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5 in. of the cross-sectional area of the
Determine the moment of inertia
T-beam
shown
Fig. A–7a
x¿ axis. area of the
Determine
theinmoment
of about
inertiathe
of centroidal
the cross-sectional
T-beam shown in Fig. A–7a about
the
centroidal
x¿ axis.
8 in.
2 in.
8 in.
(a)
1.5 in.
1.5
1.5 in.
in.
4.45 in.
4.45 in.
1.5 in.
C A–7
Fig.
A.2
MOMENT OF INERTIA
AN AREA
789
A.2 FOR
MOMENT
OF INERTIA FOR
AN AREA
789
x¿
x¿
EXAMPLE A.2
C
in.
! EXAMPLE 10A.2
Şekildeki
T kiriş
enkesit alanının x’ ağırlık merkezinden geçen eksenine
8.55
in.
10 in. 5 in.
SOLUTION
I momentini
8.55 in. area of the
göre
atalet
Determine the moment
of inertia
of the hesaplayınız.
cross-sectional
Determine
the moment
of inertia
of the cross-sectional area of the
The area is segmented into
as shown in Fig. A–7a, and
5 in.two rectangles
T-beam shown in Fig.
A–7a shown
about the centroidal
x¿ axis.
T-beam
A–7a
the centroidal
the distance
from theinx¿Fig.
axis
andabout
each centroidal
axisx¿is axis.
determined.
Using the table on8 in.
the inside2 front
cover,
the
moment
of
inertia of a
in.
8 in.
1
3
1.5 in. about its centroidal2axis
rectangle
is
Applying
the
parallelI
=
bh
.
in.
12
1.5 in.
4.45(a)
in.
in.
axis1.5
theorem,
Eq. A–5,1.5
toin.
each
rectangle and4.45
adding
the results, we
in.
(a)
x¿
have
x¿
Fig. A–7
C
C
A–7
I =
©(I x¿ + Ady2) 8.55Fig.
10 in.
10 in.
in.
8.55 in.
SOLUTION I
15 in.
5
in.
3
2
The
area is =segmented
into two
as shown
in in.
Fig.-A–7a,
SOLUTION
I c 12 in.2110
in.2 rectangles
+ 12 in.2110
in.218.55
5 in.2and
d
12
the
from the x¿into
axistwo
and
each centroidal
axis
determined.
Thedistance
area is segmented
rectangles
as shown
in is
Fig.
A–7a, and
Using
the table
on1the
the
inside
cover,
the moment
of determined.
inertia of a
the distance
from
axis front
and each
centroidal
axis is
x¿
2 in.
3
2
2 1in.the
+ its
c centroidal
18 in.213
in.2
in.
- parallel1.5 in.2
rectangle
axis
is+cover,
Applying
the
I18=in.213
bh3in.214.45
.moment
Using theabout
table
on
the
inside
front
of
inertia
of ad
12
12 (a)
1
axis
theorem,
Eq.
to each
rectangle
and3 adding the results, we
rectangle
about
itsA–5,
centroidal
axis
is I = (a)
12 bh . Applying the parallel4
Ans.
I
=
646
in
have
ÇözümEq.
1. A–5, to each rectangle and adding the results, we
axis theorem,
Fig. A–7
have
Fig. A–7
I = ©(I x¿ + Ady2)
SOLUTION
II
2
I = ©(I x¿ + Ady )
SOLUTION I
4.45 in.
1can be Iconsidered
The area
as in.2110
one large
rectangle
two
2
SOLUTION
12two
in.2110
in.23 + as
12
in.218.55
in. and
- less
5 in.2
d small
= c into
The area is segmented
rectangles
shown
in
Fig.
A–7a,
x¿
1shown
12
rectangles,
shaded in
Fig. A–7b.
We have as shown in Fig.
The
isand
segmented
twoaxis
rectangles
C
12 in.2110
in.23centroidal
+into
12 in.2110
in.218.55
in. - 5 in.22 d A–7a,13and
= cx¿area
the distance from the
axis
each
is determined.
in.
12
2the x¿ axis and each centroidal axis is determined.
1 x¿ front
Using the table on the
the
inside
I =distance
©(I
+from
Adcover,
10 in.
y ) 3 the moment of inertia of a
2
8.55 in.
+
c
18
in.213
in.2
+
18
in.213
in.214.45
in.
1.5
in.2
d
1
Using 12
the
tableison
inside
front cover,
the moment of inertia of a
6.5 in.
rectangle about its centroidal
the1parallelI =the12
bh3. Applying
1 axis
3
2
5 in.
3 - 1.5 in.2 d
1
+
c
18
in.213
in.2
+
18
in.213
in.214.45
in.
3and adding
rectangle
about
its centroidal
axis isthe
Applying
the2 dparallelI =results,
axis theorem, Eq. A–5,
rectangle
we- 6.5 in.2
12 bh .in.
= to
c 12each
418 in.2113 in.2 + 18 in.2113 in.218.55
Ans.
I
=
646
in
12
axis
theorem,
Eq.
A–5,
to
each
rectangle
and
adding
the
results, we
have
Ans.
I have
= 646 in4
3 in. 2 in. 3 in.
A
I = SOLUTION
©(I x¿ + AdyII2) 2 c 1 13 in.2110 in.23 + 13 in.2110 in.218.55 in. - 5 in.22 d
2
I =12©(I x¿ + Ady )
4.45
in.
The
can
as one large rectangle less two small
SOLUTION
II be
1 area
Çözüm
2.3considered
4+ 12 in.2110 in.218.55 in. - 5 in.22 d
in.2110
in.2
= rectangles,
c 12area
x¿
4.45 in.
(b)
I
=
646
in
Ans.2
shown
shaded
in
Fig.
We have
1
The
can
be
considered
asA–7b.
one large
rectangle less two small
12
C
d
= c 12 in.2110 in.23 + 12 in.2110 in.218.55 in. - 5 in.2
13 in.
x¿
rectangles,
shown shaded
12 2 in Fig. A–7b. We have
C
10 in.
1 I = ©(I x¿ + 3Ady )
13
in.
8.55
in.
2
2 in.213 in.214.45 in. - 1.5 in.2 d
+ c I18=in.213
in.2 + 18
6.5 in.
©(I
1 x¿ + +Adc y1) 18 3in.213 in.23 + 18 in.213 in.214.45 in. -2 1.5 in.22 d10 in. 5 in.
12
8.55 in.
6.5 in.
= c 18 in.2113
in.2 + 18 in.2113 in.218.55 in. - 6.5 in.2 d
5
in.
12
1
I = 646 in4 = c 12 18 in.2113 in.23 + 18 in.2113 in.218.55 in.Ans.
- 6.5 in.22 d
12 I 1= 646 in4
Ans.
3 in. 2 in. 3 in.
A
- 2 c 13 in.2110 in.23 + 13 in.2110 in.218.55 in. - 5 in.22 d
1
SOLUTION II
3 in. 2 in. 3 in.
12
A
3
2
- 2 c 13 in.2110 in.2 + 13 in.2110 in.218.55 in. - 5 in.2 d
4.45 in.
The area can be considered
SOLUTION
12
4 asII one large rectangle less two small
(b)
I = 646 in
Ans.
x¿
4.45 in.
rectangles, shown shaded
in Fig.
We have as one large rectangle less two small
The area
canA–7b.
be considered
C
(b)
I = 646 in4
Ans.
13
in.
x¿
shown shaded in Fig. A–7b. We have
I = ©(I x¿ + rectangles,
Ady2)
C
10 in.
8.55 in.
13
in.
6.5 in.
I = ©(I x¿ + Ady2)
10 in.
5 in.
1
8.55 in.
= c 18 in.2113 in.23 + 18 in.2113 in.218.55 in. - 6.5 in.22 d
6.5 in.
5 in.
12
1
= c 18 in.2113 in.23 + 18 in.2113 in.218.55 in. - 6.5 in.22 d
12
1
3 in. 2 in. 3 in.
A
- 2 c 13 in.2110 in.23 + 13 in.2110 in.218.55 in. - 5 in.22 d
12
1
3
in.
2
in.
3
in.
- 2c 13 in.2110 in.23 + 13 in.2110 in.218.55 in. - 5 in.22 d
4
12
(b)
I = 646 in
Ans.
I = 646 in4
Ans.
PAGE !90
(b)
A
400 mm
790
APPENDIX
100 mm
A P P E N D790
IX A
790
SOLUTION
The cross section can be considered
as Athree
composite
rectangular
790
PPENDIX A
GEOMETRIC PROPERTIES OF A
areas A, B, and D shown in Fig. A–8b. For the calculation, the centroid
x of each of these rectangles is located in the figure. From the table on
A G E O M E T R I C the
P R Oinside
P E R T I Efront
S OF A
N A R Ethe
A moment of inertia of a rectangle about its
cover,
EXAMPLE A.3
400 mm
1
centroidal axis is I = 12
bh3. Hence, using the parallel-axis theorem
for rectangles A and D, the calculations are as
y follows:
Determine the
100 mm
shown in Fig. A
! EXAMPLE A.3 Şekildeki
kiriş enkesitinin ağırlık
100 mm
Rectangle
A
:
600
mm
merkezinden geçen
eksenlerine
E O MxE Tve
R I Cy P
R O P E R T I E SDetermine
Ogöre
F A N atalet
A R Ethe
A
moments of inertia of the beam’s cross-sectional area
G E O MAEPTPREI CN DPI RX OAP E RyG
TIES OF AN AREA
A P P E N D I X A hesaplayınız.
G E O M E T R I C P R O P E R T I E S O F A N A R1E A
momentlerini
shown
in
Fig.
A–8a
about the x and
y centroidal axes.
2
3
2SOLUTION
100 mm (a)
1100 mm21300 mm2
400 mm+ 1100 mm21300 mm21200 mm2 The cross secti
12
EXAMPLE
A.3
areas A, B, and
9
E A.3
Çözüm:
= 1.425110
2 mm4
SOLUTION
EXAMPLE
A.3
x of each of thes
400 mm
1section
The
can beof
considered
ascross-sectional
three composite
rectangular
2moments
3
y
thecross
ofmm21100
inertia
the beam’s
area
IDetermine
+ Ad
mm2
+area
1100
mm21300 mm21250
mm22the inside fron
y
Determine
the moments
beam’s
cross-sectional
y = Iof
y¿ inertia
x of=the 1300
y
400 area
mmthe centroid
y
Determine
the
moments
of
inertia
of
the
beam’s
cross-sectional
areas
A,
B,
and
D
shown
in
Fig.
A–8b.
For
the
calculation,
12
shownthe
in Fig.
A–8a
about theaxes.
x and y centroidal
axes.
100 mm
centroidal axis
100 mm
shown in Fig. A–8a about
xofand
y centroidal
xin
9each
4about
Fig.
A–8a
xarea
and y centroidal
axes.
thesethe
rectangles
is located in
the figure. From the table on
mmmoments of inertiashown
Determine
the
of =the
beam’s
1.90110
2 cross-sectional
mmof
100 mm 100
for rectangles A
theaxes.
inside front cover, the moment of inertia of a rectangle about its
shown in Fig. A–8a about the x400
and
y centroidal
mm
100 mm
1
SOLUTION
100 mm 200 mm
centroidal axis is I = 12
theorem
bh3. Hence, using the parallel-axis
A SOLUTION
SOLUTION
400 mm
Rectangle
B
:
Rectangle A:
The
cross
section
can
be
considered
as
three
composite
rectangular
250
mm
600
mm
for
rectangles
A
and
D,
the
calculations
are
as
follows:
300
mm
The
cross
section
can
be
considered
as
three
composite
rectangular
400 mm
The
cross
section
can beinconsidered
three
composite
rectangular
1 Fig. A–8b. as
SOLUTION
areas
A,
B,
and
D
shown
For
the
calculation,
the
centroid
3
9
4
100 mm in Fig. A–8b. For
Ix the
= incalculation,
1600
mm21100
mm2
= (a)
0.05110 2centroid
mm
areas A, B, and D shown
the the
centroid
1
B,these
and
Drectangles
shown
A–8b.
For
calculation,
x areas
The cross section can be considered
as A,
three
composite
rectangular
of each
of
is located
in the
figure.
From the
the table on
12Fig.
Ix = I x¿ + Ady2 =
1
x B of each
x
Rectangle
A
:
of thesexrectangles
is located
in the figure.
From in
thethe
table
on From the table on
12
600 mm
of
of
these
rectangles
is located
figure.
areas A, B, and D shown
in Fig. A–8b.
For
the
calculation,
the
theeach
inside
front
cover,
the centroid
moment
of inertia
of a rectangle
about its
400 mmfront cover,
the inside
the
moment
of inertia
aboutofits
1 1of
3 a rectangle
the
inside
front
cover,
the
moment
ofusing
inertia
3a rectangle 9about
4 its = 1.42511092 mm4
x100 mm
400each
mm of these rectangles
of
is located
in
figure.
From
on
centroidal
axis is
I I=y1the
the
parallel-axis
bhtable
. Hence,
=
mm21600
mm2
= 1.80110
2theorem
mm
1 the
400
3
(a) mmaxis
11100
3 1100
300ismm
centroidal
Icentroidal
=Ix12
the
parallel-axis
theorem
=bhI3x¿. Hence,
+axis
Adisy2 using
= = 12
mm21300
mm2
+ parallel-axis
1100
mm21300
mm21200 mm22
100 mm
250
mm
12
I
Hence,
using
the
theorem
bh
.
the inside front cover, the momentfor
of rectangles
inertia of aArectangle
about
its
1
and D,
calculations
are as follows:
1212the
mm
200 mm
D,using
the calculations
astheorem
follows:
1 D3 A and
Iy = I y¿ + Adx2 =
1
for
rectangles
A9 and are
D,
the
calculations are as follows:
y
centroidal axis for
is Irectangles
= 12
the
parallel-axis
bhmm
. Hence,
4
100
12
= 1.425110 2 mm
100
mm
100
mm
100
mm
for rectangles A and D, the calculations
are Rectangle
as A
follows:
9
100 mm
Rectangle
: 2 D1:
= 1.90110
2 mm4
600 mm Rectangle A:
3
2
Rectangle
A
:
I
=
I
+
Ad
=
1300
mm21100
mm2
+
1100
mm21300
mm21250
mm2
600 mm
mm
100
600 mm
y
y¿
x
1
(b)y
1 y2 = 121100 mm21300
Ix = I x¿2 + Ad
mm23 + 1100
mm21300 mm21200
mm22
3
2
(a)
200 mm
Rectangle
A
:
A
1
I
=
I
+
Ad
=
1100
mm21300
mm2
+
1100
mm21300
mm21200
mm2
12
9
4
1
x
x¿
y
3 2 mm
(a)
3
Rectangle B:
(a)100 mmAd 2 =
= 21.90110
1100
+ 1100
mm21300
mm22
12 1100
250 mm
Ix =
I x¿ mm21300
+ Ad
mm21300
mm2mm21200
+ 1100
mm21300
mm21200
mm22
x = I x¿ +
y
AIdikdörtgeni:
Fig. A–8
y =mm2
9
4
300 mm
12
=
1.425110
2
mm
12
1
9
4
3
= 1.425110
2 mm
Ix = I x¿ + Ady2 =
1100
mm21300
mm2
+ 1100
mm21300
mm21200 mm22
Ix =
9
4
A 200 mm
1 B:
= 12
1.425110
2 mm4 = 1.42511092 mm
2
3
Rectangle
1
x 2
B
I
=
I
+
Ad
=
1300
mm21100
mm2
+
1100
mm21300
mm21250
mm2
250
mm
y
y¿
x
2
3
2
300 mm
I1y = I y¿ + Adx2 = 1 1300
mm21100
+ 1100 mm21300
mm21250 mm22
12
y4
1 mm2mm21250
= 1.42511092I mm
2
3
23
3
1291300
+ 1100
mm2
= I y¿ mm21100
+ Adx =mm2
mm21100
mm2
+ 1100 mm21300
mm21250
y =yI y¿ + Adx = Iy 1300
=
1600
mm21100
mm2
=
0.0511092mm2
mm4
y
4 Ixmm21300
Iy =
12
=9 1.90110
1
12
412 2 mm
300
mm
100 mm
2
3
2
250
mm
= 1.901109 2 xmm
B mm21100
Iy = I y¿ +100
Ad
1300
mm21300
mm21250 mm2
9
4 mm2 + 1100
4
x =
mm
= 12
1.90110 2 mm = 1.90110 2 mm
200 mm D
1
The moments of inertia
for the entire cross
thus4
3 section are
9 200 mm
=
1100
mm21600
mm2
=
1.8011092 mm
I
100 mm
y
= 1.90110A
2 mm4
300
mm
Rectangle D:
250 mm
129
Rectangle B: I = 1.425110
9
9
mm250 mm
00 mm
B200
dikdörtgeni:
2
+
0.05110
2
+
1.425110
2
300 mm A
x
Rectangle
B
:
Rectangle
B
:
200
mm
D
250 mm
1
250 mm
1
300 mm
3
9(b)
4
2
9
4mm2 = 0.05110 2 mm
Ix = 1= 1600
mm21100
1100 mm
Ix = IAns.
1
x¿ + Ady =
9
4
Rectangle B:
2
mm
32.90110
9
43
12
Rectangle
D
:
I
=
1600
mm21100
mm2
=
0.05110
2
mm
Ix = x 1600 mm21100
12
x mm2 = 0.05110 2 mm
B
A
12
12
1
Fig. A–8
9
9
9
3
x
B
Iy9221=mm
1.90110
2 + 1.8011092 +3 1.90110
2
= 1.425110
2 mm4
14
Ixx =
1600
(b) mm21100 mm2 = 0.05110
2
3
9 mm21300
4
I
=
I
+
Ad
=
1100
mm21300
mm2
+
1100
mm21200
mm2
12
x
x¿
y
=
1100
mm21600
mm2
=
1.80110
2
mm
I
1 3 12
1
y
x
3
9
4
300 mm
9
1
9 4mm2
250 mm
12
2
mm21600
Iy =mm2
2 mm
= 1100
5.60110
1100 mm21600
= 1.80110
2 mm4 = 1.80110 2 mm
Iy = IAns.
1
300
mmIy =
300 mm
Fig.
A–8
y¿ + Adx =
250 mm200 mm 1
9
4
12
mm
12
D
=
12
3 1.425110 2
9 mm 4
=
1100
mm21600
mm2
=
1.80110
2
mm
I
200
mm
D
y
200 mm D
mm
9
12
100 mm
1
= 1.90110
2 mm4
Rectangle D: 2
3
2
Dmm
dikdörtgeni:
100 mm
100
Iy = I y¿ +DAd
Rectangle
: x = 12 1300 mm21100 mm2 + 1100 mm21300 mm21250 mm2
Rectangle D:
1
(b)
2
100 mm
The moments o
I1x = I x¿ + Ad
= 1 1100
mm21300
mm233 + 1100 mm21300 mm21200 mm222
9
4
Rectangle(b)D:
= y21.90110
(b)
2
3 2 mm
2
12
I
=
I
+
Ad
=
1100
mm21300
mm2
+ 1100 mm21300
mm21200 mm2
Ix = I x¿ + Ady = x 1100x¿ mm21300
mm2
+
1100
mm21300
mm21200
mm2
y
12
1
Ix = 1.42
12 = 1.425110
9
4
2Fig. A–8
= A–81100 mm21300
mm23 + 1100
mm21300
mm21200
mm22 for the entire cross section are thus
9 2 mm
4
x = I x¿ + Ady Fig.
The
moments
of
inertia
Fig. IA–8
9
4
=
1.425110
2
mm
= 12
1.425110 2 mm
= 2.90
1
3
9
9
I1y = I y¿ + Adx22 = 1 1300
mm2
1100 9mm21300
mm21250
mm222
Ix mm21100
= 1.425110
2 A+3 +
0.05110
2 + 1.425110
2
= 1.42511092 mm4
2
3
2
12 1300
= I y¿ mm21100
+ Adx =mm2
mm21100
mm2mm21250
+ 1100 mm21300
mm21250 mm2
Iy = 1.90
Iy = I y¿ + Adx = Iy 1300
+ 1100
mm21300
mm2
9
1
9
412
12 = 1.90110
3
2 4
= 2.90110mm2
2 mm
Ans.
2
mm
Iy = I y¿ + Adx2 =
1300 mm21100
mm2
+
1100
mm21300
mm21250
9
4
A
= 5.60
= 12
1.9011092 mm4 = 1.90110 2 mm
9
9
9
Iinertia
2 entire
+ 1.80110
2 section
+ 1.90110
2
y = 1.90110
9
4
The
moments
of
for
the
cross
are
thus
= 1.90110 2 mm
The moments
of inertia
for
the
Sonuç olarak:
9 entire
4 cross section are thus
The moments of inertia
for the entire
cross
section
are
2 mm
=9 5.60110
Ans.
9 thus
9
I
=
1.425110
2
+
0.05110
2
x
9 + 1.4251109 2
The moments of inertia for the entire
section
are 9thus
9 cross
9
9
I
=
1.425110
2
+
0.05110
2
+
1.425110
2
x
Ix = 1.425110 2 + 0.05110 2 + 1.425110
9
4 2
= 2.90110
Ans.
9
9 2 mm4
Ix = 1.42511092 + 0.05110929 + 1.425110
2
4
=
2.90110
2
mm
Ans.
A
=
2.90110
2
mm
Ans.
9
9
A
Iy = 1.90110992 + 1.80110
9 2 + 1.901109 2
= 2.9011092 mm4
Ans.
I
=
1.90110
2
+
1.80110
2
+
1.90110
2
9
9
9
y
Iy = 1.90110 2 + 1.80110 2 + 1.90110
2
=92 5.60110992 mm44
Ans.
Iy = 1.9011092 + 1.8011092 9+ 1.90110
Ans.
= 5.60110 2 mm4 = 5.60110 2 mm
Ans.
= 5.6011092 mm4
Ans.
GEOMETRIC PROPERTIES
OF AN
AREA
Ix = I x¿ + Ady =
PAGE !91
4
A.3
PRODUCT OFA.3
INERTIA
FOR AN OF
AREA
9REA
3
PRODUCT
INERTIA FOR AN7A
A.3
PRODUCT OF INERTIA FOR AN AREA
793
793
EXAMPLE A.4
EXAMPLE A.4
EXAMPLE EXAMPLE
A.4
A.4
A.3 Determine
PRODUCT OF Ithe
NERTIA FOR AN AREA
793
product of inertia of the
beam’s cro
Determine the
product of inertia of the beam’s cross-sectional
area,
EXAMPLE
A.4
Determine the
product
of
inertia
of
the
beam’s
cross-sectional
area,
shownarea,
in Fig. A–12a, about the x and y centroidal ax
Determine the product of inertia of the beam’s cross-sectional
shown in Fig. A–12a, about the x and y centroidal axes.
shown in Fig. A–12a,
about
the x and about
y centroidal
axes.
shown
in Fig.
the x and
y centroidal
axes.
y
Determine
theA–12a,
product
of inertia
of the
beam’s cross-sectional
area,
y
EXAMPLE
A.4
y
y
!
Şekilde
verilen
kiriş
enkesit
alanının
PRODUCT
NERTIA
FOR AN AREAaxes.
793 100 mm
yA.3 about
y x
shown in Fig. A–12a,
theOFx Iand
y centroidal
100 mm
100 mm
100Determine
ve
ymm
eksenlerine
göre
atalet
hesaplayınız.
100
mm
mm çarpım
the100
product
ofmomentini
the beam’s 100
cross-sectional
area,
y of inertia
mm y
y
y
100 mm
shown in Fig.100
A–12a,
about the x and y centroidal axes.
A 200 mm
mm
A 200 mm
100 mm
400 mm
A 200 mm
250
mm
200
mm
300
mm
A
300 mm
mm y
250
300 mm
250 mm
300 mm
area,
A 200 mm
100 mm
x
B
x
x
250 mm
B
300 mm
x
B
x
x
B
400 mm
400 mm
A 200 mm
100 mm
300
mm
250 mm
x
250
mm
250
mm
x
B
400 300
mmmm
300 mm
20
250 mm
250 mm200 mm D 300 mm
200 mm D
400 mm
100
mm
200
mm
D 300 mm
100 mm
100 mm
100 mm
x
250 100
mmmm
x
B
100 mm
A 200 mm
100 mm
100
mm 600 mm
200
mm
D
600 mm
300 mm 600 mm 250 mm
400 mm
100 mm
100 mm
600 mm
100 mm
300 mm
(a)
(b
250 mm
(a)
(b)
(a)
200
mm
(b)
D
600(a)
mm
x
x
B
(b)
Fig. A–12
Fig. A–12
100 mm
100 mm
Fig. A–12
(a)
400 mm
(b)
Fig.
A–12
SOLUTION
300 mm
SOLUTION
250
mm600 mm
SOLUTION
Asasinthree
Example A.3, the cross section can be co
200
mm
Fig.
A–12
As
in
Example
A.3,
the
cross
section
can
be
considered
D
SOLUTION
As in Example composite
A.3, the
sectionareas
can A,
beB,considered
as
threeThe coordinates
(a) cross
(b)
composite
rectangular
and
D,
Fig.
A–12b.
100 mm
As in Example 100
A.3,
mmthe cross section can be considered as three rectangular areas A, B, and D, Fig. A–12
SOLUTION
400 mm y
400 mm
400 mm
Çözüm:
oduct of inertia of the beam’s
100 mmcross-sectional
400 mm
2a, about the x and y centroidal
axes.
x
y
100 mm 400 mm
100 mm 400 mm
100 mm100 mm
composite rectangular
areas
A, B,
and D,
A–12b.
The are
coordinates
for theDue
centroid of each of these rectangles are shown
for in
theExample
centroid
of
each
of Fig.
these
rectangles
shown
inThe
thecoordinates
figure.
composite
rectangular
areas
A, B,
and D,can
Fig.be
A–12b.
A.3,
the
cross
section
considered
as three
Fig.
A–12
for the centroidAs
of
each
of
these
rectangles
are
shown
in
the
figure.
Due is zero
to
symmetry,
to
symmetry,
the
product
of
inertia
of
each
rectangle
about
for the centroid
of eachareas
of these
rectangles
are A–12b.
shown in
thecoordinates
figure. Duea the product of inertia of each rectang
composite
rectangular
A,
B,
and D,
Fig.
to symmetry,
the
product
of
inertia
of each
rectangle
is zero
about aThe
SOLUTION
set about
of
x¿,ay¿ axes that pass through the rectangle’s
(b)
set
of
axes
that
pass
through
the
rectangle’s
centroid.
Hence,
x¿,
y¿
to the
symmetry,kendi
the
product
of
inertia
of each
rectangle
is zero
bir dikdörtgenin
ağırlık
merkezinden
geçen
eksenlere
göre
çarpım atalet momenti
for
of each
of these
rectangles
are
shown
in the
figure.
Due
set ofHer
thatcentroid
pass
through
the
rectangle’s
centroid.
Hence,
x¿,
Asy¿inaxes
Example
A.3,
the
cross
section
can
be
considered
as
three
application
of the parallel-axis theorem to each of th
application
of
the
parallel-axis
theorem
to
each
of
the
rectangles
yields
set
of
axes
that
pass
through
the
rectangle’s
centroid.
Hence,
x¿,
y¿
to
symmetry,
the
product
of
inertia
of
each
rectangle
is
zero
about
a
simetriden
sıfırdır.
Fig. A–12
application
of the dolayı
parallel-axis
theorem
to
each
of the
rectangles
yields
composite
rectangular
areas
A,
B,
and
D,
Fig.
A–12b.
The
coordinates
application
the parallel-axis
theorem
each of thecentroid.
rectangles
yields
set
of x¿, y¿of
that pass through
thetorectangle’s
Hence,
Rectangle
A:
Rectangle
Aaxes
: of these
for the centroid
of each
rectangles are shown in the figure. Due
application
of
the
parallel-axis
theorem
to
each
of
the
rectangles
yields
Rectangle
A
:
to symmetry,
theIproduct
of+inertia
A dikdörgeni:
A.3, the cross section
can be
considered
asx¿y¿three
Ixy = I x¿y¿ + Adxdy
Rectangle
Adxdyof each rectangle is zero about a
xyA=: I
ID,
= Ix¿,
Ad
that
pass
through
the
rectangle’s
centroid.
Hence,
xy of
x¿y¿y¿+ axes
xd
ycoordinates
ular areas A, B, andset
Fig.
A–12b.
The
Rectangle
= 0 + 1300 mm21100 mm21- 250 mm21
IxyA:= I0x¿y¿
+ Admm21100
+ 1300
mm21 - 250 mm21200 mm2
xdy
application
the
parallel-axis
theorem
the rectangles yields
each of these rectangles
are
in mm21100
the
figure.mm21
Due
= 0shown
+ of
1300
9- 250 mm21200
4to each ofmm2
Ixy == I40Admm21100
dy
= - 1.5011092 mm4
1.50110
+ +1300
mm21-250 mm21200 mm2
product of inertia of each
zero x¿y¿
about
a x2 mm
= rectangle
- 1.501109is2 mm
9
= 0 +
1300 mm21100
Rectangle
A: centroid.
2 mm4 mm21-250 mm21200 mm2
that pass through the
rectangle’s
Hence,
Rectangle B:
Rectangle
B=: -1.50110
9
4
: Iof
parallel-axis Rectangle
theorem to Beach
the
yields
= Ad
-1.50110
2Imm
= Irectangles
+
dy
Ixy = I x¿y¿ + Adxdy
=
I
+
Ad
d
xy
x¿y¿
x
xy
x¿y¿
x
y
Rectangle
B=: I
B dikdörgeni:
I+xy1300
+ Adxmm21-250
dy
x¿y¿
=
0
mm21100
mm21200
mm2
= 0 + 0
+ 0+ Adxdy
Ixy = I0x¿y¿
Rectangle B:
= 0 + 0
= I00 + +0 Ad d
= 0
= -1.5011092 mm4 Ixy ==
x¿y¿
x y
= 0
¿ + Adxdy
Rectangle
D:
Rectangle D:
== 00 + 0
+ 1300 mm21100
mm21-250
mm2
Rectangle
B:
Rectangle
D: mm21200
=
0
IxyD=: I x¿y¿
Ixy = I x¿y¿ + Adxdy
Rectangle
Ixy += Ad
I x¿y¿
xdy+ Adxdy
.5011092 mm4
Ixy = I x¿y¿ + Adxdy
=
0
+
1300
mm21100
mm21250
mm21
200
mm2
= 0 + 1300 mm21100 mm21250 mm21Rectangle
: I x¿y¿ += Ad
D dikdörgeni:
IxyD=
0 +
xdy0
= 0 + 1300 mm21100 mm21250
mm21
- 200 mm2
9
4
=
0
2
mm
=I01.50110
= - 1.5011092 mm4
I9xy ==
+
Ad
d
mm21250 mm21-200 mm2
x y
Ixy = I x¿y¿ + Adx=
dy - 1.50110 2 mm4 x¿y¿+ 1300 mm21100
9 for the
== 0-1.50110
1300 mm21100
mm21250
mm2 The product of inertia for the entireAcross section
of+ inertia
crossmm21-200
section is thus
2 mm4 entire
RectangleThe
D: product
= 0The
+ 0product
of inertia for the entire cross
is thus
9 9 section
4 4
9
2for
mm
= Ad
=-1.50110
2 the
mmentire
] + 0 cross
+ [ -section
1.50110is2 thus
mm4]
Ixy = [-1.5011092 mm4] + 0 + [- 1.50110
IxyThe
= Iproduct
+
d1.50110
= 0
of[x-inertia
x¿y¿Ixy
y
9
4
Ixy = [ - 1.5011092 mm4] + 0 + 9[9- 1.50110
2
mm
]
44
=of[-1.50110
-inertia
3.00110
Ans. = -3.0011092 mm4
22the
mm
The
formm21250
section
is92thus
= product
0 +I 1300
mm21100
mm2
=
mmentire
] +mm21-200
0cross
+ [-1.50110
mm4]
= - 3.0011092xymm49
Ans.
Sonuç olarak:
Ixy == [-1.50110
2 mm4 9922 mm44] + 0 + [-1.5011092 mm4]
= -1.50110
-3.00110
Ans.
¿ + Adxdy
A
9
4
= -3.00110
Ans.
2 mmcross section is thus
The mm21-200
product of inertia
for the entire
+ 1300 mm21100 mm21250
mm2
Ixy = [-1.5011092 mm4] + 0 + [-1.5011092 mm4]
.5011092 mm4
m
-3.0011092 mm4
inertia for the entire cross section =is thus
-1.5011092 mm4] + 0 + [-1.5011092 mm4]
3.0011092 mm4
Ans.
Ans.
PAGE !92
A
C
796
GEOMETRIC PROPERTIES OF AN AREA
7 9 6 principal
APPENDIX A GEOMETRIC PROPERTIES OF AN AREA
moments of inertia for the beam’s crossP R O P E RDetermine
T I E S O F A N the
AREA
APPENDIX A
GEOMETRIC PROPERTIES
OF
APPENDIX A
LE
sectional area shown in Fig. A–15 with respect to an axis passing
A.5
through the centroid C.
EXAMPLE A.5
Determine the
y
A.5A RŞekilde
EA
x¿
sectional area
! EXAMPLE
gösterilen
kiriş
enkesitinin
C
100
mm
Determine the principal moments of inertia for the beam’s crossy
through the ce
geçen
x’ veiny’ Fig.
asal
eksenlerine
göre
y¿ axis passing
x¿ ağırlık merkezinden
sectional
area
shown
A–15
withthe
respect
to an
Determine
principal
moments
of inertia for the beam’s crossy
SOLUTION
mm Determine
theasal
principal
moments
of
inertia
for
the
beam’s
crossmomentlerini
x¿ hesaplayınız.
area
shown in Fig. A–15 with respect to an axis passing
through
centroid
The moments atalet
and
of the
inertia
of theC.cross sectional
section with
respect
5
100product
mm
E O MG
E TEROI C
I EPSE ROTFI EASNOA
I XGA
M EPTRRO
I CP EPRRTO
F RAENA
sectional
area shown in Fig. A–15 with respect to an axis passing
x
400 mm
SOLUTION
through the centroid C.
to the
yy¿axes have
up1 ! 57.1#
through
the x,
centroid
C. been determined in Examples A.3 and A.4. The
The moments
the principal
moments
of inertia
for the
beam’s
cross-crossthe
principal
moments
of
inertia
for
the
beam’s
resultsDetermine
are Determine
x
SOLUTION
up1 ! 57.1#
to the x, y axe
x¿
sectional
area
shown
in
Fig.
A–15
with
respect
to
an
axis
passing
x¿
sectional
area
shown
in
Fig.
A–15
with
respect
to
an
axis
passing
C
400 mm
The moments
and product ofSOLUTION
inertia of the cross section with respect
400 mm
57.1#
SOLUTION through
results are
100 mm u ! "32.9#
the
C.up !C.
9 through
4x centroid
9
4
9
4
the
centroid
p2 the cross section with respect
The
moments
inertia
Iofythe
= 5.60110
2 have
mm
I determined
= -3.00110
2and
mmproduct
x, y axes
been
in Examples
A.3ofand
A.4. of
The
m CIx = 2.90110 2 mm
The moments and productto
inertia
of the
cross
section
x xy with respect
400 mm
toand
the A.4.
x, y axes
and A.4. The
u p2 !
C are in Examples A.3
to"32.9#
the x, y axesÇözüm:
have beenresults
determined
The have been determined in Examples A.3
Ix = 2.9011092 m
100 mm
400 mm
results
are
100
mm the
SOLUTION
x¿
Using
Eq.
A–11,
angles
of
inclination
of
the
principal
axes
up ! "32.9#
SOLUTION
results
are
7.1#
up1 ! 57.1#
9 of inertia
9section
4 with respect
9
4 mm
600
moments
product
the
y¿ The
and
are
Ix and
=x 2.90110
2 mm4 of göre
Iof
5.60110
2 mmsection
Ixywith
= -3.00110
100 mm
Theönce
moments
product
inertia
ofcross
themomentleri
cross
respect 2 mm
y =
x
ve and
y eksenlerine
atalet
Using Eq. A
9
4
9
4
xDaha
x,
y
axes
have
been
determined
in
Examples
A.3
and
A.4.
The
I
=
2.90110
2
mm
I
=
5.60110
2
mm
Ixy = - 3.0011092 mm4
9 to the
4 to
9
4
9
4
100 determined
mm
the
axes have
been
in Examples
Ix 600
= 2.90110
2 mm
Iy x,= y5.60110
2 mm
Ixy = x- 3.00110
2 mm A.3 andyA.4. The
mm
Fig.
A–15
y¿ are
and
hesaplanmıştı:
400 mm400 mm
9
resultsresults
are are
Using
the angles
-Ixy
2 of inclination of the principal axes x¿
600
mm Eq. A–11,3.00110
9#
tanA–11,
2up =the angles
=
=
Using
Eq.-2.22
A–11, the angles of inclination of the principal axes x¿
Fig.
A–15Eq.
and y¿
9
x¿
Using
of are
inclination
principal
axes
1I
[2.901109of
29 -the
5.60110
y2>2A–15
9 x - I
4 Fig.
4 and 2]>2
9
4
y¿
are
9
4 y = 5.60110 2 mm9
4 xy = -3.00110 2 mm9
I
=
2.90110
2
mm
I
I
100
mmy¿
and
are
x
tan 2up =
Ix = 2.90110 2 mm
Iy = 5.60110 2 mm
Ixy = -3.00110 2 mm4
100 mm
9
3.00110 2
2up1 = 114.2°
and -I
2uxyp2 = -65.8°
tan 2u
= of inclination
= of the principal
= -2.22 3.0011092
9
- Ixy9x¿
p 3.00110
UsingUsing
the
angles
axes
9
-Eq.
Ixy A–11,
Eq. A–11,
the
angles
inclination
of2uthe
axes=x¿
1Ix - 2of
Iy2>2
[2.90110
2 =-principal
5.60110
2]>2
tan
= - 2.22
p
tan 2upand
=Asal
= konumu9 belirleyelim:
= - 2.22
y¿ are
1Ix - Iy2>2
eksenlerin
–15
[2.9011092 - 5.6011092]>2
1Iand
2>2A–15,
Thus, as shown
iny¿IFig.
[2.90110 2 - 5.6011092]>2
x yare
2up1 = 114.2°
and
2up2 = -65.8°
Thus, as shown
9
2up1 = 114.2°
and
2up2 = - 65.8°
2up1 = 114.2° -Iand
2u
=
65.8°
3.00110
2
9
xy -I
p2
3.00110
2
xy
u
=
57.1°
and
u
=
-32.9°
tan 2utan
=
= -2.22
pp1=2u =
9
9
= p2
Thus,
in[2.90110
Fig.
A–15,
1Ip x as
- 1I
Ishown
2 - 5.60110
2]>2 92]>2 = -2.22
9
y2>2
[2.90110
2
5.60110
x - Iy2>2
Thus, as shown in Fig. A–15,
Thus, as shown in Fig. A–15,
The principal moments
inertia
with and
respect to
the x¿-65.8°
and y¿ axes
2up1of=2u
114.2°
2u
up1 = 57.1°
up2 = -32.9°
and p2 =and
2up2 = -65.8°
p1 = 114.2°
are determined by using Eq. A–12.
The principa
up1 = 57.1°
and
up2 = - 32.9°
up1 = 57.1°
and
up2 = - 32.9°
are determined
Thus, Thus,
as shown
inThe
Fig.in
A–15,
principal
moments of inertia with respect to the x¿ and y¿ axes
as shown
Fig. A–15,
2
x¿The
y¿ axes moments of inertia with respect to the x¿ and y¿ axes
The
moments
with respect
to Eq.
the A–12.
andprincipal
Iy
Ix - Iy of
Ix +principal
areinertia
determined
by
using
2
max =
¢ using
≤momentleri:
; Asal
+ Ixy
are
determined
by using Eq. A–12.
Ix
Ix + Iy
by
Eq.
A–12.
atalet
u
=
57.1°
and
u
=
-32.9°
min are determined
p1
2
C
2
up1 = 57.1°
and p2 up2 = -32.9°
¢
;
Imax
=
min
2
C
I9x - Iy 2 9
Ix + Iy
9
22
2axes
292
x¿
y¿
principal
moments
of
inertia
with
respect
to
the
and
2 I+xThe
5.60110
2.90110
2
5.60110
2
¢
≤
;
=
+
I
Iy
-Imax
I
Ix + 2.90110
+
I
I
I
I
y
xy
x
y
x
y
minThe principal
with
axes
9 x¿
2 B moments
2
CEq. A–12.
2 ofIinertia
max =
R +respect
=
[-3.00110
2]2 ≤and+y¿Ixy
¢ determined
≤ + ;I2xy
¢to the
;
;
2.9011092 + 5.60
by
min
2 2 are determined
C using
2 A–12.
2
C are
by using Eq.
2
C
2
=
2
9
9
2.9011092 - 5.6011092 2
9
9 2.90110 2 + 5.60110 2
9
9
= 4.25110
2 ; 3.29110
2 2.90110292 - 5.60110;92 2 B2.90110
9
9
9 92]2
9
2
R
=
+
[-3.00110
2.90110
2 + I5.60110
2
2 + 5.60110
2
2.90110 2 - 5.60110 2
9
I - Iy 2
x + I
= 4.25110
2 ; 3.291
9 2
C
2 92]2
R= + [ - 3.00110
+ I;y ¢ Bx
Ix ≤- +
Iy I 2 2
Iyx ;
B
R + [ - 3.00110
;
2]
Imax
=2 max
2
xy
C
2
2
C
2
min
¢
≤
;
I
=
+
I
2
C
2
xy
or min
2 = 4.25110
C 92 2; 3.2911092
or
4.2511092 ; 3.2911092
= 4.25110992 ;2 3.2911092
9
2.9011092 + 5.60110
2 9 9 4 2.9011092 - 5.60110
92
9
9
94 2
Imax = 27.54110
2; mm
Imin
= 0.960110
2 Rmm+2 [-3.00110
Ans. 92]2 9 2
2.90110
+or5.60110
2B
2.90110
2 - 5.60110
=
Ima
B
R + [-3.00110
=
2]
2
C;
2
or
or
2
C
2
9
4
9
9 moment
= 47.54110
,
Specifically,
7.54110I9max
29 mm
Imin2 =mm
0.960110
2 mm44 Ans.
= 4.2511092the
; maximum
3.29110
2 4 9 Imaxof= inertia,
9
4
9
Specifically, th
9
I
=
7.54110
2
mm
I
=
0.960110
2
mm
Ans.
=
4.25110
2
;
3.29110
2
max
min
I
=
7.54110
2 mm
Imin = 0.96011092 mm4
Ans.
occurs with respect to the x¿ axis (major axis), since by inspection
maxmost
A
occurs with res
of the cross-sectional
area is farthest
from this
axis. To
show
this,
Specifically,
the away
maximum
moment
Imax = 7.5411092 mm4,
9of inertia,
4
or
of
9
2 mm
Specifically, the maximum
moment of inertia, Imax = 7.54110
Specifically,
the, maximum
moment of
inertia, Imax = 7.54110 2 mm4,the cross-sec
substitute theordata with
u = with
57.1°respect
into thetofirst
occurs
theof
x¿ Eq.
axisA–10.
(major axis),
since by inspection
most
substitute the d
occurs withArespect to the x¿ axis (major axis), since by inspection
most
with
x¿ To
axisshow
(major
9
4
9respect
4 to
the
cross-sectional
area
is occurs
farthest
away
from
thisthe
axis.
this,axis), since by inspection most
Imaxof
7.54110
2 mm
I
=
0.960110
2
mm
Ans.
of the cross-sectional area
is =
farthest
away
from
this
axis.
To
show
this,
9
9
4
of
theinto
area
isAns.
farthest away from this axis. To show this,
Imax = 7.54110
2 mmu4 min
Imin
=cross-sectional
0.960110
the
data
57.1°
the first2ofmm
Eq.
A–10.
substitute the data with u =substitute
57.1° into
the
firstwith
of Eq.=A–10.
substitute the data with
u
=
57.1°
into the first of Eq. A–10.
Specifically, the maximum moment of inertia, Imax = 7.5411092 mm94,
4
= 7.54110 2 mm ,
Specifically, the maximum moment of inertia, I
occurs with respect to the x¿ axis (major axis), since bymax
inspection most
occurs with respect to the x¿ axis (major axis), since by inspection most
of the cross-sectional area is farthest away from this axis. To show this,
of the cross-sectional area is farthest away from this axis. To show this,
substitute the data with u = 57.1° into the first of Eq. A–10.
substitute the data with u = 57.1° into the first of Eq. A–10.
1
2
PAGE !93
308
CHAPTER 6
BENDING
08
CHAPTER 6
BIENNGD I N G 3EXAMPLE
B6E N D
6.16
BENDING
G
ER
A.4
shown,
such
A.4 or
or A.5),
A.5), the
the principal
principal axes
axes yy and
and zz are
are oriented
oriented as
asthat
shown,
theysuch
represent the m
that they represent the minimum and maximum principal
principal moments
moments
of
inertia,
I = 0.960110-32 m44 and I = 7.54110-3
-32 m44, respectively.
of inertia, Iyy = 0.960110-3
respectively.y
zz
Determine
the normal s
Determine the normal stress at point P and the orientation
orientation of
of the
the
neutral
axis.
neutral axis.
SOLUTION
SOLUTION
The Z-section shown in Fig. 6–34a is subjected to the bending
moment
The Z-section
shown in Fig.6.16
6–34a is subjected to the bending moment
For use of
Eq. 6–19, it
EXAMPLE
#
For
use
of
Eq.
6–19,
it
is
important
that
the
z
axis
represent
the
!
Şekilde
gösterilen
Z
enkesite
M=20
kNm
eğilme
momenti
tesir
represent
the
of
Using
the
methods
of
Appendix
A
(see
Example
M
=
20
kN
m.
of M = 20 kN # m. Using the methods of Appendix A (see Example
principal
axis
for the max
principal
axisy for
maximum
moment
of
inertia.
(Note
most
of
that
most
of
A.4
A.5),
the
principal
axes
y and
zasal
are
oriented
as that
shown,
such
etmektedir.
veshown
zthe
asal
atalet
momentleri
The Z-section
shown
inprincipal
Fig. 6–34aaxes
is subjected
the
bending
moment
The or
Z-section
ineksenlerdir
Fig.
6–34a
isve
subjected
to the
bending
moment
A.4 or A.5),
the
yFig.
and
z to
are
oriented
as
shown,
such
the
area
is
located
furthes
The
Z-section
shown
in
Fig.
6–34a
is
subjected
to
the
bending
moment
The
Z-section
shown
in
6–34a
is
subjected
to
the
bending
moment
z
area
is represent
located
furthest
from this
axis.)
z¿ and
that
they
the Example
minimum
maximum
principal
of M that
Using
the
methods
ofand
Athe
(see
= 20they
kN #represent
m. of
#(see
z Appendix
#Using
z¿
of
Using
the methods
of Appendix
A (seemoments
Example
M
= 20
kN
m.
maximum
principal
moments
#
the
methods
ofExample
Appendix
A (see
M20=the
20minimum
kN
m. Using
of
the
methods
of
Appendix
A
Example
M
=
kN
m.
4
-3
100-3mm
TheA.4
Z-section
shown
in 100
Fig.mm
6–34a
subjected
the bending
moment
4 z aretooriented
-3
4 A.5),
or
theIprincipal
axes-3is2ythe
and
as
such
of
inertia,
and
respectively.
Iy the
= 0.960110
2axes
m
7.54110
2 m4as
,Internal
Iz z= are
Comp
A.4
or
principal
y and
oriented
shown, Moment
such
ofA.5),
inertia,
and Iaxes
=or0.960110
m
2shown,
, respectively.
or A.5),
principal
y and
zmare
oriented
as shown,
such
y A.4
z =axes
Moment
Components.
From
Fig. 6–34a,
A.4
A.5),
the principal
y7.54110
and
zInternal
are
oriented
as shown,
such
#
of that
Using
the
methods
of
Appendix
A
(see
Example
M =they
20 kN
m.
M
P
represent
the
minimum
and
maximum
principal
moments
z
Determine
the
normal
stress
at
point
P
and
the
orientation
of
the
32.9#
that
they
represent
the
minimum
and
maximum
principal
moments
Determine
the
normal
stress
atMpoint
P and
the
orientation
of
the moments
P
that
they
represent
minimum
and
maximum
principal
moments
zthe
that
they
represent
minimum
and
principal
32.9#the
-3
-3
4maximum
A.4oforinertia,
A.5), the
axes
zIzare
oriented
such
-3
4
-3
4
and
Iy axis.
=principal
0.960110
2 myI4 and
=-37.54110
mshown,
,I respectively.
-34
4 2as
4
neutral
axis.
-3Iy -3
42 0.960110
of
inertia,
and
=M
2# m
m
Iz == 7.54110
neutral
of inertia,
and
respectively.
=
0.960110
2
m
=
7.54110
m
,
=
20
mnoktasındaki
sin
57.1°
16.79
kN2# gerilmeyi
m , respectively. My = 20 k
of
inertia,
and
respectively.
I
=
0.960110
2
m
=
7.54110
2
m
,
I
y
z
400 mm kN P
y
olarak
verilmiştir.
normal
y
z
hat
they
represent
the
minimum
and
maximum
principal
moments
u
!
57.1#
mm stress at point
Determine the400
normal
P and
theatorientation
of the
the
u ! 57.1#
theorientation
normalofstress
at point P and theM orientation
of the M = 20 k
! 20 kN"m
Determine
the
normal
stress
point
P and
of
the
-3
4
Determine
normal
stress
at-3M2point
PDetermine
and
the
orientation
the
kN"m
!
#
#
z
of neutral
inertia, axis.
and
Iy = 0.960110
2 mthe
m420
, respectively.
Iz = 7.54110
ve
tarafsız
eksenin
yönelimini
belirleyiniz.
y¿
M
=
20
kN
m
cos
57.1°
=
10.86
kN
m
z
y¿
neutral
axis.
neutral
neutral
axis.axis.
SOLUTION
SOLUTION
Determine
the normal
stress
at point P and the orientation
of the
100 mm
100 mm
For use of Eq. 6–19,
it is important that the zFor
axisuse
represent
the it is important that the z axis represent
ofStress.
Eq. 6–19,
the
neutral axis.
Bending
Stress. The
point
P must
must
Bending
The y300and
z coordinates of point
P
SOLUTION
mmmoment of inertia. (Notebe
6
SOLUTION
principal
axis
for
the
maximum
moment
of
inertia.
(Note
that
most
of
principal
axis
for
the
maximum
that
most
of first. No
300
mm
SOLUTION
determined
6
SOLUTION
of P
P are
are
determined
first. Note that (a)the y¿, z¿Mycoordinates
of
For use
Eq.
6–19, it furthest
is important
thatM
the yz axisbe
represent
the
y axis 1-0.2
For
use
of
Eq.
6–19,
it
is
important
that
the
z
represent
the
yis important
(a)
the of
area
isFor
located
from
this
axis.)
For
use
of
Eq.
6–19,
it
that
the
z
axis
represent
the
m,
0.35
the
area
is
located
furthest
from
this
axis.)
use of Eq. 6–19,
it isz important 1that
the 0.35
z axis
represent
thecolored triangles from the construction
construction m2. Using
Using the
- 0.2
z¿
SOLUTION
principal axis for the maximum moment
of inertia. (Note
thatm,
most
ofm2.
principal
axis
for
the
maximum
moment
of
inertia.
(Note
that most
principal
axis
for
the
maximum
moment
of
inertia.
(Note
that
most
of
shown
in Fig.of6–34b, we h
principal
axis
for the
maximum
of inertia.
(Note
that
shown
in the
Fig.
6–34b,
wemost
haveof
Forthe
use
of isEq.
6–19,
it 100
is Components.
important
the z moment
axis6–34a,
represent
mm
area
located
furthest
from
this that
axis.)From
Internal
Moment
Fig.
the
area
is
located
furthest
from
this
axis.)
Internal
Moment
Components.
From
Fig.
6–34a,
the
area
is
located
furthest
from
this
axis.)
zfrom this axis.)
themaximum
area is located
z¿furthest
zaxis for the
z
principal
of inertia.
(Note that most of
yP = - 0.35 sin
P moment
m
32.9# Mz
Çözüm:
yP = - 0.35 sin 32.9° - 0.2 cos 32.9° = - 0.3580 m
100
mm
Internal
Moment
Components.
From
Fig.
6–34a,
#
#
6.5
U
NSYMMETRIC BENDING
he area is located furthest
from
this
axis.)
Internal
Moment
Components.
From
Fig.
6–34a,
M
=
20
kN
m
sin
57.1°
=
16.79
kN
m
Internal
Moment
Components.
From
6–34a,My = 20 kN # m sin 57.1° = 16.79 kN # m
y Moment
Internal
Components.
From
Fig. Fig.
6–34a,
zP = 0.35 cos
Mz
P
zP = 0.35 cos 32.9° - 0.2 sin 32.9° = 0.1852 m
400 mm
32.9#
z kN"m
Mz !M
20
M
u
!
57.1#
2.9#
#
#
M
=
20
kN
m
cos
57.1°
=
10.86
kN
m
#
#
M
!
20
kN"m
z
nternal
Components.
From
Fig.
6–34a,
=
20
kN
m
sin
57.1°
=
16.79
kN
m
y¿ MomentM
#
#
#
y
# mform
#zym== 2020and
M
kN the
mcos
sin 57.1°
57.1°algebraic
16.79 kN
kN m
mApplying Eq.
kN
m
== 10.86
#kN
M
=x,kN
20
siny¿57.1°
= 16.79
y 6–17,
y, z#kN
a=right-handed
system,
proper
My The
= yu20!
maxes
sin
57.1°
16.79
kN6–17,
mM
400 mm
Applying
Eq.
57.1#
20
u ! 57.1#
#
#
! kN"m
57.1#
M
!
20
kN"m
M
=
20
kN
m
cos
57.1°
=
10.86
kN
m
#
#
M
!
20
kN"m
signs
must
be
assigned
to
the
moment
components
and
the
coordinates
z
y¿
#
#
Mzm= 20 kN m cos 57.1° = 10.86 kN m
M !M
20 kN"m
#The
#kN
M
20
m #57.1°
cos
= 10.86
kN
= 20 kN100
m
sin
16.79
m y¿57.1°
mm
Bending
Stress.
y57.1°
coordinates
of= M
point
P #M
must
M z
Mz yP
M
= z20=zkN
m kN
cos
10.86
kN
m
z and
zPPthe
y¿ y y¿
yis
P Stress.
y
zz yP
Bending
Thecase,
y the
andresulting
z coordinates
P must+ y P
when
applying
this
equation.
When
this
stress ofsPpoint
=
s
=
+
be
determined
first.
Note
that
the
coordinates
of
P
are
y¿,
z¿
kN"m
P
#ymmm
#m
P
300
I are Iy
6 Mz = 20
kN100
coswill
57.1°
=
10.86ifkN
tensile
it isof
positive
and
compressive
ifNote
it isy negative.
be the
determined
that
y¿, z¿ coordinates
Iconstruction
Ifirst.
yBending
y
The
and
z be
coordinates
point
P
[(sx)max
zz must
Stress.
The
and the
z coordinates
of pointof PPz must
Using
the
triangles
from
1 - 0.2Stress.
m, 0.35
m2.
M
Bending
Stress.
The
z Bending
coordinates
of y point
P must
y y and
(a) colored
Bending
Stress.
The
y
and
z
coordinates
of
point
P
must
y
3
3
3
# m21300
mm
P
noktasının
y-z
eksenlerindeki
koordinatları:
z¿
#
#
3
3
Using
the
colored
triangles
from
the
construction
1-0.2
m,
0.35
m2.
be determined
first.
Note
that
the
coordinates
of
P
are
y¿,
z¿
110.86(10
)
N
z¿
z
[(s
)
"
(s
¿
)
]
116.79(10
110.86(10
m21
m210.1852
m2
z
z¿
be
determined
first.
Note
that
the
coordinates
of
P
are
y¿,
z¿
shown6in Fig.
6–34b,
we
have
x
max
x
max
116.79(10
110.86(10
)
N
m210.3580
m2
)
N
m210.1852
m2
z that
be determined
first.
Note
the
of
P0.200
arem +
y¿,
z¿ coordinates
Mthat
be
determined
first.
Note
the
coordinates
of
P
are
y¿,
z¿
a
Orientation
of
the
Neutral
Axis.
The
angle
of
the
=
y
(a)
=
M
0.200
m
0.350
m
y
Using
the
colored
triangles
from
the
construction
1
0.2
m,
0.35
m2.
Bending
The y 0.200
and mz coordinates
point
P m,
must
y
0.350
shown
in Fig.
6–34b,
we
have
My Stress.
y
0.350 m
mthe of
-3
4
-3 construction
colored triangles
from the
10.2
0.35
m2. Using
-3
4the
y
7.54110-32 m
Using
colored
triangles
from
construction
1-0.2
m, 0.35
m2.
7.54110
m44 N
Using
the
colored
triangles
from
the the
construction
1-0.2
m,
0.35
m2.the
22 m
neutral
in 32.9°
Fig.
6–32d
can
be
by
applying
Eq. 6–17 0.960110-322 m
32.9#
yP =
-32.9#
0.35
sin
32.9°
- axis
0.2
cos
= shown
-of0.3580
mdetermined
32.9#
in Fig. 6–34b,
we
have
be shown
determined
first.
Note
that
coordinates
P
are
y¿,
z¿
P
P
in
Fig.
6–34b,
we
have
P Fig.
shown
in
6–34b,
we
in
6–34b,
we
N stress
s have
= 0,have
with
since the
by definition
noyPnormal
acts on
A
= -0.35
sin 32.9°
- the
0.2 neutral
cos 32.9° = -0.3580
m MPa
N
= 3.76
= 3.76
3.76 MPa
MPa
Ans.
N
Using
theFig.
colored
triangles
construction
-0.2 m, 0.35 m2. shown
=
Ans.
x
z-P0.35
= 0.35
cos 32.9°
0.2
sinfrom
32.9°
=0.3580
0.1852
m
z
y
=
sin
32.9°
0.2
cos
32.9°
=
m
axis.
We
have
P
y
=
0.35
sin
32.9°
0.2
cos
32.9°
=
0.3580
m
[(sx)max
32.9#
32.9#
hown in Fig. 6–34b, we havey =yP-=0.35
-0.35
sin 32.9°
- cos
0.2 32.9°
cos 32.9°
=P -0.3580
32.9#
a
sin 32.9°
- 0.2
= z-0.3580
mcosm32.9°
0.35
- 0.2The
sin 32.9°
P
P =of
Orientation
of
Neutral
Axis.
The
angle =uu 0.1852
isOrientation
shown
in of Neut
= 57.1°
57.1°mis
Orientation
Neutral
Axis.
angle
shown
in
=
Applying
6–17,
z =Eq.
0.35
cos 32.9° - 0.2 sin 32.9° = 0.1852 mFig. 6–34a.zPThus,
[(sx)max !
¿x)max6–34a.
]
0.35 cos 32.9° - 0.2 sin 32.9°
= 0.1852
m(sFig.
Thus,
a ! 85.3#
I=
MThus,
!
= cos
0.35
cos
32.9°
- sin
0.2 32.9°
sin
32.9°
=
0.1852
yP =P -0.35 sin 32.9°
0.2
=- -0.3580
m
Fig.
6–34a.
y
zm m
! 85.3#
85.3#
zP =-zP0.35
cosaa32.9°
32.9°
0.2
= 0.1852
Applying
Eq. 6–17,
y¿
y¿
z
y
=
M
z
y¿
M
y
y P
PP noktasındaki
normal
gerilme:
-3
Applying
Eq.z 6–17,
MzI6–17,
7.54110-3
m44
y
32.9°
0.26–17,
sin 32.9° = 0.1852 m Applying Eq.
sP = z-P = 0.35
+cos
7.54110
22 m
Applying
(d)
Applying
Eq.-Eq.
6–17,
M
z
M
y
tan
a
=
tan 57.1°
57.1°
tan a =
B
y
P
Iz M
I
z
P
tan a = B
y
RR tan
-3
-3
44
Mz yP
y zP
0.960110
2
m
M
z
s
=
+
M
y
y P
P3
0.960110
2
m
z
P
M
z
M
y
3
y
P
Applying
Eq.
6–17,
M
z
z P- 0.3580
+
M
IzM sin
Ithen
y P M
Since
M cos usand
z y# Pm21
110.86(10
)N
m2z = 116.79(10
)=NM
m2
zP = -# m210.1852
+ u,
y
y =
P
A
A
sP =sIPy-= - I+ + (b)
z
Iz 4 3 Iy a
A
+
= Ia =
= 85.3°
85.3°
Ans.
a =
I
y
y
-3
4
-3
Ans.
z
y
Iz
Iy (b)z¿
My3 zP 7.54110
y
Mz yP
110.86(10
) N # m21-0.3580
m2 (b) 116.79(1033) N # m210.1852
m2
z3 0.960110
2
m
2
m
#
#
3
116.79(10
0.3580
m2
)
N
m210.1852
m2
#
#
z¿
= -z¿ 110.86(10
+ z ) N m213
3
116.79(10
110.86(10
)
N
m21
0.3580
m2
)
N
m210.1852
m2
z
#
#
+
=
0.200
3 m #) N 0.350
3 I )
m
The
neutral
axis
is oriented
oriented
as
shown
in Fig. 6–34b. -3 The
116.79(10
110.86(10
N m210.1852
-3 m2
4 m2
+ m21-0.3580
= - Iz= z3.76 MPa
110.86(10
)m
N
m21-0.3580
m2 m2116.79(10
) Nz#axis
m210.1852
Iy
The
is
as
shown
Fig.
6–34
Fig.
6–34+ in Fig. 6–34b.
- 4neutral
0.200
0.960110 -32 m44 neutral axis(sis) orie
Ans.
m 0.960110-3=
4
tan
u -3
z -3224m
= ¢ 7.54110
≤
m4 PFig. 6–34-3 0.350
+ +2 my
= - = -3-232.9#
x max
-34
43
7.54110
m
0.960110 2 m
0.3500.350
m m 3 7.54110
7.54110
Iy -32 m42 m
32.9# 7.54110
2 m 2 m ) N # m210.1852
0.960110
116.79(10
110.86(10 ) N N# m21-0.3580
m20.960110
P m2
=
3.76
MPa
Ans.
Orientation
Axis. The angle u == 57.1°
isAns.
shown in
+
= -= 3.76 MPa
N -3 of4 Neutral
(6–18)
3.76 MPa
Ans.
=2 m
3.76
0.960110-32 m4
= 3.76
MPaMPa
Ans.Ans.
Fig.7.54110
6–34a.
Thus,
32.9#
32.9#
Orientation
of
Neutral
Axis.
The
angle
is
shown
u
=
57.1°
Orientation
of Neutral Axis. The angle u = 57.1° isOrientation
shown in of Neutral Axis. The angle u = 57.1° is showninin
y¿ 32.9#
-3 Axis.
4Axis.
= 3.7632.9#
MPa
Ans.
Orientation
of Neutral
The
angle
is shown
= 57.1°
Orientation
of
Neutral
The
angle
is shown
in in
u = u57.1°
7.54110
2
m
Fig.
6–34a.
Thus,
a ! 85.3#
Fig. 6–34a. Thus,
Fig.
6–34a.
Thus,
z
a
!
85.3#
This
equation
defines
the
neutral
axis
for
the
cross
section. Since the
Fig.
6–34a.
Thus,
tan
a
=
tan
57.1°
B
R y¿
a ! 85.3#
Fig. 6–34a.
Thus,
y¿ a ! 85.3#
-3
y¿a
Orientation y¿
ofTarafsız
Neutral
Axis.
The
angle
is =
shown
in
57.1°
0.960110
m=4 is
-3
44
-3of
4 2u
y¿
slope
this
line
tan
then
y>z,
eksenin
yönelimi:
-3
7.54110
7.54110 2 m
7.54110 22m
m
-3 -342 m4
(sx)max
Fig. 6–34a.
Thus, tan a = B
tan
aa == BB
tan7.54110
57.1°
R tan 57.1°
R7.54110
2m
A
tan
-3
4
-3
44 R tan 57.1°
a
=
85.3°
Ans.
tan
a
=
tan
57.1°
-3
B
R
y
0.960110
2=mB
0.960110
2
m
tan
a
tan
57.1°
R
-3
4
0.960110
2
m
-3
0.960110
0.960110
2 m42 m
7.54110-32 m4
A
(e)
The neutral
axis
is
oriented
as
shown
in
Fig.
6–34b.
Iz
AA
tan
a
=
tan
57.1°
R
a B= 85.3° -3(b)
85.3°
Ans.
y
85.3°
Ans.
4 a = 85.3°
yy
(6–19)
tan aAns.
=
tan u aa ==Ans.
A yA
Ans.
0.960110 (b)
2am=
(b)
85.3°
y
Iy
The neutral axis is oriented as shown in Fig. 6–34b.
!
The
neutral
axis
The
neutral
axisisisoriented
orientedas
asshown
shownin
inFig.
Fig.6–34b.
6–34b.
Fig.6–34
6–34
Fig.
The
neutral
axis
is oriented
as shown
in Fig.
6–34b.
a =
85.3°
Ans.
The
neutral
axis
is oriented
as shown
in Fig.
6–34b.
6–34
y
6.16
"
The neutral axis is oriented as shown in Fig. 6–34b.
Here it can be seen that unless Iz = Iy the angle u, defining the direction
of the moment M, Fig. 6–32a, will not equal a, the angle defining the
inclination of the neutral axis, Fig. 6–32d.
(s¿x)max
Important Points
(s¿x)max
• The flexure formula can be applied only when bending occurs
•
about axes that represent the principal axes of inertia for the
cross section. These axes have their origin at the centroid and
are oriented along an axis of symmetry, if there is one, and
perpendicular to it.
!94 some arbitrary axis, then the
If the moment is appliedPAGE
about
moment must be resolved into components along each of the
principal axes, and the stress at a point is determined by
superposition of the stress caused by each of the moment
(f)
Fig. 6–32 (cont.)
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