izmir institute of technology department of architecture ar231 fall12/13 18.04.2015 dr. engin aktaş...
Post on 14-Jan-2016
215 Views
Preview:
TRANSCRIPT
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 1
8.0 8.0 SECOND MOMENT OR MOMENT OF INERTIA SECOND MOMENT OR MOMENT OF INERTIA OF AN AREAOF 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
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 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 second moment 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
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 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 of from 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.
moment second
momentfirst 022
dAydAykM
QdAydAykR
AkyF
x
• Example: Consider the net hydrostatic force on a submerged circular gate.
dAyM
dAyR
AyApF
x2
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 4
8.3.Moment of Inertia of an Area by Integration• Second moments or moments of inertia of
an area with respect to the x and y axes,
dAxIdAyI yx22
• Evaluation of the integrals is simplified by choosing dto 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
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 5
8.4. Polar Moment of Inertia
• The polar moment of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs.
dArJ 20
• The polar moment of inertia is related to the rectangular moments of inertia,
xy II
dAydAxdAyxdArJ
222220
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 6
8.5. Radius of Gyration of an Area• Consider area A with moment of inertia
Ix. Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent Ix.
A
IkAkI x
xxx 2
kx = radius of gyration with respect to the x axis
• Similarly,
A
JkAkJ
A
IkAkI
OOOO
yyyy
2
2
222yxO kkk
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 7
Examples
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.
dyldAdAydIx 2
• For similar triangles,
dyh
yhbdA
h
yhbl
h
yh
b
l
• Integrating dIx from y = 0 to y = h,
h
hh
x
yyh
h
b
dyyhyh
bdy
h
yhbydAyI
0
43
0
32
0
22
43
12
3bhI x
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 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
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 9
8.6. Parallel Axis Theorem
• Consider moment of inertia I of an area A with respect to the axis AA’
dAyI 2
• The axis BB’ passes through the area centroid and is called a centroidal axis.
dAddAyddAy
dAdydAyI
22
22
2
2AdII parallel axis theorem
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 10
• Moment of inertia IT of a circular area with respect to a tangent to the circle,
4
45
224412
r
rrrAdIIT
• Moment of inertia of a triangle with respect to a centroidal axis,
3
361
2
31
213
1212
2
bh
hbhbhAdII
AdII
AABB
BBAA
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 11
8.7. Moments of Inertia of Composite Areas• The moment of inertia of a composite area A about a given axis is
obtained by adding the moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 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 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.
• Calculate the radius of gyration from the moment of inertia of the composite section.
225 mm
358 mm
20 mm
172 mm
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 13
SOLUTION:
• Determine location of the centroid of composite section with respect to a coordinate system with origin at the centroid of the beam section.
mm. 51.7211730
10 50.58 3
A
AyYAyAY
225 mm
358 mm
20 mm
172 mm
189 mm
12.5095.17
0011.20Section Beam
12.50425.76.75Plate
in ,in. ,in ,Section 32
AyA
AyyA mm2mm mm3
45007230
11730
189 850.5 x 103
850.5 x 103
2450020225 mmmmmmA
mmmmmmy 189202
1358
2
1
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 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
262sectionbeam,
mm 102.61
51.72189450020225
102.198
51.727230102.160
AdII
YAII
xx
xx
• Calculate the radius of gyration from the moment of inertia of the composite section.
2
46
mm 11730
mm 1053.82
A
Ik x
x
mm. 1.147xk
66plate,section beam, 102.61 106.192 xxx III
4610254 mmI x
225 mm
358 mm
20 mm
172 mm
189 mm
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 15
Example
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.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 16
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’,
464814
81 mm1076.2590 rI AA
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
moment of inertia with respect to x,
46
2362
mm103.92
8.811072.121020.7
AbII xx
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 17
• 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.
46mm109.45 xI
xI 46mm102.138 46mm103.92
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 18
8.8. Product of Inertia• Product of Inertia:
dAxyI xy
• When the x axis, the y axis, or both are an axis of symmetry, the product of inertia is zero.
• Parallel axis theorem for products of inertia:
AyxII xyxy
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 19
8.9. Principal Axes and Principal Moments of Inertia
Given
dAxyI
dAxIdAyI
xy
yx22
we wish to determine moments and product of inertia with respect to new axes x’ and y’.
2cos2sin2
2sin2cos22
2sin2cos22
xyyx
yx
xyyxyx
y
xyyxyx
x
III
I
IIIII
I
IIIII
I
• The change of axes yields
• The equations for Ix’ and Ix’y’ are the parametric equations for a circle,
2
222
22 xyyxyx
ave
yxavex
III
RII
I
RIII
• The equations for Iy’ and Ix’y’ lead to the same circle.
sincos
sincos
xyy
yxx
Note:
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 20
2
222
22 xyyxyx
ave
yxavex
III
RII
I
RIII
• At the points A and B, Ix’y’ = 0 and Ix’ is a maximum and minimum, respectively. RII ave minmax,
yx
xym II
I
22tan
• Imax and Imin are the principal moments of inertia of the area about O.
• The equation for m defines two angles, 90o apart which correspond to the principal axes of the area about O.
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 21
Example
Determine the product of inertia of the right triangle (a) with respect to the x and y axes and (b) with respect to centroidal axes parallel to the x and y axes.
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 TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 22
SOLUTION:
• Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips
b
xhyyxx
dxb
xhdxydA
b
xhy
elel 1
11
21
21
Integrating dIx from x = 0 to x = b,
bb
b
elelxyxy
b
x
b
xxhdx
b
x
b
xxh
dxb
xhxdAyxdII
02
4322
02
322
0
22
21
83422
1
22241 hbIxy
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 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
22721 hbI yx
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 24
For the section shown, the moments of inertia with respect to the x and y axes are Ix = 10.38 in4 and Iy = 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 the xy axes by dividing the section into three rectangles and applying the parallel axis theorem to each.
• Determine the orientation of the principal axes (Eq. 9.25) and the principal moments of inertia (Eq. 9. 27).
Example
IZMIR INSTITUTE OF TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 25
SOLUTION:
• Compute the product of inertia with respect to the xy axes by dividing the section into three rectangles.
2769858
13849295.445.31988
000988
13849295.445.31988
m,mm. ,mm. ,mm Area,Rectangle 42
Ayx
III
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 the xy axes is zero for each rectangle.
42770000mmAyxI xy
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 TECHNOLOGYIZMIR INSTITUTE OF TECHNOLOGY Department of ArchitectureDepartment of Architecture AR23AR2311 Fall12/13Fall12/13
21.04.23 Dr. Engin Aktaş 26
• Determine the orientation of the principal axes by
46
46
46
mm10 77.2
mm1093.2
mm1042.4
xy
y
x
I
I
I
254.9 and 9.742
72.31093.21042.4
1077.2222tan
66
6
m
yx
xym II
I
5.127 and 5.37 mm
26
26666
2
2
minmax,
1077.22
1093.21042.4
2
1093.21042.4
22
xy
yxyx IIIII
I
4min
4max
mm 07.8
mm 54.6
II
II
b
a
yx
xym II
I
22tan
RII ave minmax,
the principal moments of inertia by
2
222
22 xyyxyx
ave
yxavex
III
RII
I
RIII
m = 127.5o
m = 37.5o
top related