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Problem 10.16
The strength of long bones, as well as slender structural members in general,
is directly related to the area moments of inertia, such that if the moments of
inertia increase, then strength increases. From the point of view of the ratio of
strength to weight, discuss why most long bones in humans and animals, such
as the human femur shown, are hollow.
Problem 10.17
(a) Determine Ix.
(b) Determine Iy.
0in:
0in:D3
(2)
Part (b)
To determine the moment of inertia
Iy
, we will carry out integration with respect
0in:
0in:D3
(4)
Statics 2e 1461
Problem 10.18
(a) Determine Ix.
(b) Determine Iy.
Problem 10.19
(a) Determine Ix.
(b) Determine Iy.
Problem 10.20
(a) Determine Ix.
(b) Determine Iy.
0mm
105.1 y/3=2.8 C12y C15y2/ˇˇˇ
0mm D32
105 mm4
)IxD0:305 mm4:(7)
Part (b)
Problem 10.21
(a) Determine Ix.
(b) Determine Iy.
Problem 10.22
(a) Determine Ix.
(b) Determine Iy.
Problem 10.23
For the triangle shown, having base
b
and height
h
, show that the area moment of
inertia about the xaxis is IxDbh3=12.
0
0)IxDbh3
Alternate determination of w:
We determine expressions for
y1
and
y2
so that they intersect at the point
.x; y/ D.c; h/
; this results in
y1Dh
and
y2D.h
. We solve
y1
for
x
, calling the result
xr
,
Problem 10.24
Determine constants
c1
and
c2
so that the curves intersect at
xDa
and
yDb
.
Express your answers in terms of
a
and
b
. Then determine the area moment of
inertia indicated.
Determine Ix.
To determine the moment of inertia
Ix
, we will carry out integration
Problem 10.25
Determine constants
c1
and
c2
so that the curves intersect at
xDa
and
yDb
.
Express your answers in terms of
a
and
b
. Then determine the area moment of
inertia indicated.
Determine Iy.
yDc2x3)bDc2a3)c2Db
a3)yDb
a3x3:(2)
To determine the moment of inertia
Iy
, we will carry out integration with respect
Problem 10.26
Use integration with an area element perpendicular to the
x
axis to determine the area moment of inertia
Ixfor the shape indicated. Figure P10.20 on p. 584.
ybD0)
2.ytCyb/D1
22x x2C0;(2)
hDytybD2x x20: (3)
The moment of inertia of the area element about the xaxis is
dIxD1
Problem 10.27
Use integration with an area element perpendicular to the
x
axis to determine the area moment of inertia
Ixfor the shape indicated. Figure P10.21 on p. 584.
2.ytCyb/D1
21Cx
2Cx2;(2)
hDytybD1Cx
2x2:(3)
The moment of inertia of the area element about the xaxis is
Problem 10.28
The beam cross sections shown are symmetric about the
x
and
y
axes. Determine
the area moments of inertia Ixand Iyand the radii of gyration kxand ky.
12.6 mm/.230 mm/3C21
12.140 mm/.10 mm/3C.120 mm/2.140 mm/.10 mm/
12.230 mm/.6 mm/3C21
12.10 mm/.140 mm/3)IyD4:58 106mm4:(2)
The area of the cross section is
AD.6 mm/.230 mm/C2.10 mm/.140 mm/D4180 mm2:(3)
Problem 10.29
The beam cross sections shown are symmetric about the
x
and
y
axes. Deter-
mine the area moments of inertia
Ix
and
Iy
and the radii of gyration
kx
and
ky.
12.2:4 in:/.1:2 in:/34"1
12.0:9 in:/.0:5 in:/3C0:1 in:C1
20:5 in:2
)IyD0:0876 in:4:(2)
The area of the cross section is
Problem 10.30
The cross-sectional dimensions for a
W10 22
wide-flange I beam are shown. By
idealizing the cross section to consist of rectangular shapes, determine the area
moment of inertia
Ix
, and compare this to the value of
118 in:4
reported in the AISC
Steel Construction Manual. Note: The value reported by AISC accounts for the
effects of small fillets where the web and flanges join.
Solution
Problem 10.31
Using Eq. (10.6) on p. 585, show that the parallel axis theorem for the radius of gyration about the
x
axis
is k2
xDk2
x0Cd2
x.
A)IxDk2
xA(2)
kx0DrIx0
x0A: (3)
Problem 10.32
For the T shape shown, determine the radius of gyration about the xaxis.
Solution
12.2 in:/.3 in:/3C24 in:0:5 in:3in:
22
The area of the shape is
Problem 10.33
For the channel shown, determine the radii of gyration about the
x
and
y
axes.
Solution
We use the composite shapes shown with the parallel axis theorem.
The area of the shape is
AD.1 in:/.1 in:/ .0:75 in:/.0:75 in:/ D0:4375 in:2:(3)
Problem 10.34
Let each of the beams shown in Fig. 10.2 on p. 574 be constructed of identical planks of wood with
2in:
by
6in:
cross-sectional dimensions. Determine the area moments of inertia for each beam’s cross-sectional
area about the xand yaxes.
Solution
Rectangular cross section:
Problem 10.35
The cross sections of two beams are constructed by arranging the
2cm
by
16 cm
strips of wood as shown. Determine the area moments of inertia for each beam
about the horizontal and vertical axes passing through the centroid of each area.
Solution
12.16 cm/.2 cm/3C1
12.2 cm/.16 cm/3)IyD693 cm4:(3)
Prior to determining
Ix
, we first determine the
y
location of the centroid using the
nt
12.2 cm/.16 cm/3C.12:5 cm 8cm/2.2 cm/.16 cm/
C1
12.16 cm/.2 cm/3C.5:5 cm 1cm/2.16 cm/.2 cm/
