Mos 06-moment of inertia
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Transcript of Mos 06-moment of inertia
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HJD InstituteCivil 4th
Group member:(1) Dhwani Pathak
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Content:1. Definition2. Standard shape derivation3. Moments of Inertia of
Composite Areas4. Parallel Axis Theorem5. Some examples of M.I
Moment of Inertia
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Moment of inertia is second moment of area.
1. Definition
• Consider a lamina of area A• Let this lamina is split up in
to an infinite of small elements even of area da..
• x1,x2,x3……are distance of small element from OY axis.
• y1,y2,y3……are distance of small elements from OX axis.
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Taking secound moment of all the small elements about OX axis
Thus secound moment of area is called moment of inertia.
UNIT : mm4, cm4
dAxIdAyI yx22
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•Standerd shape darivation
Determine the moment of inertia of a triangle with respect to its base.
SOLUTION:
• A differential strip parallel to the x axis is chosen for dA.
dyldA
dAyI x2
• For similar triangles,
dyh
yhbdA
h
yhbl
h
yh
b
l
h
hh
x
yyh
h
b
dyyhyh
bdy
h
yhbydAyI
0
43
0
32
0
22
43
Integrating dIx from y = 0 to y = h
12
3bhI x
Same as rectangle..
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•Parallel Axis TheoremConsider moment of inertia I with respect to the axis AA’.
dAyI 2
This axis BB’ passes through the area centroid & is called a centroidal axis..
dAddAyddAy
dAdydAyI
22
22
2
2AdII
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The strength of a W14x38 rolled steel beam is increased by attaching a plate to its upper flange.
Determine the moment of inertia..
SOLUTION:• Determine location of the
centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.
• Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis.
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• Apply the parallel axis theorem to determine moments of inertia of beam section and plate with respect to composite section centroidal axis.
4
23
43
1212
plate,
4
22sectionbeam,
in 2.145
792.2425.775.69
in3.472
792.220.11385
AdII
YAII
xx
xx
2.145 3.472plate,section beam, xxx III
4in 618xI
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Determine the moment of inertia of the shaded area with respect to the x axis.
SOLUTION:
• Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis.
• The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.
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SOLUTION:• Compute the moments of inertia of the
bounding rectangle and half-circle with respect to the x axis.
Rectangle:
46313
31 mm102.138120240 bhIx
Half-circle: moment of inertia with respect to AA’,
464
814
81 mm1076.2590 rI AA
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23
2212
21
mm1072.12
90
mm 81.8a-120b
mm 2.383
904
3
4
rA
ra
moment of inertia with respect to x’,
46
362
mm1020.7
1072.121076.25
AaII AAx
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• The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.
xI 46mm102.138 46mm103.92 46mm109.45 xI
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Thank you