Transcript ENGR-36_Lec-26_Area_Moment_of_Inertia

Engineering 36
Chp10:
Moment of Inertia
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
[email protected]
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Bruce Mayer, PE
[email protected] • ENGR-36_Lec-26_Area_Moment_of_Inertia.pptx
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
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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 distribution
with respect to the
given axis
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Bruce Mayer, PE
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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-ofMaterials Analysis Shows that forces
vary linearly with distance from the
F  kyA
M  yF NEUTRAL AXIS which passes through
 k  Matl/GeomCONST the section centroid. In This Case
R  k  y dA  0  y dA  Qy  first moment(ResultantForce)
M  k  y 2 dA
2
y
 dA  secondmoment(ResultantMoment)
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Bruce Mayer, PE
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Example → 2nd Moment of Area
 Find the Resultant hydrostatic force on a submerged
CIRCULAR gate, and its Point of Application
• 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 = pA
• Locate Point of Resultant Application, yP, by
Equating Moments about the x-Axis
F  pA  yA
Let A  dA  dF  pdA  ydA
taking dF R    y dA   y A also dM x  ydF  y ydA
A
2
dM

M

y
R


y
 x x P
 dA  
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  y 2 dA
I x k x2
yP 


  y dA y A y
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More on
this Later
Second Moment
 The Areal Moments of Inertia Take
2
2
I

u
dA
I

u
these forms v 
 i  Ai
v
 In both the Integral and the Sum
Observe that since u is a distance, so
the expressions are of the form
2
I v  LeverArm  Intensity
 Thus “moments of Inertia”
are the SECOND
Moment of Area or Mass
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Bruce Mayer, PE
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Moment of Inertia by Integration
 SECOND MOMENTS or MOMENTS OF
INERTIA of an area with respect to the x
and y axes:
2
2
I x   y dA
I y   x dA
 Evaluation of the integrals is simplified by
choosing dA to be a thin strip parallel to
one of the coordinate axes.
I x   y dA   y a  x dy
2
2
I y   x 2 dA   x 2 ydx
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Bruce Mayer, PE
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Base Case → Rectangle
 Find Moment of Inertia for a Rectangular
h
Area
1 3
2
2
I x   y dA   y bdy  bh
3
0
 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
dIx  y 2 dA  y 2 a  x dy
dI y  dI y Strip of Lengtha   dI y Strip of Length x


1 3
1 3
1 3 3
dI y  a dy  x dy  a  x dy
3
3
Bruce Mayer, PE
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Use BaseLine Case for Iy
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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
J O   r dA
2
 Relate JO to Ix & Iy Using The
Pythagorean Theorem


J O   r 2 dA   x 2  y 2 dA   x 2 dA   y 2 dA
 Iy  Ix
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Bruce Mayer, PE
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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
I x  r A  rx 
2
x
rx
ry
rO
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Ix
A
I y  r A  ry 
2
y
 Similarly in the Polar Case the
Radius of Gyration
JO
2
J O  rO A
rO 
A
rO2  rx2  ry2
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Iy
A
Example 1
 SOLUTION:
• Choose dA as a differential
strip parallel to the x axis
dIx  y 2dA
dA  l dy
 By Similar Triangles See
that l in linear in y
l h y

b
h
l b
h y
h
dA  b
h y
dy
h
 Integrating dIx from
y = 0 to y = h
h y
b
I x   y 2 dA   y 2b
dy   hy2  y 3 dy
h
h0
0
h
 Determine the moment
of inertia of a triangle
with respect to its base.
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h
h
b  y3 y4 
 h   
h 3
4 0

bh3
I x
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
Example 2
 SOLUTION:
• Choose dA as a differential
annulus of width du
dA  2 u du
dJO  u 2 dA
r
r
J O   dJO   u 2 u du  2  u 3 du
2
0
JO 

2
0
r4
 By symmetry, Ix = Iy, so
a) Determine the centroidal
POLAR moment of inertia
 4
of a circular area by direct J  I  I  2 I

r
O
x
y
x
integration.
2
b) Using the result of part a,
 4
determine the moment of
I diameter  I x  r
inertia of a circular area
4
with respect to a diameter.
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Bruce Mayer, PE
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 2I x
Parallel Axis Theorem
 With Respect to The
OffSet Axis AA’
I AA'   y 2 dA    y  d  dA
2
  y2 dA  2d  ydA  d 2  dA
 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
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 Now Consider the
Middle Integral
2d  ydA  2d  1st Moment WRT BB'
 But this is a 1st Moment
about an Axis Thru the
Centroid, Which By
Definition = 0
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Parallel Axis Theorem cont.
I AA'   y2 dA  2d  ydA  d 2  dA
I AA'  I BB '  0  d 2 A
but BB' is a Cent roidalAxis, so
I BB'  I
 Next Consider the Last
Integral
d
2
 dA  d
2
A
 The Relation for I w.r.t.
the Centroidal Axis and
a parallel Axis:
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 Thus The Formal
Statement of the
Parallel Axis Theorem
I  I d A
2
• Note: This Theorem
Applies ONLY to a
CENTROIDAL Axis and
an Axis PARALLEL to it
Bruce Mayer, PE
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Parallel Axis Theorem Exmpls
 Find the Moment of inertia IT of a
circular area with respect to a
tangent to the circle
 
IT  I  Ad 2  14  r 4   r 2 r 2  54  r 4
 Find the Moment of inertia of a triangle
with respect to a centroidal axis
I AA  I BB  Ad  I  Ad
2
I  I AA  Ad 
2
1
12
2
bh  bh h 
3
1
2
 I  361 bh3
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1
3
2
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
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Standard Shapes are Tabulated
7230 mm 2  11.20 in 2
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160.2 106 mm 4  385 in 4
Bruce Mayer, PE
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Example 3
 SOLUTION PLAN
Abeam  11.20in 2
I beam  385in 4


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.
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• 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
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Bruce Mayer, PE
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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
Section
A, in 2
y , in.
P late
6.75
7.425 50.12
Beam Section 11.20
0
 A  17.95
2.792 in.
0
 yA  50.12
 Calc the Centroid
Y  A   yA
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yA, in 3
yA 50.12 in 3

Y 

 2.792 in.
2
 A 17.95 in
Bruce Mayer, PE
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Example 3 cont.2
2.792 in.
 Apply the parallel axis theorem to
determine moments of inertia of beam
section and plate with respect to
COMPOSITE SECTION centroidal axis.
I x,beam section  I x  AY 2  385 11.202.792
2
 472.3 in 4
I x,plate  I x  Ad 2  121 9 34   6.757.425 2.792
3
2
 145.2 in 4
 Then the Composite Moment Of Inertia About the
Composite Centroidal Axis, at x’
I x  I x,beamsection  I x,plate  472.3 145.2  I x  617.5 in 4
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Bruce Mayer, PE
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Example 3 cont.3
 Finally Calculate the Radius of
Gyration for the Composite From
Quantities Previously Determined
4
I x
617.5 in
k x 

A
17.95in 2
 So Finally
k x  5.87 in.
2.792 in.
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Bruce Mayer, PE
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WhiteBoard Work
Let’s Work
This Nice
Problem
 Find the Area Moment of Inertia of the
Parallelogram about its Centroidal axis y’
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Engineering 36
Appendix
dy
µx µs
 sinh

dx
T0 T0
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
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