. decreasing
Equation
6–1then becomes negative
, Vdecreases
Fig.moment
6–9b starts
at +VA
to zero, and
"wC
wB . Specific slopes wA = 0, - wC, - wD, and
VA
w = negative
increasing
slope
=moment
positive
decreasing
increasing
These
two
equations
a convenient
means
for have
quickly
two
equations
provide
convenient
means
for
quickly
w = negative
increa
states that These
at a point
the
slope
of
the
shear
diagram
equals
the
intensity
and
decreases
to -Vprovide
The
diagram
will
then
anVinitial
These
two
equations
a convenient
means
for
quickly
V
V w = negative
Ba. provide
"w
D0
VA
M
slope slope
= negative
increasing
(c)
slope
= negative
increasing
0
= negative
incr
0
"w
obtaining
the
shear
and
moment
diagrams
for
a
beam.
Equation
6–1
obtaining
the slope
shear
and
moment
diagrams
for
a
beam.
Equation
6–1
B
of the distributed
loading.
For
example,
consider
the
beam
in
Fig.
6–9a.
obtaining
and
moment
diagrams
for
a
beam.
Equation
6–1
+Vshear
ofthe
which
decreases
to
zero,
then
the
slope
becomes
negative
"wC "wCx
"wC
6–2 states that at a point the slope of theA
(b)
states
that
atnegative
aatpoint
the
slope
of
the
shear
diagram
equals
intensity
states that
at aand
point
the
ofthe
the
equals
the
The distributed
loading
isdecreases
and
increases
from
zero
states
that
aslope
point
slope
ofdiagram
the
shear
equals
-V
Vto
V
VDthe
, 0,the
to
Specific
andintensity
-V are
VDslopes
0 diagram
BC., intensity
B .shear
A, w
"wD "wD
"wD
o the shear. Notice that the
diagram inloading. For
VA B VA VA
Vhas
C
ofshear
thediagram
distributed
example,
the
beam
in Fig.
6–9a.
"VFig.
Therefore,
a example,
curve For
that
a consider
negative
slope,
of the shear
distributed
loading.
For
consider
the
beam
inthe
Fig.
6–9a.
of
the
distributed
loading.
example,
consider
beam
in Fig.
6–9a.
6–9
shown
inwill
Fig.be
6–9c.
B
x
(b)
x
creases to zero, and thenThe
becomes
negative
(b)V = positive(b)
wtoB . wB . decreasing
distributed
loading
isVAnegative
increases
from
-loading
wB . Specific
0,and
-w
, wD, and
increasing
from
zero
to distributed
slopes
wBzero
. toslope
The
distributed
is loading
negative
and
from
zero
tozero
The
isw
negative
and
increases
from
A =increases
C
=
positive
decreasing
"V
e moment diagram will then
have anthe
initial
B
"VB
Therefore,
shear
be be
athat
curve
has has
a negative
- wB are Therefore,
shown in
Fig.
Therefore,
the
shear
diagram
will
a curve
that
ax negative
slope,
"VB
the6–9b.
shear diagram
will
be will
a curve
hasthat
a negative
slope,
M slope,
(c)diagram
V =decreasing
positive
B decreasing
ses to zero, then
the slopeincreasing
becomes negative
V"w
= positive
decreasing
V = positive
w
.
w
=
0,
w
,
w
,
from
zero
to
Specific
slopes
and
In a similar
manner,
that
at
a
point
the
slope
of
the
B
A
C
D
w
.
w
=
0,
w
,
w
,
increasing
from
to
Specific
slopes
and
-wzero
.
w
=
0,
-w
,
-w
,
increasing
fromEq.
zero6–2
to states
Specific
slopes
and
B
A
C
D
B
A
C
D
slopeslope
=decreasing
positive
decreasing
= positive
decreasing
slope
ecific slopes VA, VC, VD-, w
0, and
- VB are
VD M= positive
0
in
Fig.
6–9b.
moment -w
diagram
is-Bequal
toFig.
the6–9b.
shear.
Notice
ware
shown
in Fig.
6–9b.that the shear diagram in
inshown
M
MC
V
B are
"wB "w
B are shown
"w
Fig.that
6–9 at a point the slope of the
B
aAsimilar
manner,
Eq.
6–26–2
states
+V
,manner,
Fig. 6–9b starts
at In
to6–2
zero,
and
then
becomes
In
adecreases
similar
manner,
Eq.
states
thatnegative
at
pointof
the
slope of the
In a similar
Eq.
states
that
at
a point
thea slope
the
V
0
V
D
VD
0
VD V
0 "V
diagram
istoequal
to the
shear.
Notice
that
thediagram
shear
diagram
in in
-moment
VB . The
and decreases
tomoment
moment
diagram
then
have
initial
isthe
equal
towill
the
shear.
Notice
that
the
shear
moment
diagram
isdiagram
equal
shear.
Notice
that
thean
shear
inAdiagram
C
VC B
VC
(c)
x
+
V
,
Fig.
6–9b
starts
at
decreases
to
zero,
and
then
becomes
negative
PAGE
!
6
6
+ VA6–9b
slope of Fig.
which
decreases
to
zero,
slopeand
becomes
negative
VA, the
Fig.
6–9b
starts
at +Athen
decreases
to then
zero,becomes
and thennegative
becomes negative V
starts
at +V
to
zero,
A, decreases
A VA
"VB
VA
- V-moment
. The
then
have
an initial
Vdecreases
V. AThe
, moment
Vdiagram
, VD, diagram
and decreases
toand
slopes
0,will
andthen
-will
Vhave
arean
BV
and
to
diagram
then
have
an initial
"VB
B .decreases
Cmoment
Bwill
-VBto
. The
and decreases
toSpecific
initial
B
(c) (c)
x
(c)
+
V
slope
of
which
decreases
to
zero,
then
the
slope
becomes
negative
x
Fig.
6–9
shown inslope
Fig. 6–9c.
slope
of +AVA which to
decreases
to the
zero,
thenbecomes
the slopenegative
becomes negative
of +V
zero, then
slope
A which decreases
- V-BV
. Specific
VA,VV,C,VVD
and
decreases
to to
slopes
- VB- V
are are
. Specific
, ,V0, , and
and
decreases
slopes
0, and
264
V CHAPTER 6
=
w1x2 dx
C H A P T E R 6andBdM
E N D I=
N GV dx. Noting that w(x) dx and V dx¢V
represent
L differential
264
CHAPTER 6 BEND
ING
Equationsloading
6–1 andand
6–2 shear
may also
be rewritten
in the form
areas
diagram,
respectively,
we dV = w1x2 dx (6–3)
C
E6R B
6BEunder
BD"V
EI N
N
DGIthe
N G distributed
2266442 6 4
CCHHAA
PP
THT
EAE
RPRT6
NN
G
E
D
IN
change
in dx area
under
C(e)
HAPTER 6
B E N D I can
N G integrate these
andareas
= V dx. any
Noting
that
w(x)
and
Vthe
dxbeam,
represent differential
x dM between
two
points
C
and
D
on
=
Equations 6–1 and 6–2 may shear
also be rewritten
in the
form dV = w1x2 dx
C H A P T ED
loading
264
C
R 6 and
B E Nwrite
D Iareas
NG
under the
distributed
shear
respectively,
we = w1x2 dx
Fig. 6–9d,
Equations
6–1loading
and 6–2 and
maydistributed
also bediagram,
rewritten
in the form dV
264
264
BENDING
Equations 6–1 and 6–2 may also be rewritten in the form dV = w1x2 dx
andEquations
dM
= V6–1
dx.
Noting
that
w(x)
dx
and Vinin
dx
represent
differential
Equations
6–1
and
6–2
may
also
rewritten
inthe
the
form
dV
w1x2
dx
and
6–2
may
also
be
rewritten
the
form
dV
===
w1x2
dx
6–1
and
6–2
may
also
bebe
rewritten
form
dV
w1x2
dx
can Equations
integrate
these
areas
any
two
points
and
D
on
the
beam,
and
dM
=
Vbetween
dx.
Noting
that
w(x)
dxC
and
V
dx
represent
differential
Equations
6–1
and
6–2
may
also
be
rewritten
in
the
form
dV
=
w1x2
dx
areas
under
the
distributed
loading
and
shear
diagram,
respectively,
we
and
dM
=
V
dx.
Noting
that
w(x)
dx
and
V
dx
represent
differential
and
dM
==Vwrite
Noting
that
w(x)
and
dx
differential
and
dM
Vdx.
dx.
Noting
that
w(x)dx
dx
andVVand
dxrepresent
represent
differential
6–9d,
and
areas
under
the
distributed
loading
shear
diagram,
respectively,
we
C
D
and dMFig.
= V
dx.
Noting
that
w(x)
dx
and
V
dx
represent
differential
can
integrate
these
areas
between
any
two
points
C
and
D
on
thedV
beam,
Equations
6–1
and
6–2
may
also
be
rewritten
in the
form
=we
w1x2
areas
under
the
distributed
loading
and
shear
diagram,
respectively,
we dx
areas
under
the
distributed
loading
and
shear
diagram,
respectively,
areas
under
the
distributed
loading
and
shear
diagram,
respectively,
we
¢V
=
w1x2
dx
(d)
¢M
=
V1x2
dx
can
integrate
these
areas
between
any
two
points
C
and
D
on
the
beam,
areas
under
the
distributed
loading
and
shear
diagram,
respectively,
we
M (d) C
Fig.
6–9d,
and
write
and
dM
=
V
dx.
Noting
that
w(x)
dxpoints
and
V
dxDD
represent
differential
integrate
these
areas
between
any
two
Cand
and
D
on
the
beam,
can
integrate
these
areas
between
any
two
points
CCand
on
the
beam,
L
D
cancan
integrate
these
areas
between
any
two
points
on
the
beam,
V
L
(d)
Fig.
6–9d,
and
write
can D
integrate
these
areas
between
any two points
C and
D
on the
beam,respectively,
(d) (d) "V
(6–4)we
(6–3)
C
areas
under
the
distributed
loading
and
shear
diagram,
Fig.
6–9d,
and
write
Fig.
6–9d,
and
write
d)
= in
w1x2area
dx under
Fig.
6–9d,inand write
C"M
DD D
change
area under¢V
C Cx
change
VC
Fig.
6–9d,
and
write
can integrate
these areas between
D
=
L = any two points C and D on the beam,
V
C
D (d)
shear
distributed
loading
¢V
=
w1x2
dx diagram
(6–3)
"V
VV V
moment
shear
Fig.
6–9d,
and
write
C
D
change in
area
under
¢V
= dxw1x2 dx
L
V (e)
x
(f)
¢V
=
w1x2
¢V
=
w1x2
dx
x
= =
¢V
w1x2 dx
(6–3)
L
"VD
C
CD
¢V
=shearin
w1x2
dx
distributed
loading
L
change
area
under
V
LL
Equation
6–3
states
that
the
change
in
shear
between
C
and
D
is
equal
"V
x
(e)
L
(6–3) to(6–3)
(6–3)
=
"V
change
in
area
under
(6–3)
"V "V
D
x
change
in
area
under
¢V
=
w1x2
dx
(e) C"V
change
in
area
under
(6–3)
shear
distributed
loading
the area
the distributed-loading
¢M under
=
V1x2
dx
= curve between these two points,
under
(e)
(e)
Fig. 6–9 (cont.)
== =area
change
in change
areainunder
C
D xx x
(e)
L
shear
distributed
Lthis
e)
C C D Dx
shear
distributed
loading
Fig.
6–9d.
In
case
the
change
is
negative
sinceloading
the distributed load
shear
distributed
loading
=
(6–4)
C
D
(6–3)
shear
distributed
loading
"V
C
D
shear
distributed
loading
change
area
under
¢M
=frominV1x2
dx the
"M
in
areaSimilarly,
under
acts downward.
Eq. 6–4,
change in moment between
M (e)
xchange
=
L = area
6
C
D
Cmoment
and D, Fig.shear
6–9f,diagram
is ¢M
equalshear
shear diagram
distributed
loading
(6–4)within
=to the
V1x2
dxunder the
M
"M
¢M
=dxdx
V1x2
dx
change
in
area
under
L
the
region
from
C
to
D.
Here
the
change
is
positive.
M x
¢M
=
V1x2
¢M
=
V1x2
(6–4)
M
M
==dx V1x2 dx
L
¢M
D
¢M moment
=
V1x2
M
L
L
M C
Since
the above
equations
not
apply
at points
shear
"M
change
in do
area
under
Equation 6–3 states that the
change
in shear
between
C
anddiagram
D is equal
to where a concentrated
(6–4) (6–4)
(6–4)
L
L
F
"M
=
(6–4)
change
in
area
under
(6–4)
(f)
"M "M
force or couple moment
acts,
we
will
now
consider
x the distributed-loading
change
area
under
change
in in¢M
area
the area under
curve
between
these
two
points,
=under
dx each of these cases.
moment
diagram
=V1x2
Fig. 6–9
(cont.) M
"MC
"M
D
change
in
under
change
in the
area
under
=area
= inshear
shearload
diagram
L
states
change
shear
between
C and D is equal to (6–4)
Fig. 6–9d. In this
case the 6–3
change
is that
negative
since moment
the
distributed
=
(f)
= moment
shear
diagram
moment
shear
diagram
x Equation
moment
shear
diagram
moment
shear
diagram
(f)
C
D
x
the
area
under
the
distributed-loading
curve
between
two
points,
(f) (f)
"M
acts Cdownward.
Similarly,
from
Eq.
6–4,
the
change
in
moment
between
Regions
of
Concentrated
Force
and
Moment.
Atofreex
x
change
in
area
under
Equation
6–3
states
that
the
change
in
shear
between
Cthese
and D
is equal
Fig.
6–9
(cont.)
D
(f)
(f)
C C D Dx
xFig. 6–9d. InEquation
=shear
6–3
states
that
the
change
in
shear
between
Cequal
and
D
this
case
the
change
is
negative
since
the
distributed
load
C
and
D,
Fig.
6–9f,
is
equal
to
the
area
under
the
shear
diagram
within
Equation
6–3
states
that
the
change
in
shear
between
C
and
D
is
tois
Equation
6–3
states
that
the
change
in
between
C
and
D
is
equal
toequal to
body
diagram
of
a
small
segment
of
the
beam
in
Fig.
6–10a
taken
from
C
D VC
D
the
area
under
the
distributed-loading
curve
between
these
two
points,
moment
shear
diagram
6–3
states
that
the
change
in
shear
between
C
and
D
is
equal
to
MFig. 6–9 (cont.)
MEquation
! "M acts
Equation
6–3
states
that
the
change
in
shear
between
C
and
D
is
equal
to
the
area
under
the
distributed-loading
curve
between
these
two
points,
Similarly,
from
Eq.
6–4,
the
change
initthe
moment
between
the
area
under
the
distributed-loading
curve
between
these
points,
from CFig.
to downward.
D.
Here
the
change
isshown
positive.
the
area
under
the
distributed-loading
curve
between
these
two
points,
Fig.region
6–9 (cont.)
under
the
force
isthe
in Fig.
6–10a.
Here
can
be two
seen
that
(f)
In
this
case
change
is negative
since
distributed
loadforce
x6–9d.
Fig. 6–9
Fig. (cont.)
6–9the
(cont.)
the
area
under
the
distributed-loading
curve
between
these
two
points,
the
area
under
the
distributed-loading
curve
between
these
two
points,
6
Fig.
6–9d.
In
this
case
the
change
is
negative
since
the
distributed
load
C
D
Fig. 6–9 (cont.)
Fig. 6–9 (cont.)
CFig.
and
D,Equation
Fig.
is equal
to
the
area
the
shear
diagram
within
6–9d.
In
this
case
the
change
ischange
negative
since
thethe
distributed
Fig.
6–9d.
In
this
case
change
isunder
since
distributed
Since the above
equations
do6–9f,
not
apply
atthe
points
where
anegative
concentrated
equilibrium
requires
downward.
Similarly,
from
Eq.
6–4,
the
change
in
moment
between
6–3
states
that
the
in
shear
between
C and
Dload
isload
equal to
Fig. 6–9d.acts
In
this
case
the
change
is
negative
since
the
distributed
load
Fig.
6–9d.
In
this
case
the
change
is
negative
since
the
distributed
load
F
acts
downward.
Similarly,
from
Eq.
6–4,
the
change
in
moment
between
the
region
from
C
tonow
D.equal
Here
the
change
is
positive.
6
acts
downward.
Similarly,
from
Eq.
6–4,
the
change
inbetween
moment
between
acts
downward.
Similarly,
from
Eq.
6–4,
the
change
in
moment
between
force or couple
moment
acts,
we6–9f,
will
consider
each
of
these
cases.
C
and
D,
Fig.
is
to
the
area
under
the
shear
diagram
within
the
area
under
the
distributed-loading
curve
these
two points,
acts downward.
from
Eq.
6–4,
the
change
in¢V2
moment
between
6
cSimilarly,
Fig. 6–9 (cont.)
+downward.
©F
=6–9f,
0;
V
+the
Fthe
-6–4,
1V
+points
=under
0in
acts
Similarly,
from
Eq.
the
change
moment
between
Cy and
D,
Fig.
6–9f,
is
equal
tounder
area
the
shear
diagram
within
6
6
the
above
equations
do
not
apply
at
where
a
concentrated
CSince
and
D,
Fig.
is
equal
to
area
the
shear
diagram
within
C
and
D,
Fig.
6–9f,
is
equal
to
area
under
the
shear
diagram
within load
the
region
from
C
to
D.
Here
the
change
is
positive.
Fig.
6–9d.
In
this
case
the
change
is
negative
since
the
distributed
6
Fig.
6–9f,
is
equal
to
the
area
under
the
shear
diagram
within
Tekil yük
bölgeleri:C and D,
F
C
and
D,
Fig.
6–9f,
is
equal
to
the
area
under
the
shear
diagram
within
the
region
from
C
to
D.
Here
the
change
is
positive.
force
or couple
moment
we
will
now
consider
each
of athese
cases. between
theSince
region
from
CForce
to
Here
the
change
is
positive.
the
region
from
C D.
toacts,
D.and
Here
the
change
is
positive.
¢VEq.
=
F
(6–5)
the
above
equations
do
not
apply
at6–4,
points
where
concentrated
Regions
Concentrated
Moment.
Athe
freeacts
Similarly,
from
change
inwhere
moment
theof
region
from
C
to downward.
D.
Here
change
is
positive.
the
region
from
Cequations
tothe
D.
Here
the
change
is points
positive.
Since
the
above
equations
doat
not
apply
atwhere
points
a concentrated
F
Since
the
above
do
not
apply
where
a
concentrated
Since
the
above
equations
do
not
apply
at
points
a
concentrated
V
!
"V
6
V
force
or
couple
moment
acts,
we
will
now
consider
each
of
these
cases.
"x
of the
a small
segment
the
beam
in
Fig.
taken
C
and
D,of
is equal
to6–10a
the
area
under
the shear
diagram
Since
above
equations
do6–9f,
not
apply
at the
points
where
afrom
F
F bodyFdiagram
Thus,
when
FFig.
acts
upward
beam,
¢V
isconcentrated
positive
so
the
shear
will
Since
the
above
equations
doon
not
apply
at
points
where
athese
concentrated
force
or
couple
moment
acts,
weconsider
will
now
consider
each
of
thesewithin
cases.
force
orthe
couple
moment
acts,
we
will
now
each
of of
force
or
couple
moment
acts,
we
will
now
consider
each
these
cases.
FM ! "M F
of
Concentrated
Force
and
Moment.
Acases.
freeunder the
force
is
shown
in
Fig.
6–10a.
Here
it
can
be
seen
that
force
region
from
C
to
D.
Here
the
change
is
positive.
force
orRegions
couple
moment
acts,
we
will
now
consider
each
of
these
cases.
“jump”
upward.
Likewise,
if
F
acts
downward,
the
jump
1¢V2
will
be
force
or
couple
moment
acts,
we
will
now
consider
each
of
these
cases.
(a)
V
body
diagram
a small
ofForce
the
beam
in Fig.
6–10awhere
takenA
equilibrium
requires
Regions
ofofthe
Concentrated
and
freeSince
abovesegment
equations
do not
apply
atMoment.
points
a from
concentrated
M
downward.
Regions
of
Concentrated
Force
and
Moment.
A freeFM ! "M
Regions
of
Concentrated
Force
and
Moment.
A
freeRegions
of
Concentrated
Force
and
Moment.
A
freeunder
the
force
is
shown
in
Fig.
6–10a.
Here
it
can
be
seen
that
force
V
body
diagram
of
a
small
segment
of
the
beam
in
Fig.
6–10a
taken
from
force
or
couple
moment
acts,
we
will
now
consider
each
of
these
cases.
V
Regions
of
Concentrated
Force
and
Moment.
A
freeWhen
the
beam
segment
includes
the
couple
moment
M
,
Fig.
6–10b,
c
V
+
©F
=
0;
V
+
F
1V
+
¢V2
=
0
M
Regions
of
Concentrated
Force
and
Moment.
A
free0
M
!
"M
body
diagram
of
a
small
segment
of
the
beam
in
Fig.
6–10a
taken
M
!
"M
M
V
V
y
body
diagram
ofsegment
small
segment
of the
beam
Fig.
6–10a
taken
from
body
a small
ofinthe
beam
in
Fig.
6–10a
taken
from from
equilibrium
requires
under
the
force
isaofequilibrium
shown
insegment
Fig.
6–10a.
Here
itincan
be
seen
that
force
V
M
!
"M then
body
diagram
of
adiagram
small
of
the
beam
Fig.
6–10a
taken
from
M MV
M !M
"M
! "M M
moment
requires
the
change
in
moment
to
be
body
diagram
of
a
small
segment
of
the
beam
in
Fig.
6–10a
taken
from
under
the
force
is
shown
in
Fig.
6–10a.
Here
it
can
be
seen
that force
M
M ! "M
under
the
force
isin=shown
in Fig.
6–10a.
Here
can
be
seen
that
force
under
the
force
is
shown
in Here
Fig.
6–10a.
Here
it can
be
seen
that
force
¢V
FV
(6–5)
requires
M
M ! "M
under the
force
is 0;
shown
Fig.
6–10a.
it¢V2
can
be0it seen
that
force
Regions
of
Concentrated
Force
and
Moment.
A freec
+equilibrium
©F
=
+
F
1V
+
=
y
equilibrium
requires
under
the
force
is
shown
in
Fig.
6–10a.
Here
it
can
be
seen
that
force
equilibrium
equilibrium
V ! "V
d+ ©MOrequires
=requires
0;
Ma +small
¢Msegment
- M0 -ofVthe
¢x - M in
= 0
O
equilibrium
"x
V
diagram
taken from
crequires
Thus, when
F
acts+upward
on0;the
beam,ofV
¢V+isFpositive
so¢V2
the shear
will Fig. 6–10a
©Fybody
=
- 1V
+
= 0beam
equilibrium
requires
M
M
!
"M
¢V
=
F
(6–5)
c
M0
+
©F
=
0;
V
+
F
1V
+
¢V2
=
0
c
c
+
©F
=
0;
V
+
F
1V
+
¢V2
=
0
+
©F
=
0;
V
+
F
1V
+
¢V2
=
0
y
under
the
force
is
shown
in
Fig.
6–10a.
Here
it
can
be
seen
that force
y ify F¢x
0,
get + the
upward.
acts
+ c ©Fy Likewise,
= 0; Letting
V:
+ downward,
F we
- 1V
¢V2jump
= 0 1¢V2 will be
(a)
V ! “jump”
"V
"x
c
¢V
=
F
(6–5)
+
©F
=
0;
V
+
F
1V
+
¢V2
=
0
equilibrium
requires
y
Thus, when F acts upward on the¢V
beam,
is positive so the shear (6–5)
will
downward.
= F
=¢V
(6–5)
¢M
=F¢V
M0 = F
(6–6)(6–5)
V ! "V
¢V = F ¢V
(6–5)
V ! "VV ! "V “jump” upward. Likewise,
"x
"x
V
if
F
acts
downward,
the
jump
1¢V2
will
be
V
!
"V
V
!
"V
Thus,
when
F
acts
upward
on
the
beam,
¢V
is
positive
so
the
shear
will
When
the
beam
segment
includes
the
couple
moment
M
,
Fig.
6–10b,
¢V
=
F
(6–5)
(a)
"x
c
"x "x
0 + ¢V2 = 0
+ when
©F
0;
V the
+beam,
Fbeam,
- the
1V
! "V
y F=when
Thus,
F acts
on
beam,
is
positive
so the
shear will
"xM !V"M
Thus,
when
Fcase,
acts
upward
onupward
the
¢V
is¢M
positive
sokuvveti
the
shear
acts
upward
¢V
isthe
positive
so
the
shear
will
In
this
if
M
applied
clockwise,
is¢V
positive
so
the
moment
Vthen
! "Vmoment
Thus,equilibrium
when
FThus,
acts
upward
on
the
¢V
is
positive
so
the
shear
will
F downward.
yukarı
doğru
uygulandığında
pozitiftir
ve
kesme
“jump”
upward.
Likewise,
ifinFon
acts
downward,
jump
1¢V2
willwill
be
requires
the
change
moment
to be
0 isbeam,
"x (b)
(a)
“jump”
upward.
Likewise,
if
F
acts
downward,
the
jump
1¢V2
will be
“jump”
upward.
Likewise,
if
F
acts
downward,
the
jump
1¢V2
will
be
Thus,
when
F
acts
upward
on
the
beam,
¢V
is
positive
so
the
shear
will
“jump”
upward.
Likewise,
if
F
acts
downward,
the
jump
1¢V2
will
be
V
(a) “jump” upward.
¢V
=
F
(a) (a)
diagram
will segment
“jump”
upward.
Likewise,
when
M0 acts
When
the
beam
includes
the couple
moment
M
Fig.
Likewise,
if çıkış
F acts
downward,
the jump
1¢V2
will
be 6–10b, (6–5)
downward.
0,counterclockwise,
(a)
M ! "M
M
diyagramında
yukarı
olacaktır.
downward.
+
©M
=
0;
M
+
¢M
M
V
¢x
M
=
0
downward.
“jump”
upward.
Likewise,
if
F
acts
downward,
the
jump
1¢V2
will
be
V
!
"V
downward.
O
O
0
V
(a) Fig.d
"x
downward.
theThus,
jump
1¢M2
will
be
downward.
thenWhen
moment
requires
the
change
in moment
to
the equilibrium
beam
segment
includes
couple
moment
Mbe
0, Fig.
M ! "M
V6–10
M
when
Fthe
acts
upward
onthe
the
beam,
¢Vmoment
is positive
so 6–10b,
the
shear will
V
V
When
beam
segment
includes
the
couple
moment
M
When
the
beam
segment
includes
the
couple
moment
, Fig.
6–10b,
When
the
beam
segment
includes
the
couple
6–10b,
downward.
M0 V
0, Fig. 6–10b,
0M
M ! the
"M
M"M M !M
0, Fig.
"M
M M
! "M
When
beam
segment
includes
the
couple
moment
M
, Fig. M
6–10b,
then
moment
equilibrium
requires
the
change
in
moment
to
be
0
M
!
Letting
¢x
:
0,
we
get
M
“jump”
upward.
Likewise,
if
F
acts
downward,
the
jump
1¢V2
V
d
+
©M
=
0;
M
+
¢M
M
V
¢x
M
=
0
(a)
then
moment
equilibrium
requires
the
change
in
moment
to
bewill be
then
moment
equilibrium
requires
the
change
in
moment
to
be
O
then
equilibrium
requires
themoment
change
be 6–10b,
O moment
0
the beam
segment
couple in
Mto
0, Fig.
M ! "M
then momentWhen
equilibrium
requires
theincludes
change
in
tomoment
be
M
downward.
¢M
=
M
(6–6)
d
+
©M
=
0;
M
+
¢M
M
V
¢x
M
=
0
M
O
moment
requires
the -change
inVmoment
0
0
O : 0, equilibrium
V ! "V
"x
¢x
we
get
=M
M
+ 00M
¢M
M¢x
Mto=be
0 M , Fig. 6–10b,
dthen
+ ©M
=
0;+
M
+0;
=¢x0= -0moment
V O
d + ©M
=©M
0;
+0 segment
¢M
- ¢x
V=
-couple
M
O d + ©M Letting
O
O
0- -M
O M
Od+
When
the
beam
includes
the
¢M
M¢M
VM
¢x
-0 V
M
0
O
0
O = 0;
M ! "M
M
M0
Letting
¢x
:
0,
we
get
In
this
case,
if
M
is
applied
clockwise,
¢M
is
positive
so
the
moment
M
M
M
d
+
©M
=
0;
M
+
¢M
M
V
¢x
M
=
0
0
0
0
O
0
then
moment
equilibrium
requires
be
O Letting
0 M0 the change in moment to
M0
¢M
=
(6–6)
Letting
¢x
:
0,
we
get
¢x
:
0,
we
get
(b)
Letting
¢x
:
0,
we
get
V ! "V
¢x :
0, we get
"x
diagram Letting
will “jump”
upward.
Likewise, when M0¢M
acts=counterclockwise,
M0
M
(6–6)
0¢M
0, we get M clockwise,
V ! "V
d +¢x
©M
¢M
M
V=positive
¢x
0 moment
"x
=- M
(6–6)
M0 - Mso=(6–6)
In Letting
this
case,
if:M
¢M
the
¢M
=
(6–6) (6–6)
O 0= is0;applied
0 0M
1¢M2
be downward.
Fig. 6–10
V the
! "V
0 is
! "V will
¢M = +M¢M
VOjump
!"x"V V
"x "x
0
V !(b)
"V
"x
M0
diagram
will
“jump”
upward.
Likewise,
when
M
acts
counterclockwise,
In
case,
if
M
is
applied
clockwise,
¢M
is
positive
so
the
moment
¢M
= Mclockwise,
0 ¢M
V ! "V
Letting
¢x
:
0,
In this
this
case,
ifthis
M
is0clockwise,
applied
clockwise,
¢M
is0 so
positive
sopositive
the
moment
Inapplied
Mget
applied
¢M
isolur
so(6–6)
the moment
In
this
if00case,
M
isifwe
applied
clockwise,
is
positive
so
the
moment
(b)
Burada
yönünde
uygulandığında
pozitif
ve
"x
o0 saat
0 is
if M
M
iscase,
¢M
is positive
thecounterclockwise,
moment
(b)6–10
(b) In this case,
(b)
the
jump
1¢M2
will
be
downward.
diagram
will
“jump”
upward.
Likewise,
when
M
acts
Fig.
(b)
0
diagram
will
“jump”
upward.
Likewise,
when
M
acts
counterclockwise,
diagram
will
“jump”
upward.
Likewise,
when
M
acts
counterclockwise,
diagram
will
“jump”
upward.
Likewise,
when
M
acts
counterclockwise,
0
0
0
diagram
will
“jump”
upward.
Likewise,
when
M
acts
counterclockwise,
In
this
case,
if
M
is
applied
clockwise,
¢M
is
positive
so
the
moment
0
¢M
= M0
(6–6)
0 be downward.
the
jump
1¢M2
will
Fig.
V ! "V
(b)6–10
diyagramda
artış,
aksi
taktirde
azalış
olacaktır.
the
jump
1¢M2
will
be
downward.
the
jump
1¢M2
will Likewise,
be downward.
the
jump
will
be downward.
Fig. "x
6–10Fig. 6–10
the jump
1¢M2
will
be1¢M2
downward.
Fig. 6–10 Fig. 6–10
diagram
will
“jump”
upward.
when M0 acts counterclockwise,
In this
case,
Mdownward.
0 is applied clockwise, ¢M is positive so the moment
the jump
1¢M2
willifbe
Fig. 6–10
(b)
diagram will “jump” upward. Likewise, when M0 acts counterclockwise,
the jump 1¢M2 will be downward.
Fig. 6–10
C
D
(d)
PAGE !67
266
CHAPTER 6 BENDING
266
C Hdiagrams
A P T E R 6 for
B Ethe
N D Ibeam
NG
Draw the shear
and moment
shown in Fig. 6–11a.
SOLUTION
! EXAMPLE 6.5 Ölçü ve yükleme durumu şekilde verilen konsol kirişin kesme kuvveti ve
EXAMPLE
6.5 at the fixed support is shown on
Support Reactions.
The reaction
Draw the shear and moment diagrams for the beam shown in Fig. 6–11a.
the free-body diagram,
6–11b.
momentFig.
diyagramlarını
çiziniz. Draw
the shear and moment diagrams for the beam shown in Fig. 6–11a.
Shear Diagram. The shear at each end of the beam is plotted first,
SOLUTION
Fig. 6–11c. Since there is no distributed
loading Pon theSOLUTION
beam,
the slope
P
P how theSupport
Reactions.
The reaction at the fixed support is shown on
of the shear diagram is zero as Pindicated. Note
force P at
the
Support
Reactions.
The
the
free-body
Fig.reaction
6–11b. at the fixed support is shown on
center of the beam causes the shear diagram to jump
downward
andiagram,
the
free-body
diagram,
Fig.
6–11b.
amount P, since this force acts downward.
Shear Diagram. The shear at each end of the beam is plotted first,
Diagram.
The shear
at each endloading
of the beam
plotted
L the ends ofShear
L moments at
Fig.
Since
is no distributed
on theisbeam,
thefirst,
slope
Moment Diagram. L
The
the 6–11c.
beam
are there
L
Fig.
6–11c.
Since
there
is
no
distributed
loading
on
the
beam,
the
diagram is zero as indicated. Note how the force Pslope
at the
plotted, Fig. 6–11d. Here the moment diagram consists of
of the
twoshear
sloping
ofcenter
the
diagram
zero as
indicated.
Note how
the force
P at thean
(a)
of the
beamiscauses
the
shear diagram
to jump
downward
lines, one with a slope of !2P and
of shear
!P.
(a) the other with a slope
center
of
the
beam
causes
the
shear
diagram
to
jump
downward
an
P, since
The value of the moment in the center of the amount
beam can
be this force acts downward.
amount
P,
since
this
force
acts
downward.
determined byÇözüm:
the method of sections, or from the area
underDiagram.
the
Moment
The moments at the ends of the beam are
Diagram.
The the
moments
the ends
of the
are
shear diagram. If we choose the left half of the shear Moment
diagram,
plotted, Fig. 6–11d. Here
momentatdiagram
consists
of beam
two sloping
plotted,
Fig.
6–11d.
Here
the
moment
diagram
consists
of
two
sloping
lines, one with a slope of !2P and the other with a slope of !P.
M ƒ x = L = M ƒ x = 0 + ¢M
lines,The
onevalue
with a of
slope
!2P andinthethe
other
with of
a slope
!P. can be
theofmoment
center
the of
beam
The value of the moment in the center of the beam can be
determined by the method of sections, or from the area under the
determined by the method of sections, or from the area under the
shear diagram. If we choose the left half of the shear diagram,
M ƒ x = L = -3PL + (2P)(L) = -PL
shear diagram. If we choose the left half of the shear diagram,
Mƒ
= Mƒ
+ ¢M
M ƒ x =xL= L= M ƒ x =x0= 0+ ¢M
6
P
6
P
Mƒ
= -3PL + (2P)(L) = -PL
M ƒ x =xL= L= -3PL + (2P)(L) = -PL
2P
3PL
V
2P
w#0
slope # 0
(b)
P
downward force P
downward jump P
x
P
3PL
3PL
(b)
w#0
wslope
# 0 # 0 (b)
V slope # 0
downward force P
V
downward
force
PP
downward
jump
downward jump P
2P
2P
P
P
x
x
(c)
(c)
(c)
V # positive constant
slope # positive constant
x
"PL
"3PL
P
2P
2P
P
M
P
M
(d)
M
V # positive constant
# positive
constant
Vslope
# positive
constant
slope # positive constant
x
Fig. 6–11
"PL
"PL
"3PL
"3PL
(d)
(d)
Fig. 6–11
Fig. 6–11
PAGE !68
x
M0
LETHOD
6.2 L GRAPHICAL
RAPHICAL M
ETHOD FOR
FOR C
CONSTRUCTING
ONSTRUCTING SSHEAR
HEAR AND
ANDM
MOMENT
OMENTD
DIAGRAMS
IAGRAMS
267
267
(a)
EXAMPLE
SOLUTION
EXAMPLE
6.6
$
Ölçü ve yükleme durumu şekilde verilen basit mesnetli kirişin kesme
Support Reactions. The reactions are shown on the free-body
Drawdiagram
the shear
and
moment diagrams for the
Draw
the beam
beam shown
shown in
in
in Fig.
6–12b.
kuvveti vethe
moment
diyagramlarını
çiziniz.
Fig.
6–12a.
Fig. 6–12a.
Shear Diagram. The shear at each end is plotted first, Fig. 6–12c.
M00
Since there is no distributed load
on the beam, the shear diagram has
zero slope and is therefore a horizontal line.
L moment L
L
L
Moment Diagram. The
is zero at each end, Fig. 6–12d. The
moment diagram has a constant
negative
slope of !M0!2L since this
(a)
(a)
is
the
shear
in
the
beam
at
each
point.
Note
that the couple moment
Çözüm:
SOLUTION
SOLUTION
M0 causes a jump in the moment diagram at the beam’s center, but it
does not
affect the The
shearreactions
diagram at
this
point.on
Support
Reactions.
are
shown
Support
shown
on the
the free-body
free-body
diagram
in
Fig.
6–12b.
diagram
M0
6
Shear Diagram.
Diagram. The
The shear
shear at
Shear
at each
each end
end is
is plotted
plotted first,
first, Fig.
Fig. 6–12c.
6–12c.
Since there
there is
is no
no distributed
distributedLload
Since
load on
on the
the Lbeam,
beam, the
the shear
shear diagram
diagram has
has
zero
slope
and
is
therefore
a
horizontal
line.
M
/2L
M
/2L
0
zero slope and is therefore a horizontal
line. 0
(b)
Moment Diagram.
Diagram. The
The moment
moment is
Moment
is zero
zero at
at each
each end,
end, Fig.
Fig.6–12d.
6–12d.The
The
V
moment
diagram
has
a
constant
negative
slope
of
!M
!2L
w "negative
0
moment diagram has a constant
slope of !M00!2L since
since this
this
slopepoint.
"0
is the
the shear
shear in
in the
the beam
beam at
at each
each
is
point. Note
Note that
that the
the couple
couple moment
moment
M00 causes
causes aa jump
jump in
in the
the moment
moment diagram
M
diagram at
at the
the beam’s
beam’s center,
center, but
but itit
does not
not affect
affect the
the shear
shear diagram
diagram at
does
at this
this point.
point.
x
M
M00
!M0 /2L
(c)
L clockwise moment
L
L
L
M0
positive
jump
M0M0 /2L
M
M
/2L
(b)
M00/2L
M
/2L
(b)
V 0"
negative constant
V
V
w"
" 00
w
slope "
slope
" 00
M0 /2
slope " negative constant
x
– M0 /2
!M
M0/2L
/2L
!
0
M
M
(c)
(c)
xx
(d)
Fig. 6–12
clockwise moment
clockwise
moment M
M00
positive jump
positive
jump M
M00
V
V"
" negative
negative constant
constant
slope
"
negative
slope
"
negative constant
constant
M
/2
0
M /2
0
xx
–– M
/2
M00 /2
(d)
(d)
Fig.
Fig. 6–12
6–12
PAGE !69
66
Shear
Diagram. The shear at each end point is plotted first, Figs. 6–13c
SOLUTION
and 6–14c. The distributed loading on each beam indicates the slope
and 6–14c.
The distributed
loading
on
each beam
indicates
the slope
Support
Reactions.
The
reactions
atthe
the
fixed support
are shown
of
the
shear
diagram
thus
produces
shapes
Support
Reactions.
Theand
reactions
at the fixed
supportshown.
are shown
of the shear
diagram
and thus
produces
the
shapes
shown.
on
each
free-body
diagram,
Figs.
6–13b
and
Fig.
6–14b.
on each
free-body
diagram, Figs.
andatFig.
6–14b.
Moment
Diagram.
The 6–13b
moment
each
end point is plotted first,
MomentShear
Diagram.
The moment
atateach
end point
pointisisplotted
plottedfirst,
first,
Diagram.
The
shear
each
Figs.on
6–13c
Figs.
6–13d
and
6–14d.
Various
values
of
the
shear
at each
the
Shear
Diagram.
The
shear
at
each
end
point
is
plotted
Figs. point
6–13c
268
C H A P T E R 6 B E N D I N G Figs. 6–13d and 6–14d. Various values of the shear at eachfirst,
point
on thethe slope
and
6–14c.
The
distributed
loading
on
each
beam
indicates
beamThe
indicate
the slope
of the
diagram
at the
and 6–14c.
distributed
loading
onmoment
each beam
indicates
thepoint.
slopeNotice
beam indicate
the
slope
of theand
moment
diagramthe
at the
point. Notice
of the
shear
diagram
thus
shapes
this
variation
theproduces
curves
shown.
of the how
shear
diagram
andproduces
thus produces
the shapes
shown.shown.
how this variation produces the curves shown.
Moment
Diagram.
The
moment
atend
each
end ispoint
iswplotted
NOTE:
Observe
how
the
degree
of
the
curves
from
to V tofirst,
M
Moment
Diagram.
The
moment
each
point
plotted
first,
$ EXAMPLE 6.7 Ölçü ve yükleme
durumu
şekilde
verilen
konsol
kirişlerin
kesme
NOTE:
Observe
how
the
degree
of at
the
curves
from
w kuvveti
to
V topoint
M on the
Figs.
6–13d
and
6–14d.
Various
values
of
the
shear
at
each
increases
exponentially
due toofthe
integration
ofpoint
dV on
= w
dx and
Figs.
6–13d
and
6–14d.
Various
values
the
shear
at
each
the
increases
exponentially
dueslope
to the
integration
of ofdV
dx point.
and Notice
beam
indicate
of
the
moment
diagram
atwthe
Draw
the= shear
moment
diagrams
for
each
the=beams
shown
inload
dM
Vdx
.and
Forthe
example,
in
Fig.
6–14,
the
linear
distributed
beam
indicate
the
slope
of
the
moment
diagram
atdistributed
the
point.
Notice
ve moment diyagramlarını çiziniz.
dM
=
Vdx
.
For
example,
in
Fig.
6–14,
the
linear
load
how
this
variation
produces
the
curves
shown.
Figs.
6–13a
and
6–14a.
produces
a parabolic
shear
diagram
and cubic moment diagram.
how
this
variation
produces
the
curves
shown.
produces a parabolic shear diagram and cubic moment diagram.
NOTE:
Observe
how
the
the curves
w0 degree
NOTE:
Observe
how wthe
degree
of the of
curves
from wfrom
to VwtotoMV to M
SOLUTION
0
increases
exponentially
due
to
the
integration
of
dV = w dx and
increases
w0 exponentially due to the integration of dV = w dx and
w0
6
Support
Reactions.
The reactions
at6–14,
the fixed
support
are shown
dM
=
Vdx
.
For
example,
in
Fig.
the
linear
distributed
dM = Vdx. For example, in Fig. 6–14, the linear distributed load load
on each
free-body
diagram,
Figs.diagram
6–13b and
Fig.
6–14b.
produces
a
parabolic
shear
and
cubic
moment
diagram.
produces a parabolic shear diagram and cubic moment diagram.
6
6
L
(a)
L
(a)
6
w0
L is plotted first, Figs. 6–13c
Shear Diagram. The shear
atweach
end point
0
w0
L
(a)
and w6–14c.
The
distributed
loading
on
each
beam indicates the slope
0
(a)
of the shear diagram and thus
w0 produces the shapes shown.
w0
w0
w0Lmoment at each end point is plotted first,
Moment
Diagram.
The
w0L
2
w0 L
2
Figs. 6–13d and 6–14d.
Various
valuesLof theLshear at each point on the
w0 L
Çözüm:
L
L
(a)
(a) diagram at the point. Notice
beam indicate the slope of the moment
(a)
(a)
w0 curves shown.
2
how this variation
produces
the
ww0L
(b)
0
w0 L2
w0L2
(b)
w0
w0L
(b)
w0 L2
w0 NOTE: Observe whow
(b)
0L 6the V
2 w L
6
2 degree of the curves from w to V to M
0
2
2V
w0 L
w ! negative constant ("w
wdue
w ! negative of
decreasing
0)
0L
increases
exponentially
to
the
integration
w ! negative constant ("w0 )
w
L
w!
negative
decreasing dV = w dx and
0
V
slope = negative constant ("w0 )
slope ! negative decreasing
2
V
slope = negative constant ("w0 )
slope
!
negative
decreasing
2
dM = Vdx. For example, in Fig. 6–14, the linear distributed load
w L
w0L2 diagram and cubic
(b) moment diagram.
w0 L w L20
produces
a parabolic
w0L2 shear
(b)
0
2
w0
(b)
(b)
w ! negative
w ! negative
constantconstant
("w0x) ("w0 ) x
V
w0 0 )
slope = negative
constant
V
slope = negative
constant
("w0 ) ("w
6
(c)
(c)
w
L
0
w0 L
V!
positive decreasing
VM
! positive
decreasing
M
slope =
positive decreasing
slope = positive
decreasing
L
xx
x x
(a) (c)
(c)
V
! positive
decreasing
w0
V
decreasing
M! positive
ML
slope = positive
decreasing
2
w
w
L
slope
=
positive
decreasing
w0 L2 "0 0
"
x
2
2
(d)
x
(d)
w0 L
2
2
6–13
w0 L2
Fig. 6–13 Fig.
(b)
2
2
w0 L2 " w0 L
w ! negative constant ("w0 )
"
2
V
slope = negative(d)
constant ("w0 )
2
(d)
w0 L
6
6
0
V wwL0 V
0 decreasing
x
w ! negative
0
w0L
x
w ! negative
decreasing
slope
! negative
decreasing
2
(c)
slope
!
negative
decreasing
2
(c)
V!
positive decreasing
V!
positive
decreasing
M slopeM! positive
slope !
positive decreasing
decreasing
0
L
0
xx
x x
(a)
(c)
(c)
w0
V ! positive
decreasing
VM! positive
decreasing
w0L
M slope ! slope
! positive
decreasing
positive
decreasing
2
x
2
x
w0 L2 " w0 L
"
(d)
6
(d)
6
w0L2
(b)
Fig. 6–14 Fig. 6–14
6
V
w0L
w ! negative decreasing
2
slope ! negative decreasing
2 w0 L
w0 L2 "
(d)
"
6
(d)
6
Fig. 6–14
Fig. 6–14
Fig. 6–13
Fig. 6–13
0
x
(c)
M
V ! positive decreasing
slope = positive decreasing
M
V ! positive decreasing
slope ! positive decreasing
x
x
"
w0 L2
2
x
(c)
(d)
"
Fig. 6–13
PAGE !70
w0 L2
6
(d)
Fig. 6–14
6.2
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT
EXAMPLE 6.8
6.2
269
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS
Draw the shear and moment diagrams for the cantilever beam
in Fig. 6–15a.
2 kN
$ EXAMPLE 6.8 Ölçü ve yükleme durumu şekilde verilen konsol kirişin kesme kuvveti ve
moment
diyagramlarını
çiziniz.
MB " 11 kN#m
Draw the
shear and moment
diagrams for the cantilever beam
2 kN
2 kN
in Fig. 6–15a.
1.5 kN/m
1.5 kN/m
2m
OD FOR
269
CONSTRUCTING SHEAR AND MOMENT2 D
IAGRAMS
kN
Çözüm:
2 kN
B
By " 5 kN
2m
w"0
slope " 0
(b)
2V6(kN)
9
w " negative constant
slope " negative constant
B CONSTRUCTING SHEAR
6.2
(a) AND MOMENT DIAGRAMS
11RAPHICAL
kN#m METHOD FOR
MB "G
1.5 kN/m
2m
w"0
slope " 0
2m
2
EXAMPLE 6.8
B
2m
2m
2m
A
ntilever beam
(b)
1.5 kN/m A
V (kN)
(a)
!2
SOLUTION
MB " 11 kN#m
Draw the shear and moment diagrams
for the cantilever beam
2 kN
2
4
By " 5 kN
x (m)
Support Reactions. The support reactions at the1.5
fixed
kN/msupport
in Fig. 6–15a.
2m
2m
(c)
B are shown in Fig. 6–15b.
!2
SOLUTION (b)
Shear
Diagram.
The
shear
at end
is M
!2
kN. D
This
value is
G
RAPHICAL
METHOD
FOR
CONSTRUCTING
SHEARAAND
OMENT
IAGRAMS
269
Support Reactions.
The 6.2
support
reactions
at the
fixed
support
w "2 0kN w " negative constant
plotted
at
x
"
0,
Fig.
6–15c.
Notice
how
the
shear
diagram
is
(c)
1.5
kN/m
B are shown
in0Fig.
6–15b.
slope "
By " !5
5VkN
slope
" negative constant
" negative constant
constructed by following the slopes defined
by the 2loading
w.
2m
m
slope " negative constant
Shear
Diagram.
The
shear
at
end
A
is
!2
kN.
This
value
is
The shear at x = 4 m is !5 kN, the reaction on the beam. This
EXAMPLE
6.8
6
V (kN)
plotted at A
x " 0, Fig. 6–15c. Notice
theverified
shear diagram
is the area under
(b) the distributed
value how
can be
by finding
M
(k
N#m
)
V " negative constant
constructed
following
slopes
defined
by the loading
MB " 11 kN#m
B
Draw
the shearbyand
momentthe
diagrams
for
the
beam w. 2slope
kN " negative
loading,
Eq.cantilever
6–3.
V " negative
increasing
w " 0 constant
w " negative
constant
2
4
shear at x = 4 m
is
!5
kN,
the
reaction
on
the
beam.
This
1.5
kN/m
in The
Fig. 6–15a.
x
(m)
slope
" negative
increasing
2
slope " 0 slope "
2m
2m
negative
constant
0
value can be verified by finding the area under the distributed M (kN#m)
V ƒ x = 4 m = V ƒ x = 2 m + ¢V = -2 VkN
!2
loading, Eq. 6–3.
(kN)- (1.5 kN>m)(2 m) = -5 kN
he fixed support
2 kN
(c)
(a)
1.5 kN/m
2
0
4
x (m)
Moment
Diagram.
of zero at x2 = 0 is plotted4 in By " 5 kN
V ƒ x = 4 m = V ƒ x = 2 m + ¢V = - 2!5
kN
- (1.5 kN>m)(2
m) The
= - 5moment
kN
2m
2m
x (m)
Fig. 6–15d. Notice how the moment diagram is constructed
based
!4
N. This value is
6 its slope, which is equal to the shear
on knowing
(b) at each point.
A
SOLUTION
2 kN
Moment Diagram.
The moment of zero at x = 0 is plotted !2
in
hear diagram is
The
change
of
moment
from
to
m is determined !11
x
=
0
x
=
2
B
VFig.
" negative
constant
6–15d.
Notice The
how support
the moment
diagram
is constructed
based
w " 0 (d) w " negative constant
the loading w.
Support
Reactions.
reactions
at
the
fixed
support
slope " negative constant V " negative increasing
from themarea
under the shear diagram. Hence,
moment
slope(c)
" 0 the
2m
slope
" negativeat
constant
knowing
its6–15b.
slope,
which
is equal to2the
shear at each point.
n the beam. This
B areon
shown
in Fig.
slope " negative increasing
!5
2 kN
m
is
x
=
2
change
of moment from x = 0 to x = 2 m is determined
the distributed MThe
(kN#m
)
Shear
Diagram.
shear
end
A is !2Hence,
kN. This
is at V (kN)
(a)
from
the area The
under
the at
shear
diagram.
thevalue
moment
6
V " 2 kN
plotted
6–15c. Notice
how
is = 0 + [-2 kN(2 m)] = -4 kN # m
2m
M ƒ xthe
= M ƒ xdiagram
is 0, Fig.
x =at2 xm "
= 2 m shear
= 0 + ¢M
2
4
M " 4 kN#m
0
x (m)
constructed
by following the slopes defined
by the loading w. V " negative constant 2
4
(e)
(2 m) = - 5 kNThe shear at x = 4 m is !5 kN, the reaction on the beam. This# slope " negative constant V " negative increasing x (m)
slope "of
negative
increasing
value
determined
from the2 mmethod
sections,
M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 This
+ [ -same
2 kN(2
m)] can
= -be
4 kN
m
!4 by finding the area under the distributed
value
can be verified
M !2
(kN#m)
Fig. 6–15e.
SOLUTION
(e)
loading, Eq. 6–3.
= 0 is plotted in
This same
value can The
be determined
from the
sections,
Support
Reactions.
support reactions
at method
the fixedofsupport
2 (c)
4
!11
onstructed based B are
(d) 6–15b.
Fig.shown
6–15e. in Fig.
0
x (m)
Fig. 6–15
!5
ar at each point.V ƒ x = 4 m = V ƒ x = 2 m + ¢V = - 2 kN - (1.5 kN>m)(2 m) = - 5 kN
2 kN
m is determined Shear Diagram. The shear at end A is !2 kN. This value is
6
!4
Fig.
6–15c. Notice how the shear diagram is
the moment at plotted at x " 0,V "
2 kN
V
"
negative
constant
Moment
Diagram.
The moment
of zero
at x =by0 the
is plotted
constructed
by following
the slopes
defined
loadingin w.
slope " negative constant V " negative increasing
!11
M
"!5
4 kN#m
Fig.
6–15d.
Notice
the
is constructed
based
(d)
The
shear
at x how
ismoment
kN, diagram
the reaction
on the beam.
This
= 4m
slope " negative increasing
on
knowing
its
slope,
which
is
equal
to
the
shear
at
each
point.
value
can
be
verified
by
finding
the
area
under
the
distributed
2M
kN(kN#m)
2m
= -4 kN # m
The
change
of6–3.
moment from x = 0 to x = 2 m is determined
loading,
Eq.
from the area
(e) under the shear diagram. Hence, the moment at
4
V "22 kN
0
x (m)
thod of sections,
x = 2 m is
V ƒ x = 4 m = V ƒ x =Fig.
+ ¢V = - 2 kN - (1.5 kN>m)(2 m) = - 5 kN
M " 4 kN#m
2 m 6–15
M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(2 m)] = - 4 kN # m
Moment Diagram. The moment of zero at x = 0 is plotted in
Fig. 6–15d. Notice how the moment diagram is constructed based
This same value can be determined from the method of sections,
on knowing its slope, which is equal to the shear at each point.
Fig. 6–15e.
The change of moment from x = 0 to x = 2 m is determined
from the area under the shear diagram. Hence, the moment at
x = 2 m is
M ƒ x = 2 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(2 m)] = - 4 kN # m
PAGE !71
This same value can be determined from the method of sections,
Fig. 6–15e.
2m
(e)
!4
!11
(d)
2 kN
Fig. 6–15
V " 2 kN
M " 4 kN#m
2m
(e)
Fig. 6–15
!4
(d)
V"
Fig.
270
DING
$
CHAPTER 6
EXAMPLE 6.9
BENDING
Ölçü ve yükleme durumu şekilde verilen çıkmalı kirişin kesme kuvveti ve
Draw the shear and moment diagrams for the overhang beam in
moment
çiziniz.for the
Draw the
shear diyagramlarını
and moment diagrams
Fig.overhang
6–16a. beam in
Fig. 6–16a.
4 kN/m
4 kN/m
A 270
A
270
CHAPTER 6
CHAPTER 6
" 10 kN
B
B
BENDING
2m
(a)
(a)
slope " 0
w " negative constant
slope " negative constant
gative constant
NDING
egative constant
2m
4m
2m
4m
A
BENDING
4m
EXAMPLE (b)6.9
Ay " 2 kN
B " 10 kN
EXAMPLE
w " 0 6.9 y
2m
4 kN/m
4 kN/m
Draw the shear and moment diagrams for the overhang beam in
Fig.
6–16a.
Draw
the shear and moment diagrams for the overhang beam in
SOLUTION
Fig.
6–16a.
Support Reactions. The support reactions are shown in
V (kN)
SOLUTION
4 kN/m
8
Support Reactions. The
support4 kN/m
reactions are
Fig. shown
6–16b. in
4 kN/m
4 kN/m
Fig. 6–16b.
Shear Diagram. The
shear of !2 kN at end A of the beam
A
Shear Diagram.
The shear of !2 kN at end Aisofplotted
the beam
at x " 0, Fig. 6–16c. The slopes
are determined
A
B
A
x (m)
Draw the
moment
diagrams
for
the overhang
beamloading
in
0shear and
is plotted
at
x
"
0,
Fig.
6–16c.
The
slopes
are
determined
from
the
and
from
this
the
shear
diagram is
4
6
A
B
x (m) Fig. 6–16a.
from the loading4 m
and !2
from this2 m
the shearconstructed,
diagram isas indicated in the
figure.
In
particular,
notice
6
2
m
4m
6
the
positive
jump
of
10
kN
at
m
due
to
the
force
By, as
x
=
4
constructed, as indicated in the figure.
In
particular,
notice
2m
4 kN/m
4m
2m
4m
(b)
Ay "jump
2 kN of(c)
indicated
4 kN/m
the positive
10 kN at x =By4"m10due
force Byin
, asthe figure.
kN to the
(a)
w " 0 (b)
indicatedAin
the
figure.
2 kN
y"
By " 10 kN
(a)
slope
"0"
0 negative
V
decreasing
w"
V " negative constant
w " negative
constant Moment Diagram. The moment of zero at x " 0 is
A
slope
"
negative
decreasing
plotted,
Fig.
6–16d.
Then
following
the
behavior of the slope
slope
"
0
slope
"
negative
constant
SOLUTION
Moment
Diagram.
The
moment
of
zero
at
x
"
0
is
slope
negative
constant
negative decreasing
B
w ""negative
constant
V (kN)
" negative decreasing plotted, Fig.
found
from
the
shear
diagram,
the
moment
diagram
is
6–16d. Then following
the
behavior
of
the
slope
SOLUTION
slope " negative constant
Support
Reactions. The support
reactions
are shown
in
M (kN#mV) (kN)
constructed.
The
moment
at
x
"
4
m
is
found
from
the
area
found from
the shear diagram,
the
moment
diagram
is
Fig.
6–16b. Reactions. The support reactions are shown in
Support
4 m 8 slope " 0 2 m
2m
under
thearea
shear diagram.
constructed. The moment at x "
the
8 4 m is found from
Fig. 6–16b.
4
slope " 0
Shear
Diagram. The shear of !2 kN at end A# of the beam
under
the
shear
diagram.
0
By " 10 kN
(a) x (m)
M ƒ x = 4 is
= M ƒ x = 0at+ x¢M
= 0Fig.
+ [6–16c.
- 2 kN(4
m)]slopes
= - 8 kN
m plotted
" 0,
areA m
determined
6
Shear Diagram.
The shear
of The
!2 kN
at end
of the beam
x (m)
#
x
(m)
M
=
kN(4
m)]
=
8
kN
m
0 M ƒ x = 0 + ¢M = 0 + [-2
ƒ
x
=
4
m
from
the
loading
and
from
this
the
shear
diagram
is
We canisalso
obtain
by using
method
sections,
plotted
atthis
x "value
0, Fig.
6–16c.the
The
slopesofare
determined
4
6
6 constant
negative
x
(m)
SOLUTION
0
" negative constant
constructed,
as
indicated
in
the
figure.
In
particular,
notice
!24
as of
shown
in Fig.
from
the6–16e.
loading and from this the shear diagram is
We
the6 method
sections,
!8 by using
6 can also obtain this value
the
positive
of 10 kN in
at x
the force B
4 m due
constructed,
as indicated
the= figure.
Intoparticular,
notice
The support
reactions are shown
in jump
!2
y, as
as6Support
shown in Reactions.
Fig. 6–16e.
(d)
indicated
in
the
figure.
the positive jump of 10 kN at x = 4 m due to the force By, as
Fig. 6–16b.
(c)
indicated in the figure. V " 2 kN
(c)
Moment
Shear
Diagram.
The shearVVof
!2 kN decreasing
at end A of the
beam Diagram. The moment of zero at x " 0 is
" negative
V"
negative constant
" 2 kN
M " 8 the
kN#m
slope
"The
negative
decreasing
plotted,
Fig.Diagram.
6–16d. ThenThe
following
behavior
of xthe"slope
A
Moment
moment
of
zero at
0 is
slope
"
negative
constant
is plotted
at
x
"
0,
Fig.
6–16c.
slopes
are
determined
V
"
negative
decreasing
V " negative constant
x (m)
slope
"
negative
decreasing
found
from
the
shear
diagram,
the
moment
diagram
is
M
"
8
k
N#m
plotted,
Fig.
6–16d.
Then
following
the
behavior
of
the
slope
from
the
loading
and
from
this
the
shear
diagram
is
slope
"
negative
constant
4
6
A
4m
M (kN#m)
constructed.
at x " 4 m
from
the areais
found
themoment
shear diagram,
theis found
moment
diagram
constructed, as indicated in the figure. In particular,
noticefromThe
2
M (kN#m) 4 m
under
constructed.
Thediagram.
moment at x " 4 m is found from the area
the positive jump of 10 kN at x = 4 mslope
due "to0 the force
By,the
as shear
slope
"0
4
2 kN the shear diagram.
under
indicated
in
the
figure.
x (m)
0
M ƒ x = 4 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m
4
6
(e)
20kN
x
(m)
Moment Diagram. The moment of
zero at x M
"ƒ x =04 mis = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m
" negative decreasing
6
We
can
also
obtain
this value by using the method of sections,
(e)
" negative decreasing plotted, Fig. 6–16d. Then
Fig. 6–16
following the behavior of the slope
as
shown
in Fig.
6–16e.
We
can
also
obtain
this value by using the method of sections,
!8
found from the shear
the moment diagram is
Fig. diagram,
6–16
as
shown
in
Fig.
6–16e.
(d) at x!8
constructed. The moment
" 4 m is found from the area
slope " 0
under the shear diagram.(d)
V " 2 kN
8
6
x (m)
M ƒ x = 4 m = M ƒ x = 0 + ¢M = 0 + [ - 2 kN(4 m)] = - 8 kN # m
A
We can also obtain this value by using the method of sections,A
as shown in Fig. 6–16e.
2 kN
2 kN
V " 2 kN
M " 8 kN#m
A
4m
2 kN
(e)
Fig. 6–16
PAGE !72
V " 2 kN
M " 8 kN#m
M " 8 kN#m
4m
4m
(e)
(e)
Fig. 6–16
Fig. 6–16
6.2
EXAMPLE 6.10
$ EXAMPLE 6.10 Ölçü ve yükleme durumu şekilde verilen şaft için kesme kuvveti ve
The shaft in çiziniz.
Fig. 6–17a
supported byyatay
a thrust
bearing
at A and a
moment
diyagramlarını
A ismesnedinde
hareket
The shaft
injournal
Fig. 6–17a
is supported
by athe
thrust
at
A and
aengellenmiş, B mesnedi ise
bearing
at B. Draw
shearbearing
and moment
diagrams.
kayıcı
mesnettir.
journal
bearing at B. Draw the shear and moment diagrams.
120 lb/ft
A
A
D FOR
CAL
CONSTRUCTING SHEAR AND MOMENT DIAGRAMS
120 lb/ft
A
B
271
METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS
12 ft
Çözüm:
2
6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS
271
B
12 ft
271
(a)
(a)
120 lb/ft
A
B
120 lb/ft
B
12 ft
12 ft
(b)
Ay = 240
lb
(b)w
# negative increasing By # 480 lb
Ay = 240 lb
# negative
increasing
w # negativeslope
increasing
By #
480 lb
V (lb)
slope # negative increasing
V (lb)
240
SOLUTION
12
6.93
240
aring at A and a
120 lb/ft
x (f
SOLUTIONSupport Reactions.
0
12
The
support
reactions
are
shown
in
6.93
diagrams.
thrust
bearing at A and a
x IAGRAMS
(ft)
120 lb/ft
6.2
G
RAPHICAL
M
ETHOD
FOR
C
ONSTRUCTING
S
HEAR
AND
M
OMENT
D
0
Support Reactions.
Fig. 6–17b. The support reactions are shown in
(c)
moment diagrams.
/ft
Fig. 6–17b. Shear Diagram.
(c)
As
shown
in
Fig.
6–17c,
the
shear
at
is
x
=
0
6.2 B
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT
IAGRAMS
271
V #D
positive
decreasing
A
120 lb/ft
" 480
Shear Diagram.
As shown the
in Fig.
6–17c,
the shear
at xloading,
= 0 is the
Following
slope
defined
by
the
shear
slope
# positive decreasing
V
#
positive
decreasing
A !240.
EXAMPLE 6.10 B
V # negative
" 480 increasing
!240. Following
the
slope
defined
by
the
loading,
the
shear
slope
#
positive
decreasing
diagram
is
constructed,
where
at
B
its
value
is
"480
lb.
Since
the
12 ft
# negative increasing
B
V # negativeslope
increasing
diagram is constructed,
where
at
Bpoint
its value
is "480
shear
changes
sign,
the6–17a
bethe
located.
doand M
V = by
0lb.must
EXAMPLE
6.10
(b)The
12 ft
shaft
in Fig.
iswhere
supported
aSince
thrust
bearing To
at A
a (lb$ft)
120 lb
slope
#
negative
increasing
V#0
B
Ashear
lb
y = 240 changes
thissign,
we
will
use
the
method
free-body
of
the point
where
must
beGThe
located.
To
do diagram
VDraw
=of0 sections.
M (lb$ft)
6.2shear
RAPHICAL
METHOD
FOR
CONSTRUCTING
SHEARslope
AND M
DIAGRAMS
# OMENT
0
(b)
journal
bearing
at lb
B.
the
and
moment
diagrams.
w # negative increasing
B
#
480
y
V#0
AyFig.
= the
2406–17a
lb
left
of the
shaft,
sectioned
at at
andiagram
arbitrary
this
we
will
use
the segment
method
sections.
The
free-body
The
shaft
in
is supported
by
a thrust
bearing
A
and a ofposition x, is slope
120 lb/ft
slope
# negative
#0
w # increasing
negative increasing By # 480 lb
V (lb)
120
lb/ft
1109
6
shown
in
Fig.
6–17e.
Notice
that
the
intensity
of
the
distributed
the
left
segment
of
the
shaft,
sectioned
at
an
arbitrary
position
x,
is
journal bearing at
B. #
Draw
the
shear
and
moment
diagrams.
slope
negative
increasing
A
6.2 GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS
271
(lb)
load
at
x
is
w
#
10x,
which
has
been
found
by
proportional
1109
shown inV Fig.
6–17e.
Notice
that
the
intensity
of
the
distributed
EXAMPLE 6.10
120 lb/ft by proportional
i.e.,which
120!12has
# w!x.
load at x istriangles,
w # 10x,
been found
240
A
B 12 ft
x(
0
A
12 6–17a
Thus,
V #in0,Fig.
B
6.93
Thefor
shaft
is
supported
by
a
thrust
bearing
at
A
and
a
triangles,
i.e., 120!12
#
w!x.
240
6.93
12 120 lb
x (ft)
(b)
EXAMPLE
6.10
0
are shown in
x (ft)
0
12
6.93 at B. Draw
journal
bearing
the
A = 240 lb
Thus, for V
# 0,
1 shear and moment diagrams.
12 ft y 6.93w # negative
(ft)2 (10x)x = 0
240 lbx = 0;
(d) 12 increasing B #
0 + c ©F
12 ft
A y
eactions are shown in
y
B
(c)
The+shaft
Fig. 6–17a is supported
thrust bearing
at A and a 120 lb/ft
120
lb/ft
slope
# negative increasing
(b)(d)
c ©Fin
240 lb by
- 12a(10x)x
= 0
y = 0;
V
(lb)
x
=
6.93
ft
1
A
=
240
lb
(c)
A
hear at x = 0 is journal
y
[10 x ] x
V # positive
decreasing
bearing
at B. Draw the shear and moment diagrams.
w # negative increasing By # 480 lb2
12
ft
x
" 480
7c, the
shear
at xslope
= 0#ispositive
x = 6.93 ft (a)
ding,
the
shear
1 [10 x ] x
V #decreasing
positive decreasing
slope
#
negative
increasing
Moment
Diagram.
diagram starts at 0 V
since
3 10 x
2
(lb) there
V # negative
increasingThe moment
"120
480lb/ft
decreasing
12
ft
80the
lb. loading,
Since thethe shear slope # positive
240
Bx
A it is constructed based on Athe
is slope
no SOLUTION
moment
atincreasing
A; then
(a)
# negative
Moment
Diagram.
The
diagram starts at 0 since there B slope as
3 10 (b)
V #moment
negative
increasing
12
alue
is "480
the
e located.
To lb.
doSince
6.93 V
x
M (lb$ft)
determined
from
the shear
diagram.
The
maximum
moment
Ay0= 240 lb
slope
#
negative
increasing
is no
at
A;
then
it
is
constructed
based
on
the
slope
as
V
#
0
Support
Reactions.
The
support
reactions
are
shown
in
A
be located.
To
do moment
0 must
ody
diagram
of
V
M
(lb$ft)
w # negative increasing By #
12 ft is equal to 240
slope
#0
M
occurs
at
the shear
zero, since 12 ft
= 6.93diagram.
ft, whereThe
A Fig.
V6–17b.
#x0shear
determined
from
the
slope #12negative
The
free-body
of
6 Bmaximum moment
ary position
x, diagram
is SOLUTION
(c) increasing
A(b)
6.93
slope
#
0
V
(lb)
Fig.
6–17d,
dM>dx
=
V
=
0,
x
(ft)
occurs
where
the shear
is equal
zero, since
x = 6.93
ft,The
0 lbat x = 0 is
= 240
x M
6to(a)
an
position
x, is atReactions.
1109
thearbitrary
distributed
Support
support
reactions
shown
in
Shear
Diagram.
As
shown
inare
Fig.
6–17c,
theAyshear
V # positive
decreasing
w # negative increasing
By # 480
lb
12 ft
Fig.
6–17d,
dM>dx
=
V
=
0,
1109
ensity
of the distributed
d
+
©M
=
0;
by
proportional
Aypositive
# 240 lbdecreasing
Fig. 6–17b.
"
!240. Following the slope defined by the loading,
shear(c)
slope x#
slopethe
# negative
increasing
1
1
V (lb)
240
n found by proportional
M
+
[(10)(6.93)]
6.93
(6.93)
240(6.93)
=
0
V # negative increa
A
B
diagram
is
constructed,
where
at
B
its
value
is
"480
lb.
Since
the
(e)
max
SOLUTION
2
3
d+
©M
=
0;
A
#
240
lb
Shear Diagram. As shown in Fig.
(a) 6–17c, the shear at x = 0 is
y
12
V # positive decreasing
slope #6.93
negative incre
x (ft)
0
1 changes sign,
1 point where V = 0 must be located. To do
shear
the
M0 (lb$ft)
#
Mmax
+6.93
[(10)(6.93)]
6.93
(6.93)
240(6.93)
= 0 slope
A
B
"6–17
480
!240. Following
the
slope
defined
by
the
loading,
the
shear
#
positive
decreasing
(e)
Support
Reactions.
The
support
reactions
are
shown
in
12
M
=
1109
lb
ft
2
3
Fig.
max
x (ft)
0
V#0
240
this we
will at
use
the
method
of sections.
Thethe
free-body
diagram ofV # negative increasing
6.93
12
diagram is constructed,
where
B
its value
is "480
lb. Since
slope #(c)
0
SOLUTION
#
(d)Fig. 6–17b.
M
=
1109
lb
ft
12
6.93
Fig.
6–17
max
Finally,
notice
how
integration,
first
of
the
loading
w
which
is
linear,
slope # negative increasing
the
left
segment
of
the
shaft,
sectioned
atTo
an do
arbitrary
position x, is
x (ft)
x = 0
shear
changes
sign,
the
point
where
must
be
located.
V
=
0
0
(d)
M
(lb$ft)
Shear
Diagram.
As
shown
in
Fig.
6–17c,
the
shear
at
is
x
=
0
Support Reactions.
The
support
reactions
shown and
in then a moment
V # positive decreasing 1109
produces
shear
diagram
which
isare
parabolic,
ft
1a[10
shown
inxof
6–17e.
Notice
that
intensity
of the distributed
V#0
notice
how
integration,
first
of
the
loading
w the
which
linear,
] Fig.
x sections.
thisFinally,
we will
use the
method
free-body
diagram
of loading,
"
!240.
theThe
slope
defined
by isthe
the slope
shear
slope # positive decreasing
Fig.
6–17b.
2 Following
#
0
(c)
x = 6.93 ft
diagram
which
is
cubic.
1
load
at xxsectioned
is w[10is
#x]parabolic,
which
has
been
found
by proportional
x10x,
produces
a shear
diagram
which
and
then
avalue
moment
6
the
left
segment
of
the
shaft,
at
an
arbitrary
position
x,
is
V
#
negative
increas
2
diagram
is
constructed,
where
at
B
its
is
"480
lb.
Since
the
at 0 since there Shear Diagram. As
shown 3in
shear at x = 0 is V # positive decreasing
x # the
10
x 6–17c,
triangles,
i.e.,Fig.
120!12
w!x.
diagram
which
is shear
cubic.
slope # negative incre
NOTE:
Having
studied
these
examples,
test= yourself
bylocated.
coveringTo do 1109
shown
in Fig.
6–17e.
that
the
of
thethe
distributed
am the
starts
at 0 as
since
there
changes
sign,
the
point
where
be
V
0 must
3intensity
M0(lb$ft) " 480
on
slope
!240.
Following
the Notice
slope
defined
by
the
loading,
shear
slope
# positive decreasing
10
x
V# 0,
Thus,
for
V
6.93
the
shear
and
moment
diagrams
in Examples
6–1 through 6–4
V#0
12
load
at isx constructed,
isHaving
wover
# this
10x,
which
has
been
found
by
ed
based
on the slope
as
we
will
the
method
of
sections.
# negative increasing
imum
moment
NOTE:
studied
these
yourself
byThe
covering
diagram
atuse
Bexamples,
its
value
"480
lb.proportional
Since
the free-body diagramVof
V istest
slope # 0
A and see ifwhere
1 concepts discussed here.
you
can
construct
them
using
the
c
slope
#
negative
increasing
triangles,
i.e.,
120!12
#
w!x.
240
lb
+
©F
=
0;
(10x)x
=
0
The
maximum
moment
the
left
segment
of
the
shaft,
sectioned
at
an
arbitrary
position
x,
is
(d)
M= 0
to zero, since shear
over
the shear
diagrams
inmust
Examples
6–1
2 through
changes
sign,and
theAmoment
pointy where
be located.
To do 6–4
V
M (lb$ft)
x (ft)1109
0 distributed
M
V#0
forifuse
Vyou
#the
0, method
is equal to zero,
since
shown
in
Fig.
6–17e.
Notice
that diagram
the
intensity
of the
and
can
construct
them
using
the
concepts
discussed
here.
thisThus,
we see
will
of
sections.
The
free-body
of
6.93
12
x = 6.93 ft
x
slope # 0
1 [10 x ] x
load
at
x
is
w
#
10x,
which
has
been
found
by
proportional
6 2
the
of the shaft,
sectioned
at an arbitrary
position x, is
x - 1 (10x)x
240
lb
+ cleft
©Fsegment
=
0
x
(d)
y =A0;
#
240
lb
y
i.e., 120!12
# w!x.
1109
shown in Fig.
6–17e.triangles,
Notice
that
the2 intensity
of the distributed
Moment
3 10
A
# 240 lb Diagram. The moment diagram starts at 0 since there
240(6.93) = 0
y(e)
0
forhas
V #at
0,A; then
x = 6.93
1 [10 x ] x
load= at
been
found
proportionalbased on the slope as
is Thus,
nowhich
moment
it by
is ft
constructed
6.93
12
6.93) B - 240(6.93)
0 x is w # 10x,
V
(e)
2
1
triangles, i.e., 120!12
#
w!x.
from
the
shear
diagram.
The
maximum
moment
x
Fig.determined
6–17
c
240 lb - 2(10x)x = 0
+ ©Fy = 0;
(d)
A
Moment
TheFig.
moment
diagram
starts the
at 0 shear
since there
3 10x x(ft)
0 to zero, since
t
Thus, forDiagram.
V # 0, occurs
at 6–17
is equal
x = 6.93
ft, where
6.93
12
w which is linear, is no moment at A; then it is constructed based on the slope
as
x
=
6.93
ft
1
V
dM>dx
=
V
=
1 0, Fig. 6–17d,
loading
w which is
linear,
c ©Fy = 0; from the
x 2 [10 x ] x
240
lb
+determined
(10x)x
=
0
then
a moment
(d)
2
shear diagram.
The maximum moment
x
bolic, and then a moment
Theequal
moment
diagram starts at 0 sinceAthere
dft,
+ ©M
= Diagram.
0;the shear
3 10
Ay # 240
M lb
occurs at x = 6.93Moment
where
x 1 =is 6.93
ft to zero,1 since
1 [10 x ] x
is no moment
at A;
it is constructed
based
the slope
Mmax
+ then
6.93 A 3(6.93)
240(6.93)
= 0as 2
B - on
(e)
V
2 [(10)(6.93)]
self by covering dM>dx = V = 0, Fig. 6–17d,
x
x
determined
from
the
shear
diagram.
The
maximum
moment
Moment
Diagram.
The
moment
diagram
starts
at
0
since
there
3
test
yourself
by
covering
#
A
10 x
6–1 through 6–4 d+ ©M = 0;
Mmax = 1109 lb ft
Fig. 6–17
Ay #
240 lb
whereonthe
since
x = 6.93 ft, based
is no6–4
moment at A; occurs
then
it at
is constructed
the shear
slope is
as equal to zero,
Examples
through
V
s discussed6–1
here.
1
1
Mmax the
+dM>dx
6.93
240(6.93)
= 0
AThe
B - first
6–17d,
= diagram.
V =how
0, Fig.
Finally,
integration,
ofmoment
the loading
w which is linear,
(e)
determined
from
shearnotice
maximum
2 [(10)(6.93)]
3 (6.93)
e concepts discussed
here.
x
A
produces
a
shear
diagram
which
is
parabolic,
and
then
a
moment
M
occurs at x = 6.93 ft,
where
the
shear
is
equal
to
zero,
since
#
Mmax= =0;1109 lb ft
d + ©M
Ay # 240 lb
Fig. 6–17
diagram
is cubic.
1
6–17d, which
dM>dx = V = 0, Fig.
Mmax + 12[(10)(6.93)]
6.93
(6.93) B - 240(6.93) = 0 x
A
(e)
PAGE
!
7
3
3
Finally, notice how integration, first of the loading w which is linear,
NOTE: Having studied these examples,
test
yourself
by
covering
d+ ©M = 0;
A
#
240
lb
# ft
y
Mmax
= then
1109 albmoment
produces a shear diagram
which is parabolic,
and
Fig. 6–17
1
the shear6.93
andA 13moment
6–1 through 6–4 (e)
M
+ over
(6.93) B -diagrams
240(6.93)in=Examples
0
2 [(10)(6.93)]
diagram whichmax
is cubic.
Finally,
howconstruct
first
of the
w which
is linear,
and seenotice
if you can
them
using
theloading
concepts
discussed
here.
# integration,
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