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 "73
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 "74
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
5
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,
is applied about an axis perpendicular to this axis of symmetry as
shown in Fig. 6–18. The behavior of members that have unsymmetrical
6.3 orBare
ENDING DEFORMATION OF A STRAIGHT MEMBER
281
cross sections,
made from several different materials, is based on
similar observations and will be discussed separately in later sections of
this chapter.
DeformationBy using a highly deformable material such as rubber, we can illustrate
what happens when a straight prismatic member is subjected to a bending
6.10
Prizmatik bir kirişin eğilme deformasyonu
ight Member
moment. Consider, for example, the undeformed bar in Fig. 6–19a, which
has a square cross section and is marked with longitudinal and transverse
discuss the deformations
that occur
when
a
of
grid lines. When
a bending
moment
is Axis
applied,
it tends to distort these
y
symmetry
made of homogeneous
material,
is subjected
to Fig. 6–19b.
lines into
the pattern
shown in
Notice
that the longitudinal
will be limited to beams
havingcurved
a cross-sectional
lines become
and the vertical transverse lines remain straight and
with respect to an yet
axis,undergo
and thea bending
rotation. moment
M
is perpendicular toThe
thisbending
axis ofmoment
symmetry
z within the bottom portion of
causesasthe material
the bar
stretch
and the material within the top portion to compress.
behavior of members
thattohave
unsymmetrical
Neutral
x
Consequently,
between
these on
two regions there must be a surface,
called
de from several different
materials,
is based
surface
the
neutral
surface,
in
which
longitudinal
fibers
of
the
material
will
not
d will be discussed separately in later sections of
Longitudinal
undergo a change in length, Fig. 6–18. axis
rmable material such as rubber, we can illustrate
aight prismatic member is subjected to a bending
xample, the undeformed bar in Fig. 6–19a, which
n and is marked with longitudinal and transverse
ing moment is applied, it tends to distort these
own in Fig. 6–19b. Notice that the longitudinal
d the vertical transverse lines remain straight and
causes the material within the bottom portion of
he material within the top portion to compress.
hese two regions there must be a surface, called
hich longitudinal fibers of the material will not
gth, Fig. 6–18.
rmation
M
z
Neutral
surface
x
Longitudinal
axis
Fig. 6–18
6
Fig. 6–18
M
6
Horizontal lines
become curved
M
Vertical lines remain
straight, yet rotate
Before deformation
M
After deformation
(a)
(b)
Şekildeki moment kirişin alt tarafında çekme üst tarafında basınç oluşturmaktadır. Bu iki
Fig. 6–19
bölge arası geçiş sırasında bir “tarafsız
yüzey” bulunacağı aşikardır. x ekseninin
konumlandırıldığı boyuna eksende boy değişimi olmadığı görülmektedir. Tüm kiriş
kesitlerinin deformasyon sırasında düzlem kaldığı ve boyuna eksene dik olduğu
gözlemlenmektedir.
282
CHAPTER 6 BENDING
From these observations we will make
assumptions regarding the way the stress deform
longitudinal axis x, which lies within the neutral
Vertical lines remain
not experience any change in length. Rather t
straight, yet rotate
deform
the beam so that this line becomes a c
After deformation
plane of symmetry, Fig. 6–20b. Second, all cro
(b)
remain plane and perpendicular to the longi
deformation. And third, any deformation of th
Fig. 6–19
own plane, as noticed in Fig. 6–19b, will be neg
z axis, lying in the plane of the cross section an
section rotates, is called the neutral axis, Fig. 6–2
In order to show how this distortion will str
isolate a small segment of the beam located a dis
Note the distortion of the lines due to length and having an undeformed thickness ¢x,
from the beam,
is shown in profile view
bending oföncesi
this rubber
bar. Thegösterilmiştir.
top line taken
Şekildeki lastik elemanın deformasyon
ve sonrası
Deformasyon
sonrası
Horizontal lines
become curved
M
theorta
bottom
line compresses,
üst çizginin uzadığı alt çizgininstretches,
kısaldığı
çizginin
ise aynı and
uzunlukta kaldığı görülmektedir.
the center line remains the same length.
Furthermore the vertical lines rotate and yet
remain straight.
6
z
PAGE "76
Note the distortion of the lines due to
bending of this rubber bar. The top line
stretches, the bottom line compresses, and
the center line remains the same length.
Furthermore the vertical lines rotate and yet
remain straight.
y
6
z
6.3
283
BENDING DEFORMATION OF A STRAIGHT MEMBER
6.3
x
BENDING DEFORMATION OF A STRAIGHT MEMBER
283
6.3 !xB
BENDING
6.3
ENDING D
DEFORMATION
EFORMATION OF
OF A
AS
STRAIGHT
TRAIGHT M
MEMBER
EMBER
O¿
x
6.3
O¿
BENDING DEFORMATION OF A STRAIGHT MEMBER
O¿
O¿
y
(a)
O¿
r
6.3
z
longitudinal
axis
!s " !x
!x y
!x
d element
M
r
r
r
O¿
!u
!u
!u
!s¿
!x
!x
longitudinal
!x
!s¿
axis Undeformed
element
(b) y Deformed
element
!x
Undeformed
element
Undeformed
element
(a)
(b)(a)
(a)
neutral
surface
longitudinal
6.3
!u
EFORMATION
STRAIGHT MEM
BENDING6.3
D!EFORMATION
OF
A STRAIGHTOF
MAEMBER
s¿ BENDING D
longitudinal
axis longitudinal
y
axis
axis y
y
(b)
!s¿
!x
!xO¿
! s¿
!x
O¿
Fig. 6–20
Deformed element
!x
!s " !x
) Undeformed element
y
!xFig. 6–21
(a)
r
neutral
axis
y longitudinal
longitudinal
axis!x r
! s "!s¿
!x
axis
!s " !x ! s " ! x
x y
!x
longitudinal y
y
longitudinal
longitudinal y! x
axis
!u!x
axis
axis
!x
!s " !x
z
283
BENDING DEFORMATION OF A STRAIGHT MEMBER
!u
gitudinal
axis
!x
dinal
isolate a small segment of the beam located a distance x along the beam’s
length and having an undeformed thickness ¢x, Fig. 6–20a. This element,
taken from the beam, is shown in profile view in the undeformed and
Fig. 6–21
Fig. 6–21
Fig. 6–21
Fig. 6–21
!x
Deformedrelement
r
Deformed
Deformed
elementelement
(b)
(b)
(b)
!u
6
!u
6
ndeformed
element
element
ions
in Fig.
6–21. Notice that any line segmentDeformed
¢x, located
longitudinal longitudinal! s¿
! s¿
!s " !x
!s " !x
surface, does not change its length,
whereas
any
linein Fig. 6–21.
axis
deformed
positions
Notice
any
lineany
segment
¢x, located
y located
y axis
deformed
in Fig.
6–21. that
Notice
that
line
segment
¢x,
(b) positions
ormed (a)
positions in Fig. 6–21. Noticelongitudinal
that any line
segment
located
¢x,
!
x
y
!x
y surface,
longitudinal
onthe
the
neutral
surface,
does
not
change
its
whereas
line
cated at the arbitrary distance y above
neutral
deformed
positions
in Fig.
6–21.
Notice
thatchange
anylength,
line
located
¢x,any
on
the
neutral
surface,
does
not
itssegment
length,
whereas
any line
he neutral surface, does
not
change
its
length,
whereas
any
line
6–21
axis
axis
segment
located
the
arbitrary
distance
y above
neutral
surface,
¢s,
! x arbitrary
and become ¢s¿ afterFig.
deformation.
By
definition,
the
!at
xlocated
segment
atnot
the
distance
y the
above
the neutral
surface,
¢s,
on
the
neutral
surface,
does
change
its
length,
whereas
any
line
ment ¢s, located at the arbitrary distance
y above
the
neutral
surface,
will
contract
and
become
after
deformation.
By
definition,
the the
¢s¿
long ¢s is determined from Eq. 2–2,segment
namely,
willlocated
contract
and
become distance
By definition,
¢s¿ afterydeformation.
at
the
arbitrary
above
the
neutral
surface,
¢s,
contract and become ¢s¿ after deformation.
By
definition,
the
normal
strain
along
is determined
from Eq.
2–2,
namely,
¢salong
normal
strain
is determined
from
Eq.
2–2, namely, 6
¢s
will
contract
and
¢s¿
- ¢s
! x after deformation. By definition, the
! xbecome ¢s¿
mal strain along
is
determined
from
Eq.
2–2,
namely,
¢s
P = lim
¢s¿
- ¢s
¢s¿
- ¢snamely,
normal
strain along
from
Eq. 2–2,
¢s is determined
¢s :
0
¢s that
ed positions in Fig.
6–21.
Notice
segment
located
¢x,
P = limP = lim
¢s¿any
- line
¢s
Undeformed element
Deformed element
Undeformed element
Deformed element
¢s
:
0
¢s
:
0
¢s
¢s
Pin=terms
lim its length,
¢s¿ - ¢s
surface,
does not
change
whereas any line
w neutral
represent
this strain
¢s : 0 of the
¢s location y of the (a) P = lim
Weof
will
now represent
strain
in of
terms
the
y(b)of the
nt
located
the arbitrary
y above
theaxis
neutral
surface,
We will
now
represent
this ¢s
strain
in
the of
location
y of the
(a)
(b) location
:this
0
¢sterms
he¢s,
radius
of at
curvature
the longitudinal
the
r ofdistance
We
will and
now become
represent
strain
insegment
terms of
the
location
yofof
the of curvature
segment
and
the
radius
the longitudinal
axis
of the
r of
ntract
after
deformation.
By
definition,
the
¢s¿this
and
the
radius
curvature
of
the
longitudinal
axis
of
the
r
e deformation, ¢s = ¢x, Fig. 6–21a. After
deformation
now represent
this
strain
in
ofFig.
the6–21a.
location
y deformation
of the
Fig.
6–21
Fig.
6–21
element.
Before
After
¢sterms
= ¢x,
ment
the¢sradius
of curvature
ofWe
thewill
longitudinal
axis deformation,
of
the
relement.
strainand
along
is determined
from Eq.
2–2,
namely,
Before
deformation,
Fig.
6–21a.
After
deformation
¢s
=
¢x,
us of curvature r, with center of curvature
at
point
segment
and
theaO¿,
radius
of
thecenter
longitudinal
axis at
of point
the O¿,
r of
has
of curvature
curvature
with
of curvature
ment. Before deformation, ¢s
Fig.
6–21a.
After
deformation
¢x,
has
a¢x
radius
ofradius
curvature
center
of curvature
at point
¢x
r, with r,
O¿,
¢s¿ =-the
¢ssides
ce ¢u defines the angle
between
of
the
element,
element.
Before
deformation,
Fig.
6–21a.
After
deformation
¢s
=
¢x,
P = lim r, with center
Fig.
6–21b.
Since
defines
thebetween
angle between
theof
sides
of
the element,
¢u the
has a radius of curvature
of
curvature
at
point
O¿,
Fig.
6–21b.
Since
defines
angle
the
sides
the
element,
¢u
¢s : 0 the¢s
deformed
length
¢u. In the same manner,
a ¢x
radius
of¢s
curvature
with
curvature
at point
¢x
r, manner,
In the
samecenter
manner,
the deformed
length
= of
¢sof
=the
r¢u.
defines
the angle between
the
element,
¢u
In
the
same
the ofdeformed
length
of O¿,
¢x has
= ¢s
=sides
r¢u.
¢sof ¢s
H6–21b.
A P T E R Since
6
B
END
ING
Substituting
intointhe
above
equation,
we¢u
get
1r now
- y2¢u.
will
represent
this strain
terms
of6–21b.
thepositions
location
y¢s¿
of
the
deformed
positions
in
Fig.
6–21.
Notice
that
any
line
segment
located
¢x,
Fig.
Since
defines
the
angle
between
the
sides
of
the
element,
becomes
Substituting
into
the
above
equation,
we get
=
1r
y2¢u.
deformed
in
Fig.
6–21.
Notice
that
any
line
segment
located
¢x,
becomes
into the above equation, we get
¢s¿ = 1rlength
- y2¢u.
the deformed
ofSubstituting
= ¢s = r¢u. In the same manner,
¢s
nt and the radius of curvature r ofonthe
longitudinal
axis
of the
the surface,
on
the
neutral
does not
change
its length,
whereas
any line
In
same
manner,
the
deformed
length
of
¢x
=
¢s
=
r¢u.
¢s
the
neutral
surface,
does
not
change
its
length,
whereas
any
line
omes ¢s¿ = 1r -1ry2¢u.
Substituting
into the above equation, we get
1r - y2¢u - r¢u
- y2¢u
r¢u
t. Before deformation,
Fig.
6–21a.¢s,
After
deformation
¢s = -¢x,
1r
-lim
y2¢u
r¢u
segment
located
arbitrary
distance
yequation,
abovesurface,
thewe
neutral
¢s,
segment
located
arbitrary
distance
y-above
the neutral
becomes
Substituting
into
the
above
get surface,
¢s¿
=
1rat
-the
y2¢u.
Pat
=Pthe
#
P = lim
max
P
=
lim
¢u
:
0
r¢udeformation.
a radius of¢ucurvature
with
center
of
curvature
at
point
r,r¢u
O¿,
:0
become
after
By
definition,
the
¢s¿
1r - will
y2¢ucontract
- r¢uwill
andcontract
becomeand
By
definition,
the
¢s¿
¢u :
0after deformation.
r¢u
y
1r
-#
y2¢u
- 2–2,
r¢u
21b. Since ¢u defines
thelim
angle between
thestrain
sides
of
the element,
P =
P is
P
or
normal
strain
along
determined
from
Eq.
2–2,
namely,
¢s
!
normal
along
is
determined
from
Eq.
namely,
¢s
max
c
0
r¢u
c
= lim
ordeformed length ofP ¢s
manner,
the
¢s = r¢u. In the same¢u :
r¢u
y ¢u : 0 ¢s¿ y - ¢s
¢s= -¢s¿
y
es ¢s¿ = 1r - y2¢u. Substituting into the above equation, we get
(6–7)
P =yPlim
P = lim
P = (6–7)
or
- ¢s : 0 r ¢s
(6–7)
¢s : 0P = ¢s
r
r
yr¢u
1r - y2¢u
P = --We
(6–7)
y strain
This
important
result
indicates
that of
thein
longitudinal
normal
We
will nowthis
represent
terms
of the
location
of the
will now
represent
strain
inthis
terms
the
location
y of
thestrainy of
P = lim
" x Pthat
r This
=
(6–7)
nt result indicates
that
the
longitudinal
normal
strain
of
important
result
indicates
the
longitudinal
normal
ofaxis
r¢usegment
! ¢u : 0
any
element
within
the
beam
depends
on
its the
location
the
cross
andofthe
radius ofr curvature
longitudinal
of the
r of
andsegment
the radius
curvature
of
axisystrain
ofonthe
r the longitudinal
within
the beam
depends
on its
location
y onelement.
theNormal
cross
anylongitudinal
element
thebirim
beam
on
its
location
ydeformation
on the
cross
deformasyon
Before
deformation,
Fig. 6–21a.
After
deformation
¢s 6–21a.
=dağılımı
¢x,
his important
result
indicates
that
the
normal
strain
ofdepends
element.
Beforewithin
deformation,
After
¢s
= ¢x, Fig.
Normal
strain
distribution
This
result
indicates
that
ther,longitudinal
strain
of point O¿,
a on
radius
of
with
center normal
ofatcurvature
¢x has
element within the beam depends
its
location
the
cross
has
a important
radius
of ycurvature
with
center
of curvature
point
r,curvature
O¿,at
y ¢x on
any
element
within
the
beam
depends
on
its
location
y
on
the
cross
P = - Fig.
(6–7)
Fig.
6–21b.
Since
defines
the
angle
between
the
sides
of
the
¢u
Fig.angle
6–22 between the sides of
6–21b. Since ¢u defines the
element, element,
r
In
the
same
manner,
the
deformed
length of ¢s
¢x
=
¢s
=
r¢u.
¢x = ¢s = r¢u. In the same manner, the deformed length of ¢s
PAGE
"
7
7
important result indicates that thebecomes
longitudinal
normal
strain
of
becomes
Substituting
into
the
above
equation,
we get
¢s¿
=
1r
y2¢u.
¢s¿ = 1r - y2¢u. Substituting into the above equation, we get
ement within the beam depends on its location y on the cross
P = lim
1rr¢u
- y2¢u - r¢u
1r - y2¢u P = lim
¢u : 0
r¢u
6
- P2
willoccurs
occur atinthe
fibers
locatedfiber,
above
the neutral
axis
by1division,
Here
the then
maximum
strain
outermost
located
a
1+y2, whereas elongation 1+P2 will6–22.
occur
in contraction
fibers
located
below
the
1
+
y2,
1
+
P2
whereas
elongation
will
occur
in
fibers
located
below
the
distance
of yat
!the
c from
the neutral
on and the radius
of curvature
of the beam’s
longitudinal
axis
axis 1-y2.
This variation
in strain
over
the cross
section
is shown
in Fig.axis. Using Eq. 6–7, since Pmax = c>r,
6
1
y2.
axis
This
variation
in
strain
over
the
cross
section
is
shown
in
Fig.
then
6–22. for
Here
thespecific
maximum
occurs
thedivision,
outermost fiber, located a
.ature
In other
any
cross
section,
theatby
longitudinal
y>r
of thewords,
beam’s
longitudinal
axis
atstrain
the
P
6–22.
Here
thePaxis
maximum
fiber, located a
section
andoflinearly
the
radius
of the
curvature
ofaxis.
the
beam’s
longitudinal
at the strain occurs
=at-the
a outermost
b
distance
y
!
c
from
neutral
Using
Eq.
6–7,
al
strain
will
vary
with
y
from
the
neutral
axis.
A since
max = c>r,
ny specific point.
cross section,
the
longitudinal
Pmax
c>r
Pmax = c>r,
distance
of
y
!
c
from
the
neutral
axis.
Using
Eq.
6–7,
since
In
other
words,
for
any
specific
cross
section,
the
longitudinal
then
by
division,
action
will
occur
fibers located
the neutral axis
y>r
P A
nearly 1-P2
with
y6from
thein neutral
axis.
Aabovewith
thenthe
by division,
normal
strain
will
vary
linearly
y
from
neutral
axis.
= -a
b
, in
1+P2the
whereas
willneutral
occur in
fibers located below the
So the
that neutral
fibers elongation
located
above
axis
Pmaxaxis
c>r
1 - P2
contraction
will deformasyonlar
occur
in section
fibers located
above
Boyuna
birim
tarafsız
eksenden
gösterilen
y
mesafesine
bağlı olarak doğrusal
-y2. This variation
in strain
over
the
cross
is
shown
in
Fig.
y>r in fibers located below the
+P2 will occur
in whereas
fibers located
below1 +the
P will occur
1 + y2,
P2
elongation
y>r
Here the maximum
strain
occurs
at
the
outermost
fiber,
located
a
=
a
b
P
birsection
değişim
that
train over the
is gösterecektir.
shownininstrain
Fig.
Pmax overSothe
1 - y2.
axiscross
This variation
cross section is shown in Fig.
c>r
= P-=a - a by b P
(6–8)
Pmax = c>r,
nce of y ! c from the neutral axis. Using Eq.
6–7, since
c>r c max
ain occurs at
the Here
outermost
fiber, located
6–22.
the maximum
straina occurs at the outermost fiber, located a Pmax
by division,
y
So that
distance
y !since
c from
the=neutral
Pmax
c>r, axis. Using Eq. 6–7, since Pmax =P c>r,
utral axis. Using
Eq.of6–7,
= - a bPmax
(6–8)
So
that
c
then by division,
This normal strain depends only on the assumptions made with regards
y>r
y
P
to the deformation.
y is applied to the beam, therefore, it
(6–8)When a moment
= -a
b P = - a bPmax
cThis normal strain
Pstress
= - ain the
b P longitudinal
(6–8)
P
c>r
y>r
max
will
only
cause
a
normal
or x direction.
All the
depends
only
on
the
assumptions
made with regards
P
y>r
P
c max
y= - a
b
= -a
b
other
components
of
normal
and
shear
stress
will
be
zero.
It
is this
to
the
deformation.
When
a
moment
is
applied
to
the
beam,
therefore,
it
Pmax
c>r
c>r
max
at
uniaxial
state
of
stress
that
causes
the
material
to
have
the
longitudinal
will
only
cause
a
normal
stress
in
the
longitudinal
or
x
direction.
All
the
This
normaletkisi:
strain
depends only on the assumptions made with regards
Poisson
y
normal
strain component
depends
only
the
with
regards by
Pon
normal
strain
Eq.
6–8.
Furthermore,
components
of normal
anditshear
stress
willassumptions
be by
zero.
Itmade
is this
Sotothat
x, defined
the deformation. When a momentother
is applied
toThis
the
beam,
therefore,
y
to
the
deformation.
When
a
moment
is
applied
to
the
beam,
therefore,
it
there
must also
be the
associated
strain components
state(6–8)
ofPoisson’s
stress
thatratio,
causes
the material
to have
longitudinal
will only cause
stress in theuniaxial
longitudinal
or
x direction.
All the
P = a-normal
a bPmax
will
only
cause
a
normal
stress
in
the
longitudinal
or
x
direction.
All
the
c normal
Pcomponent
= nPx and
= - nPxby
, which
the plane of by
the cross-sectional
normal
strain
Eq. deform
6–8. Furthermore,
y
y be
other components of
and shear
stress
will
zero.
ItPxis,Pzdefined
this
y
y
P = - a Poisson’s
b Pmax other
(6–8)
components
of
normal
and
shear
stress
will
be
zero.
It is this Such
area,
although
here
we
have
neglected
these
deformations.
ratio,
there
must
also
be
associated
strain
components
state of stress that
causes the
c material to have the longitudinal
P = - a bPuniaxial
(6–8)
max
uniaxial
state
of
stress
that
causes
the
material
to
have
the
longitudinal
M Eq.
c normal strain component Px, defined
deformations
will,deform
causeofthe
cross-sectional dimensions to
Py = by
-nP
-nPx, which
and
Pz =Furthermore,
the plane
the cross-sectional
x regards
6–8.
byhowever,
is normal strain
depends only on the assumptions made
with
P
,
normal
strain
component
defined
by
Eq.
6–8. Furthermore,
by axis.
z
x
become
smaller
below
the
neutral
axis
and
larger
above
area,
although
here
we
have
neglected
these
deformations.
Such the neutral
Poisson’s
ratio, there
must toalso
be associated
strain
components
x
e deformation.
When
a moment
is applied
the
beam,
therefore,
it ratio,
This
normal
strain depends
only
on
the
assumptions
made
with
regards
Poisson’s
there
must
also
be
associated
strain
components
M
For
ifcause
the beam
has a square dimensions
cross section,
deformations
will,
however,
the cross-sectional
to it will actually
Py = -nP
=
-nP
, which
deform
the plane
of
theexample,
cross-sectional
x and
z When
nly
atonormal
stress
in Pthe
longitudinal
or
xisdirection.
All
the
onlycause
on the
assumptions
made
with
regards
the deformation.
ax moment
appliedsmaller
to
the
beam,
therefore,
it
P
=
nP
=
nP
,
and
P
which
deform
the
plane
of
the
cross-sectional
z
y
x
z
x
deform
as
shown
in
Fig.
6–23.
become
below
the
neutral
axis
and
larger
above
the
neutral
axis.
Fig.
6–23
area,
although
here
we
have
deformations.
x neglected
components
of
normal
and
shear
stress
zero. these
Itarea,
isorthis
moment
is applied
to
the beam,
therefore,
it
will
only
cause
a normal
stress
inwill
the be
longitudinal
x direction.here
AllSuch
the have neglected these deformations. Such
For
example,
if although
the beam
has a we
square cross section, it will actually
M
deformations
will,
however,
cause
the
cross-sectional
dimensions
ial
state
of
stress
that
causes
the
material
to
have
the
longitudinal
other components
of normal
and shear
will be zero.will,
It is however,
thisto
ss in the longitudinal
or x direction.
All the
M stressdeformations
cause the cross-sectional dimensions to
deform
as
shown
in
Fig.
6–23.
Fig.
6–23
becomestate
smaller
below the
neutral
axisFurthermore,
and larger
above
thelongitudinal
neutral axis.
x component
,ofdefined
al
strain
Eq.
by smaller
uniaxial
the
material
to
have
the
zstress
l and
shear
stress willPxbe
zero. that
Itbyis causes
this 6–8.
become
below
the
neutral
axis and larger above the neutral axis.
x
For6.4
example,
if thebe
a square
section,
it will actually
Thave
HE Falso
LEXURE
Fbeam
ORMULA
285
on’s ratio,
there
must
associated
strain
components
Phas
normal
strain
component
bycross
Eq.For
6–8.
Furthermore,
by has a square cross section, it will actually
auses
the material
to
the longitudinal
x, defined
example,
if the beam
deform
aswhich
shown
in Fig.
6–23.
-nPx and PPoisson’s
= -nP
deform
the
plane
theassociated
cross-sectional
ratio,
there
must
alsoofbe
strain
components
zby
Eq. x,6–8.
Furthermore,
by
deform
as shown
in Fig. 6–23.
Fig.
6–23
x, defined
nPx and
= components
- nPx, which
Pz neglected
the plane Such
of the cross-sectional
although
here
have
thesedeform
deformations.
y = -we
also be Passociated
strain
6.4 THE FLEXURE FORMULA
6.4.
Eğilme
Formülü
area,
although
here
wecross-sectional
have neglected
these deformations.
Such
mations
will,
cause
the
dimensions
to
hich
deform
thehowever,
plane
of the
cross-sectional
M
deformations
will,axis
however,
cause
thethe
cross-sectional
me
belowthese
the neutral
and larger
above
neutral axis. dimensions to
ave smaller
neglected
deformations.
Such
6.4 THE FLEXURE FORMULA
285
y
become
smaller
below
the
neutral
axis
and
larger
above the neutral axis.
xample,
if the
beam has a dimensions
square crosstosection, it will actually
tes
the
stress
x the
cause
cross-sectional
Flexure
Formula
For
example,
if the beamPmax
hasThe
a square
cross section,
it will actually
m as axis
shown
inlarger
Fig. 6–23.
moment
acting
utral
and
above
the
neutral
axis.
deform
as
shown
in
Fig.
6–23.
at the material
y
as a square cross section,
it In
willthis
actually
section,
we will develop an equation that relates the stress
The Flexure
Formula
r variation of
c
3.
Pdistribution
in a beam to the internal resultant bending moment acting
Pmax
y
ar variation in
x
on
the
beam’s
cross
section.
To
do
this
we
will
assume
that
the
material
y
this section, we will develop an equation that relates the stress
ariation, sInwill
behaves in a linear-elastic manner and therefore a linear variation of
c
m value, smax
,a
distribution
in a beam
to the internal resultant bending moment acting
Pmax
P
y
normal
strain,
Fig.
6–24a,
must
then
be
the
result
of
a
linear
variation
in
roportionality
on the beam’s cross section. To do this we will assume that the material
x
normal
stress,
Fig. 6–24b. Hence, like the normal strain variation, s will
Normal strain
variation
nd Eq. 6–8,
we
behaves
in a linear-elastic
manner
and
therefore
a
linear
variation
of
c
(profilezero
view)at the member’s neutral axis to a maximum value, smaxP, a
vary from
y
normal strain, Fig. 6–24a,
must
then
be
the
result
of
a
linear
variation
in
distance c farthest from the neutral axis. Because of the proportionality
x
(a) like the normal strain variation, s will
normal stress, Fig. 6–24b.
Hence,
Normal strain variation
of triangles,
Fig. 6–23b, or by using Hooke’s law, s = EP, and Eq. 6–8, we
(profile view)
y
vary from zero at thecan
member’s
neutral
axis
to
a
maximum
value,
s
,
a
max
write
6.4
6.4
smaxfrom the neutral axis. Because of the proportionality
distance c farthest
(a)
(6–9)
Normal strain variation
of triangles, Fig. 6–23b, or by using Hooke’s law, s = EP, and Eq. 6–8, we
y
(profile view)
can write 2 8 6
C H A P T E R 6c B E N D I N G
s
M
max
y
s
y
(a)
s = - a b smax
(6–9)
x
c
y
s
c
cross-sectional
M
s
y
smax s
t. For positive
y
x
= -variation
a bsmax
(6–9)
smax
Bendingsstress
give negative
y
c
6
This(profile
equation
view) describes the stress distribution over the cross-sectional
the negative x
c
area. The sign convention established here is significant.
For positive
s
dAM
ive or tensile
y
(b)
z
Bending stress variation
M, which acts in the + z direction, positive values of xy give negative
d at a specific
(profile view)
values
forstress
s, that
is, a compressive
since it acts in the negative x
Fig. 6–24
This equation describes
the
distribution
over thestress
cross-sectional
essive normal
direction. Similarly, negative y values will give
(b)
s or tensile
M positive
dF
area.
ated at +y
is The sign convention established here is significant. For positive
values for s. If a volume element of material is selected at a specific
Bending stress variationFig. 6–24
M, which acts in the +z direction, positive values of y give negative
6
point on the cross section, only these tensile
normal (profile
y c view)
x or compressive
oss section
by for s, that is, a compressive stress since it acts in the negative
values
x
stresses will act on it. For example, the element located at + y is
d by the stress
direction. Similarly,shown
negative
y values
(b)
in Fig.
6–24c. will give positive or tensile
o zero. Noting
values for s. If a volume
element
material
selected
at a axis
specific
We can
locateofthe
positionisof
the neutral
on the cross section by
Fig. 6–24
A in Fig. 6–24c,
stress
variation
point on the cross section,
tensile
compressive
normal
satisfyingonly
the these
condition
thatorthe
resultant force
produced byBending
the stress
stresses will act ondistribution
it. For example,
element located
at +y
is
over thethe
cross-sectional
area must
be equal
to zero. Noting(c)
shown in Fig. 6–24c.that the force dF = s dA acts on the arbitrary element dA in Fig. 6–24c,
Fig. 6–24 (cont.)
We can locate the we
position
of the neutral axis on the cross section by
require
satisfying the condition that the resultant force produced by the stress
PAGE
"78 Noting
distribution over the cross-sectional area must be equal
to zero.
Since smax>c is not equal to zero, then
that the force dF = s dA acts on the arbitrary element dA in Fig. 6–24c,
we require
FR = ©Fx;
0 =
dF =
s dA
y dA = 0
(6–10)
smax
s
z
ydA
dF
s
s
distributionBending
about
neutral
axis. T
stressthe
variation
M, which acts in the +z sdirection,
valuesby
of ythe
givestress
negative
y positive
max
6
shown
incan
Fig.s6–24c.
s
(profile
view)
We
determine
the
stress
in
the
beam
from
the
requirement
s
y
values for s, that is, a compressiveWe
stress
it the
acts
inmaxthe
negative
x
Fig.
6–24c
about
the
neutral
isbydM = y dF. that
Sin
cansince
locate
position
of the
neutral
axis on
the crossaxis
section
s
smax
y
dA
direction.
Similarly,
negative
y
values
will
give
positive
or
tensile
the
resultant
internal
moment
M
must
be
equal
to
the
moment
produced
(b)
z
s
satisfying the condition
the resultant
by thecross
stress section,
Eq.
6–9,
haveforce
for produced
the entire
dA that
y
values
for s. If a volume element
of material
is the
selected
at a we
specific
zdistribution
distribution
over
cross-sectional
area must
equal Fig.
to axis.
zero.
Noting
by
about
thebeneutral
The
moment of dF in
dA the stress
6–24
zthese tensile or compressive normal
point on the cross section,
onlythat
s
the
force
dF
=
s
dA
acts
on
the
arbitrary
element
dA
in
Fig.
6–24c,
s
M
F
Fig. 6–24c
aboutlocated
the neutral
axis is dM = y dF. Since dF = s dA, using
s ond it.
stresses will act
For example,
the element
at +y is
we require
s
M
dF
smaxin Fig.
Eq.s 6–9,
we haveiçin
forgerilmelerin
the entirebileşkesi
cross section,
shown
y 6–24c.
dF
Tarafsız
ekseninM konumunu
belirlemek
(Resultant)
y c
x
M
=
yolan
dF =
y1s dA2 =
R2
z = ©M
z;
We can locate the position of the neutral axis on the1M
cross
section
by
c
y
dA kuvvetler için denge denkleminix yazarsak:
LA
LA
satisfying
the condition that the resultant force producedc by the stress
z
x
M
x
y
distribution over
area must be equal to zero. Noting
y c the cross-sectional 1M
=dF = y dF
2x;z =element
©MzdA
; 0 in=M
Rvariation
Bending
stress
F
=
©F
s dA=
R
that the
Fig. 6–24c,
s force dF = s dA acts on the arbitrary
dF
L
A
L
A
L
A
or Bending stress variation
c
we yrequire
(c)
Bending stress variation
y
c
y
=
-(c)
a bsmax dA
Bending stress variation
Fig. 6–24 (cont.)
c
(c)
or
LA
(c)
Fig. 6–24 (cont.)
ending stress variation
Bending stress variation
is
not
equal
to
zero,
then
s
>c
Fig.Since
6–24
(cont.)
F =max
©Fx;
0 =
dF =
s dA
(c) R
(c)
LA
LA
Fig. 6–24 (cont.)
LA
y1s dA2 =
y
y ¢ smax ≤ dA
LA c
M =
smax
y2 dA
Thisc wood
LAspecimen f
-smax
smax
2
to its fibers being crush
y TdA
M
=
(6–11)
HE
F
LEXURE
FORMULA y dA 2 8 7 apart at its bottom.
c 6.4
LA
c LA
Since smax>c is not equal to zero, then
Since
is
not
equal
to
zero,
then
s
>c
max the moment
6–24 (cont.)
y 0 of
The integral
inertia
the cross-sectional area
Fig. represents
6–24 (cont.)
(6–10)
yof
dA
=
bs
=
a
dA
al to zero,
then
max
Kesit ağırlık merkezinin
konumu:
about the neutral axis. We will symbolize
its
value
as I. Hence,
Eq.
6–11
*Recall
that
the
centroid of the cross-sect
LALA
c
y dA
= 0 the location y for (6–10)
can be solved for smax and written as
L
A
= centroid
y dA>
dA.
y dA
= 0,
the
equation
If 1
then
This
wood
failed
in bending
due y = 0, and
(6–10)
y dAthat
= 0the
1
1specimen
-smax *Recall
y fory the
location
of
the cross-sectional
area
sthen
not equal
to zero,
then the first moment
6.4
THE
FLEXURE
2 8torn
7 is defined from
Iny dA
other
of
the L
member’s
cross-sectional
area
AdA
to its
fibers
beingFORMULA
crushed at its top and
=
y
(6–10)
= words,
0
reference
(neutral)
axis.
See
Appendix
A.
y
=
y
dA>
dA.
y
dA
=
0,
=
0,
the
equation
If
then
y
and
so
the
centroid
lies on the
cThis
at its bottom. area
1 be satisfied
1 the member’s
LA
LAabout the neutral axis must beInzero.
condition
can only
if 1 apartcross-sectional
other
words,
the first
moment
of
Mc
reference
(neutral)
axis.
See
Appendix
A.area
smoment
= the
(6–12)
the neutralIn
is=also
thethe
horizontal
centroidal
axis
for
the
cross
section.*
about
neutral
must
be
zero.
This
condition
can only be satisfied if
(6–10)
yaxis
dA
0words,
max
other
the
first
moment
member’s
cross-sectional
represents
the
cross-sectional
area
(6–10)
y dA = 0The integral
Iof inertiaofofthe
L
A
Consequently,
once
the
centroid
for
the
member’s
cross-sectional
area
is
st moment
of the
cross-sectional
thesymbolize
neutral
also the
horizontal
centroidal
axis for
about
the
neutral
axis
mustarea
beaxis
zero.
This
condition
canEq.
only
be satisfied
if the cross section.*
about
the member’s
neutral
axis.
We will
itsisvalue
as I.
Hence,
6–11
Yani,
tarafsız
eksene
göre
alan
momentinin
gerekiyor.
Bu şartarea
ancak
determined,
location
of
the
neutral
axis
is known
must be zero.
condition
only
be
satisfied
if atalet
Consequently,
once
the
centroid
member’s
cross-sectional
is tarafsız
thethe
neutral
axis
is written
also
theashorizontal
centroidal
axis forsıfır
the olması
cross
section.*
scan
can This
be solved
for
and
max
ds,
the
first
moment
of
the
member’s
cross-sectional
area
We
can
determine
the
stress
in
the
beam
from
the
requirement
that
he
horizontal
centroidal
axis
for the
cross
determined,
the
of the neutral
axis ağırlık
is known
eksenin
ağırlık
merkezinden
geçmesi
sağlanabilir.
Yani
hesaplayarak
Consequently,
once
thesection.*
centroid
forlocation
theile
member’s
cross-sectional
areamerkezini
is
t of
the Here
member’s
cross-sectional
area
tral
axis
must
be zero.
This condition
can
only
be
satisfied
the
resultant
internal
moment
M
must
be
equal
toifthe
moment
produced
We
can
determine
the
stress
in
the
beam
from
the
requirement
that
centroid
for
the
member’s
cross-sectional
area
is
determined,
the
location
of
the
neutral
axis
is
known
tarafsız
eksenin
konumunu
elde
edebiliriz.
ro.
can only
be satisfied
if the cross
s isThis
alsocondition
the
horizontal
centroidal
axis
for
Mcsection.*
by the
stress
distribution
about
the stress
neutral
axis.
The
moment
ofrequirement
dF
the
resultant
internal
moment
M the
must
be in
equal to the
moment produced
n
of
the
neutral
axis
is
known
s
=
(6–12)
We
can
determine
the
in
the
beam
from
that
max
=
the
s
maximum
normal
stress
in
the
member,
which
occurs
at
ntal
centroidal
section.*
once
the centroid
forthe
thecross
member’s
cross-sectional
area is
max axis for
6–24c
about
thethe
neutral
axis
is dM
using produced
=IM
ydistribution
dF. Since
dF
=tosthe
dA,
by the
stress
about
the
neutral
axis. The moment of dF in
he
stress Fig.
in of
the
beam
requirement
that
the from
resultant
internal
moment
must
beaway
equal
moment
aKesitte
point
on
theiscross-sectional
area
farthest
from
the
he
the
neutral
axis
known
forlocation
the member’s
cross-sectional
area
is
oluşacak
moment,
kesitteki
gerilmelerin
bileşkesi
kuvvetlerin
tarafsız eksene göre
Eq.
6–9,
we
have
for
the
entire
cross
section,
Fig.
6–24c
about
the
neutralaxis.
axisThe
is dM
= y dF.
oment M
must
be
equal
to
the moment
produced
by
the
stress
distribution
about
the
neutral
moment
of Since
dF indF = s dA, using
neutral
axis
ermine
the
stress
in
the
beam
from
the
requirement
that
eutral axis is known
momentleri
toplamı
olarak
elde
edilebilir.
Eq.
6–9,
we
for the
crossdF
section,
on
about
the neutral
axis.
moment
of dF
inhave
Fig.
6–24c
about
neutral
axis
is dM
Since
=from
yentire
dF.the
nternal
moment
must
beThe
equal
tothe
the
moment
produced
n
the beam
from
the
requirement
that
M
=M
the
resultant
internal
moment,
determined
method = s dA, using
Here
y
utral
axis is dM
Since
using cross
= Eq.
ythe
dF.
dF
=for
s the
dA,
distribution
about
neutral
axis.
The
moment
of
dF
in
6–9,
we
have
entire
section,
the
must be equal
the
moment
produced
M
= equations
1MRto
2zof
=sections
©M
; and
y dF = of equilibrium,
y1s dA2 = and
y ¢calculated
s ≤ dA
6.4 THE FLEXURE FORMULA
287
eutentire
cross section,
the neutral
axis
is dMz = y dF. Since
dF = L
s
dA, using L
y
c max
L
A in
Ain
A which
about
the
neutral
axis
of
the
cross
section
=
the
s
maximum
normal
stress
the
member,
at dA2 =
the
neutral
axis.
The
moment
of
dF
max
1MR2z = ©Mz ; M =
y dFoccurs
=
y1s
y ¢ smax ≤ dA
ave for the entire
cross section,
point
the cross-sectional
areaneutral
farthest
away
L
A afrom
LAy
LA c
is dM = y dF.
dF
= on
s dA,
c =Since
theaperpendicular
distance
axis
to
pointthe
1M
©Musing
; M =fromythe
dF =
y1s
dA2
=
y ¢ smax ≤ dA
R2the
z = moment
zy
or
The
integral
represents
of
inertia
of
the
cross-sectional
area
neutral
axis
c
oss
section,
away
axis. This is
A
LAwhere smax acts
LA
=
y dF = farthest
y1s dA2
= from
y ¢ the
s neutral
≤ dA
yL
c=ormaxmoment,
axis.
We
will
symbolize
itsssvalue
Eq.the
6–11
max
= neutral
Mzabout
;L
yA=
dFthe
= resultant
y1s
dA2
y¢
dA
≤ as
A Mthe
L
LA
2 I. Hence,
maxdetermined
M
internal
from
method
M
=
y
dA
(6–11)
IL
=
the
moment
of
inertia
of
the
cross-sectional
area
about
the
c
A for s
L
A
L
A
can be solved
and
written
asequations
smax
c ofL
A
ofmax
sections
and the
equilibrium,
and M
calculated
or
y
neutral
axis
=
y2 dA
(6–11)
F =
y1s dA2 = about
y ¢ the
smax
dA
≤
c
neutral axis of the cross section
s
A
L
max
2
c
LA
L
A
M =
y dA
(6–11)
smax
6
Mc(6–11)
2s
distance
fromstress
the neutral
to a point
c the
I=M
=
>cthe=yperpendicular
-dA
s>y, Eq.
Since
the
normal
at
intermediate
max
LAaxis
M = scmax
2 6–9,
s
=
(6–12)
max
=
y
dA
(6–11)
c
y
*Recall
that
the
location
for
the
centroid
of
the
cross-sectional
area
is
defined
from
s
farthest
away
from
the
neutral
axis.
This
is
where
acts
A
L
I
c L
distance y can be determined
from an equation similar to Eq.max
6–12.
A
the equation y = 1 y dA> 1 dA. If 1 y dA = 0, then y = 0, and so the centroid lies on the
*Recall
thatcross-sectional
the location y for area
the centroid
of
the cross-sectional area is defined from
have
I
=
the
moment
of
inertia
of
the
about
the
sWe
reference
(neutral) axis. See Appendix A.
max
the equation y = 1 y dA> 1 dA. If 1 y dA = 0, then y = 0, and so the centroid lies on the
M =
y2 dA neutral
(6–11)
axis
y for(neutral)
*Recall that the location
the centroid
of the
cross-sectional
area is defined from
c LA
reference
axis. See
Appendix
A.
Here
yefor
the
centroid
of centroid
thethe
cross-sectional
is defined
y for the
location
of the y
cross-sectional
area
is1defined
= area
y dA = from
0, then y = 0, and so the centroid lies on the
equation
Iffrom
1 y dA>
1 dA.
My
6
dA.Since
y dA
0,
=the
0,centroid
ys>y,
and
centroid
lies onstress
the
dA.
=If0,1then
0,then
y ==>c
and
lies
on the
reference
(neutral)
axis.
See
Appendix
A.
sso
=the
(6–13)
1 yIfdA>
1 y1dA
smax
=
-so
Eq.
6–9,
the
normal
at the intermediate
I member,
l) axis. See
Appendix
= the A.
smax
maximum normal stress in the
which
occurs at
Appendix
A.
(Eğilme
formülü)
distance y can be determined from an equation similar to Eq. 6–12.
a point onarea
theiscross-sectional
area farthest away from the
ntroid of the
cross-sectional
defined from
We
have
neutral
axiscentroid lies on the
A = 0, then y = 0,
and so the
Note
thatresultant
the negative
is necessary
sincefrom
it agrees
with the
.
M
= the
internalsign
moment,
determined
the method
My
established
x,
y,
z
axes.
By
the
right-hand
rule,
M
is
positive
along
of sections and the equations
and calculated the
s =ofs
-equilibrium,
(6–13)
+ z axis,
y
is
positive
upward,
and
therefore
must be negative
I
about the neutral axis of the cross section
(compressive) since it acts in the negative x direction, Fig. 6–24c.
c = the perpendicular distance from the neutral axis to a point
Either of the above two equations is often referred to as the flexure
farthest away from the neutral axis. This is where smax acts
formula.
is used
determinesign
the is
normal
stress since
in a straight
member,
NoteIt that
theto negative
necessary
it agrees
with the
I
=
moment
inertia
ofthe
theright-hand
cross-sectional
about
having
athe
cross
section
isBy
symmetrical
with respect
axis,the
and
the the
established
x, y, zofthat
axes.
rule, area
Mto
is an
positive
along
neutral
axispositive
moment
is applied
perpendicular
to this
have be
assumed
+ z axis,
y is
upward,
andaxis.
s Although
thereforewemust
negative
that(compressive)
the member issince
prismatic,
cannegative
in most cases
of engineering
design
it acts we
in the
x direction,
Fig. 6–24c.
6
s
>c flexure
= the
- s>y,
Since
Eq. two
6–9,
the
normalisthe
stress
at
the
intermediate
also use
the
formula
toequations
determine
normal
stress
members
Either
of
above
often
referred
toinas
the flexure
max
distance
y can
beisdetermined
from an
equation
similarin to
Eq. 6–12.
thatformula.
have
a slight
taper.to
For
example,
using
a mathematical
based
It
used
determine
the
normal
stress
aanalysis
straight
member,
Weon
have
the theory
of elasticity,
a member
having with
a rectangular
section
having
a cross
section that
is symmetrical
respect tocross
an axis,
and the
andmoment
a lengthis that
is tapered
15° willtohave
an actual
maximum
normal
applied
perpendicular
this axis.
Although
we have
assumed
stress
is about is5.4%
less than
that
calculated
using
the flexure
thatthat
the member
prismatic,
we
can
in most
cases of
engineering
design
My
s = to
- determine the normal stress(6–13)
formula.
also use the flexure formula
in members
I
that have a slight taper. For example, using a mathematical analysis based
on the theory of elasticity, a member having a rectangular cross section
and a length that is tapered 15° will have an actual maximum normal
Note that the negative sign is necessary since it agrees with the
PAGEthe
"79 flexure
stress that is about 5.4% less than that calculated using
established x, y, z axes. By the right-hand rule, M is positive along the
formula.
+ z axis, y is positive upward, and s therefore must be negative
(compressive) since it acts in the negative x direction, Fig. 6–24c.
=
6.11
6.4
6.4
THE FLEXURE FORMULA
T6.4
HE FLEXURE
FORMULA
THE FLEXURE
FORMULA
289 2 8 9
289
EXAMPLE
6.11 6.11 Dikdörtgen kesitli bir kiriş gösterilen gerilme dağılımına maruzdur.
EXAMPLE
Kesit momenti M değerini (a) Eğilme
ile
(b) TGerilme
6.4
HE FLEXURE FORMULA
6.4 formülü
THE FLEXURE
FORMULA
289
dağılımının
bileşkesi
ve
temel
prensipleri
kullanarak
hesaplayınız.
beam
has a rectangular
cross section
is subjected
the stress
A beamAhas
a rectangular
cross section
and is and
subjected
to the to
stress
289
6.4 THE M
289
THE FLEXURE
FORMULA
289
distribution
in6.11
Fig. 6–25a.
Determine
the internal
moment
at6.4 FORMULA
distribution
shown shown
in Fig. 6–25a.
Determine
the internal
moment
M at FLEXURE
6 in. 6 in.
EXAMPLE
EXAMPLE
6.11
6.4
T
HE
F
LEXURE
F
ORMULA
2
8
9
ectangular
cross
section
and
is
subjected
to
the
stress
the section
the distribution
stress distribution
(a) the
using
the flexure
the section
causedcaused
by the by
stress
(a) using
flexure
wn in Fig. 6–25a.
Determine
internal
Mof
at the of
6 in.distribution
formula,
(b)
by finding
the resultant
the distribution
stress
formula,
(b) bythe
finding
themoment
resultant
stress
using using
EXAMPLE
6.11
ksi 2 ksi
EXAMPLE
6.11
used
by the stress
distribution
(a)
using
the flexurecross section and is subjected to the2 stress
basic
principles.
A beam
has
a rectangular
basic
principles.
A
beam
has
a
rectangular
cross
section
and
is
subjected
to
the
stress
N
finding EXAMPLE
the resultant of 6.11
the
stress distribution
using
distribution
shown in Fig.
6–25a. Determine the internal moment M
6 in.
N at
distribution shown in Fig. 6–25a. Determine the internal
moment M at
6 in.
2 ksi
the
section
caused
by
the
stress
distribution
(a)
using
the
flexure
A beam
has
a
rectangular
cross
section
and
is
subjected
to
the
stress
6 in. 6 in.
A SOLUTION
beam
has by
a rectangular
section (a)
and using
is subjected
to the stress
theSOLUTION
section
caused
the stress cross
distribution
the flexure
N of the stress distribution using
formula,
(b)
by finding
the
resultant
distribution
shown
in
Fig.
6–25a.
Determine
the
internal
moment
M
at
6
in.
shown
in
Fig.section
6–25a.
the
internal
moment
at
in.
A
beamdistribution
has
abyrectangular
cross
and
to the
stress
formula,
(b)
finding
resultant
ofDetermine
theisisstress
distribution
using
26 ksi
Part
(a).
Thethe
flexure
formula
From
Fig.6 M
6–25a,
ssubjected
max = Mc>I.
basic
principles.
in. 2 ksi
Partcaused
(a).
The
flexure
formula
is smax
Fig.M
6–25a,
=using
Mc>I.
A
the section
by
the
stress
distribution
(a)
theFrom
flexure
the
section
caused
by
the
stress
distribution
(a)
using
the
flexure
distribution
shown
in
Fig.
6–25a.
Determine
the
internal
moment
at
A
6 in.
basic principles.
and
The
neutral
axis
is
defined
as
line
NA,
c
=
6
in.
s
=
2
ksi.
max
N
6 in.
and
The
neutral
axis
is
defined
as
line
NA,
c
=
6
in.
s
=
2
ksi.
max
formula,
(b)
by
finding
the
resultant
of
the
stress
distribution
using
6 in.
formula,
bystress
finding
the
resultant
the
stress
distribution
using
the section
caused
by
the
stress
distribution
(a)
using
the
flexure
the
is zero
along
this of
line.
Since
the
cross
section
has Na
flexure formula
isbecause
From
Fig.
6–25a,
sthe
=(b)
Mc>I.
2 ksi
maxstress
2 ksi
6 in.
because
is zero
along
this
line.
Since
the
cross section
has
a A
basic principles.
basic
principles.
SOLUTION
formula,
(b)
by
finding
the
resultant
of
the
stress
distribution
using
6
in.
rectangular
shape,
the
moment
of
inertia
for
the
area
about
NA
is
The
neutral
axis
is
defined
as
line
NA,
smax = 2 ksi.
2
ksi
2
ksi
6
in.
SOLUTION
rectangular shape, the moment of inertia for the area about NA isN
2 ksi
Part the
(a).
Thesection
flexurehas
formula
is smax
From front
Fig. 6–25a,N
= Mc>I.
(a)
from
the
formula
for
given
on the inside
ess is zero basic
alongprinciples.
thisdetermined
line. Since
cross
aa rectangle
A
determined
the
formula
for
a
rectangle
given
on
the
inside
front
Part
(a). cover;
Thefrom
flexure
formula
is
From
Fig.
6–25a,
s
=
Mc>I.
6 in. (a) A 6 in.
N
max
and
The
neutral
axis
is
defined
as
line
NA,
c
=
6
in.
s
=
2
ksi.
i.e.,
pe, the
moment of
inertia for the area about
NA is
max
SOLUTION
6 in.
SOLUTION
2
ksi
and sbecause
The neutral
axis isthis
defined
as line
c =cover;
6 in. i.e.,
= 2 ksi.
max given
the
stress
is zero
along
line. Since
theNA,
cross
6 in.
(a) section has a
m the
formula
for
a rectangle
on is
thesinside
SOLUTION
Part
(a).
The
formula
From
Fig. section
6–25a,
= front
Mc>I.
Part
(a). is
The
flexure
formula
is smax
Fromhas
Fig.
=cross
Mc>I.
max line.
because
theflexure
stress
zero
along
this
Since
the
a 6–25a,
A
rectangular shape,
the
moment
of
inertia
for
the
area
about
NA
is
1
1
2 ksi A
3
4
neutral
is16
as
NA,
c = 6 Part
in. and
= flexure
2and
ksi.
6 in.
The
axis
isline
defined
line
NA,
cs=max
6shape,
in.
2 ksi.
(a).
The
formula
is3 axis
From
6–25a,
s
=defined
Mc>I.
I ==
bh
=formula
in.2112
in.2
=Fig.
864
inas
1max
1the
6 in.
maxneutral
rectangular
thesThe
moment
inertia
for
the
area
about
NA
is
6 in. (a)
3 12 of
3
4
A
determined
from
for
a
rectangle
given
on
the
inside
front
2
ksi
12
6 in.
I stress
= 2 ksi.
bh
=line.
16
in.2112
in.2
= 864
because
stress
is szero
along
this
Since
the
cross
section
has
a section
because
the
is
zero
along
this
line.
thein
cross
and
The
neutral
axis
is Since
defined
as
line
NA, has a
cdetermined
=the
6 in.
=
max
(a) 6 in.
from
the
formula
for
a
rectangle
given
on
the
inside
front
12
12
cover;
i.e.,
1
1 shape,
thein.2
moment
ofininertia
forof
the
area
about
NA
3zero along
4moment
rectangular
inertia
the
the
stress
isshape,
this line.
Since
the for
cross
section
has a NA is
Çözüm:
2 ksi
Therefore,
6area
in. is about
2 ksi
I = rectangular
bh3because
=
16
in.2112
= 864the
cover;
i.e.,
12 rectangular
12from
(a)
determined
the
formula
for
a
rectangle
given
on
the
inside
front
Therefore,
(a)
determined
frommoment
the formula
for
a
rectangle
given
on
the
inside
front
shape, the
of inertia
for
the
area
about
NA
is
2
ksi
1
Mc
3 2 1 M16 in.2
3
4
6
N
6 in.
cover;determined
i.e.,
I
=
bh
=
16
in.2112
in.2
=
864
in
cover;
i.e.,
(a)
from
the
formula
for
a
rectangle
given
on
the
inside
front
s
=
;
2
kip>in
=
M1612
max
1 3
1
(a) Mc
4 4
12
3in.2
6
"F
2
N
I
4
in.
864
in
6
in.
I
=
bh
=
16
in.2112
in.2
=
864
in
6
in.
cover;
smaxi.e.,
=
;
2 kip>in =
12 1
12
"F
IM16
4 in. 6 in.
in.2
864# in4 4 N
1Therefore,
6
# ft in4
1
1 3kip
2
3
Ans.6 in.
M3 =
288
in. in
=in.2
243 kip
2 kip>in I= =
bh4 = I = 16 in.2112
=in.2112
864
6 in.
bh
= in.2
16
=
864
4 in.
"F
A
#
#
1 kip12in. = 24
Therefore, 864
12 in 1 12
4 Ans.
in. 6 in.
M
288
3 12
3 kip ftM16
4 in.2
Mc
6 in.
Is = =bh
==
16
in.2112
in.2
=
2 864 in
N
4
in.
6
in.
A
; 12
2for
kip>in =
force
two triangular stress
12The
# Part
# ftIresultant
M16
in.2each of the
"F
Ans.
M = 288 kipTherefore,
in. = (b).
24max
kip
Therefore,
Mc
4 in. 66 in.
864 in4
2
N
6 in.
in. volume
F
distributions
in 2Fig.
6–25b
graphically
equivalent
to 4stress
the
smax
= (b).
; The resultant
kip>in
= iseach
A
Part
force
for
of
the
two
triangular
4
"F
Therefore,
Icontained within each M16
4 in. 6 in.
864
in M16
#Thus,
# ftvolume
in.2
Mc
=2 288
kip in.2
in. = each
24
stressM
distribution.
volume
is6Nin.
6
Mc
(b)
FAns.N
2
distributions
in Fig.
6–25b
equivalent
to kip
the
6
resultant
each
two
triangular
stress
4 in.
smax =force
;for
= is 2graphically
A
sMc
= of the
;2 kip>in
kip>in
=
4
max
M16
in.2
"F
#
#
4
I
"F
4
in.
864
in
Ans.
M
=
288
kip
in.
=
24
kip
ft
6
contained
within
each
stress
distribution.
Thus,
each
volume
is
Iequivalent
4 in. 6 in.
6 in.(b)
2 volume
864 in
N
F
Fig. 6–25b
is graphically
to the
6 in.
smax
=
; Part
kip>in
=
4 in.Fig. 6–25
(b). 2 The
force
each of the two triangular
A
4 for
1resultant
"F stress
I
2
4
in.
864
in
#
#
6
in.
n each stress distribution.
Thus,
each
volume
is
(b)
#
#
Ans.
M
= 288 kip
in.=Fig.
=288
24
kipkip>in
ft
F =
16
in.212
216
in.2 ft
= equivalent
36 kip
Ans.
M
kip
in.
=
24
kip
F
distributions
in
6–25b
is
graphically
to
the
volume
6
in.
Fig. 6–25A 4 in.
4 in.
Part (b). The resultant1 force2for# each 2of the #two triangular stress
A
Ans.volume is
288
kip
in.stress
= 24
kip
FM
= =16
in.212
kip>in
216
in.2ft= 36tokip
contained
within
each
distribution.
Thus,
each
(b)
Fig.
6–25
F
distributions
in
Fig.
6–25b
is
graphically
equivalent
the
volume
6 4in.
in.
6
in.
A
1 (b). The
2
Part
resultant
force
for
each
of
the
two
triangular
stress
2(b). The resultant force for each of the two triangular stress
Part
F = 16 contained
in.212 kip>in
216forces,
in.2
36 kip
These
which
form a couple,
in volume
the same
as the
each=is
stress
distribution.
Thus,act
each
is direction
(b)
6 in. Fig. 6–25
F
distributions
in within
Fig.
6–25b
graphically
toequivalent
the
volume
2 Part (b).
The
force
for is
each
of Fig.
the two
triangular
F
distributions
in Fig.
6–25b
graphically
tostress
the volume
1equivalent
2 Furthermore, they act
stressesresultant
within
each
distribution,
6–25b.
F
=
16
in.212
kip>in
216
in.2
=
36
kip
These
forces,
which
form
a
couple,
act
in
the
same
direction
as
the
contained
within
each
stress
distribution.
Thus,
each
volume
is
(b)
F
distributions
in Fig.
is stress
graphically
equivalent
to
the
volume
contained
within
each
distribution.
Thus,
volume
(b)
2 volume,
Fig. 6–25
through
the6–25b
centroid
of each
i.e., 23each
16 in.2
= 4 in.is from the
1same
2 the
within
each
distribution,
Fig.
6–25b.
Furthermore,
they act
hich formcontained
a stresses
couple, within
act
inFeach
the
direction
as
stress
distribution.
Thus,
each
volume
is
(b)
=
16
in.212
kip>in
216
in.2
=
36
kip
neutral axis 2of the beam. Hence the distance
between them is 8 in. as Fig. 6–25
2
Fig. 6–25
1These
through
centroid
of each
volume,
i.e.,
from direction
the
=in4the
in. same
each distribution,
Fig.the
6–25b.
Furthermore,
they
act
1which
2the
3 16 in.2
forces,
form
a
couple,
act
as the
2therefore
shown.
The
moment
of
couple
is
F
=
16
in.212
kip>in
216
in.2
=
36
kip
2
F
=
16
in.212
kip>in
216
in.2
=
36
kip
Fig.
6–25
(b)
neutral
axisi.e.,
of
the
the
distance
between
themFurthermore,
is 8 in. as
ntroid of each
volume,
from
the
161beam.
in.2
= Hence
2stresses
within
each
distribution,
Fig.
6–25b.
they act
24 in.
2
F 3=between
16
in.212
kip>in
216
in.2
= 36
kip
2
These
forces,
which
form
a couple,
act
in
the volume,
same
direction
as
the= 4 in. from the
shown.
moment
ofthe
the
couple
is
therefore
he beam. Hence
theThe
distance
them
is
8
in.
as
#
#
2M
through
centroid
of
each
i.e.,
16
in.2
= 36 kip 18 in.2 = 288 kip in. = 24
Ans.
3 kip ft
within
each awhich
distribution,
Fig.
6–25b.
Furthermore,
they
act them
mentThese
of thestresses
coupleThese
is therefore
forces,
which
form
couple,
act abeam.
in
the
same
direction
as the
neutral
axis of
the
Hence
thethe
distance
between
forces,
form
couple,
act
in
same
direction
as theis 8 in. as
2
# in.
through
the
centroid
of
each
volume,
i.e.,
in.2
=24
4they
in. # from
stresses
within
each
distribution,
Fig.
6–25b.
Furthermore,
actas the
M
=
36
kip18
in.2
=of
288
kip
= therefore
kip
ft
3 166–25b.
shown.
The
the
couple
is
These
forces,
which
form
amoment
couple,
act
in
the
same
direction
the
stresses
within
each
distribution,
Fig.
Furthermore,
they
act
NOTE:
This
result
can
also
be
obtained
by
choosing
a Ans.
horizontal
2
#
#
2
neutral
axis
of
the
beam.
Hence
the
distance
between
them
is
8
in.
as
= 36
kip18
in.2
288
kip
Ans.
in.
=
24
kip
ft
through
the =centroid
of
each
volume,
i.e.,
from
the
16
in.2
=
4
in.
stresses
within
each
distribution,
Fig.
6–25b.
Furthermore,
they
act
through
the
centroid
of
each
volume,
i.e.,
from
the
16
in.2
=
4
in.
3
strip of area dA ! (6 in.) dy and using
3 integration by applying
2
shown.
moment
the
couple
is
# ftas Ans.
neutral
axis
ofThe
beam.
Hence
the
is
8 #in.
as
NOTE:
This
result
can
also
be
obtained
by=them
choosing
afrom
horizontal
Mdistance
=Hence
36therefore
kipbetween
18 in.2
288
kip
in.
= 24
through
the
centroid
each
volume,
i.e.,
the
=between
4 in.
neutral
axis
ofofof
the
beam.
the
distance
them
iskip
8 in.
Eq.
6–11.
3 16 in.2
esultshown.
can neutral
also
be
obtained
by
choosing
a dy
horizontal
The
moment
of
the
couple
is therefore
strip
of area
! (6
in.)
and
using
integration
axis
of The
thedA
beam.
Hence
the
distance
them by
is 8 applying
in. as
shown.
moment
of
the
couple
is therefore
# in.between
M NOTE:
= 36
in.2result
= isapplying
288
kipalso
Ans.
= obtained
24 kip # ft by choosing
A ! (6 in.)
dy 6–11.
and moment
using
integration
by
can
be
a horizontal
Eq.
shown.
The
ofkip18
theThis
couple
therefore
#
#
M = 36 strip
kip 18M
in.2
= 288
kip
in.=in.)
=288
24dy
kip
ft using
# in.
# ft
of
dA18
!in.2
(6
and
integration
by applying
=area
36
kip
kip
Ans.
= 24Ans.
kip
NOTE: ThisMresult
can
also
by choosing
Eq.
6–11.
= 36
kip 18
in.2be=obtained
288 kip # in.
Ans.
= 24 kip # afthorizontal
NOTE:
result
can
be
choosing
a horizontal
stripThis
ofNOTE:
area dA
!also
(6
in.)obtained
dy also
and by
using
integration
by applying
This
result
can
be
obtained
by
choosing
a horizontal
strip of
dA
(6
in.)
dyalso
by
applying
Eq.area
6–11.
NOTE:
This!
can
obtained
choosing
a horizontal
strip
ofresult
area
dA
!and
(6beusing
in.)
dyintegration
andbyusing
integration
by applying
Eq. 6–11.
strip ofEq.
area
dA ! (6 in.) dy and using integration by applying
6–11.
Eq. 6–11.
PAGE "80
6
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