Post on 23-Dec-2015
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx1
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering 36
Chp10:Moment of
Inertia
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx2
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Introduction Previously we considered distributed forces
which were proportional to the area or volume over which they act; i.e., Centroids• The resultant Force was obtained by summing or
integrating over the areas or volumes• The moment of the resultant about any axis was
determined by Calculating the FIRST MOMENTS of the areas or volumes about that axis– FIRST MOMENT = An Area/Volume/Mass INCREMENT
(or INTENSITY) times its LEVER ARM (kind of like F•d for Torque) Lever Arm = the ┴ Distance to Axis of Interest
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx3
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Introduction cont. Next consider forces which are proportional
to the Area or Volume over which they act, but also VARY LINEARLY WITH DISTANCE from a given axis• It will be shown that the magnitude of the
resultant depends on the FIRST MOMENT of the force distribution with respect to the axis of interest
• The Equivalent point of application of the resultant depends on the SECOND moment of the Force distributionwith respect to the given axis
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx4
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Area Moment of Inertia Consider distributed forces, F ,
whose magnitudes are proportional to the elemental areas, A, on which they act, and also vary LINEARLY with the Distance of A from a given axis.
Example: Consider a beam subjected to pure bending. A Mechanics-of-Materials Analysis Shows that forces vary linearly with distance from the NEUTRAL AXIS which passes through the section centroid. In This Case
Moment) (Resultantmoment second
Force) (Resultantmoment first 0
CONST Matl/Geom
22
dAydAykM
QdAydAykR
k
FyMAkyF
y
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx5
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example → 2nd Moment of Area Find the Resultant hydrostatic force on a submerged
CIRCULAR gate, and its Point of Application
y
k
Ay
I
dAy
dAyydAyRyMdM
ydAyydFdMAydAyRdF
ydApdAdFdAAAyApF
xxPPxx
xA
22
2
also taking
Let
• Recall That the gage Pressure, p, is Proportional to the Depth, y, by p = γy
• Consider an Increment of Area, A• Then The Associated Force Increment
F = pA• Locate Point of Resultant Application, yP, by
Equating Moments about the x-Axis
More onthis Later
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx6
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Second Moment
The Areal Moments of Inertia Take these forms
In both the Integral and the Sum Observe that since u is a distance, so the expressions are of the form
Thus “moments of Inertia” are the SECOND Moment of Area or Mass
iivv AuIdAuI 22
IntensityArmLever 2 vI
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx7
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Moment of Inertia by Integration SECOND MOMENTS or MOMENTS OF
INERTIA of an area with respect to the x and y axes: dAxIdAyI yx
22
Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes.
ydxxdAxI
dyxaydAyI
y
x
22
22
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx8
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Base Case → Rectangle Find Moment of Inertia for a Rectangular
Area
Apply This Basic formula to strip-like rectangular areas That are Most conveniently Parallel to the Axes
Consider a Horizontal Strip• dIx: b∙y → [a-x]∙y
• dIy → Subtract strips: b+ → a; b− → x
3
0
22
3
1bhbdyydAyI
h
x
dyxadyxdyadI
dIdIdI
dyxaydAydI
y
yyy
x
3333
22
3
1
3
1
3
1
Length x of StripaLength of Strip
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx9
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Use BaseLine Case for Iy
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx10
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx11
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Polar Moment of Inertia The polar moment of inertia is an important parameter in
problems involving torsion of cylindrical shafts, Torsion in Welded Joints, and the rotation of slabs
In Torsion Problems, Define a Moment of Inertia Relative to the Pivot-Point, or “Pole”, at O
dArJO2
Relate JO to Ix & Iy Using The Pythagorean Theorem
xy
O
II
dAydAxdAyxdArJ
22222
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx12
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Radius of Gyration (k↔r) 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, Then Define the RADIUS OF GYRATION, r
A
IrArI yyyy 2
Similarly in the Polar Case the Radius of Gyration
222
2
yxO
OOOO
rrr
A
JrArJ
xr A
IrArI xxxx 2
yrOr
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx13
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example 1
Determine the moment of inertia of a triangle with respect to its base.
SOLUTION:• Choose dA as a differential
strip parallel to the x axis
dyldAdAydI x 2
By Similar Triangles See that l in linear in y
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
4312
3bhI x
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx14
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example 2
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:• Choose dA as a differential
annulus of width du
By symmetry, Ix = Iy, so
rr
OO
O
duuduuudJJ
duudAdAudJ
0
3
0
2
2
22
2
4
2rJO
xxyxO IrIIIJ 22
2 4
4
4rII xdiameter
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx15
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Parallel Axis Theorem With Respect to The
OffSet Axis AA’
dAddAyddAy
dAdydAyI AA22
22'
2
Now Consider the Middle Integral
BB' Moment WRT1st 22 ddAyd
But this is a 1st Moment about an Axis Thru the Centroid, Which By Definition = 0
Consider the moment of inertia I of an area A with respect to the axis AA’
Next Consider Axis BB’ That Passes thru The Area Centriod
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx16
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Parallel Axis Theorem cont.
Next Consider the Last Integral
II
AdII
dAddAyddAyI
BB
BBAA
AA
'
2''
22'
so Axis, Centroidal a is BB'but
0
2
Thus The Formal Statement of the Parallel Axis Theorem
AdII 2• Note: This Theorem
Applies ONLY to a CENTROIDAL Axis and an Axis PARALLEL to it
AddAd 22 The Relation for I w.r.t.
the Centroidal Axis and a parallel Axis:
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx17
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Parallel Axis Theorem Exmpls Find the Moment of inertia IT of a
circular area with respect to a tangent to the circle
445224
412 rrrrAdIIT
Find the Moment of inertia of a triangle with respect to a centroidal axis
3
361
2
31
213
1212
22
bhI
hbhbhAdII
AdIAdII
AA
BBAA
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx18
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Composite Moments of Inertia 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.• This Analogous to Finding the Centroid of a Composite
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx19
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Standard Shapes are Tabulated
44622 in 385mm 102.160in 11.20mm 7230
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx20
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example 3
The strength of a W14x38 rolled steel beam is increased by welding a ¾” 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 NEW section.
SOLUTION PLAN• Determine location of the
Centroid for the 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 the 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
4
2
in 385
in 11.20
beam
beam
I
A
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx21
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx22
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example 3 cont. Determine location of the CENTROID
of the COMPOSITE SECTION with respect to a coordinate system with origin at the CENTROID of the BEAM section. The First Moment Tabulation
12.5095.17
0011.20Section Beam
12.50425.76.75Plate
in ,in. ,in ,Section 32
AyA
AyyA
in. 792.2in 17.95
in 12.502
3
A
AyYAyAY
Calc the Centroidin. 792.2
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx23
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example 3 cont.2 Apply the parallel axis theorem to
determine moments of inertia of beam section and plate with respect to COMPOSITE SECTION centroidal axis.
Then the Composite Moment Of Inertia About the Composite Centroidal Axis, at x’
4
23
43
1212
plate,
4
22sectionbeam,
in 2.145
792.2425.775.69
in3.472
792.220.11385
AdII
YAII
xx
xx
4plate,section beam, in5.6172.145 3.472 xxxx IIII
in. 792.2
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx24
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Example 3 cont.3 Finally Calculate the Radius of
Gyration for the Composite From Quantities Previously Determined
So Finally
2
4
in 17.95
in 5.617
A
Ik xx
in. 87.5xk
in. 792.2
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx25
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
WhiteBoard Work
Let’s WorkThis NiceProblem
Find the Area Moment of Inertia of the Parallelogram about its Centroidal axis y’
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx26
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Engineering 36
Appendix 00
sinhT
µs
T
µx
dx
dy
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx27
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics
BMayer@ChabotCollege.edu • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx28
Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics