ME 101: Engineering · PDF fileME 101: Engineering Mechanics Rajib Kumar Bhattacharjya...
Transcript of ME 101: Engineering · PDF fileME 101: Engineering Mechanics Rajib Kumar Bhattacharjya...
ME 101: Engineering MechanicsRajib Kumar Bhattacharjya
Department of Civil Engineering
Indian Institute of Technology Guwahati
M Block : Room No 005 : Tel: 2428
www.iitg.ernet.in/rkbc
Area Moments of Inertia
Parallel Axis Theorem• Consider moment of inertia I of an area A
with respect to the axis AA’
∫= dAyI2
• The axis BB’ passes through the area centroid
and is called a centroidal axis.
• Second term = 0 since centroid lies on BB’
(∫y’dA = ycA, and yc = 0
( )
∫∫∫
∫∫
+′+′=
+′==
dAddAyddAy
dAdydAyI
22
22
2
2AdII += Parallel Axis theorem
Parallel Axis theorem:
MI @ any axis =
MI @ centroidal axis + Ad2
The two axes should be
parallel to each other.
Area Moments of InertiaParallel Axis Theorem
• Moment of inertia IT of a circular area with
respect to a tangent to the circle,
( )4
45
224
412
r
rrrAdIIT
π
ππ
=
+=+=
• Moment of inertia of a triangle with respect to a
centroidal axis,
( )3
361
2
31
213
1212
2
bh
hbhbhAdII
AdII
AABB
BBAA
=
−=−=
+=
′′
′′
• 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.
Area Moments of Inertia: Standard MIs
Moment of inertia about �-axis �� =1
3�ℎ
� =1
3�ℎ
Answer
Moment of inertia about �-axis
Moment of inertia about ��-axis �� =1
12�ℎ
� =1
12�ℎMoment of inertia about ��-axis
Moment of inertia about �-axis passing through C �� =1
12�ℎ �� + ℎ�
Moment of inertia about �-axis �� =1
12�ℎ
Moment of inertia about ��-axis �� =1
36�ℎ
Moment of inertia about ��-axis �� =1
4���
Moment of inertia about �-axis passing through O �� =1
2���
Moment of inertia about ��-axis �� = � =1
8���
Moment of inertia about �-axis passing through O �� =1
4���
Moment of inertia about ��-axis �� = � =1
16���
Moment of inertia about �-axis passing through O �� =1
8���
Moment of inertia about �-axis �� =1
4���
Moment of inertia about �-axis passing through O �� =1
4��� �� + ��
Moment of inertia about �-axis � =1
4���
Area Moments of InertiaExample:
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.
Area Moments of InertiaExample: Solution
SOLUTION:
• Compute the moments of inertia of the bounding
rectangle and half-circle with respect to the x axis.
Rectangle:
( )( ) 463
3
13
3
1 102.138120240 mmbhI x ×===
Half-circle:
moment of inertia with respect to AA’,
( ) 464
814
81 mm1076.2590 ×===′ ππrI AA
( )( )
( )2
mm
rA
mm 81.8a-120b
mm r
a
3
2
2
12
2
1
1072.12
90
2.383
904
3
4
×=
==
==
===
ππ
ππ
Moment of inertia with respect to x’,
( ) ( )( )4
AAx
mm
AaII
6
2362
1020.7
2.381072.121076.25
×=
×−×=−= ′′
moment of inertia with respect to x,
( )( )4
xx
mm
AbII
6
2362
103.92
8.811072.121020.7
×=
×+×=+= ′
Area Moments of Inertia
Example: Solution
• 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.
46 mm109.45 ×=xI
xI = 46 mm102.138 × − 46 mm103.92 ×
Consider area (1)
�� =1
3�ℎ =
1
3× 80 × 60 = 5.76 × 10 ""�
Consider area (2)
�� =1
4
���
4=
�
1630 � = 0.1590 × 10 ""�
��$ = 0.1590 × 10 −
�
430 � × 12.73 � = 0.0445 × 10 ""�
�� = 0.0445 × 10 +�
430 � 60 − 12.73 = 1.624 × 10 ""�
Consider area (3)
�� =1
12�ℎ =
1
12× 40 × 30 = 0.09 × 10 ""�
�� = 5.76 × 10 − 1.624 × 10 − 0.09 × 10 = &. '( × )'* ++&
, = 60 × 80 −1
4� 30 � −
1
240 × 30 = 3490""�
-� =��
,=
4.05 × 10
3490= .&. ''++
Determine the moment of inertia and the radius of gyration of
the area shown in the fig.
�1� �1
12�/ �
1
12� 24 � 6 � 432 ""�
�2� �1
12�/ �
1
12� 8 � 48 � 73728 ""�
�3� �1
12�/ �
1
12� 48 � 6 � 864 ""�
�1� � �1� � ,�� � 432 � 24 � 6 � 24 � 3 � � 105408 ""�
�3� � �3� � ,�� � 864 � 48 � 6 � 24 � 3 � � 210816 ""�
�� � 105408 � 73728 � 210816 � 390 � 10""�
-� ���
,�
390 � 10
816� 0). 1 ++
Area Moments of Inertia
Products of Inertia: for problems involving unsymmetrical cross-sections
and in calculation of MI about rotated axes. It may be +ve, -ve, or zero
• Product of Inertia of area A w.r.t. x-y axes:
x and y are the coordinates of the element of area dA=xy
∫= 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 +=
+ Ixy
+ Ixy - Ixy
- Ixy
Quadrants
Area Moments of InertiaRotation of Axes Product of inertia is useful in calculating MI @ inclined axes.� Determination of axes about which the MI is a maximum and a minimum
∫
∫∫=
==
dAxyI
dAxIdAyI
xy
yx22
Moments and product of inertia
w.r.t. new axes x’ and y’ ?
θθ
θθ
sincos
sincos
xyy
yxx
−=′
+=′Note:
( )
( )
( )( )dAxyyxdAyxI
dAyxdAxI
dAxydAyI
yx
y
x
∫∫
∫∫
∫∫
−+==
+==
−==
θθθθ
θθ
θθ
sincossincos''
sincos'
sincos'
''
22
'
22
'
θθθθθθ
θθ
θθ
2cossincos2sin2/1cossin
2
2cos1cos
2
2cos1sin
22
22
=−=
+=
−=
θθ
θθ
θθ
2cos2sin2
2sin2cos22
2sin2cos22
' xy
yx
yx
xy
yxyx
y
xy
yxyx
x
III
I
IIIII
I
IIIII
I
+−
=
+−
−+
=
−−
++
=
′
′
′
Area Moments of Inertia
Rotation of Axes Adding first two eqns:
Ix’ + Iy’ = Ix + Iy = Iz � The Polar MI @ O
θθ
θθ
θθ
2cos2sin2
2sin2cos22
2sin2cos22
' xyyx
yx
xyyxyx
y
xyyxyx
x
III
I
IIIII
I
IIIII
I
+−
=
+−
−+
=
−−
++
=
′
′
′
Angle which makes Ix’ and Iy’ either max or
min can be found by setting the derivative
of either Ix’ or Iy’ w.r.t. θ equal to zero:
Denoting this critical angle by α
( ) 02cos22sin' =−−= θθ
θxyxy
x IIId
dI
xy
xy
II
I
−=2
2tan α
� two values of 2α which differ by π since tan2α = tan(2α+π)� two solutions for α will differ by π/2� one value of α will define the axis of maximum MI and the other defines the
axis of minimum MI
�These two rectangular axes are called the principal axes of inertia
Area Moments of Inertia
Rotation of Axes
θθ
θθ
θθ
2cos2sin2
2sin2cos22
2sin2cos22
' xyyx
yx
xyyxyx
y
xyyxyx
x
III
I
IIIII
I
IIIII
I
+−
=
+−
−+
=
−−
++
=
′
′
′
xy
xy
xy
xy
II
I
II
I
−=⇒
−=
22cos2sin
22tan ααα
Substituting in the third eqn for critical value
of 2θ: Ix’y’ = 0
� Product of Inertia Ix’y’ is zero for the
Principal Axes of inertia
Substituting sin2α and cos2α in first two eqns
for Principal Moments of Inertia:
( )
( )0
42
1
2
42
1
2
@
22
min
22
max
=
+−−+
=
+−++
=
αxy
xyyxyx
xyyxyx
I
IIIII
I
IIIII
I
�� =�� + �
2+
�� − �
223425 − ��46725
�� =�� − �
246725 + ��23425
�� −�� + �
2=
�� − �
223425 − ��46725
Squaring both the equation and adding
�� −�� + �
2
�
+ �� �
=�� − �
223425 − ��46725
�
+�� − �
246725 + ��23425
�
�� −89:8;
�
�
+ �� �
=89<8;
�
�
234�25 − 289<8;
��� 2342546725 + ��
�467�25
+89<8;
�
�
467�25 + 289<8;
��� 2342546725 + ��
�467�25
�� −�� + �
2
�
+ �� �
=�� − �
2
�
+ ��
�
�� −�� + �
2
�
+ �� �
=�� − �
2
�
+ ��
�
Defining �� + �
2= �=>?
And @ =�� − �
2
�
+ ��
�
�� − �=>?
�+ ��
�= @� Which is a equation of circle with center �=>? , 0 and radius @
�
��
�=>?
@
25
Area Moments of InertiaMohr’s Circle of Inertia: Following relations can be represented
graphically by a diagram called Mohr’s CircleFor given values of Ix, Iy, & Ixy, corresponding values of Ix’, Iy’, & Ix’y’ may be
determined from the diagram for any desired angle θ.
θθ
θθ
θθ
2cos2sin2
2sin2cos22
2sin2cos22
' xyyx
yx
xyyxyx
y
xyyxyx
x
III
I
IIIII
I
IIIII
I
+−
=
+−
−+
=
−−
++
=
′
′
′
xy
xy
II
I
−=2
2tan α
( )
( )0
42
1
2
42
1
2
@
22
min
22
max
=
+−−+
=
+−++
=
αxy
xyyxyx
xyyxyx
I
IIIII
I
IIIII
I
( )
2
2
222
22xy
yxyxave
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,
I
Ixy
Area Moments of InertiaMohr’s Circle of Inertia: Construction
θθ
θθ
θθ
2cos2sin2
2sin2cos22
2sin2cos22
' xyyx
yx
xyyxyx
y
xyyxyx
x
III
I
IIIII
I
IIIII
I
+−
=
+−
−+
=
−−
++
=
′
′
′
xy
xy
II
I
−=2
2tan α
( )
( )0
42
1
2
42
1
2
@
22
min
22
max
=
+−−+
=
+−++
=
αxy
xyyxyx
xyyxyx
I
IIIII
I
IIIII
I
Choose horz axis � MI
Choose vert axis � PIPoint A – known {Ix, Ixy}
Point B – known {Iy, -Ixy} Circle with dia AB
Angle α for Area� Angle 2α to horz (same
sense) � Imax, Imin
Angle x to x’ = θ
� Angle OA to OC = 2θ� Same sense
Point C � Ix’, Ix’y’
Point D � Iy’
Area Moments of Inertia
Example: Product of Inertia
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.
Area Moments of InertiaExamples
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
+−=
+−=
−===
∫
∫∫∫
22
241 hbIxy =
Area Moments of Inertia
Examples
• 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 −=′′′′
SOLUTION
Area Moments of Inertia
Example: Mohr’s Circle of Inertia
The moments and product of inertia
with respect to the x and y axes are Ix =
7.24x106 mm4, Iy = 2.61x106 mm4, and
Ixy = -2.54x106 mm4.
Using Mohr’s circle, determine (a) the
principal axes about O, (b) the values of
the principal moments about O, and (c)
the values of the moments and product
of inertia about the x’ and y’ axes
SOLUTION:
• Plot the points (Ix , Ixy) and (Iy ,-Ixy).
Construct Mohr’s circle based on the
circle diameter between the points.
• Based on the circle, determine the
orientation of the principal axes and the
principal moments of inertia.
• Based on the circle, evaluate the
moments and product of inertia with
respect to the x’y’ axes.
Area Moments of Inertia
Example: Mohr’s Circle of Inertia
46
46
46
mm1054.2
mm1061.2
mm1024.7
×−=
×=
×=
xy
y
x
I
I
I
SOLUTION:
• Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s
circle based on the circle diameter between the points.
( )
( )
( ) ( ) 4622
46
21
46
21
mm10437.3
mm10315.2
mm10925.4
×=+=
×=−=
×=+==
DXCDR
IICD
IIIOC
yx
yxave
• Based on the circle, determine the orientation of the
principal axes and the principal moments of inertia.
°=== 6.472097.12tan mmCD
DXθθ °= 8.23mθ
Area Moments of InertiaExample: Mohr’s Circle of Inertia
46
46
mm10437.3
mm10925.4
×=
×==
R
IOC ave
• Based on the circle, evaluate the moments and product
of inertia with respect to the x’y’ axes.
The points X’ and Y’ corresponding to the x’ and y’ axes
are obtained by rotating CX and CY counterclockwise
through an angle θ= 2(60o) = 120o. The angle that CX’
forms with the horz is φ = 120o - 47.6o = 72.4o.
oavey RIYCOCOGI 4.72coscos' −=′−== ϕ
46mm1089.3 ×=′yI
oavex RIXCOCOFI 4.72coscos' +=′+== ϕ
46mm1096.5 ×=′xI
oyx RYCXFI 4.72sinsin' =′=′=′ ϕ
46mm1028.3 ×=′′yxI
Determine the product of the inertia of the shaded area shown below about the x-y axes.
Solution:
Parallel axis theorem: Ixy = Īxy +dx dy A
Both areas (1) and (2) are symmetric @ their
centroidal
Total Ixy = Ixy1 + Ixy2 = 18.40×106 mm 4
Similarly, for Area (2): Ixy2 = dx2 dy2 A2
Ixy2 = 60×20×130×30 = 4.68×106 mm 4
Therefore, for Area (1): Ixy1 = dx1 dy1 A1
Ixy1 = 20×140×70×70 = 13.72×106 mm 4
Axis � Īxy = 0 for both area.
Ix < Iy , Ixy (+)
Ix >Iy , Ixy (+)
Ix > Iy , Ixy (-)
Ix < Iy , Ixy (-)
Area Moments of InertiaMass Moments of Inertia (I): Important in Rigid Body Dynamics- I is a measure of distribution of mass of a rigid body w.r.t. the
axis in question (constant property for that axis).- Units are (mass)(length)2 � kg.m2
Consider a three dimensional body of mass m
Mass moment of inertia of this body about axis O-O:
I = ∫ r2 dm
Integration is over the entire body.
r = perpendicular distance of the mass element dm
from the axis O-O
Area Moments of InertiaMoments of Inertia of Thin Plates
• For a thin plate of uniform thickness t and homogeneous
material of density ρ, the mass moment of inertia with
respect to axis AA’ contained in the plate is
areaAA
AA
It
dArtdmrI
,
22
′
′
=
== ∫∫ρ
ρ
• Similarly, for perpendicular axis BB’
which is also contained in the plate,
areaBBBB ItI ,′′ = ρ
• For the axis CC’ which is perpendicular to the plate,
( )
BBAA
areaBBareaAAareaCCC
II
IItJtI
′′
′′′
+=
+== ,,, ρρ
Area Moments of InertiaMoments of Inertia of Thin Plates
• For the principal centroidal axes on a rectangular plate,
( ) 2
1213
121
, mabatItI areaAAAA === ′′ ρρ
( ) 2
1213
121
, mbabtItI areaBBBB === ′′ ρρ
( )22121
,, bamIII massBBmassAACC +=+= ′′′
• For centroidal axes on a circular plate,
( ) 2
414
41
, mrrtItII areaAABBAA ==== ′′′ πρρ
Area Moments of InertiaMoments of Inertia of a 3D Body by Integration
• Moment of inertia of a homogeneous body
is obtained from double or triple
integrations of the form
∫= dVrI2ρ
• For bodies with two planes of symmetry,
the moment of inertia may be obtained
from a single integration by choosing thin
slabs perpendicular to the planes of
symmetry for dm.
• The moment of inertia with respect to a
particular axis for a composite body may
be obtained by adding the moments of
inertia with respect to the same axis of the
components.
Q. No. Determine the moment of inertia of a slender rod of
length L and mass m with respect to an axis which is
perpendicular to the rod and passes through one end of the rod.
Solution
Q. No. For the homogeneous rectangular prism shown,
determine the moment of inertia with respect to z axis
Solution
Area Moments of InertiaMI of some common geometric shapes
Q. Determine the angle α which locates the principal axes of inertia through point O for the
rectangular area (Figure 5). Construct the Mohr’s circle of inertia and specify the
corresponding values of Imax and Imin.
x
y
y’
x’
O
2b
b
α
Ix = B
�(2�) �
E
��
Iy = B
2�(�) �
�
��
Ixy = 2��(F
�)(�) � ��
With this data, plot the Mohr’s circle, and using trigonometry
calculate the angle 2 α
2 α =tan-1(b4/b4) = 45o
Therefore, α =22.5o (clockwise w.r.t. x)
From Mohr’s Circle:
Imax = G
�� + �� 2 � 3.08��
Imin = G
�� − �� 2 � 0.252��
Q. No. Determine the moments and product of inertia of the area of the square with respect to the x'- y' axes.
Ix =B
� � =
B
��; Iy =
B
��
Ixy = 0 + (F
�×
F
�× ��) =
B
���
With θ = 30o, using the equation of moment of inertia about any inclined axes
���
= �� + �
2+
�� − �
2H34 25 − �� I67 25
��
= �� + �
2−
�� − �
2H34 25 + �� I67 25
����
= �� − �
2I67 25 + �� H34 25
we get,
�Ix' =FJ
+ 0 −
B
���I67 60° = (
B
−
E)�� = 0.1168��
�Iy' =FJ
+ 0 +
B
���I67 60° = (
B
+
E)�� = 0.5498��
�Ixy' = 0 + (B
��� ×
B
�) = (
B
E× ��) = 0.1250��
Alternatively, Mohr’s Circle may be
used to determine the three
quantities.