2E NBDEI NNGD I N G
290
CHAPTER 6
M (kN"m)
BENDING
The simply supported beam in Fig. 6–26a has the cross-sectional area
22.5
shown in Fig. 6–26b. Determine the absolute maximum bending stress
The simply supported
beam
in
Fig.
6–26a
has
the
cross-sectional
area
in the beam and draw the stress distribution over the cross section at
EXAMPLE
6.12
2
shown
2
9 0 in Fig.C6–26b.
H A P T E RDetermine
6 B E N D I Nthe
G absolute maximum bending stress
this location.
in the beam and draw the stress distribution over the cross section at
The simply supported beam in Fig. 6–26a has the cross-sectional area
6m
this
location.
TheThe
simply
supported
beam
inmFig.
6–26a
has
the
cross-sectional
areaarea
5beam
kN/
Thehas
simply
supported
in Fig. 6–26a has the cross-sectional area
simply
supported
in Fig.
6–26a
the
cross-sectional
shown in Fig.
6–26b.
Determine
the
absolute
maximum
bending
stress beam
(a)
EXAMPLE
6.12
Mthe
(kN"m)
shown
in Fig.
6–26b.
Determine
the
absolute
maximum
bending
stress
shown
in Fig.
6–26b.
the maruz
absolute
maximum
bending
stress
shown
in Fig.
6–26b.
Determine
absolute
maximum
bending
stress
! and
Şekilde
enkesiti
üniform
yayılı
yüke
basit
mesnetli
kirişte
in the beam
stress
distribution
over
the verilen
cross
section
atDetermine
5draw
kN/
mthe
in the
beam
and
draw
the
stress
distribution
over
the
cross
section
at
in
the
beam
and
draw
the
stress
distribution
over
the
cross
section
at
in
the
beam
and
draw
the
stress
distribution
over
the
cross
section
at
SOLUTION
this location.
M (kN"m)
oluşacak
mutlak
değerce maksimum
eğilme gerilmesini belirleyerek enkesit üzerindeki
22.5
thisthis
location.
this
location.
location.
The simply supported beam in Fig. 6–26a has the cross-sectional
area Momen
Maximum Internal
dağılımını gösteriniz.
290
C H A P T E22.5
R 6
BENDING
20
mm Determine the absolute maximum
5 kN/m
shown
in
Fig.
6–26b.
bending
stress
the
beam,
M
=
22.5 kN # m, o
5 kN/
m
kN/ m
5 kN/m
M 5(kN"m)
in the beamMand
draw the stress
distribution
x (m) over the cross section at
B
M (kN"m)
M (kN"m)
(kN"m)
3
6150 mm
Section Property. By reaso
C
290
C
H
A
P
T
E
R
6
B
E
N
D
I
N
G
this
location.
6m
6
x (m)
N
A through the centroid C at the
EXAMPLE
6.12
22.5
22.5 22.5
3
620 mm
(a)
22.5(c)
area is subdivided into the t
5 kN/ m
6m
150 mm
M
(kN"m)
inertia of each part is calcu
SOLUTION
(a)
EXAMPLE 6.12
(c)
20 mm
The simply supported
beam
in Fig. 6–26a has the cross-sectional
area theorem. (See E
x (m)
x
(m)
parallel-axis
xinternal
(m)
x (m)
Maximum Internal
The
maximum
moment
in
D
3
6
3 Moment.
6
SOLUTION
shown
in
Fig.
6–26b.
Determine
the
absolute
maximum
bending
stress
3#
6
22.5
3
work in6 meters, we have
m 6m
the beam,
occurs
at
the
center.
M
=
22.5
kN
m,
250
mm
6m
in the
beam
andinternal
draw the
stress
section at area
The
simply
supported
beamdistribution
in Fig. 6–26aover
has the
the cross
cross-sectional
Maximum Internal Moment.
The
maximum
moment
(c)
a) (a)
B
(c) (c)
I = ©1I
+ Ad22
(a)
150
mm
(c)
#
this
location.
shown
in Fig.
6–26b. Determine
absolute
maximum bending
stress
the
center.
22.5 kN Property.
m, occurs atBy
reasons
of symmetry,
the neutral
passes
(b)the axis
C the beam, M = Section
x (m)
SOLUTION
in
and draw
stressFig.
distribution
over the cross section
SOLUTION
N 150 mm
SOLUTION
Çözüm: A through the centroid CSOLUTION
at the
the beam
midheight
of thethe
beam,
6–26b.
3 The
6 1 at
20 mm
6m
5 kN/
m the neutral axis passes
Section Property.
Bysubdivided
reasons
of into
symmetry,
= 2c 10.25 m210.020
this
area
is
thelocation.
three
parts
shown,
moment of
Maximum
Moment.
TheThe
maximum
internal
moment
inand
Maximum
Internal
Moment.
The maximum
internal
moment
inmoment
Moment.
maximum
internal
in the
12 in
150
mm Internal
M
(kN"m)
Maximum
Internal
Moment.
The
maximum
internal moment
A Maximum
12.7 MPa
through
theInternal
centroid
C
at
the
midheight
of
the
beam,
Fig.
6–26b.
The
(a)
(c)
#kN
#
#
inertia
each
is
calculated
about the neutral
axis using the
the the
beam,
occurs
atpart
the
center.
M =M
22.5
kN
m,of
the beam,
occurs
at
the
center.
M
=
22.5
kN
m,
beam,
occurs
at
the
center.
=
22.5
m,
#
20
mm
the shown,
beam,
at the center.
M
kN m, occurs
1
area is subdivided
into the theorem.
three parts
and
the
moment
of
5 kN/
m = 22.5
150 mm
parallel-axis
(See
Eq.
A–5
of Appendix
A.) Choosing to
+ c 10.020 m210.300
SOLUTION
D
22.5 MB
0 mm
(Maksimum
moment)
11.2
MPa
150 mm
B
(kN"m)
inertia
of
each
part
is
calculated
about
the
neutral
axis
using
the
12
150
mmof we
Section
Property.
By reasons
symmetry,
the the
neutral
axisaxis
passes
Section
Property.
By
reasons
of have
symmetry,
neutral
passes
Section
Property.
By reasons
symmetry,
the neutral
axis
passes
work
inof
meters,
Property.
ByMoment.
reasons
the neutral
axis
passes in
C
250 mm parallel-axis theorem.
Maximum
Internal
The maximum
internal
moment
(See
Eq.
A–5ofSection
ofthe
Appendix
A.)
Choosing
to of symmetry,
A 6A
through
the
centroid
C
at
the
midheight
beam,
Fig.
6–26b.
The
AD through
through
the
centroid
C
at
the
midheight
of
the
beam,
Fig.
6–26b.
The
the
centroid
C
at
the
midheight
of
the
beam,
Fig.
6–26b.
The
N
A
2 through the centroid C at the
=
301.3110
midheight
of
the
beam,
Fig.
6–26b.
The -62 m4
#
20
mm
the
beam,
occurs
at
the
center.
M
=
22.5
kN
m,
I = ©1I + Ad 2
work
meters,
we into
have
22.5
20
mm the
M
!
22.5
kN"m
is subdivided
into
the the
three
parts
shown,
and
the
moment
of
area
isinsubdivided
three
parts
shown,
and
the
moment
of
area(b)isarea
subdivided
into
three
parts
shown,
and
the
moment
of
0 mm
x
(m)
150 mm
D
area is subdivided into the three parts
shown, and the moment
150150
mmmm
B
22.5
Mc of
3
6
of
is 2calculated
the
neutral
axispart
using
the
of©1I
each
is
calculated
the
neutral
axis
the
1 about
I part
=each
+ part
Ad
inertiainertia
of inertia
each
ispart
calculated
theabout
neutral
axis
using
the using
Section
Property.
By
reasons
of
symmetry,
the
neutral
axis
passes
3 each
2the neutral
s
=
;
smax =
inertia
of
is
calculated
about
axis
using
the
C= 2about
62m
max
c
10.25
m210.020
m2
+
10.25
m210.020
m210.160
m2
d
20
mm
I The
theorem.
(See
Eq.
A–5
of through
Appendix
A.)
Choosing
to midheight
parallel-axis
(SeeA–5
Eq.of
A–5
ofAAppendix
A.)
Choosing
6 parallel-axis
N theorem.
theorem.
(See
Eq.
Appendix
A.) Choosing
to (See
thetheorem.
centroid
Ctoat Eq.
the
of the beam, Fig. 6–26b.
D D parallel-axis
(m)
parallel-axis
A–5
(a) D 12
12.7 MPa
(c) of Appendix A.) Choosingx to
201
mmhave
3
2
work
in
meters,
we
have
work
in
meters,
we
3
6
=
2
c
10.25
m210.020
m2
+
10.25
m210.020
m210.160
m2
d
area
is
subdivided
into
the
three
parts
shown,
and
the
moment
of
work in meters, we have
work
in
meters,
we
have
A
three-dimensional
view
150
mm
1
6
m
mm + c
12 250
10.020SOLUTION
m210.300
m23ofd each part
.7 MPa
2
2
inertia
is
calculated
about
the
neutral
axis
using
the
12.7
MPa
Fig.
6–26d.
Notice
how
the
I
=
©1I
+
Ad
2
2
11.2
MPa
I
=
©1I
+
Ad
2
2
12
mmAd 2
(a)
(c)
I = ©1I20 +
I = ©1I + theorem.
Ad 2
1
parallel-axis
(See
Eq.
A–5
of
Appendix
A.)
Choosing
to
3 Maximum
Internal
Moment.
The
maximum
internal
moment
in
section
develops
a
force
th
(b)
D
+1 c 1 10.020 m210.300 m2 -6
d
4
11.2 MPa 20 mm
= 301.3110
2 mbeam,
#
12 10.25
work
inm210.160
meters,
we
have
2
SOLUTION
3m23 +the
2m2
occurs
at
the
center.
M
=
22.5
kN
m,
1= 2 =
1
neutral
axis
that
has
the
same
2
c
m210.020
10.25
m210.020
m210.160
d
250
mm
c
10.25
m210.020
m2
+
10.25
m210.020
m2
d
2
! 22.5 m210.020
kN"m
= D
2c M 10.25
m23 4+ 10.25 m210.020 m210.160
d m210.020
= 2c 3 m2
10.25
m23 + 10.25 m210.020 m210.160 m22 d
-6
B12
7 MPa
22.5(10
) Internal
N # +m10.170
yB = moment
150 mm, and
Maximum
The maximum internal
in so
Pa
(d)
Mc
12 =12301.3110
2 mmm
I = 12
©1I
Ad2Moment.
2 m2
150
12.7
MPa
Property.
By reasons
of
symmetry,
the neutral axis passes
Ans.
;
sSection
=
12.7
MPa
Cs(b)
#
max =
max =
20
mm
1
M
!
22.5
kN"m
1
-6
4
the
beam,
occurs
at
the
center.
M
=
22.5
kN
m,
3
D
3m2 d3
I22.5(10
301.3110
m the midheight
1 1 C2 at
6
1 N+ c + c10.020
10.020
m210.300
m210.300
m2
d A )through
N # m10.170
22.
the
centroid
the beam, Fig. 6–26b. My
TheB 2
3
Mc
Fig.3 6–26
+=
c m2
10.020
m210.300
m2
d 3 +of10.25
11.2 11.2
MPaMPa + c s10.020
m210.300
d=150
12 ; B sm2
2012
mm
Ans.
=
=
12.7
MPa
2
c
10.25
m210.020
m2
m210.020
m2; d sB = sm210.160
- of
mm
11.2
MPa
max
max
B
a
B = axis
12
-6
4
12
area
is
subdivided
into
the
three
parts
shown,
and
the
moment
A
three-dimensional
view
of
the
stress
distribution
is
shown
in
Section
Property.
By
reasons
of
symmetry,
the
neutral
passes
I
12
150
mm
C12.7
I
4 MPa 301.3110 2 m
-6 -642 m
=N-6301.3110
6
2.7 MPa
= 301.3110
m 6–26d. Noticeinertia
of =stress
each
part
is2 calculated
about
the cross
neutral
axis Fig.
using
the The
A
-6
4CBatand
howthrough
the
at
points
D
on the
the
centroid
the
midheight
of
the
beam,
6–26b.
4 2Fig.
1
301.3110
m
20
mm
= 301.3110
220mmm
! 22.5
kN"m
A three-dimensional
view3 of 3 the
distribution
is m210.300
shown A–5
inm23ofd Appendix A.) Choosing to
M
22.5
kN"m
D !M
ccontributes
10.020
theorem.
(See
# stress
section
am10.170
that
moment
aboutshown,
the and the moment of
aream2
is +subdivided
intoaEq.
the
three
parts
Mdevelops
!11.2
22.5
kN"m
22.5(10
N force
m10.170
m2
kN"m
)MPa
N )# parallel-axis
mm
D22.5(10
3
Mc Mc Notice
B6–26d.
12
# m10.170
3 150
Fig.
how
the
stress
at
points
B
and
D
on
the
cross
#
22.5(10
) Nat
Ans.
s
=
;
s
=
=
12.7
MPa
Mc
22.5(10
)
N
m10.170
m2
work
in
meters,
we
have
Ans.
sMc
=
;
s
=
=
12.7
MPa
neutral
axis
that
has
the
same
direction
as
M.
Specifically,
point
B, m2
max
max
inertia
of
each
part
is
calculated
about
the
neutral
axis
using
the
max
max
-64
4
-6 s
250=mm
I
Ans.
=
;
s
=
= 12.7
MPa
20 ;mm I sdevelops
301.3110
2
m
a
force
that
contributes
a
moment
about
the
Ans.
smax = section
=
12.7
MPa
max
max
-6
4
301.3110
2
m
-6
4
max
-6
4
=
301.3110
2
m
and
so
y
=
150
mm,
I
(d)
parallel-axis
theorem.
(See
Eq.
A–5
of
Appendix
A.)
Choosing
to
301.3110 2 m
Ineutral axis that Bhas301.3110
2direction
mkN"m I =
D
+ Ad22
the
same
as ©1I
M.
Specifically,
at
point in
B,
M ! 22.5
A three-dimensional
of
the
stress
distribution
is
shown
D view
3
(b)
A three-dimensional
view
of
the
stress
distribution
is
shown
in
work
in
meters,
we
have
#
3
m10.170
m2
Mc
250 mm
and
so
yB =6–26d.
150 mm,
A 22.5(10
three-dimensional
view
of the) N
stress
distribution
is shown in
) N;D# m10.150
m222.5(10
Fig.
6–26 Fig.
My
A three-dimensional
view
of the
stress
distribution
shown
insthe
B stress
Notice
at =points
B
on
1=isand
sBmaxand
=cross
= 12.7 2MPa Ans.
Fig. 6–26d.
Notice
how
stress
points
D
on
the
cross
max
3
2
sB how
=the
- the
; at s
=
11.2
MPa
-6
4
12.7 MPa
B Fig.
=
2c
10.25
m210.020
m2
+
10.25
m210.020
m210.160
m2the
d cross
6–26d.
Notice
how
the
stress
at
points
B
and
D
on
I
I
=
©1I
+
Ad
2
-6
4
301.3110
2
m
Fig. 6–26d.
Notice
how the
at
points
on
the cross
3 B and
I22.5(10
section
develops
astress
force
that
contributes
a 301.3110
moment
2about
m the the
section
develops
a (b)
force
that
contributes
aDmoment
about
12
) N # m10.150
m2
6
My
B 12.7
MPa
section
develops
a
force
that
contributes
a
moment
about
the
sectionneutral
develops
athat
force
that
aas moment
about
s the
sB =
-axis
; has
=contributes
11.2
MPa
neutral
that
same
direction
as
M. three-dimensional
Specifically,
at point
B, of the stress distribution is shown
A
axis
M.
point
B,view
1 = at-the
-6Specifically,
3
2 B, in
1=4axis
I has theBsame direction
301.3110
2
m
neutral
that
has
the
same
direction
as
M.
Specifically,
at
point
3
2c
10.25
m210.020
m2
+
10.25
m210.020
m210.160
m2
12.7
neutralyBaxis
hasMPa
the
as M. Specifically,
at12
point
B, howm2thed stress at points B and D on the dcross
and
yBthat
= 150
mm,
+ c 6–26d.
10.020
m210.300
Fig.
Notice
andsame
so sodirection
=
150
mm,
11.2MPa
MPa
B
12
and
so
y
=
150
mm,
12.7
(d)
B
yB = 150 mm, and so
section develops a force that contributes a moment about the
3
3
22.5(10
)# m10.150
N # m10.150
m2
1 -6
6
My
22.5(10
)
N
m2
4
B
33
My
neutral
that
has
the22.5(10
same
as M.m2
Specifically, at point B,
= 301.3110
2 -m11.2
+ caxis
10.020
m210.300
m2direction
d) N # m10.150
s
s
= - B ; ; Fig.
MPa
6–26
B-= 3)-N
#
sB22.5(10
sBMy
=B =M
=12
11.2
MPa
B-=
-64
4 My
11.2
MPa
m10.150
m2
B
-6
I
!
22.5
kN"m
301.3110
2B=m
B
s
s
;
=
= -11.2 MPa
and
so
y
=
150
mm,
I
(d)
D
301.3110
2
m
3
B
B
-6 m2
sB = sB = ;
=Mc
-11.2
) 301.3110
N # m10.170
I MPa -6 22.5(10
2 m4
4
4
I
301.3110-62 ms
=
s
=2 m
= 12.7 MPa Ans.
=; 301.3110
3
#
22.5(10-6
)2 N
Fig. 6–26 M ! 22.5 kN"mmax
I MyB max
301.3110
m4 m10.150 m2
D
3
sB = 22.5(10
sB = ;
= - 11.2 MPa
#
) N m10.170
-6
McI
2 m4m2 =is12.7
A three-dimensional
view= of the 301.3110
stress -6
distribution
shown
smax =
;
smax
MPain Ans.
4
I
2 mand
12.7 MPa
Fig. 6–26d. Notice
how the stress 301.3110
at points B
D on the cross
12.7 MPa
(d)
Fig. 6–26
(d)
Fig. 6–26
sectionAdevelops
a force that
a moment
about
the in
three-dimensional
viewcontributes
of the stress
distribution
is shown
neutral
that Notice
has the how
samethe
direction
Specifically,
at on
point
Fig.axis
6–26d.
stress as
at M.
points
B and D
theB,cross
and so a force that contributes a moment about the
yB =section
150 mm,develops
neutral axis that has the same3 direction
as M. Specifically, at point B,
22.5(10 ) N # m10.150 m2
MyB
and
so
y
=
150
mm,
sB = sB = B;
= - 11.2 MPa
I
301.3110-62 m4
3
#
22.5(10 ) N m10.150 m2
MyB
sB = sB = ;
= - 11.2 MPa
I
301.3110-62 m4
PAGE !82
6.4
THE FLEXURE FORMULA
291
6.4
THE
FLEXURE
FORMULA 2 9 1
6.4F
THE
FFLEXURE
FORMULA
6.4 THE
LEXURE
ORMULA
6.4 THE FLEXURE FORMULA
291
6.4
THE FLEXURE FORMULA
6.4 THE FLEXURE FORMULA
291
EXAMPLE 6.13
EXAMPLE
6.13 6.13
EXAMPLE
6.13 EXAMPLE
6.13 6.13
! EXAMPLE
Şekilde enkesiti verilen konsol kirişin a-a kesitinde oluşacak maksimum
eğilme gerilmesini
EXAMPLE
6.13hesaplayınız. The beam shown in Fig. 6–27a2.6has
kN a cross-sectional area in the shape
The beam shown in Fig. 6–27a has a cross-sectional
area in the shape
291
2 96.4
1 THE FLEXU
291
2.6 kN
2.62.6
kNkN
The
shown
in in
Fig.
6–27a
hashas
a cross-sectional
area
in in
thethe
shape
beam
shown
Fig.
a cross-sectional
area
shape
2.6
kN in the
wn in Fig.The
6–27a
hasThe
abeam
cross-sectional
area
the
shape
2.6
kN
beam
shown
in Fig.
6–27a
has 6–27a
aincross-sectional
area
shape
of
a
channel,
Fig.
6–27b.
Determine
the
maximum
bending
stress
that
of a channel,
Fig.
6–27b.
Determine
the
maximum
bending
stress
that
13
13
12
12
of of
a channel,
Fig.Fig.
6–27b.
Determine
the
maximum
bending
stress
that
athe
channel,
6–27b.
Determine
the
maximum
bending
stress
that
13 13
12 12
ig. 6–27b.ofDetermine
maximum
bending
stress
that
13 12
a channel,
Fig.
6–27b.
Determine
the
maximum
bending
stress
that
13 12
5
5
occurs
in5 the beam
atin
section
a–a.
occurs occurs
in occurs
the
beam
at
section
a–a.
5 kN
5
a
2.6
in
the
beam
at
section
a–a.
The
beam
shown
in
Fig.
6–27a
has
a
cross-sectional
area
the
shape
in
the
beam
at
section
a–a.
a
5
a
eaminatthe
section
occurs
in the
beam at section a–a.
2.6 kN
a
a
rea
shapea–a.
of a channel, Fig. 6–27b. Determine the maximum bending stress that 13 12
SOLUTION
SOLUTION
nding stress that SOLUTION
13SOLUTION
12
5
occurs
in the beam at section a–a.
SOLUTION
a
5
Internal
Moment.
Here the
support
reactions
do not do
have
a beam’s
Internal
Moment.
Here
the
beam’s 2support
reactions1 m
do not
m
2m
Internal
Moment.
Here
the
beam’s
support
reactions
not
have
2
m
1 m1have
Internal
Moment.
Here
the
beam’s
support
reactions
do
not
have
2
m
m
Internal
Moment.
Here
thethe
beam’s
support
reactions
do not
ma
ment. Here
thedetermined.
beam’s
support
reactions
do not
have
2Instead,
m have
1 m 2 mof sections,athe1segment
to be
Instead,
by
method
of
sections,
the
segment
to
the
to
be
determined.
by
the
method
to
the
a
SOLUTION
a
to be
determined.
Instead,
byby
thethe
method
of of
sections,
thethe
segment
to ato
thethe
(a)
to be
determined.
method
sections,
toleft
be
Instead,
byInstead,
the
ofto
sections,
the
segment
tosegment
the
(a)(a)
ed. Instead,
by determined.
the
method
of sections,
themethod
segment
the
of left
section
a–a
can
be
used,
Fig.
6–27c.
Inofparticular,
note
that
(a) In particular,
left
section
a–a
can
be
used,
Fig. 6–27c.
note
that
(a)
of
section
a–a
can
be
used,
Fig.
6–27c.
In
particular,
note
that
Internal
Moment.
Here
the
beam’s
support
reactions
do
not
have
left
of
section
a–a
can
be
used,
Fig.
6–27c.
In
particular,
note
that
2
m
1m
250
mm
left
of
section
a–a
can
be
used,
Fig.
6–27c.
In
particular,
note
that
a–adocan
used,
Fig.internal
6–27c.
In
particular,
note through
that
the
resultant
axial
force N1passes
the centroid
ofaxial
the force N passes 250
ons
notbe
have
250
mm
2
m
m
250
mm
the
resultant
internal
through
the
centroid
of
the
a
mm
the
internal
axial
force
Nthrough
passes
the
centroid
thethe
_toof
to the
beresultant
determined.
Instead,
by
the
method
ofthrough
sections,
the
segment
resultant
internal
force
N
passes
through
the
250
mmof
_
the
resultant
internal
axial
force
N
the
centroid
ofcentroid
thebe
aaxial
20 mm
(a)moment
_ mm
cross
section.
Also,through
realize
that
thepasses
resultant
internal
moment
must
y ! 59.09
_
e segment
to
the
nternal
axial
force
N
passes
the
centroid
of
the
_
cross
section.
Also,
realize
that
the
resultant
internal
must
be y ! 59.09 mm
20
mm
(a)
20
mm
cross
section.
Also,
realize
that
the
resultant
internal
moment
must
be
y
20
mm
!
59.09
mm
_
left
of
section
a–a
can
be
used,
Fig.
6–27c.
In
particular,
note
that
cross
section.
Also,
realize
that
the
resultant
internal
moment
must
be
y
A
cross
section.
Also,
that
the resultant
internal
moment must be y ! 59.09
! 59.09 mm
N mm
N
20 mm
calculated
therealize
beam’s
neutral
axis at
section
cular,realize
note
that
C a–a.
Also,
thatcalculated
theabout
resultant
internal
moment
must
be
ya–a.
250 A
mm 200Amm
calculated
about
at Nsection
! 59.09
mmthe beam’s neutral
A
N axis N
themm
beam’s
neutral
ata–a.
section
a–a.
the
resultant
internal
axial
force
Naxis
passes
through
the centroid
of the
C C
calculated
about
the
beam’s
neutral
axis
at
section
a–a.
calculated
about
theabout
beam’s
neutral
axis
ataxis,
section
250
200
mm
C
A
N
To
find
the
location
of
the
neutral
the
cross-sectional
area
is
200
mm
200
mm
_
of the
utcentroid
the beam’s
neutral
axis
at the
section
a–a.
To find
the
location
of
the
neutral
axis,mm
the cross-sectional
15
15 mm area
20 mmis
To
find
location
of of
the
neutral
axis,
thethe
cross-sectional
area
is mm
200
_cross
section.
Also,
realize
that
the
resultant
moment
be
find
the
location
the
neutral
axis,
cross-sectional
area
isy ! 59.09
To find
theTo
location
of
the
neutral
axis,
the
cross-sectional
area
isCmust
15 mm
mm
subdivided
into
three
composite
parts
shown
in internal
Fig.into
6–27b.
Using
15as
mm
20
mmas
mm
moment
be neutral
y ! 59.09
15 mm
15
mm
156–27b.
mm A15
15
mm
mm
subdivided
three
composite
parts
shown
in
Fig.
Using
locationmust
of
the
axis,
the
cross-sectional
area
is
N
subdivided
into
three
composite
parts
as
shown
in
Fig.
6–27b.
Using
calculated
about
the
beam’s
neutral
axis
at
section
a–a.
subdivided
into
three
composite
parts
as
shown
in
Fig.
6–27b.
Using
subdivided
into
three
composite
parts
as
shown
in
Fig.
6–27b.
Using
15 mm
C
15 mm
NAppendix A, we have A
Eq. A–2 of
(b)
200 mm
Eq. A–2 of Appendix A, we have
Cin A,
o three composite
parts
asthe
shown
Fig.
6–27b.
200Using
mm
A–2
of
Appendix
wethe
have
(b)(b)
Eq. A–2Eq.
ofTo
Appendix
A,
we
have
find
location
of
neutral
axis, the cross-sectional area is
(b)
Eq.
A–2
of
Appendix
A,
we
have
15 mm
ectionalA,area
is©yA
15 mm
pendix
we have
2[0.100
m]10.200
m210.015
m2 + as
[0.010
m]10.02
m210.250
m2m210.015
(b)Using
15 mm
15 mm parts
subdivided
into
threem]10.200
composite
shown
in
Fig.
6–27b.
2[0.100
m]10.200
m2 + [0.010 m]10.02 m210.250 m2
Çözüm:
©yA
2[0.100
m210.015
m2
+
[0.010
m]10.02
m210.250
m2
y =©yA
=©yA
2[0.100
m]10.200
m210.015
m2
+
[0.010
m]10.02
m210.250
m2
ig. 6–27b. Using
2[0.100
m]10.200
m210.015
m2
+
[0.010
m]10.02
m210.250
m2
©yA
=
m210.015
m2y += 0.020
m2 210.200 m210.015 m2 + 0.020 m10.250
A,m]10.02
we havem210.250
(b)
yEq.
=y=A–2
=Appendix
y = ©A
[0.100 m]10.200
m210.015
[0.010
m2+ m10.250
=©Aofm2
=+ 210.200
©A
m2
(b) 210.200
m210.015
m2
0.020
m10.250
m2m2
©A
210.200
m210.015
m2
+ 0.020
m2
©A
210.200
m210.015
m2m10.250
+
0.020
m10.250
=
0.05909
m
=
59.09
mm
m2 + [0.010
m]10.02
©yA m2 2[0.100
210.200 m210.015
+ 0.020m]10.200
m10.250m210.015
m2
= 0.05909
m =m210.250
59.09 mmm2
m]10.02 m210.250
== 0.05909
mm
=mm
59.09
mm
m 0.05909
= 59.09
ym2
== 0.05909
=
= 59.09
mm
This
dimension
6–27c. m210.015 m2 + 0.020 m10.250 m22.4 kN
6 2.4 kN
.05909 mm2=
59.09
mm©Ais shown in Fig.210.200
m10.250
This dimension is shown in2.4
Fig.
6–27c.
ThisApplying
dimension
is moment
shownisinshown
Fig.
6–27c.
This
dimension
in in
Fig.
6–27c.
kN
2.42.4
kNkN
V
the
ofFig.
equilibrium
about the neutral
6
This
dimension
isequation
shown
6–27c.
66
0.05909 m
1.0 kN
Applying
the
moment
equation
of
equilibrium
about
the
neutral
=
0.05909
m
=
59.09
mm
Vm M V V
Applying
the
moment
equation
of
equilibrium
about
neutral
Applying
thethe
moment
equation
of of
equilibrium
about
thethe
neutral
1.0 kN 0.0
n is shown axis,
in
Fig.
we6–27c.
have
2.4
kN the
0.05909
m0.05909
Applying
moment
equation
equilibrium
about
neutral
1.0
kN
6
1.0
kN
N M
1.0 kN 0.05909Mm
axis, we have
M
axis,
we axis,
have
wewe
have is shown
V 2.4 kN
e moment
equation
of
equilibrium
about the
neutral
Fig.
6–27c.
N
axis,
NN
0.05909 m
6
1.0M
kN = 0
d + ©MThis
=dimension
0; have
2.4 kN12 m2 +in1.0
kN10.05909
m2
2.4
C
NAkN
M
6
d
+
©M
=
0;
2.4
kN12
m2
+
1.0
kN10.05909
m2
M
=
0
V
2
m
Applying
the
moment
equation
of
equilibrium
about
the
neutral
NA
0.05909
mC
d + ©MNA
= 0;
2.4
kN12
m2
+
1.0
kN10.05909
m2
M
=
0
d +d+
©M
=
0;
2.4
kN12
m2
+
1.0
kN10.05909
m2
M
=
0
N
C
1.0
kN
#
V
NA
out the neutral
2
©MNA0.05909
= 0; M
= kN12
4.859m2
kN +m1.0 kN10.05909 m2 - M = 0
M
m2.4
C
axis,1.0
wekN
have
M #
m (c) 2 m2 m
M = 4.859 kN2# m
N
#
m kN
Mm2
= 4.859
=kN
; 2.4 kN12
m2 + 1.0
kN10.05909
-M
MM
= 4.859
0N4.859
(c)
(c)(c)
Section
Property.
The
moment
of
about
axis is C
=inertia
kNm# mthe neutral2 m
Fig.
6–27the neutral axis is
Section
Property.
The
about
d+
©M
= The
0;parallel-axis
2.4
kN12
m2
+ 1.0
kN10.05909
m2of
-the
M
=is0moment
# mthe
Property.
moment
of
inertia
about
the
neutral
axis
C
NA
determined
using
theorem
toabout
each
three
Section
Property.
The
moment
ofapplied
inertia
the
neutral
is isof inertia
M = 4.859
kN
- M = 0 Section
(c) axis
Section
Property.
of
inertia
about
the
neutral
axis
C The momentdetermined
Fig.
6–27
2 mFig.
6–27
using
the
parallel-axis
theorem
applied
to Fig.
each
of the three
#
6–27
2parallel-axis
mcross-sectional
determined
using
theusing
theorem
applied
to
each
of
the
three
composite
parts
of
the
area.
Working
in
meters,
we
have
m
M
=
4.859
kN
determined
the
parallel-axis
theorem
applied
to
each
of
the
three
the
theorem
applied
to the
each
of the three area. Working(c)
erty. The momentdetermined
of inertia using
about
theparallel-axis
neutral axis
is
composite
parts
of
cross-sectional
in
meters,
we have
(c)
composite
parts of parts
the cross-sectional
area. Working
in meters,
we
have
Fig.we
6–27
composite
of of
thethe
cross-sectional
Working
meters,
have
1Section
Property.
The
moment
of area.
inertia
about in
thein
neutral
axis
is
composite
parts
cross-sectional
area.
Working
meters,
we
have
the parallel-axis
theorem
applied
to
of
the
three
3 each
2
engneutral
axis
is
Fig. 6–27
I = c1 10.250 m210.020 m2 + 10.250 m210.020 m210.05909
m - 0.010 m2 d
1
3 2
2
Fig.
6–27
determined
using
the
parallel-axis
tom210.020
each of the
three
1area.
s of the
cross-sectional
wetheorem
have
I = m210.05909
c applied
10.250
m2+
10.250
3 m210.020
2m210.020 m210.05909 m - 0.010 m2 d
1 Working
I three
= c 12
10.250
m210.020
m23 in
+ meters,
10.250
m
0.010
m2
d
each
of the
I
=
c
10.250
m210.020
m2
+
m210.05909
m
0.010
m2
d
3 10.250 m210.020
2
12
I
=
c
10.250
m210.020
m2
+
10.250
m210.020
m210.05909
m
0.010
m2
d
12composite
parts
of
the
cross-sectional
area.
Working
in
meters,
we
have
12
1
meters, we have
+32+
c1 10.250
10.01512
m210.200 m23 + 10.015
m210.100
m - 0.05909 m22 d
1 2d
0 m210.020 m2
mm210.200
- 0.010 m2
3 2
11 m210.020 m210.05909
12
3
3
22m210.200 m210.100 m - 0.05909 m22 d
+
2
c
10.015
m210.200
m2-m2
+ d10.015
3
+ 2 c I 10.015
m210.200
m2
+
10.015
m210.200
m210.100
m
- 0.05909
1
10.250
m210.020
m2
++3 10.250
m210.020
m210.05909
0.010
m2
dd 2
+= 22+cdc -6
10.015
m210.200
m2m2
10.015
m210.200
m210.100
mm
-0.05909
m2m2
12
4
12
0.05909 m - 0.010
m2
2c
10.015
m210.200
+
10.015
m210.200
m210.100
m
0.05909
d
m
= 42.2611012
12212
-6
4
-6
4
3
2
2
m
=
42.26110
-6
21m
42.26110
5 m210.200=m2
+ =10.015
m210.200
m210.100
m - 0.05909 m2 d
2-6m24m
42.26110
4
Maximum
Bending
bending stress
occurs
=
+ 2c
10.015Stress.
m210.200The
m23maximum
+ 10.015 m210.200
m210.100
m -at0.05909 m22 d
2 42.26110
0.100 m - points
0.05909
m2
d
12
Maximum
Bending
Stress.
The maximum bending stress occurs at
farthest
away
from the
neutral
axis.maximum
This
is at the
bottom
of the
Maximum
Bending
Stress.
The
maximum
bending
stress
occurs
at
6
Maximum
Bending
Stress.
The
bending
stress
occurs
at
2 m4
-6
4
Maximum
Bending
Stress.
The
maximum
bending
stress
occurs
points
farthest
away
from
the
neutral
beam,
Thus,
c
=
0.200
m
0.05909
m
=
0.1409
m.
points farthest
away from
neutral
axis. This
is at
theisbottom
of the of the at axis. This is at the bottom of the
2 m thefrom
= 42.26110
points
farthest
away
the
neutral
axis.
This
at
the
bottom
points
farthest
awaymfrom
the neutral
axis.
at -the
bottom
beam,
c = This
0.200is m
0.05909
m of= the
0.1409 m. Thus,
3 stress
beam, c beam,
Thus,
=
0.200
- 0.05909
=
m.
nding Stress.
The
maximum
bending
at
4.859(10
) 0.1409
N #m
m10.1409
m2
c =mMc
0.200
m
- 0.05909
=occurs
0.1409
m. Thus,
Maximum
Bending
Stress.
The
maximum
bending
stress
occurs
at
beam,
Thus,
c
=
0.200
m
0.05909
m
=
0.1409
m.
3
smax =axis. This
Ans.
=4.859(10
3the bottom
# m10.1409
ngaway
stress
occurs
from
the at
neutral
is at42.26110
the = 16.2 MPa
-6
4 m2
4.859(10 ) N # m10.1409 m2
) Nthe
3 2 m#of
McI away
points=farthest
neutral
axis. =This
is=MPa
atMc
the=bottom
of the
4.859(10
)N
m2
3 m10.1409
Mc from
#
s
= 16.2 MPa Ans.
Ans.
= Thus,
16.2
the
of the
max
4.859(10
)
N
m10.1409
m2
00
m bottom
- 0.05909
msmax
= 0.1409
m.
-6
4
-6
4
Ans.
= m Mc
=42.26110
= I16.2 MPa 42.26110
2
m
beam,
Thus,
cs
=Imax
0.200
0.05909
m
=
0.1409
m.
-6
4
Show that
at the
top
of
the
beam
the
bending
stress
is
s¿
=
6.79
MPa.
s
Ans. 2 m
=
=
=
16.2
MPa
max I
42.26110
2
m
-6
4
3
I
# m10.1409
42.26110
2 is
ms¿
4.859(10
Nthe
m2
Mc Show
#bending
that)at
top offorce
the
beam
the
bending
stress
=V 6.79
MPa.
that
atm2
the
ofkN
thewill
beam the bending stress is s¿ = 6.79 MPa.
The
normal
of
N
=beam
1MPa
kN3)the
and
shear
force
=top
2.4
4.859(10
NShow
m10.1409
Mc
that
at
the
top
of
the
stress
is
s¿
= MPa
6.79 MPa.
Ans.
=
= NOTE:Show
=
16.2
-6
4 =the top
sadditional
Ans.
=Nof=the
=
16.2
Show
that
at
beam
the
bending
stress
is
s¿
=will
6.79
MPa.
max
also
contribute
stress
on
the
cross
section.
The
superposition
I MPaNOTE:
-6
4
42.26110
2
m
The
normal
force
of
1
kN
and
shear
force
V
=
2.4
kN
NOTE:
The
normal
force
of
N
= 1 kN and
shear force
V = 2.4 kN will
Ans.
16.2
Bu
çözümde,
a-aI kesitinde
varolan
normal
kuvvet
ve
kesme
kuvvetinin
katkıları
dikkate
42.26110
2and
m shear
NOTE:
The
normal
force
of
N
=
1
kN
force
V
=
2.4
kN
will
of all
these
effects
will be
discussed
in
Chapter
8.
also
contribute
additional
stress
on
the
cross
section.
The
superposition
NOTE:
The
normal
force
of
N
=
1
kN
and
shear
force
V
=
2.4
kN
will
also
contribute
additional
stress
on
the
cross
section.
The
superposition
alınmamıştır.
8.
bölümde
etkiler
alınacaktır.
he top of the beam
the
bending
stress
s¿ stress
=bu6.79
MPa.
also
contribute
additional
on
theele
cross
section.
Show
that
at the
ofis the
beam
the
stress
is The
s¿
=superposition
6.79
MPa. in Chapter 8.
ofMPa.
all these
effects
will
betop
discussed
instress
Chapter
8. these
also
contribute
additional
on
the
cross
section.
The
superposition
ofbending
all
effects
will
be
discussed
s¿ = 6.79
all1these
effects
will
be discussed
Chapter
8. force V = 2.4 kN will
normal force of Nof
=of
kN
and
shear
force
=N 2.4
kN
NOTE:
The
normal
force
= 1in
kN
and shear
all
these
effects
will
beVof
discussed
inwill
Chapter
8.
V = 2.4 kN
will
eceadditional
stress
on
the
cross
section.
The
superposition
also
contribute
additional
stress
on
the
cross
section.
The superposition
The superposition
ects will be discussed
Chapter
8. will be discussed in Chapter 8.
of allin
these
effects
PAGE !83
6.5 Unsymmetric
Bending
6.5 Unsymmetric
Bending
6.5 Unsymmetric Bending
When developing the
flexure
formula
weflexure
imposed
a condition
that a condition that
When
formula
we imposed
302
C Hdeveloping
APTER 6
B Ethe
NDING
the cross-sectional
area
be
about
anwe
axisimposed
perpendicular
the
When developing
thesymmetric
flexure formula
a condition
that
the cross-sectional
area be
symmetric
about
an to
axis
perpendicular
to the
neutral
axis;
furthermore,
the
resultant
internal
moment
M
acts
along
the
cross-sectional
area
be
symmetric
about
an
axis
perpendicular
to
the
neutral
axis;
furthermore,
the
resultant
internal
moment
M
acts
along
Axis of symmetry
Axis of symmetry
axis.
Such
is neutral
the caseaxis.
for Such
the
“T”
orinternal
channel
sections
shown
neutral
axis;
furthermore,
the
resultant
Mchannel
actsinalong
the
is the
case formoment
the “T” or
sections shown in
Axis of symmetry the neutral
Unsymmetric
6–29.
These conditions,
are
unnecessary,
and
insections
this
section
the
axis.
Such
ishowever,
the
caseconditions,
for the
“T”
or channel
shown
in in this Bendin
6.5
SimetrikC HOlmayan
Fig.
6–29.
These
however,
are
unnecessary,
and
section
302
A P T E R 6 Fig.
BEğilme
EN
D
I N Gneutral
we will
the
flexure
formula
can
also
be
applied
either
to
aapplied either to a
Fig.show
6–29.that
These
however,
are
unnecessary,
and
in
this
weconditions,
will show
that the
flexure
formula
can
also
besection
Neutral axis
Neutral
y area
When
developing
formula
beamwe
having
ashow
cross-sectional
of
any shape
or
tobe
a beam
having
atotoaflexure
will axis
that
flexure
formula
can also
applied
either
abeam having
beamthe
having
a cross-sectional
area
of
any
shape
orthe
a w
Neutral axis
the
cross-sectional
area
be
symmetric
abou
resultant
internal
that
acts in
anyofdirection.
beam
havingmoment
aresultant
cross-sectional
area
any
shape
to adirection.
beam having a
internal
moment
that
acts or
in any
Unsymmetric
Bending
neutral axis; furthermore, the resultant in
M
x
Axis
of
symmetry
resultant
internal
moment
that
acts
in
any
direction.
M
x
z
z
Moment Applied
About Applied
PrincipalAbout
Axis. Principal
Consider
beam’s
the
neutralthe
axis.
Such
is the case
the “T”
Moment
Axis.
Consider
thefor
beam’s
M
x
z
y
developing
the
flexure
formula
we
imposed
a condition
that
cross Moment
section to When
have
the
unsymmetrical
shape
shown
in 6–29.
Fig.
6–30a.
As
in
Applied
About
Principal
Axis.
Consider
the
beam’s
Fig.
These
conditions,
cross
section
to have
the unsymmetrical
shape
shown
in Fig. however,
6–30a. Asare
in un
thetocross-sectional
area be symmetric
about
axis
perpendicular
toformula
the
y
y Sec. 6.4,
thesection
right-handed
x,the
y,unsymmetrical
zright-handed
coordinate
system
is
established
such
cross
have
the
shape
shown
inan
Fig.
6–30a.
As
in
will
show
that
the
flexure
Sec.
6.4,
x, y, zwe
coordinate
system
is established
such can
axis
axis;
furthermore,
resultant
moment
M acts along
y
Axis ofthat
symmetry
the
isneutral
located
at the
centroid
Cthe
onat
the
cross
section,
the
Sec.origin
6.4, the
right-handed
x, y,
z Neutral
coordinate
is C
established
such
beaminternal
having
athe
cross-sectional
that the
origin
is located
thesystem
centroid
onand
cross
section,area
andof
theany
the
neutral
axis.
Such
is
the
case
for
the
“T”
or
channel
sections
shown
resultant
moment
M internal
acts
along
the + zM
require
theaxis.
stress
thatinternal
the origin
is located
at the
centroid
Caxis.
on
the
cross
section,
and
resultant
internal
moment
that
actsin
in any d
resultant
moment
actsWe
along
the
+z
We the
require
the
stress
Axis of symmetry
Axis
of symmetry
Fig. over
6–29.
These
conditions,
however,
are We
unnecessary,
and
in this section
M
distribution
acting
the
entire
cross-sectional
to require
have a the
zero
resultant
internal
moment
M
acts
+ z area
axis.
stress
x along
distribution
acting
over the entire
cross-sectional
area
to have a zero
z
Axis of symmetry
we
will
show
that
the
flexure
formula
can
also
be
applied
either to
a
Moment
Applied
About
force
resultant,
the
resultant
internal
moment
about
the
y
axis
to
be
zero,
distribution
acting
over
the
entire
cross-sectional
area
have
a
zero
force
resultant,
the
resultant
internal
moment
about
the
y
axis toPrincipal
be zero,
Neutral axis
beam
having
a
cross-sectional
area
of
any
shape
or
to
a
beam
having
a
cross
section
to
have
the
unsymmetrical
and the
resultant
internal
moment
about
the moment
z axis
to about
equal
M.*
force
resultant,
the the
resultant
internal
moment
about
the
ythe
axis
to
beto
zero,
and
resultant
internal
zThese
axis
equal
M.* These sha
y moment that acts in
resultant
internal
anymathematically
direction.
6.4,
the right-handed
x, y, z coordina
conditions
be
expressed
mathematically
by
considering
the
the resultant
internal
moment
the zSec.
axis
to equal
M.*
These
Neutral axis three and
three
conditions
can about
be expressed
by considering
the
Neutral
axiscan
M
x
that
the
origin
is located
atz),the
centroid
C
z
acting
on
the
differential
element
dA
located
at
(0,
y,
z),
Fig.
6–30a.
conditions
can
be
expressed
mathematically
by
considering
the
Neutralforce
axis three
force
acting
on
the
differential
element
dA
located
at
(0,
y,
Fig.
6–30a.
Asal eksenlere göre döndüren moment
etkisi
Moment
Applied About Principal
Axis.
Consider
the
resultant
internal
moment
M beam’s
acts along the
This
force
is
dF
=
s
dA,
and
therefore
we
have
force
acting
on
the
differential
element
dA
located
at
(0,
y,
z),
Fig.
6–30a.
This
force
is
dF
=
s
dA,
and
therefore
we
have
M
M
symmetry
z
x
cross section to Axis
haveofthe
unsymmetrical
shape shown
in over
Fig. 6–30a.
As incross-s
zx
6
distribution
acting
the entire
This force is dF = s dA, and therefore we have
M
y
x
z
Sec.
6.4,
the
right-handed
x,
y,
z
coordinate
system
is
established
such mome
6
force
resultant,
the resultant internal
FR = ©Fx ; that the F
0©F
= located
-;
s dA
(6–14)
=
0
=
s
dA
(6–14)
R
x
origin
is
at
the
centroid
C
on
the
cross
section,
and
the
Fig. 6–29
Fig. 6–29
and L
the
about t
A resultant internal
FR = resultant
©Fx ;
0L
=A - Msacts
dAalong
(6–14) moment
internal
moment
the
+
z
axis.
We
require
the
stress
Fig. 6–29
Neutral
LA axis three conditions can be expressed mathe
Axis of symmetry
overzsthe
areadifferential
to have aelement
zero
1MR2y = ©My ;distribution
0©M
= dAentire
1MR2y =acting
0 =cross-sectional
- acting
zs dAon(6–15)
(6–15)dA
y;
force
the
L
A
L
A
force
resultant,
the
resultant
internal
moment
about
the
y
axis
to
be
zero,
1MR2y = ©My ;
0 = - zs dA This force is dF = s
(6–15)
dA, and therefore we
x L
z the resultantMinternal
A
6
and
moment
about the z axis to equal M.* These
1MR2z = ©Mz ;
= z ; ys dA
1MR2z =M©M
M =
ys dA (6–16)
(6–16)
mathematically
considering the
Neutral axis three conditions can
LA=be expressed
LAFR = ©Fx ; by(6–16)
0 = - s dA
1MR2z = ©Mz ;
M
ys dA
force actingFig.
on6–29
the differential
element dA located at (0, y, z), Fig. 6–30a.LA
LA
This
force
is
dF
=
s
dA,
and
therefore
we have
M
x
z
6
y R2y = ©My ;
1M
0 = - zs d
y
y
y
LA
y
s
smax (6–14)
F y= ©Fx ;
0 = - s dAmax
z dFR!
z dF ! sdA
Fig. 6–29
sdA
L
A
smax303
6.5 UNSYMMETRIC BENDING
1M
ys dA
s M =
R2zs = ©M
z;
z dF ! sdA
cs
L
A
c
dA 1M 2 = ©M ; dA
M
M
y
0 = - zs dA
y (6–15)
R y
y
c
6.5 L
UANSYMMETRIC ByENDINGx M
303
x
As shown in Sec. 6.4, Eq. 6–14 is satisfied since the 6.5
z dA
axisUNSYMMETRIC
passes
BENDING
303
6.5 UxNSYMMETRIC BENDING
through the centroid of the area. Also, since the z axis represents
the
C
C
1MR2z = M
©M
MM=
ys dA
(6–16)
y
y
y z;
neutral axis for the As
crossshown
section,
stress
willis vary
linearly
from
C
L
A
in the
Sec.normal
6.4, Eq.
6–14
satisfied
since
the
z
axis
passes
x
x
z
y
M
z
z dF distribution
Bending-stress
Bending-stress
! sdA
nzero
Sec.at6.4,
6–14axis,
is satisfied
since the
axis
passes
yzshown
=
c, Fig.
the Eq.
neutral
Hence
As
in 6–30b.
Sec. 6.4,
Eq. the
6–14 isx satisfiedthe
since the zdistribution
axis passes
through to
thea maximum
centroid ofat
the
area.
Also,
(profile
view)
z since the z axis represents
(profile
view)
Bending-stress
distribution
entroid
of the area.isAlso,
since
the
z -axis
represents
the
s
=
(y>c)s
.
stress distribution
defined
by
When
this
equation
is
through
the
of thewill
area.
Also,
since the z axis(b)
represents the
max
neutral axis for the cross
section,
thecentroid
normal
vary
linearly
(a)stress
(a) from
(profile view) dA y (b)
the cross section,
the6–16
normal
stress
willneutral
vary
linearly
from
substituted
into Eq.
and
integrated,
it
leads
to
the
flexure
formula
y
axis
for
the
cross
section,
the
normal
stress
will
vary
linearly
from
c, Fig. 6–30b. Hence the
zero at the neutral axis, to a maximum at y = (a)
(b)
smax
y = c, Fig.
utral
to a maximum
at
6–30b.
Hence
the
smax axis,
= Mc>I.
When
it
is
substituted
into
Eq.
6–15,
we
get
Fig.
6–30
y
=
c,
zero
at
the
neutral
axis,
to
a
maximum
at
Fig.
6–30b.
Hence
the
Fig.
6–30
z
stress distribution is defined by s = -(y>c)s max . When this equation
dF ! sdAis
C
ion is defined by s = -(y>c)s max . When
this
equation
is
Fig.
6–30
s
s flexure
= -(y>c)s
stress
distribution
is
max . When this equation
y
M
substituted into Eq. 6–16
and
integrated,isitdefined
leads toby
the
formula
c x
o Eq. 6–16 and integrated, it leads
to the
flexure formula
-smax
dA
substituted
into
Eq.
6–16
and
integrated,
it
leads
to
the
flexure
formula
M
z
y
smax = Mc>I.
When it isyzsubstituted
intothat
Eq.moments
6–15,
get thethat
dA
= 6–15,
*The condition
about
y axis
be equal
to zero
in was not considered in B
*Thewe
condition
moments
about
the ywas
axisnot
be considered
equal to zero
When it is substituted
into0Eq.
sget
= 6.4,
Mc>I.
When
it
is
substituted
into
Eq.
6–15,
we
get
c we
x
maxSec.
LA
since the bending-stress distribution was symmetric with respect to the y axis and such
0 =
which requires
y
y
y
6.5
6.5
Sec. 6.4,
since the
bending-stress
symmetric
respect to
*The condition that
moments
about
the y axisdistribution
be equal towas
zero
was notwith
considered
inthe y axis and such
(a)
a distribution
stress
produces
moment
about
therespect
yzero
axis.moment
See
Fig.
6–24c.
a distribution
ofCstresszero
automatically
produces
thesuch
y axis. See Fig. 6–24c.
-s
Sec.
6.4,ofsince
theautomatically
bending-stress
distribution
was
symmetric
with
to the
y about
axis
and
max
-smax
y -smax
M
yz of
dA
= a distribution
stress automatically
produces zero moment about the y axis. See Fig. 6–24c.
Bunu 6.15
denkleminde0 yerine
yz dA
c koyarsak
yz dA x
0z =
A
L
Fig. 6–30
c LA
Bending-stress distribution
c LA
(profile view)
(a)
(b)
Buradan,
which requires yz dA = 0
*The condition that moments about the y axis be
which requires
LA
Sec. 6.4, since the bending-stress distribution was symme
Fig. 6–30
a distribution of stress automatically produces zero mo
yz dA = 0
This integral isBu
called
the
of inertia
for
the area. Asdenilmektedir.
indicated
yz
dA
= 0 product
yz dA =y ve
0 z eksenleri kesitin asal
integrale
çarpım
atalet
L
A momenti
L
A
L
A that moments about the y axis be equal to zero was not considered in
*The
condition
in Appendix A, eksenleri
it will indeed
be
zero
provided
the
y
and
z
axes
are
olarak seçildiğinde çarpım Sec.
atalet
momenti
sıfır olacaktır.
6.4, since
the bending-stress distribution was symmetric with respect to the y axis and such
chosen as principal axes of inertia for the area. For an arbitrarily
shaped
Kesitin
asal eksenlerinin
EkaA
da
atalet
dönüşüm
denklemleri
This integral
is called theyönelimi
product of inertia
forverilen
theofarea.
As
indicated
stress
automatically
produces
zero momentile
about the y axis. See Fig. 6–24c.
area,
the the
orientation
of inertia
the principal
axes
can
always
be distribution
determined,
is called
product of
for the area.
indicated
ThisAs
integral
is called
the product of inertia for the area. As indicated
in Appendix A, Kesitte
it will indeed
be zero
provided
the
y eksenlerin
and z axes yönelimini
are
hesaplanabilir.
bir
simetri
ekseni
varsa
asal
belirlemek
kolay
either
the inertia
transformation
equations
Mohr’s
of inertia
A,using
it will
indeed
be zero
provided the
and zoraxes
iny Appendix
A,are
it circle
will indeed
be zero provided the y and z axes are
chosen
as
principal
axes
of
inertia
for
the
area.
For
an
arbitrarily
shaped
olacaktır
daima
simetri
eksenleri
boyunca
ve for
buna
yöneleceklerdir.
as explained
in Appendix
A,
Secs.
A.4
and
A.5.
Ifshaped
the area
hasofan
axis
6
cipal
axes of inertia
for the çünkü
area.
For
anchosen
arbitrarily
as principal
axes
inertia
thedik
area.
For an arbitrarily shaped
area, thethe
orientation ofaxes
the can
principal
can
always be determined,
of symmetry,
however,
easilyaxes
be of
established
ntation
of the principal
axes principal
can always
be the
determined,
area,
orientation
the principal axes can always be determined,
using either
the inertiaalong
transformation
equations
or Mohr’s
of inertia
since
will always
be oriented
the ofaxis
of symmetry
and circle
inertiathey
transformation
equations
or Mohr’s
circle
inertia
using
either
the
inertia
transformation
equations or Mohr’s circle of inertia
as
explained
in
Appendix
A,
Secs.
A.4
and
A.5.
If
the
area
has
an axis
6
it. A.4 and A.5. If the
nperpendicular
Appendix A, to
Secs.
has aninaxis
asarea
explained
Appendix A, Secs. A.4 and A.5. If the area 6
has an axis
of
symmetry,
however,
the
principal
axes
can
easily
be
established
For example,
consideraxes
the members
in Fig.
6–31. Inthe
each
of
however,
the principal
can easily
be established
ofshown
symmetry,
however,
principal
axes can easily be established
since
they
willthe
always
be oriented
alongfor
thethe
axis
of symmetry and
these
cases,
y
and
z
must
define
principal
axes
of
inertia
cross
they willand
always be oriented along the axis of symmetry and
l always be oriented along the axis since
of symmetry
it. through 6–16. In Fig. 6–31a the
section
in order perpendicular
to satisfy Eqs.to6–14
perpendicular to it.
to
it.
For example,
consider the
members
shown
in Fig.
6–31. In each of
located
by symmetry,
Figs.
6–31b
and 6–31c
Forinexample,
consider
thetheir
members shown in Fig. 6–31. In each of
e,principal
consideraxes
theare
members
shown
in Fig.and
6–31.
In each
of
these
cases,
y
and
z
must
define
the
principal
axes
of
inertia
for the cross
orientation
is determined
using
theofmethods
of Appendix
A. Since
M !8the
is4 principal
PAGE
these cases,
and z must
define
axes of inertia for the cross
nd
z must define
the principal
axes
inertia
for
they cross
section
in
order
to
satisfy
Eqs.
6–14
through
6–16.
In
Fig.
6–31a the
applied
aboutEqs.
one 6–14
of thethrough
principal6–16.
axessection
(z Fig.
axis),in
the
stress
order
satisfy Eqs.is6–14 through
6–16. In Fig. 6–31a the
er
to satisfy
In
6–31a
thetodistribution
principal
axes are located
by
and
in shown
Figs.
, and
determined
the flexure
s and
= symmetry,
-6–31c
My>I
is
for and 6–31c
principal
axes their
located
by 6–31b
symmetry,
and intheir
Figs. 6–31b and 6–31c their
are
located byfrom
symmetry,
and in formula,
Figs. 6–31b
zare
is determined
using the
of Appendix
Since M
each case. using orientation
orientation
is methods
determined
using the A.
methods
of is
Appendix A. Since M is
determined
the methods of Appendix
A. Since
M is
of symmetry,
theaxis
principal axes can easily be established 6
ndix A, Secs. A.4 and
A.5. If the however,
area has an
sincecan
theyeasily
will be
always
be oriented along the axis of symmetry and
r, the principal axes
established
perpendicular
it.
s be oriented along
the axis oftosymmetry
and
For example, consider the members shown in Fig. 6–31. In each of
these cases,
y and
z must
define
er the members shown
in Fig.
6–31.
In each
of the principal axes of inertia for the cross
section
satisfy
Eqs. 6–14 through 6–16. In Fig. 6–31a the
st define the principal
axesinoforder
inertiatofor
the cross
304
CHAPTER 6 BENDING
principal6–16.
axes In
areFig.
located
by the
symmetry, and in Figs. 6–31b and 6–31c their M
tisfy Eqs. 6–14 through
6–31a
Aşağıdaki
kesitlerde
asalusing
eksenlere
göreofdöndüren
orientation
is determined
the methods
Appendix A.moment
Since M isetkisi
ed by symmetry, and
in Figs. 6–31b
and 6–31c
their
u
y
applied
about
oneA.
ofdağılımı
the principal
axesdenklemini
(z axis), thekullanarak
stress distribution
is
olduğunda
gerilme
eğilme
belirlenebilir.
ned using the methods
of Appendix
Since
M is
z
determined
from distribution
the flexure is
formula, s = - My>Iz, and is shown for
he principal axes (z
axis), the stress
flexure formula, seach
= -case.
My>Iz, and is shown for
Moment Arb
y
My, Mz = the resultant internal moment components directed along
the principal y and z axes. They are positive if directed
along the + y and + z axes, otherwise they are negative.
Or, stated another way, My = M sin u and Mz = M cos u,
where u is measured positive from the + z axis toward the
6
z
z
"
"
My, Mz = the resultant internal moment components directed along
Fig. 6–32
y
(c)
My ! M sin u
x
Here,
z
s = the normal stress at the point
Iz
(b)
Mz ! M cos u
y
(a)
u
M
y, z = the coordinates of the point measured from x, y, z axes
having their origin at the centroid of the cross-sectional
area and forming a right-handed coordinate system
The x axis is directed outward from the cross-section and
the y and z axes represent respectively the principal axes
of minimum and maximum moment of inertia for the
area
(6–17)
Iy
Myz
+
Mzy
s = -
x
y
x
BENDING
"
"
!
304
"
!
CHAPTER 6
!
!
!
Moment Arbitrarily Applied. Sometimes a member may be
loaded such that M does not act about one of the principal axes of the
cross section. When this occurs, the moment should first be resolved
into components directed along the principal axes, then the flexure
formula can be used to determine the normal stress caused by each
moment component. Finally, using the principle of superposition, the
resultant normal stress at the point can be determined.
To show this, consider the beam to have a rectangular cross section
and to be subjected to the moment M, Fig. 6–32a. Here M makes an
angle u with the principal z axis. We will assume u is positive when it is
directed from the + z axis toward the + y axis, as shown. Resolving
M into components along the z and y axes, we have Mz = M cos u
and My = M sin u, as shown in Figs. 6–32b and 6–32c. The normalstress distributions that produce M and its components Mz and
My are shown in Figs. 6–32d, 6–32e, and 6–32f, where it is assumed
that 1sx2max 7 1s¿x2max . By inspection, the maximum tensile and
compressive stresses [1sx2max + 1s¿ x2max] occur at two opposite corners
of the cross section, Fig. 6–32d.
Applying the flexure formula to each moment component in Figs. 6–32b
and 6–32c, and adding the results algebraically, the resultant normal stress
at any point on the cross section, Fig. 6–32d, is then
loaded such that
cross section. W
into component
formula can be
moment compon
resultant normal
sözkonusu
To show this,
Momentx Arbitrarily
Applied
and to be subjec
loaded such that M does not act ab
angle u with the
cross section. When this occurs, th
directed from th
into
(a) components directed along th
M into compon
formula can be used to determine
My = M sin
and
y
304y
CHAPTER 6 BENDING
moment component.
Finally,
using
stress
distributi
M
y
resultant
normal
stress
at
the
point
y
My are shown c
u
y
y
To
show
this,
consider
the
beam
Moment Arbitrarily Applied.
Sometime
y
that 1s
x2max 7
x
z
and
to
be
subjected
to
the
momen
loaded such that M does not act compressive
about one ofstre
the
CHAPTER 6 BENDING
M
withoccurs,
the principal
z axis.
We
cross section.angle
Whenu this
thethe
moment
sho
of
cross
secti
M
directed
from along
the +z
axis
toward
into components
directed
the
principal
M
Applying
the fa
(a)
M
y
M
into
components
along
the
z ad
as
Moment Arbitrarily
Applied. Sometimes aformula
membercan
may
be be
used to determine
normal
M
and the
6–32c,
and
z
M
=
M
sin
u,
and
as
shown
in
Fi
loaded such that M does not act aboutxone of the principal
of the y Finally, using
momentaxes
component.
the point
principle
at any
on th
x
M
z
stress
distributions
the moment should
first be normal
resolved
stress
at the point that
can beproduce
determi
z resultant
x cross section. When this occurs,
u x
z
Myconsider
are shown
in Figs.
into components
directed(b)
along the principal
then
the flexure
To show
this,
the beam
to 6–32d,
have a6–32
rec
z
(a)
yaxes, (c)
6
x
z
x
1s¿x2maxM,
. By
that
andcaused
to be by
subjected
the 7moment
Fig. inspe
6–32
formula can be
each 1sto
x2max
(c) used to determine the normal stress
(b)
z
Fig. 6–31
Mz ! Mucos
u the
[1swill
+ 1s¿u
compressive
with
principal zstresses
axis. We
assume
moment component. Finally, using the principle of angle
superposition,
the
x2max
M
Fig. 6–31
+y axis,
axis
toward
the +z
cross
section,
Fig.the
6–32d.
resultant normal stress at the point can be determined.directed fromofthe
(a)
(b)
u
M into
components
along
z and
y axes,
we
Applying
thethe
flexure
formula
to ea
To show
this,döndüren
consider themoment
beam to have
a rectangular
cross
section
Herhangi
bir
eksene
göre
etkisi
304
6 to
BEN
NG
x C H A P T E Rand
= Mand
sin
and M
as shown
in Figs.
6–32b alge
and
z
6–32c,
and adding
the results
beD I subjected
to the moment M, Fig. 6–32a. Here
My makes
anu,
that on
produce
and its
the crossMsection,
Fig.
angle u with the principal z axis. We will assume u is stress
positivedistributions
whenatitany
is point
Here,
showny in Figs. 6–32d, 6–32e,
and 6–32f,
directed
from the +z axis towardy the +y axis, as M
shown.
y are Resolving
y
Moment Arbitrarily Applied.
Sometimes a member may be
(a)
7 1s¿
that M1s
M cos
u x2max . By inspection, the m
M into components along
the z and y axes, we have
6
z x=2max
loaded such that M does not act about xone
of the principal
axes of the
[1sx2max + 1s¿ x2max
compressive
stresses
occur
s] =
the at
n
and My = M sin u, as shown
in
Figs.
6–32b
and
6–32c.
The
normalz
cross section.
When
this
occurs,
the
moment
should
first
be
resolved
Mz ! Mcos u
Mcz
of the crossMsection,
6–32d.
stress distributions that produce
M and its components
and MFig.
!
M
sin
u
z
y,
z
=
the
into components directed along the principal axes, ythen the flexure s = Applying
the flexure formula to each moment
co
My are shown in Figs.
6–32d,
andto6–32f,
where
it normal
is assumed
Iz
havin
y
(b)
formula
can 6–32e,
be used
determine
the
stress caused by each
and
6–32c,
and
adding
the
results
algebraically,
the
that 1sx2max 7 1s¿moment
inspection,
the
maximum
tensile
and
area
x2max . By
component. Finally, using the principle of superposition, the
at"
any pointcorners
on the cross section, Fig. 6–32d, is The
thenx
M
! resultant
[1sx2maxnormal
+ 1s¿ xstress
2max] occur
compressive stresses
two
opposite
at the at
point
can
be determined.
of the cross section, Fig.
u
the y
To6–32d.
show this, consider the beam to have a rectangular
Here, cross section
6 the flexure formula to each moment component
z in Figs. 6–32b
y
Applying
of mi
x
z
and to be subjected to the
x makes an
x moment M, Fig. 6–32a. Here M
and 6–32c, and zadding
the
results
algebraically,
the
resultant
normal
stress
area
u with
angle
it is
Mz ! Mcos
u the principal z axis. We will assume u is positive whenM
s = the normalzystressMatyzthe p
at any point on the cross
section,
Fig.
6–32d,
is
then
+ z = the r
= - My, M
directed from the +z axis toward the +y axis, as shown.s Resolving
Iz
Iythe
(a)
u
y, z =Mthe=coordinates
of
(b)
y ! Msin
(c) have
thepo
p
M
cos
u
M into components
along the zMand
y axes,
we
z
having
their
origin
at
the
along
M
=
M
sin
u,
and
as
shown
in
Figs.
6–32b
and
6–32c.
The
normaly
x
Fig. 6–32
area M
and and
forming a Or,
right
s
stress distributions that
z
Mz ! Mcos u
Myz produce M and its components
Mzy
The
x
axis
is
directed
ou
wher
Here,
M
are
shown
in
Figs.
6–32d,
6–32e,
and
6–32f,
where
it
is
assumed
y
(6–17)
s = - y +
y
y andand
z axes represe
Iy. By inspection, the maximum the
+y a
(b)
that 1sx2max 7Iz1s¿x2max
tensile
z
of
minimum
and maxim
compressive stresses [1sx2max + 1s¿ x2maxxs] occur
at
two
opposite
corners
Iy, Iz = the p
= the normal stress at the point
area
of the cross section, M
Fig.!6–32d.
axes,
Msin u
y, z = the coordinates of the point measured
y
My, Mz = in
theFigs.
resultant
Applying the flexure formula to each moment component
6–32b internal mo
Here,
having their origin at the centroid of t
y
(c) algebraically, the resultant normal
the principal
and 6–32c, and adding the results
stress y and z axe
area and forming a right-handed coor
+y and +z axe
along
the
at any point on the cross section, Fig. 6–32d, is then
The x axis is directed outward from t
s = the normal stress at the point Fig. 6–32
Or,
stated
another
way, M
the y and z axes represent
respective
My ! Msin u
y, z = the coordinates of the point measured from x, y, z axes
u
where
is
measured
poso
6
z
of minimum and maximum moment
having
their origin at the centroidxof the cross-sectional
x
+y
axis
z
Mz ! Mcos u
area and forming a right-handed coordinate
Myzarea
Mzy system
Iy, Iz = internal
the principal
moments
of
+
(6–17)
s
=
M
,
M
resultant
moment
compon
The x axis is directed outward from the cross-section
and
y
z = the
I
I
axes,
respectively.
See
Ap
z
y
(b)
(c)
principal y and z axes. They are po
the y and z axes represent
respectively the principalthe
axes
z
along
the +y and +z axes, otherwise t
of minimum and maximum moment of inertia for the
x
Fig. 6–32
Or,
stated
another way, My = M sin u
area
Here,
u
where
is
measured
positive from the
y y, Mz = the resultant internal moment components directed along
M
+y
axis
(c)
the principal y and z axes. They are positive if directed
s = +z
theaxes,
normal
stress atthey
the are
point
, Iz = the principal moments of inertia calcula
Iynegative.
along the +y and
otherwise
Fig. 6–32
axes,
respectively.
See Appendix A
Mz =measured
My = M sinofuthe
M cos
u, from
Or,u stated another
way,coordinates
andpoint
My ! Msin
y, z = the
x, y, z axes
where u is measured
positive
from
theat+zthe
axis
towardofthe
having
their
origin
centroid
the cross-sectional
+y axis
area and forming a right-handed coordinate system
Theofx inertia
axis is calculated
directed outward
from
thez cross-section and
Iy, Iz = the principal moments
about the
y and
the
y
and
z
axes
represent
respectively
the principal axes
axes, respectively. See Appendix A
z
PAGE
!85
of
minimum
and
maximum
moment
of
inertia
for the
x
area
(c)
x
y
(b)
z
z
6
"
"
"
"
"
"
PAGE !86
My, Mz = the result
the princi
y, z = the coord
having th
area and
The x axi
the y and
of minim
area
My ! M sin u
x
Mz ! M cos u
y
(a)
x
z
M
u
y
"
CHAPTER 6
BENDING
"
"
"
304
s = the norma
Here,
Moment Arbitra
loaded such that M d
cross section. When
into components dir
formula can be used
moment component.
resultant normal stre
To show this, consi
and to be subjected
angle u with the prin
directed from the +
M into components
and My = M sin u,
stress distributions
My are shown in Fi
that 1sx2max 7 1s¿x2
compressive stresses
of the cross section, F
Applying the flexur
and 6–32c, and adding
at any point on the cr
y
ection. When this occurs, the moment should first be resolved
[(sx)max " (s¿x)max]
stress distributions that p
(6–18)the
a of the
ion
of the
Neutral
Axis.
The angle
the flexure
mponents
directed
along
principal
axes, then
My are shown in Figs. 6–3
N
y
candetermine
be determined
by applying
6–17by each
as in
canFig.
be 6–32d
used to
the normal
stress Eq.
caused
that 1sx2max 7 1s¿x2max . B
,t since
by definition
nousing
normal
acts on
neutral
A
component.
Finally,
thestress
principle
of the
superposition,
the
x
compressive stresses [1sx2m
z
z
This
equation
defines
the
neutral
axis
for
the
cross
section.
Since
the
venormal stress at the point can be determined.
[(sx)max " (s¿x)max]
nt
a
of the cross section, Fig. 6–3
slope of this line is tan a = y>z, then
how this, consider the beam to have a rectangular cross section
6
(sx)maxApplying the flexure form
[(sx)max ! (s¿x)max]
I
M
y z
be subjected to the
moment
M, Fig. 6–32a. Here M makes an
z
y
=
and 6–32c, and adding the re
6.5 U
NSYMMETRIC
ENDING
305
Tarafsız
eksenin
yönelimi:
NSYMMETRIC
BBENDING
MzIU
u is positive 305
with the principal
z6.5
axis.
We
will assume
when it is
y
(e)at any point on the cross sect
Iz
(d)
NSYMMETRIC BENDING
305
d from the +z axis toward the +y axis, as shown.
(6–19)
tan a Resolving
=
tan u6.5 U6.5
UNSYMMETRIC
BENDING
305
I
y
y
M
=
M
cos
u
along
the
z
and
y
axes,
we
have
hecomponents
proper
algebraic
M cos u and My = M sin u, then y
z
! BENDING
6
6.5 UNSYMMETRIC
305
= the
M sin
u, as shown in Figs. 6–32b and 6–32c. The normalcoordinates
y
and
ynd
x
y
y, z axes form a right-handed system, and the proper algebraic
Mz and
distributions
that
produce
M and system,
its components
The
x,
y, z stress
axes
form
a right-handed
and the proper
algebraic z
he
resulting
he
Mz ! Mcos
6.5u y UNSYMMETRIC
305
st be
assigned
to the moment
components and the coordinates
Iz
6.5 UNSYMMETRIC
BENDING
3 0 5BENDING
y
shown
in
Figs.
6–32d,
6–32e,
and
6–32f,
where
it
is
assumed
The
x,
y,
z
axes
form
a
right-handed
system,
and
the
proper
algebraic
gns
must
be
assigned
to
the
moment
components
and
the
coordinates
egative.
s
Here
it
can
be
seen
that
unless
I
the
angle
u,
defining
the
direction
=
I
tan
u
z
y
=
¢
≤
[(s
)
!
(s
¿
)
]
[(s
)
!
(s
¿
)
]
z
y
(s
)
x
max
x
max
max
x max
plying this equation. When
this is the case, the xresulting
stress
x max
(s¿x)max
Iinspection,
(b)
y
shen
2maxapplying
7 1s¿x2this
.[(s
By
the
maximum
tensile
and
signs
must
be
assigned
to
the
moment
components
and
the
coordinates
equation.
When
this
is
the
case,
the
resulting
stress
of
the
moment
M,
Fig.
6–32a,
will
not
equal
a,
the
angle
defining
the
x
max
[(s
)
"
(s
¿
)
]
)
"
(s
¿
)
]
6.5
U
NSYMMETRIC
B
ENDING
max
nsile ifait of
is positive
and
it is negative.system, and the proper algebraic
xx max
xx max
[(s )
! (s¿x)max
y]
The
y,maxwhen
zcompressive
axes
aifright-handed
angle
the
y
+
1s¿
2form
ssive
stresses
occur
at
corners
y,
axes
form
ait right-handed
and
the
algebraic
applying
this
equation.
When
this is the case, the resulting
stressx max [(s"
illzbe
tensile
if[1s
isx2x,
positive
and
compressive
iftwo
itproper
isopposite
negative.
of] the
neutral
axis,
Fig.
6–32d.
maxinclination
xsystem,
max
!
x)max ! (s¿x)max]
N
N
[(s
)
"
(s
¿
)
]
signs
must
be
toifthe
components
and
x maxthe coordinates
x max
plying
Eq.
t be section,
assigned
to the
moment
and
the coordinates
ross
Fig.
6–32d.
willassigned
be components
tensile
it
ismoment
positive
and
if[(s
itx)ismax
negative.
a
ation
of 6–17
the
Neutral
Axis.
The
angle
ofcompressive
the
[(sx)max ! (s¿x)max]
" (s¿x)max]
y
when
applying
this
equation.
When
this
is
the
case,
the
resulting
stress
sying
AAresulting
The
x,
y,
z
axes
form
a
right-handed
system,
and the proper algebraic
on
the
neutral
a
Orientation
of
the
Neutral
Axis.
The
angle
of
the
N
lying
this
equation.
When
this
is
the
case,
the
stress
the
flexure
formula
to
each
moment
component
in
Figs.
6–32b
Here,
xis in Fig. 6–32d can be determined
by
applying
Eq.
6–17
x
[(s
)
"
(s
¿
)
]
x
max
x max
zz it for
ythe xcoordinates
z The components
N of the
on and
defines
theFig.
neutral
axis
theifsigns
cross
section.
Since
the
will
be
tensile
if
is
positive
and
compressive
if
it
is
negative.
must
be
assigned
to
the
moment
and
a
Orientation
of
the
Neutral
Axis.
angle
eutral
axis
in
6–32d
can
be
determined
by
applying
Eq.
6–17
[(s
)
"
(s
¿
)
]
Important
Points
[(s
)
!
(s
¿
)
]
[(s
)
"
(s
¿
)
]
nsile
if
it
is
positive
and
compressive
it
is
negative.
2c,
adding
the
results
algebraically,
the
resultant
normal
stress
x xmax
x max
x max
max neutral
max
[(s
0, since by definition no normal
stress acts onx xmax
the
Ax)max ! (s¿x)max]
aa
N
sith
line
is= the
tan
then
a = y>z,
x resulting
when
applying
this
equation.
When
is the [(s
case,
the
neutral
axisnormal
in
Fig.stress
6–32dacts
can
determined
by
applying
Eq.
6–17
son
0, since
by
definition
no
onbethe
neutral
z )this
oint
cross
section,
Fig.
6–32d,
is
then
(s¿x))maxA]" (s¿ )stress
x)max "[(s
have
[(s
)
"
(s
¿
]
(s
¿
)
6
x
max
x
max
]
[(s
)
!
(s
¿
)
]
x
max
s
= the normal stress
x
[(sxx)max
]
)max
x max
x max
x¿x)max
max ! (s
xz
a acts
Neutral
The angle
ofa(sthe
aAxis.
tion
of Orientation
the Neutral
Axis.
The
the
will
be
tensile
if of
it isno
positive
and
compressive
if itneutral
is negative.
sof
=maxthe
0, since
with
byangle
definition
normal
stress
on
the
A
xis.
We have
[(sx)ma
[(s
x)max " (s¿x)max]
N
x the coordinates
a
•in
The
flexure
formula
can
applied
only
when
bending
occurs
z
neutral
Fig.
6–32dby
can
be determined
by applying
Eq. N6–17
My ! Msin
u
y,¿ )z =
xis in Fig. 6–32d
canaxis
beaxis.
determined
applying
Eq.be6–17
[(s
We
have
x)max ! (s¿x)max]
I
M
[(s
)
"
(s
¿
)
[(s
)
"
(s
]
x max
x max]
y z
x max
x max
(d)
(e)
[(sneutral
)max
! (s¿x)maxThe
] Athe
Iabout
axes
that
represent
the
principal
inertia
for
(d)Orientation
of
the
Neutral
Axis.
angle a of the a A
s =y 0,
stress
acts
on axes
the
xof
having their orig
z by
0, since by with
definition
nosince
normal
stress
actsno
onnormal
the neutral
(f)
z Mdefinition
=
yIz
x
x
(6–19)
tanhave
a = Mcross
tan u
N
z
[(sx)max
! (s¿x)max][(s )
z These
y
zIy= section.
have
their originz at the centroid
and
neutral
axisaxes
in Fig.
6–32d
6–17
axis. We
ave
area and formin
M
[(sx)max "Eq.
(s¿x)max
]
IM
yIz can be determined
(d) by applying
x max " (s¿x)max]
y zy MM
Iyyz
a
a
Fig.
6–32
(cont.)
z
!
z
y =by
axis
ofdefinition
symmetry,
if
there
is
one,
s = 0,ansince
withalong
no
normal
stress
acts
on
the
neutral
(d)and
+
(6–17)
s = - are oriented
The x axisxAis dir
eksende
sıfır olacağından) z
MzIy [(sx)(Tarafsız
y
Iyaxis. to
[(sgerilme
max ! (s¿x)max]
x)max ! (s¿x)max]
IIzz u, then
Mysin
= M cos u and My = M
perpendicular
Weit.M
have
yIz
(d)
the y and z[(s
axes
x)ma
a
= My =z M sin u, then
z
y =
ince Mz = M cos uyand
z
of
minimum
and
•MIzIfIthe
is M
applied
about some arbitrary axis, then the
y the moment
zI
y direction
[(sx)max ! (s¿x)max]
be seen that unless Iz =
defining
the
Since
and
= Mu,cos
u(s
M
(d)
y Mzangle
(d) x
(sx)xmax
)max
y = M sin u, then MyIz
(s¿x)max
area
moment
must
be
resolved
intothe
components
along
I
z equal a, the angle defining
z each of the
y =
ment M, Fig. 6–32a,ywill
not
tan
u
z
=
¢
≤
,= M cos u and
I
(s
)
MzIisy determined byx max
stress at a point
z axes,
MM
sin
u, uthen
My, Mz = the
Since
and
M
cos
Muy ≤=z and
M sinthe
u, then
y Fig.
Iyprincipal
of the neutral
axis,
6–32d.
z= =M
(d) resultant int
y = ¢
tan
Iz
superposition
of the stress caused
by each of the moment(c) (sx)max
the
principal y a
Iy
y = ¢ tan u ≤ z
(sx)max
components.
Since
and
then
M
=
M
cos
u
M
=
M
sin
u,
s6–18)
= the normal stress at the
point
I
along
the +y an
z
y
y
Iz
s section.
section. Since
Since the
Iz
Fig. 6–32
z tan u ≤ z
they = (6–18)
¢
Or,
stated
anoth
portant
Points
(s
)
u ≤zz axes
y = from
x max
z = the coordinates of the
x, y,
¢ tan
(sx)max
Iy point measured
Iy
z
ation
defines
the
neutral
axis
for
the
cross
section.
Since
the
6
u
where
is
measu
having their origin at the centroid
(sx)max of the cross-sectional 6
Iz(s¿x)max
x)maxfor the cross section. Since the
(6–18)
his
defines
the
neutral (s
axis
y = ¢ tan uz ≤ z
his equation
line
tan
a = y>z,
+y axis (sx)max
areaisand
forming
athen
right-handed
coordinate system
exure
formula
can
beaThis
applied
only when
bending
occursaxis for the Icross
6
y
z
(sx)max Since the
equation
defines
the
neutral
section.
ope
of
this
line
is
tan
then
=
y>z,
(e) from the cross-section and
The x axis is directed outward
6Iy, Iz = the principal mo
(6–18)
(e)
axes that represent
the
principal
axes
for then
the
(sx)max
(f)
slope
offor
this
line
isof
taninertia
a =principal
y>z,
(6–19)
the
y and
z
axes
represent
respectively
the
axes
z
tion
defines
the
neutral
axis
the
cross
section.
Since
the
axes, respectively
(6–19)
6
z
Thisaxes
equation
defines
the at
neutral
axis for and
the cross section. Since(e)the
section. These
have their
the centroid
(sx)max
Iz origin
!
of minimum
and
maximum
moment
of
inertia
for
the
his
line
is
tan
then
a
=
y>z,
Fig.
6–32
(cont.)
(6–19)
tan
=symmetry,
riented along
an ofaxis
if
there
is
one,
and
slope
thisaof
line
is tan ua!
then
=
y>z,
(e)
y
6
(s )max
Iy a = IzThis
area
6
y uequation defines the neutral
z
the cross section.(sSince
(6–19) axis forx !
tan
tan
x)max the
ndicular to it.
(e)
I
z
I
y
slope
of
this
line
is
tan
then
a
=
y>z,
= thethe
resultant
internal
components
tan athe
=along
tan u
y ! (6–19)
zfining
direction
moment
is applied
aboutmoment
arbitrary
axis,directed
then
(e)
Izsome
(sx)max
(s¿x)max if directedIy
fining
the
direction
the
principal
y
and
z
axes.
They
are
positive
y
(e)
I
!
z
(s¿x)each
(6–19)
a = components
tan u
angle
defining
thetan into
max
nt
must
be resolved
along
of uthe
(6–19)
tan
a
=
tan
angle
defining
the
I
y
they
are
negative.
an bealong
seen the
that+y
unless
I+z
angle u,
direction
Iyy the
z =axes,
Iy the
pal
axes,
and
theand
stress
at
aotherwise
point
is defining
determined
by
(s¿x)max
(e)
Iz ! y
!
M
M
=
M
sin
u
=
M
cos
u,
Or,
stated
another
way,
and
oment
M,
Fig.
6–32a,
will
not
equal
a,
the
angle
defining
the
Here
it
can
be
seen
that
unless
I
the
angle
u,
defining
the
direction
=
I
y
z
z
y
position of the stress caused by each of the moment
(s¿x)max (6–19)
tan a =
tan u
y
u is M,
+z
measured
positive
thethat
axis
toward
Here
it will
can from
be
Iz = defining
the direction
Ithe
on
ofwhere
the
neutral
axis,
6–32d.
f the
moment
Fig.Fig.
6–32a,
notseen
equal
a, unless
the angle
the u,Idefining
y the angle
y
onents.
(sx¿ )max
!
axis
moment
M,defining
Fig. 6–32a,
will not equal a, the angle defining the
nclination
of
the
neutral
6–32d.
n be+y
seen
that
unless
Izofaxis,
the
angle u,
the direction
= the
Iy Fig.
y
(s¿x)max
Here
it
can
be
seen
that
unless
I
the
angle
u,
defining
the
direction
=
I
zangle
y defining
inclination
of calculated
the
neutral
axis,
Fig.
oment
Fig. 6–32a,
will
not
equal
a, the
(s¿x)max
principal
moments
of inertia
about
the 6–32d.
y the
and z
(s¿x)max
z = theM,
of the
moment
M, (s
Fig.
mportant
Points
¿x)max6–32a, will not equal a, the angle defining the
n of axes,
the neutral
axis,
Fig.
6–32d.
respectively.
See
Appendix
A it can be seen that unless Iz = Iy the angle u, defining the direction
n bending
occurs
inclination
of
the
neutral
axis, Fig. 6–32d.
(s¿x)max
Important
Points Here
¿x)max a, the angle defining the
nofbending
occurs
of
the moment M, Fig. 6–32a, will not(sequal
inertia for the
(f)
Important
Points
(s¿x)max
of
inertia
for and
the can be applied only
of the neutral
flexure
formula
when bending
occurs axis, Fig. 6–32d.
(f) inclination
the
centroid
mportant
Points
(s¿x)max
6–32
(cont.)
he
centroid
and
ut
that and
represent
the
axes
of when
inertiabending
for the occurs
• axes
The
flexure
formula
canprincipal
be Fig.
applied
only
here
is one,
(f)
Important
Points
Fig.
6–32
(cont.)
(s
¿
)
here
is one,
andthat
s section.
These
axes
have
their
origin
at the
centroid
andfor the
• The
flexure
formula
can
be applied
only whenx max
bending(f)occurs
about
axes
represent
the
principal
axes
of inertia
Fig. 6–32 (cont.) (sx¿ )max
Important
Points
oriented
along
an
axis
of
symmetry,
if
there
is
one,
and
flexure
formula
can
be
applied
only
when
bending
occurs
about
axes
that
represent
the
principal
axes
of
inertia for the
crossthen
section.
(f)
ry axis,
the These axes have their origin at the centroid and
Fig. 6–32 (cont.)
•
The
flexure
formula
can
be
applied
only
when
bending
occurs
t
axes
that
represent
the
principal
axes
of
inertia
for
the
pendicular
to
it.
(s¿x)max
cross
section.
These
axes
have
their
origin
at
the
ry
axis,
then
the
oriented
(f) centroid and
longare
each
of the along an axis of symmetry, if there is one, and
Fig. 6–32 (cont.)
about
axes
that
represent
the
principal
axes
of
inertia
for
the
she
section.
These
axes
have
their
origin
at
the
centroid
and
(f)
are
oriented
along
an
axis
of
symmetry,
if
there
is
one,
and
ong
each
of
the
perpendicular
to it. about some arbitrary
moment
is by
applied
axis, then
the can be applied
determined
• The
flexure
formula
only
6–32 when
(cont.)bending occurs
cross
section.
These axes
have
their
at the centroidFig.
and
oriented
along
an
axis
of perpendicular
symmetry,
if there
one,origin
to
it. is
determined
ment
must
be by
resolved
into
components
along
each
ofand
the
of
the
about
axes
that
represent
for6–32
the(cont.)
•
If
themoment
moment
is applied
about some
arbitrary
axis,
then the the principal axes of inertiaFig.
(f)
are oriented
along
an axis
of symmetry,byif there is one, and
endicular
to it.
of moment
theaxes,
moment
cipal
and
stress
a moment
point
isisdetermined
• If atthe
applied
some
arbitrary
then
thecentroid and
cross
section.
These
axes
their axis,
origin
at the
must the
be resolved
into
components
along about
each of
thehave
perpendicular
to it. arbitrary
Fig. 6–32 (cont.)
e moment
is axes,
applied
some
axis,
the components
erposition
the
stress
caused
each
the
moment
along
an by
axis of symmetry,
moment
must
beoforiented
resolved
into
along eachifofthere
the is one, and
principalof
andabout
the
stressby
at
a are
point
is then
determined
mentsuperposition
must be •
resolved
into
components
along
each
the
If of
thethe
moment
iscaused
applied
some
arbitrary
the
mponents.
perpendicular
to
it.moment
principal
axes,
and
the
stress
at a axis,
pointthen
is determined
by
stress
byabout
each
ofof
the
cipalcomponents.
axes, and moment
the stress
at be
a point
isofdetermined
by
must
resolved
into
components
along
each
ofofthe
superposition
the
stress
caused
by
each
the
moment
• If ofthethe
moment is applied about some arbitrary axis, then the
rposition of theprincipal
stress caused
by each
axes,
and
the stress
atmoment
a point is determined by
components.
moment must be resolved into components along each of the
ponents.
superposition of the stress
caused
by and
eachthe
of stress
the moment
principal axes,
at a point is determined by
components.
superposition of the stress caused by each of the moment
components.
C H A P T E R 6 306
BENDING CHAPTER 6 BENDING
306
CHAPTER 6 BENDING
MPLE 6.15 EXAMPLE 6.15
6.5eğilme
UNSYMMETRIC
BENDING
307
!
Şekilde verilen dikdörtgen enkesite 12 kNm
momenti
EXAMPLE
6.15
gösterildiği gibiThe
tesir
etmektedir.
Kesitin
meydana
gelecek
normal
rectangular
cross
section
shownköşesinde
in Fig.
is subjected
a 6–33agerilmeyi
The herbir
rectangular
cross6–33a
section
shown
in to
Fig.
is subjected to a
The
rectangular
cross
section
shown
in
Fig.
6–33a
is subjected to a
#
#
ve tarafsız eksenin
yönelimini
belirleyiniz.
bending moment of M bending
normal
stress
m. Determine
= 12 kNmoment
of M =the
Determine
the
normal stress
m.
12
kN
#
bending
of and
Determine
the normal stress
M =
12 kNthem.orientation
4.95 and
MPa
developed at each corner
of moment
the section,
specify
ofspecify
developed
at each
corner
of the section,
the orientation of
developed
at
each
corner
of
the
section,
and
specify
the orientation of
x
the neutral
axis.
the neutral axis.
the neutral axis.
A
E
2.25 MPa
SOLUTION
SOLUTION
2.25 MPa
D
E
0.2 m
SOLUTION
B it is seen that
Internal Moment Components.
By inspection
the y
Internal Moment
Components.
By inspection
it is seen that the y
Moment Components.
By inspection
it is seen that the y
0.2 m
D 5Internal
and z axes represent
the principal
of inertia
since
they
are
axes
of
and z axesaxes
represent
the
principal
axes
of
inertia
since
they are axes of
N
and
zsection.
axes represent
thezprincipal
axes
of4.95
inertia
since
they are axes of
3 cross
4is subjected
The rectangular cross sectionB shown
in Fig.
to required
a the
MPa
symmetry
for6–33a
the
As
we
have
established
the
z
symmetry
for
cross
section.
As
required
we
have
established
C
M ! 12 kN"m
symmetry
forstress
the
cross section.
As required
weThe
have established the z the z
# m.asDetermine
bending moment of M = 12 kN
the
normal
axis
the principal
axis
for
maximum
moment
of inertia.
axis
as
the
principal
axis
for
maximum
moment
of inertia.
as
for
maximum moment
of inertia.
The The
developed at each corner
and
specify axis
the
orientation
0.2 m axis
moment
is resolved
into
its the
y andprincipal
zof
components,
where
0.1 m of the section,
C
moment
is
resolved
into
its
y
and
z
components,
where
moment is resolved into its y and z components, where
the neutral axis.
z
0.1 m
4
y
4 #m
My = - 112 kN # m2 = M
- 9.60
kN
4 112 kN # m2 = -9.60 kN # m
5
My = -y =
112 kN
5 # m2 = -9.60 kN # m
SOLUTION
5
3
3# m
112 kN # m2 = 7.20
3 kN
Internal Moment Components. By inspectionM
it zis=seen
= kN
112
7.20
# m2kN=# m2
# mkN # m
5 that the yMz = Mz112
7.20=kN
5
and z axes represent the principal axes(a)
of inertia since they are axes of
5
(b)
symmetry for the cross section. As
required
we have established
the
z of inertia about
Section
Properties.
The
moments
the y andofz inertia
axes
Section
Properties.
The moments
the yz axes
and z axes
Fig. 6–33
Properties.
The moments
of inertia about about
the y and
axis as the principal
axis for are
maximum momentSection
of are
inertia.
The
Çözüm:
are
moment is resolved into its y and z components, where
1
6
-3
1
6
Iy =
10.4 m210.2 m23 I1= 0.2667110
2 m34 m23 = 0.2667110
-3
4
=
10.4
m210.2
4
y
12
Iyof= the10.4
m210.2
m2 = 0.2667110-32 m4M2!m12 kN"m
#
#
Orientation
of
Neutral
Axis.
The
location
z
neutral
axis
12
M
=
112
kN
m2
=
9.60
kN
m
y
12
5 6–33b, can be established
1 by proportion.3Along
A
(NA), Fig.
-3
1the edge
I
=
10.2
m210.4
m2
2 m43 m23 = 1.067110
-3
4
1= 1.067110
z
3
I
=
10.2
m210.4
BC,=we 112
require
Iz = z 10.212m210.4 m2 = 1.067110-32 m54 42 m
M
kN # m2 = 7.20 kN # m 12
z
12
3
5
2.25 MPa
4.95 MPa
D
E
Bending Stress. =
Thus,Bending Stress. Thus,
10.2y m
- zz2
Section Properties. The moments of inertiaz about
the
and
axes Thus,
Bending
Stress.
u ! #53.1$
are
M0.450
Mzy
yz
- 2.25z =M4.95z
M
z
z
y
y
zMyz
s = +
My
+ m
a ! #79.4$
Iz
Iy s = -s =zz- =+ 0.0625
1
3
Iy
Iz -3I2 zm
Iy4
Iy =
10.4 m210.2 m2
=# 0.2667110
a
3
3
12
7.20110
2
N
m10.2
m2
9.60110
2# ND# m1
- 0.1neutral
m2 3 axis 3
3
In the same manner this is also the 7.20110
distance
from
to
the
#
3
2
N
m10.2
m2
-9.60110
2
N
m1-0.1
m2
s1B = + 2 N # m10.2 m2-3 -9.60110
= 22.25
MPa B m2Ans.C
N # m1-0.1
4 7.20110
4 +
in
Fig.
6–33b.
s
=
2.25 MPa
3s -32
-3
4 0.2667110 +
B
1.067110
m
2
m
Ans. Ans.
=
=
2.25=MPa
-3
4
Iz =
10.2 m210.4 m2 =
B 1.067110 2 m 1.067110
4-32 m4
-3
0.2667110
1.067110-3of2 m
0.2667110
2 m4 2 m
12 can also establish
We
the orientation
the
NA
using Eq.
6–19,
3
3
#
#
7.20110 2 N m10.2 m2
- 9.60110
2# N m10.1 m2
3
3
3that #the
7.20110
2N
m10.2
m2
2 N # m10.1
m2
which
the angle+a
axis
makes
with -9.60110
the
N Ans.
= 3-2z4.95
MPa
sC =is-used to specify
-9.60110
2 N m10.2
m2
Nor# m10.1
m2
-3 s4 7.20110
-3
4
=
+
= -4.95
1.067110
2
m
0.2667110
2
m
Bending Stress. maximum
Thus,
C
y -4.95
Ans. Ans.
=
+
MPa MPa
s
principal axis.
our-3sign4-3
convention,
u
must
be
4
-3
4=
C According to 1.067110
-3
4
2m
0.2667110
2m
1.067110
2
m
0.2667110
2
m
3
3
#
#
measured 7.20110
from the
axis
toward
+
y axis.
comparison,
in 3
2 N+ zm1
- 0.2
m2 the
- 9.60110
2 NBym10.1
m2
3
(c)
# m1-0.2
Myz
7.20110
2+ N
m2
-9.60110
2 N # m10.1
sD6–33c,
= - u = - tan-1 43-3= -453.1°
+32uN=# m1-0.2
= 3-22.25
MPa
7.20110
m2
-9.60110
N # m10.1
m2Ans.m2
Thus,
Fig.
(or
306.9°
).
-3
4
s
=
+
= -2.25
+
1.067110
2
m
0.2667110
2
m
D
Ans. Ans.
sD = +
= -2.25
MPa MPa
-3
4-32 m4
-3
4-32 m4
Iy
1.067110
0.2667110
Iz
1.067110
2 UmNSYMMETRIC
0.2667110
2m
6.5
B
ENDING
307
3
3
#
#
2 Na m1
0.2
2 N m1 - 0.1 m2
7.20110
- 9.60110
= -m2
tanm2
u 7.20110
3
3
3
2 N # m1-0.2
2 N # m1-0.1
-9.60110
1032 N # m10.2 m2 sE- 9.60110
2 N #tan
m1 - 0.1
Ans. m2
= +32 N # m1-0.2
=324.95
MPa
Iy 47.20110
m2-3 m2
N # m1-0.1
m2
-9.60110
-3
4 +
s
=
= 4.95 MPa
Ans.
+
=
2.25
MPa
1.067110
2
m
0.2667110
2
m
E
Ans. Ans.
-3
4+
-3
4 = 4.95 MPa
-3
4
-3 sE 4= -3
4
-3
4
-3
4
1.067110
0.2667110
.067110 2 m
0.2667110 2 m 1.067110
2m
1.067110
2m 2m
0.2667110
2m 2m
aresultant
=m2
tan1distribution
- 53.1°2
#tan
MPa
-3
normal-stress
has been sketched using
1032 N # m10.2 m2
- 9.6011032 N The
m10.1
0.2667110
2 m4The4.95
resultant
normal-stress
distribution
been sketched
Ans.
+
=
4.95
MPa
The
resultant
normal-stress
distribution
has been
using using
x
these
Fig.
6–33b.
Since
superposition
applies,
the distribution
ishas sketched
-3
4
-3 values,
4
.067110 2 m
0.2667110 2 m
Fig. 6–33b.
Since
applies,
the distribution
is
A
a =
- 79.4° these these
Ans.superposition
values,values,
Fig. 6–33b.
Since
superposition
applies,
the distribution
is
linear as
shown.
3
#
linear
as
shown.
E
1032 N # m1 - 0.2 m2 This- 9.60110
2
N
m10.1
m2
linear
as
shown.
2.25
MPa
in Fig. 6–33c.
Using
value of z calculated
+ result is shown
= - 2.25
MPatheAns.
4 2.25 MPa
D one obtains
0.2 m -32 mE4
1.067110
0.2667110
2m
above, verify,
using -3
the
geometry
of the cross section, that
B
3
3
the
same answer.
10
m 2 N # m1 - 0.2 m2
D 5 - 9.60110 2 N # m1 - 0.1 m2
N = 4.95 MPa
Ans.
+
3
4
4.95 MPa
1.067110-32 m4
0.2667110-32 m4 z
C
M ! 12 kN"m
The
0.2 m has been sketched using
C resultant normal-stress distribution
these values, Fig. 6–33b. Since superposition applies, the distribution is
z
linear as shown.
(a)
(b)
Fig. 6–33
PAGE !87
of Neutral Axis. The location z of the neutral axis
M ! 12 kN"m
6
Fig. 6–33
Fig. 6–33
(a)
Orientation
of Neutral
The
location
z of
the neutral
Orientation
of Neutral
Axis. Axis.
The location
z of the
neutral
axis axis
Fig. 6–33
Fig. 6–33b,
be established
by proportion.
the edge
(NA), (NA),
Fig. 6–33b,
can becan
established
by proportion.
AlongAlong
the edge
we require
BC, weBC,
require
(b)
neutral axis
ng the edge
2.25 MPa
4.95 MPa
MPa
4.95
Orientation of Neutral2.25
Axis.
The
location
E
= MPaz of the neutral axis
=
z 10.2 m10.2
m - z2
Tarafsız eksenin
yönelimi:
z
- z2
(NA), Fig.
6–33b, can be established
by proportion.
Along the edge
0.450
- =
2.25z
= 4.95z
BC, we require
0.450 2.25z
4.95z
M ! 12 kN"m
z
=
0.0625
m
2.25 MPa z = 0.0625
4.95 MPa
m
E
A
=
In
the
same
manner
this
is
also
the
distance
from
D
to
the
neutral
axis
z
10.2
m
z2
5
6
4
In the
same manner this is also the distance from D to the neutral axis
E
neutral axis
g Eq. 6–19,
with the z or
n, u must be
mparison, in
a
B
A
A
5
ED
M ! 12 M
kN!
"m12 kN"m
4
3
D
5
4
3
M ! 12 kN"m
A
u ! #53.1$
u5! #53.1$
4
z
3
a ! #79.4$
D a ! #79.4$
a
a
B u ! #53.1$
C
B
C
in Fig. 6–33b.
in3 Fig. 6–33b.
0.450 - 2.25z = 4.95z
z
D
We can also establish the orientation of the NA using Eq. 6–19,
We can also establish the orientation
of
the
NA
using
Eq.
6–19,
a
!
#79.4$
= angle
0.0625am
which is used to specify zthe
that the axis makes with the z or
N
is used to specify the angle a that the axis makes with the z or
N a
u which
! #53.1$
y
maximum
principal
axis.
According
to
our
sign
convention,
u
must
be
Inmaximum
the same principal
manner this
isAccording
also the distance
D to the neutral
axisbe
y C
axis.
to our from
sign
convention,
ucomparison,
must
B
z
measured
from
the
+
z
axis
toward
the
+
y
axis.
By
in
(c)
inmeasured
Fig. 6–33b.
from the
+-ztan
axis
-1 4toward the + y axis. By comparison, in
a!
#79.4$
Fig. 6–33c,
u =-1
(c)
3 = - 53.1° (or u = + 306.9°). Thus,
4
C
N
y
(c)
E
Ans.
z calculated
one obtains
(b)
We6–33c,
can also
NA). using
Thus, Eq. 6–19,
Fig.
u = establish
- tan 3 the
= - orientation
53.1° (or u =of+the
306.9°
which is used to specify the angle Iaz that the axis makes with the z or
N
tanIza =
tan u
y
maximum principaltan
axis.
a According
=
tan uIy to our sign convention, u must be
measured from the +z axisIytoward
the
+y
-3 axis.
4 By comparison, in
1.067110 2 m
(c)
a = (or-3u2 =
tan1
- 53.1°2
Fig. 6–33c, u = -tan-1 43 =tan
-53.1°
). Thus,
1.067110
m4+306.9°
-3
4
0.2667110 tan1
2 m- 53.1°2
tan a =
Iz0.2667110-32 m4
a =u - 79.4°
Ans.
tan a =
tan
Iy- 79.4°
a
=
Ans.
This result is shown in Fig.
6–33c.
Using the value of z calculated
-3
m4of the
Thisabove,
result verify,
is shown
in 1.067110
Fig.geometry
6–33c.2 Using
the
value
of z calculated
using
the
cross
section,
that one obtains
tan a =
tan1-53.1°2
-3the cross
4
same
answer.
above,the
verify,
using
the geometry
of
section,
that
one
obtains z
0.2667110 2 m
E
D
the same answer.
a = -79.4°2.25
Ans.
D
This result is shown in Fig. 6–33c.
Using the value of z calculated
above, verify, using the geometry of the cross section, that one obtains 0.20
the same answer.
4.95
4.95
B
z
0.2-z
0.1-z
C
2.25
B
PAGE !88
C
z
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