Upload
surendran-murugesan
View
223
Download
0
Embed Size (px)
Citation preview
8/12/2019 AR231 Chap08 MomentofInertia (1)
1/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 1
8.0 SECOND MOMENT OR MOMENT OF INERTIA
OF AN AREA
8.1 Introduction
8.2 Moment of Inertia of an Area
8.3 Moment of Inertia of an Area by Integration
8.4 Polar Moment of Inertia
8.5 Radius of Gyration
8.6 Parallel Axis Theorem
8.7 Moments of Inertia of Composite Areas
8.8 Product of Inertia
8.9 Principal Axes and Principal Moments of Inertia
8/12/2019 AR231 Chap08 MomentofInertia (1)
2/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 2
8.1. Introduction
Forces which are proportional to the area or volume over which they act
but also vary linearly with distance from a given axis.
- the magnitude of the resultant depends on the first moment of the
force distribution with respect to the axis.
- The point of application of the resultant depends on the secondmoment of the distribution with respect to the axis.
Herein methods for computing the moments and products of inertia for
areas and masses will be presented
8/12/2019 AR231 Chap08 MomentofInertia (1)
3/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 3
8.2. Moment of Inertia of an Area Consider distributed forces whose magnitudes are
proportional to the elemental areas on which they
act and also vary linearly with the distance offrom a given axis.
F
A
A
Example: Consider a beam subjected to pure bending.
Internal forces vary linearly with distance from the
neutral axis which passes through the section centroid.
momentsecond
momentfirst022
dAydAykM
QdAydAykRAkyF
x
Example: Consider the net hydrostatic force on asubmerged circular gate.
dAyM
dAyR
AyApF
x
2
8/12/2019 AR231 Chap08 MomentofInertia (1)
4/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 4
8.3.Moment of Inertia of an Area by Integration Second momentsor moments of inertiaof
an area with respect to thexandyaxes,
dAxIdAyI yx22
Evaluation of the integrals is simplified by
choosing dA to be a thin strip parallel to
one of the coordinate axes.
For a rectangular area,
331
0
22 bhbdyydAyIh
x
The formula for rectangular areas may also
be applied to strips parallel to the axes,
dxyxdAxdIdxydI yx223
31
8/12/2019 AR231 Chap08 MomentofInertia (1)
5/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 5
8.4. Polar Moment of Inertia
Thepolar moment of inertiais an important
parameter in problems involving torsion of
cylindrical shafts and rotations of slabs.
dArJ 2
0
The polar moment of inertia is related to the
rectangular moments of inertia,
xy II
dAydAxdAyxdArJ
22222
0
8/12/2019 AR231 Chap08 MomentofInertia (1)
6/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 6
8.5. Radius of Gyration of an Area Consider areaA with moment of inertia
Ix. Imagine that the area isconcentrated in a thin strip parallel to
thexaxis with equivalentIx.
A
IkAkI xxxx
2
kx= radius of gyrationwith respectto thexaxis
Similarly,
A
JkAkJ
A
IkAkI
OOOO
yyyy
2
2
222yxO kkk
8/12/2019 AR231 Chap08 MomentofInertia (1)
7/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 7
Examples
Determine the moment of
inertia of a triangle with respect
to its base.
SOLUTION:
A differential strip parallel to thexaxis is chosen for
dA.
dyldAdAydIx 2
For similar triangles,
dyh
yhbdA
h
yhbl
h
yh
b
l
Integrating dIxfromy= 0 toy = h,
h
hh
x
yyh
h
b
dyyhyhbdy
hyhbydAyI
0
43
0
32
0
22
43
12
3bh
Ix
8/12/2019 AR231 Chap08 MomentofInertia (1)
8/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 8
a) Determine the centroidal polar
moment of inertia of a circular
area by direct integration.
b) Using the result of part a,
determine the moment of inertia
of a circular area with respect to a
diameter.
SOLUTION:
An annular differential area element is chosen,
rr
OO
O
duuduuudJJ
duudAdAudJ
0
3
0
2
2
22
2
4
2rJO
From symmetry,Ix=Iy,
xxyxO IrIIIJ 22
2 4
4
4rII xdiameter
8/12/2019 AR231 Chap08 MomentofInertia (1)
9/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 9
8.6. Parallel Axis Theorem
Consider moment of inertiaIof an areaA
with respect to the axisAA
dAyI 2
The axisBBpasses through the area centroidand is called a centroidal axis.
dAddAyddAy
dAdydAyI
22
22
2
2AdII parallel axis theorem
8/12/2019 AR231 Chap08 MomentofInertia (1)
10/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 10
Moment of inertiaITof a circular area with
respect to a tangent to the circle,
445
224412
r
rrrAdIIT
Moment of inertia of a triangle with respect to a
centroidal axis,
3
361
231
213
1212
2
bh
hbhbhAdII
AdII
AABB
BBAA
8/12/2019 AR231 Chap08 MomentofInertia (1)
11/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 11
8.7. Moments of Inertia of Composite Areas The moment of inertia of a composite areaAabout a given axis is
obtained by adding the moments of inertia of the component areas
A1,A2,A3, ... , with respect to the same axis.
8/12/2019 AR231 Chap08 MomentofInertia (1)
12/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 12
Example
The strength of a W360x57 rolled steel
beam is increased by attaching a
225x20 mm plate to its upper flange.
Determine the moment of inertia and
radius of gyration with respect to an
axis which is parallel to the plate and
passes through the centroid of the
section.
SOLUTION:
Determine location of the centroid ofcomposite section with respect to a
coordinate system with origin at the
centroid of the beam section.
Apply the parallel axis theorem todetermine moments of inertia of beam
section and plate with respect to
composite section centroidal axis.
Calculate the radius of gyration from the
moment of inertia of the compositesection.
225 mm
358 mm
20 mm
172 mm
f /
8/12/2019 AR231 Chap08 MomentofInertia (1)
13/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 13
SOLUTION:
Determine location of the centroid of composite sectionwith respect to a coordinate system with origin at the
centroid of the beam section.
mm.51.7211730
1050.58 3
A
AyYAyAY
225 mm
358 mm
20 mm
172 mm
189 mm
12.5095.170011.20SectionBeam
12.50425.76.75Plate
in,in.,in,Section 32
AyA
AyyA mm2 mm mm3
4500
7230
11730
189 850.5 x 103
850.5 x 103
2450020225 mmmmmmA
mmmmmmy 1892021358
21
IZMIR INSTITUTE OF TECHNOLOGY D t t f A hit t AR231 F ll12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
14/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 14
Apply the parallel axis theorem to determine moments of
inertia of beam section and plate with respect to composite
section centroidal axis.
46
23
1212
plate,
4
262
sectionbeam,
mm102.61
51.72189450020225
102.198
51.727230102.160
AdII
YAII
xx
xx
Calculate the radius of gyration from the moment of inertiaof the composite section.
2
46
mm11730
mm1053.82
A
Ik xx
mm.1.147xk
66
plate,sectionbeam, 102.61106.192 xxx III
4610254 mmIx
225 mm
358 mm
20 mm
172 mm
189 mm
IZMIR INSTITUTE OF TECHNOLOGY D t t f A hit t AR231 F ll12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
15/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 15
Example
Determine the moment of inertia
of the shaded area with respect to
thexaxis.
SOLUTION:
Compute the moments of inertia of the
bounding rectangle and half-circle with
respect to thex axis.
The moment of inertia of the shaded area is
obtained by subtracting the moment ofinertia of the half-circle from the moment
of inertia of the rectangle.
IZMIR INSTITUTE OF TECHNOLOGY D t t f A hit t AR231 F ll12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
16/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 16
SOLUTION:
Compute the moments of inertia of the bounding
rectangle and half-circle with respect to thex axis.
Rectangle:
46313
31 mm102.138120240 bhIx
Half-circle:moment of inertia with respect toAA,
464814
81 mm1076.2590 rI AA
23
2
2
12
2
1
mm1072.12
90
mm81.8a-120b
mm2.383
904
3
4
rA
ra
moment of inertia with respect tox,
46
362
mm1020.7
1072.121076.25
AaII AAx
moment of inertia with respect tox,
46
2362
mm103.92
8.811072.121020.7
AbII xx
IZMIR INSTITUTE OF TECHNOLOGY D t t f A hit t AR231 F ll12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
17/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 17
The moment of inertia of the shaded area is obtained by
subtracting the moment of inertia of the half-circle fromthe moment of inertia of the rectangle.
46mm109.45 xI
xI 46mm102.138 46mm103.92
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
18/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 18
8.8. Product of Inertia Product of Inertia:
dAxyIxy
When thexaxis, theyaxis, or both are an
axis of symmetry, the product of inertia iszero.
Parallel axis theorem for products of inertia:
AyxII xyxy
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
19/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 19
8.9. Principal Axes and Principal Moments of Inertia
Given
dAxyI
dAxIdAyI
xy
yx22
we wish to determine momentsand product of inertia with
respect to new axesxandy.
2cos2sin2
2sin2cos22
2sin2cos22
xyyxyx
xyyxyx
y
xyyxyx
x
IIII
IIIII
I
IIIII
I
The change of axes yields
The equations forIxandIxyare the
parametric equations for a circle,
2
222
22 xy
yxyxave
yxavex
III
RII
I
RIII
The equations forIyandIxy lead to the
same circle.
sincos
sincos
xyy
yxx
Note:
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
20/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 20
2
222
22 xy
yxyxave
yxavex
III
RII
I
RIII
At the pointsAandB, Ixy= 0 andIxis
a maximum and minimum, respectively.RII aveminmax,
yx
xym
II
I
2
2tan
ImaxandIminare theprincipal momentsof inertiaof the area about O.
The equation for Qmdefines two
angles, 90oapart which correspond to
theprincipal axesof the area about O.
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
21/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 21
Example
Determine the product of inertia of
the right triangle (a)with respect
to thexandyaxes and
(b)with respect to centroidal axes
parallel to thexandyaxes.
SOLUTION:
Determine the product of inertia using
direct integration with the parallel axis
theorem on vertical differential area strips
Apply the parallel axis theorem to
evaluate the product of inertia with respect
to the centroidal axes.
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
22/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 22
SOLUTION:
Determine the product of inertia using direct integrationwith the parallel axis theorem on vertical differential
area strips
b
xhyyxx
dxb
xhdxydA
b
xhy
elel 1
11
2121
Integrating dIxfromx= 0 tox= b,
bb
b
elelxyxy
b
x
b
xxhdx
b
x
b
xxh
dxb
xhxdAyxdII
02
4322
02
322
0
22
21
83422
1
22
24
1 hbIxy
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
23/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 23
Apply the parallel axis theorem to evaluate the
product of inertia with respect to the centroidal axes.
hybx31
31
With the results from part a,
bhhbhbI
AyxII
yx
yxxy
21
31
3122
241
22
721 hbI yx
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
24/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 24
For the section shown, the moments of
inertia with respect to thexandyaxes
areIx= 10.38 in4andIy= 6.97 in4.
Determine (a) the orientation of the
principal axes of the section about O,
and (b) the values of the principal
moments of inertia about O.
SOLUTION:
Compute the product of inertia with
respect to thexyaxes by dividing the
section into three rectangles and applying
the parallel axis theorem to each.
Determine the orientation of theprincipal axes (Eq. 9.25) and the
principal moments of inertia (Eq. 9. 27).
Example
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
25/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
02.06.2014 Dr. Engin Akta 25
SOLUTION:
Compute the product of inertia with respect to thexyaxesby dividing the section into three rectangles.
276985813849295.445.31988
000988
13849295.445.31988
m,mm.,mm.,mmArea,Rectangle 42
AyxIII
II
I
mAyxyx
Apply the parallel axis theorem to each rectangle,
AyxII yxxy
Note that the product of inertia with respect to centroidal
axes parallel to thexy axes is zero for each rectangle.
42770000mmAyxIxy
102 mm
76 mm
13 mm
13 mm
76 mm
31.5 mm
44.5 mm
44.5 mm
31.5 mm
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231 Fall12/13
8/12/2019 AR231 Chap08 MomentofInertia (1)
26/26
IZMIR INSTITUTE OF TECHNOLOGY Department of Architecture AR231Fall12/13
Determine the orientation of the principal axes by
46
46
46
mm1077.2
mm1093.2
mm1042.4
xy
y
x
I
I
I
254.9and9.742
72.31093.21042.4
1077.2222tan66
6
m
yx
xy
mII
I
5.127and5.37 mm
262
6666
2
2
minmax,
1077.22
1093.21042.4
2
1093.21042.4
22
xy
yxyx IIIIII
4
min
4
max
mm07.8
mm54.6
II
II
b
a
yx
xym
II
I
2
2tan
RII aveminmax,the principal moments of
inertia by
2
222
22 xy
yxyxave
yxavex
III
RII
I
RIII
m = 127.5o
m = 37.5o