k got the answer of 19.9 for the moment of inertia for the engle
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Hello kindly explain to me how did the book got the answer of 19.9 for the moment of inertia for the engle bars. Is it not 6.995 instead. Thank you
![(Chap. VIII
MOMENTS OF INERTIA
206
Taking the summations from the table and muk
stituting in Eq. (b), we obtain
I, = 454.91 + 776.8
I,=1232 in. Ana.
A girder is composed of four 6 X 6 xin
angles connected to a web plate 23.5 in. x1 in
in Fig. 8-13. The properties of the angle are
I.- 1,= 19.9 in., the area 5.75 in.2, and
2- - 1.68 in. Compute the moment of inertin
with respect to the centroidal X. axis.
Solution: The tabular computation used in Prob.
814 is well suited for cases in which there are many
elements. In the present problem, the elements
are symmetrically placed so that a table is hardly
justified.
x-
%3D
FIG. 8-13.
Direct application of the transfer formula I = 1+ Ad to each element gives
1x (23.5)3
+(23.5X 1)(0)* = 1080 in.
For web plate:
12
18 X (1)3
=D2
+(18 X 1)(12.5)2
5620 in.
For two flange plates: I 2
12
||
I = 4[19.9 + 5.75(10.32)] = 2530 in.
For four angles:
For entire figure:
I,
= 1080 + 5620+2530 =9230 in. Ans.
PROBLEMS
816. A rectangle is 3 in. by 6 in. Determine the polar moment of inertia and the
radius of gyration with respect to a polar axis through one corner.
Ans. J 270 in.; k, = 3.88 in.
817. Determine the moment of inertia and radius of gyration with respect to a
polar centroidal axis of the eross section of a hollow tube whose outside diameter in
6 in. and inside diameter is 4 in.
Bubt nollow square cross section consists of an 8 in. by 8 in. square from whic
polar radius of gyration with respect to a Z axis passing through on
tside corners:
Ans. J 2176 in.; k = 6.74 i
%3D
%3D
its e the moment of inertia of the T-section shown in Fig. P-819 wi
centroidal X. axis.
Ans. 3.5 in.; I, 291 in
%3D
et to its centroidal axes.
821. Find the moman
shown in Fir
Ans. ū = 57 in :
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1bffd075-2869-4715-8d86-fc50cf98a7c1%2F68242cc6-39c9-4dda-812d-2316cb53115d%2F6jkpe47.jpeg&w=3840&q=75)
Transcribed Image Text:(Chap. VIII
MOMENTS OF INERTIA
206
Taking the summations from the table and muk
stituting in Eq. (b), we obtain
I, = 454.91 + 776.8
I,=1232 in. Ana.
A girder is composed of four 6 X 6 xin
angles connected to a web plate 23.5 in. x1 in
in Fig. 8-13. The properties of the angle are
I.- 1,= 19.9 in., the area 5.75 in.2, and
2- - 1.68 in. Compute the moment of inertin
with respect to the centroidal X. axis.
Solution: The tabular computation used in Prob.
814 is well suited for cases in which there are many
elements. In the present problem, the elements
are symmetrically placed so that a table is hardly
justified.
x-
%3D
FIG. 8-13.
Direct application of the transfer formula I = 1+ Ad to each element gives
1x (23.5)3
+(23.5X 1)(0)* = 1080 in.
For web plate:
12
18 X (1)3
=D2
+(18 X 1)(12.5)2
5620 in.
For two flange plates: I 2
12
||
I = 4[19.9 + 5.75(10.32)] = 2530 in.
For four angles:
For entire figure:
I,
= 1080 + 5620+2530 =9230 in. Ans.
PROBLEMS
816. A rectangle is 3 in. by 6 in. Determine the polar moment of inertia and the
radius of gyration with respect to a polar axis through one corner.
Ans. J 270 in.; k, = 3.88 in.
817. Determine the moment of inertia and radius of gyration with respect to a
polar centroidal axis of the eross section of a hollow tube whose outside diameter in
6 in. and inside diameter is 4 in.
Bubt nollow square cross section consists of an 8 in. by 8 in. square from whic
polar radius of gyration with respect to a Z axis passing through on
tside corners:
Ans. J 2176 in.; k = 6.74 i
%3D
%3D
its e the moment of inertia of the T-section shown in Fig. P-819 wi
centroidal X. axis.
Ans. 3.5 in.; I, 291 in
%3D
et to its centroidal axes.
821. Find the moman
shown in Fir
Ans. ū = 57 in :
%3D
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