Transcript Structural Engineering - Indian Institute of Technology
ME101: Engineering Mechanics (3-1-0-8)
Jan-May 2014 Division I/III (Room No.: L1) Instructor
Dr. Arunasis Chakraborty
Room M-106 Department of Civil Engineering, IITG [email protected]
Phone: 258 2430 Web: www.iitg.ernet.in/arunasis shilloi.iitg.ernet.in/~arunasis 26-04-2020 1
Syllabus for 7
th
Week:
Parallel axis theorem Product of inertia Rotation of axes Principal moment of inertia Moment of inertia of simple and composite bodies
Mass moment of inertia
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MI of Standard Sections:
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Parallel Axis Theorem:
• Consider moment of inertia
I
with respect to the axis
AA’
of an area
A I
y
2
dA
• The axis
BB’
passes through the area centroid and is called a
centroidal axis.
Parallel Axis theorem: MI @ any axis = MI @ centroidal axis + Ad
2 The two axes should be parallel to each other.
Parallel Axis theorem
I
y
2
dA
y
d
2
dA
y
2
dA
2
d
y
dA
d
2
dA
• Second term = 0 since centroid lies on BB’ (∫
y’dA
=
y c A
, and
y c
= 0
I
I
Ad
2 4
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Example - Parallel Axis Theorem:
• Moment of inertia
I T
of a circular area with respect to a tangent to the circle
I T
I
Ad
2 1 4
r
4
r
2 5 4
r
4 • Moment of inertia of a triangle with respect to a centroidal axis
I A A
I B B
Ad
2
I B B
I A A
Ad
2 1 12
bh
3 1 2
bh
3
h
2 1 36
bh
3 5
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MI of Thin Plates:
• For a thin plate of uniform thickness material of density
t
and homogeneous r , the mass moment of inertia with respect to axis
AA’
contained in the plate is
I A A
r
r t
2
dm I
r
t A A
,
area
r
2
dA
• Similarly, for perpendicular axis
BB’
which is also contained in the plate,
I B B
r
t I B B
,
area
• For the axis
CC’ I C C
r
t
which is perpendicular to the plate,
J C
,
area
r
t
I A A
,
area
I B B
,
area
I A A
I B B
6
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Moments of Inertia of Thin Plates:
•
For the principal centroidal axes on a rectangular plate
I A A
r
t I A A
,
area
r
t
1 12
I B B
I C C
I
r
t I B B
,
area A A
,
mass
a
3
b
1 12
ma
2
I
r
B B
,
t
1 12
mass ab
3 1 12 1
m
12
a mb
2 2
b
2 •
For centroidal axes on a circular plate
I A A
I B B
r
t I A A
,
area
r
t
1 4 1 4
mr
2 7
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Composite Bodies and Figures:
Divide bodies or figures into several parts such that their mass centers can be conveniently determined Use Principle of Moment for all segments
m
1
m
2
m
3
X
m
1
x
1
m
2
x
2
m
3
x
3 Mass Center Coordinates can be written as:
X
m x m Y
m y m Z
m z m
m’s can be replaced by L’s, A’s, and V’s for lines, areas, and volumes 8
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Integration vs App Summation:
Reduce the problem to one of locating the centroid of area App Summation may be used instead of integration Divide the area into several strips Volume of each strip = AΔx Plot all such A against x.
Area under the plotted curve represents volume of whole body and the x-coordinate of the centroid of the area under the curve is given by:
x
A
x A
x
x c
x
Vx c V
Accuracy may be improved by reducing the width of the strip
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MI of 3D Body by Integration:
• Moment of inertia of a homogeneous body is obtained from double or triple integrations of the form
I
r
r
2
dV
• 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.
10
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Mass Moments of Inertia:
• •
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.m
2 Consider a three dimensional body of mass m Mass moment of inertia of this body about axis O-O:
I = ∫ r
2
dm
Integration is over the entire body.
r = perpendicular distance of the mass element dm from the axis O-O
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MI of Common Shapes:
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Theorems of Pappus:
- method for calculating surface area generated by revolving a plane curve about a non-intersecting axis in the plane of the curve - method for calculating volume generated by revolving an area about a non-intersecting axis in the plane of the area Surface Area Area of the ring element: circumference times dL dA = 2πy dL
If area is revolved through an angle θ<2π
Total area,
A
2
y dL
θ in radians
y L
y dL
A
2
y L
y L y
is the y-coordinate of the centroid C for the line of length L Generated area is the same as the lateral area of a right circular cylinder of length L and radius
y
Theorems of Pappus can also be used to determine centroid of plane curves if area created by revolving these figures @ a non intersecting axis is known
13
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Theorems of Pappus:
method for calculating volume generated by revolving an area about a non-intersecting axis in the plane of the area Volume Volume of the ring element: circumference times dA dV = 2πy dA Total Volume,
y A
y dA
V
V
2
y dA
2
y A
If area is revolved through an angle θ<2π θ in radians
V
y A y
is the y-coordinate of the centroid C of the revolved area A Generated volume is obtained by multiplying the generating area by the circumference of the circular path described by its centroid.
Theorems of Pappus can also be used to determine centroid of plane areas if volume created by revolving these figures @ a non-intersecting axis is known
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Example:
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Determine the volume V and surface area A of the complete torus of circular cross section The torus can be generated by revolving the circular area of radius a through 360 o about z-axis
V
r A
2 2 2
Ra
2
Similarly,
A
r L
2 2
a
4 2
Ra
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Example:
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Calculate the volume V of the solid generated by revolving the 60mm right triangular area through 180 o about z-axis The volume after revolving the area by 180 o about z-axis
V
r A
30 1 3 60 1 2 60 .
60 2 .
83
E
5
The mass of the body if density is 7800kg/m 3 (i.e. steel)
m
r
V
7830 2 .
83
E
5 1 1000 3 2 .
21
kg
16
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Example 5/94:
In order to provide sufficient support for the stone masonry arch designed as show, it is necessary to know the total weight W. Given the density is 2.40Mg/m 3
V W
r A
r
gV
r
A
8 2 2 3 1 .
5 2 * 2 * 1 .
5 1 .
5 2 2 2 3 .
5
m
2 3
7 .
05
3 2400 * 9 .
81 * 25 .
8
m
3 25 .
8 / 1000 608
kN
7 .
05
m
17
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Example:
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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.
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Solution:
a b A
120 -
1 2 4
r
3
r
2
a
3 38 .
2
mm 81.8
mm
1 2 2 12 .
72 10 3
mm 2
• • •
Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis.
Rectangle –
I x
1 3
bh
3 1 3 3 138 .
2 10 6
mm
4
Half Circle –
I A A
1 8
r
4 1 8 4 25 .
76 10 6 mm 4
Moment of Inertia w.r.t. x’ –
I x
I A A
Aa
2 25 .
76 10 6 12 .
72 10 3 38 .
2 2 7 .
20 10 6
mm 4
Moment of Inertia w.r.t. x –
I x
x
Ab
2 7 .
20 10 6 12 .
72 10 3 81 .
8 2 92 .
3 10 6
mm 4
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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
I x
I x
45 .
9 10 6 mm 4 138 .
2 10 6 mm 4 92 .
3 10 6 mm 4 21
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Product of Inertia:
PI is important for problems involving unsymmetrical cross-sections and in calculation of MI about rotated axes.
Product of Inertia of area A w.r.t. x-y axes
I xy
xy dA
x y xydxdy
-
I xy
+
I xy
It may be +ve, -ve, or zero
+
I xy
-
I xy
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Product of Inertia:
Parallel axis theorem for products of inertia
I xy
I xy
x y A
When the x axis, the y axis, or both are an axis of symmetry, the product of inertia is zero
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Rotation of Axes:
Note:
x
y
x
cos
y
cos
y
sin
x
sin
I x
'
I y
'
I x
'
y
'
y
' 2
dA
x
' 2
dA
x
'
y
'
dA
y
cos
x
cos
x
sin 2
dA y
sin 2
dA
x
cos
y
sin
y
cos
x
sin
dA
sin 2 sin cos 1 cos 2 cos 2 2 1 / 2 sin 2 1 cos 2 cos 2 sin 2 2 cos 2 24
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Rotation of Axes: Moments and product of inertia w.r.t. new axes x’ and y’ ?
I I x
y
I x
y
'
I I x x
2 2
I I y y I x
2
I y
I I
sin
x x
2 2 2
I y I y
cos cos 2 2
I xy
cos 2
I xy
sin 2
I xy
sin 2
Qs. Determination of axes about which the MI is a max or a min?
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Rotation of Axes:
I x’
+ I
y’
Adding first two eqns: = I
x
+ I
y
= I
z
The Polar MI @ O
I I x
y
I x
y
'
I x I x
2
I I y y I x
2
I y
2
I I x x
sin 2 2 2
I y I y
cos 2 cos 2
I xy
cos
I xy I xy
2 sin 2 sin 2
Angle which makes I
x’
and I
y’
either max or min can be found by setting the derivative of either I
x’
w.r.t. θ equal to zero:
dI d
x
'
I
y
I x
sin 2 2
I xy
cos 2
or I
y’
0
Denoting this critical angle by α
tan 2
I y
2
I xy
I x
26
Rotation of Axes:
I I x
y
I x
y
'
I x
I x
2
I x
2 2
I y I y I y
I x I x
I y
2
I y
2 sin 2 cos 2 cos 2
I xy
cos
I xy
I xy
2 sin 2 sin 2 tan 2
I y
2
I xy
I x
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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
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Rotation of Axes:
I I x
y
I x
y
'
I x I x
2
I I y y I x
2
I y
2
I I x x
sin 2 2 2
I y I y
cos 2 cos 2
I xy
cos
I xy I xy
2 sin 2 sin 2 tan 2
I
2
I xy y
I x
sin 2 cos 2
I
2
I xy y
I x
Substituting in the third eqn for critical value of 2θ: I
x’y’
= 0
Product of Inertia I
x’y’
is zero for the Principal Axes of inertia
Substituting sin2α and cos2α in first two eqns for Principal Moments of Inertia:
I
max
I
min
I xy
@
I x
I x
0 2
I
2
I y y
1 2 1 2
I
I x x
I y
2
I y
2 4
I
2
xy
4
I
2
xy
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Mohr’s Circle of Inertia:
Following relations can be represented graphically by a diagram called Mohr’s Circle. For given values of I
x
, I
y
, & I
xy
values of I
x’
, I
y’
desired angle θ.
I x
I ave
2
I
2
x
y
, & I
R
2
x’y’
corresponding
I xy
I I
max min
I xy
@
I x
I x
0 2
I
2
I y y
1 2 1 2
I I
x x
I y
I y
2 2
may be determined from the diagram for any
4
I
2
xy
4
I
2
xy I ave
I x
I y
2
R
I x
I y
2 2
I
2
xy
I
At the points A and B, I
x’y’
= 0 and I
x’
is a maximum and minimum, respectively.
I
max, min
I ave
R
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Construction of Mohr’s Circle of Inertia:
I
max
I
min
I x
I I xy
@
I x
0 2 2
I y y
1 2 1 2
I x
I y
2 4
I
2
xy
I x
I y
2 4
I
2
xy
Choose horz axis
Choose vert axis
MI PI Point A – known {I
x
, I
xy
} Point B – known {I
y
, -I
xy
} Circle with dia AB Angle α for Area
Angle 2α to horz (same sense)
I max
, I
min
Angle x to x’ = θ
Angle OA to OC = 2θ
Same sense Point C
Point D
I x’ I y’
, I
x’y’
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Example (Product of Inertia):
Determine the product of inertia of the right triangle with respect to the x and y axes
I xy
1 24
b
2
h
2
SOLUTION:
y
h
1
x el
x x b dA
y dx
h
1
y el
1 2
y
1 2
h
1
x b x
dx b
Integrating
dI x
from
x
= 0 to
x
=
b
,
I xy
dI xy
x el y el dA
0
b
x
2
h
2
h
2
b
0
x
2
x
2
b
x
3 2
b
2
x b
2
dx dx
h
2
x
2 4
x
3 3
b
x
4 8
b
2
b
0 32
Example:
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Determine the moments of inertia I
x
and I
y
of the area of the thin semicircular ring about the x- and y-axes.
Also find the polar moment of inertia of the combined action of the three forces on the I
C
of the ring about its centroid C
.
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Solution:
As width of ring is very small so inner and outer radius is nearly same.
Complete ring: Polar Moment of inertia at
O
,
I O
Ar
2 2
rtr
2 2
r
3
t t<
And from our previous knowledge,
I O
I x
I y
Due to symmetry of complete ring,
I x = I y
I x
I y
I O
2
r
3
t
Half ring: Evaluating MI
I x
I y
Polar Moment of inertia at
C ,
1 2
r
3
t I C
and
I O
Ad
2
I O
1 2
2
r
3
t
r
3
t
2
r
3
t
rt
2
r
I C
r
3
t
1 4 2 0 .
5947
r
3
t
1 .
8684
r
3
t x
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Example:
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Solution:
Rectangle moment of inertia
1
b
4
I x
b
b
3 3 3
I y
b
4 3
Product of inertia
I xy
b
2
b
2
b
2
b
4 4
Rotation of axes,
I x
I y
I x
2
I y I x
2
I y
I x
I y
I x
2
I y
2 cos 2
I xy
sin 2
b
4 3 cos 2
I xy
sin 2
b
4 3 0
b
4 4 sin 2 30 0 .
1168
b
4 0
b
4 4 sin 2 30 0 .
5498
b
4
I x
y
I x
I y
2 sin 2
I xy
cos 2 0
b
4 4 cos 2 30 0 .
125
b
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Example:
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Determine the orientation of the principal axes of inertia through the centroid of the angle section.
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Solution:
Product of inertia at centroid of a rectangle is zero by symmetry Product of inertia at
C I
xy
, I 0
I
xy
, I
12 .
5
d x d y
7 .
5
A
I
I
xy
, II 0
I
xy
, II
12 .
5
d x d
7 .
5
y A
II 3 .
75 10 4 mm 4 3 .
75 10 4 mm 4
I xy
I xy
, I
I xy
, II 7 .
5 10 4 mm 4
Moment of inertia at
C I x
, I
I x
, I
d x
2
A
I 1 12
12 .
5
6 .
58 10 4 mm 4 38
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Solution (contd.):
I y
, I
I y
, I
d y
2
A
I 1 12
I I x
, II
y
, II
I x
, II
d x
2
A
II
I y
, II
d y
2
A
II 1 12 1 12
I
Combined
x
I x
, I
I x
, II 18 .
16 10 4 mm 4
I y
I y
, I
I y
, II 10 .
16 10 4 mm 4 7 .
5 12 .
5 7 .
5 7 .
58 10 4 mm 4
Principal axes
tan 2
I y
2
I xy
I x
2
10 .
16 7 .
5
18 .
16 1 .
875 11 .
58 10 4 mm 4 2 .
58 10 4 mm 4 31 .
0 39
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Example (Mohr’s Circle of Inertia):
The moments and product of inertia with respect to the x and y axes are I
x
= 7.24x106 mm 4 , I
y
2.61x106 mm 4 , and I
xy
= = 2.54x10
6 mm 4 . 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.
40
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Example (Mohr’s Circle of Inertia):
I x
7 .
24 10 6 mm 4
I y
2 .
61 10 6 mm 4
I xy
2 .
54 10 6 mm 4 SOLUTION: • Plot the points (
I x , I xy
) and (
I y ,-I xy
).
Construct Mohr’s circle based on the circle diameter between the points.
OC CD
I
1 2
ave
I x
1 2
I
y I
x
I y
2 .
315 4 .
925 10 6 10 mm 4 6 mm 4
R
DX
2 3 .
437 10 6 mm 4 • Based on the circle, determine the orientation of the principal axes and the principal moments of inertia.
m
23 .
8 tan 2
m
DX CD
1 .
097 2
m
47 .
6 41
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Example (Mohr’s Circle of Inertia):
OC
I ave
4 .
925 10 6 mm 4
R
3 .
437 10 6 mm 4 • 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(60 o ) = 120 o . The angle that CX’ forms with the horz is f = 120 o - 47.6
o = 72.4
o .
I x
'
OF
OC
C X
cos
I ave
R
cos 72 .
4
o I x
y
'
F X
C Y
sin
R
sin 72 .
4
o I x
5 .
96 10 6 mm 4
I y
'
OG
OC
C Y
cos
I ave
R
cos 72 .
4
o I x
y
3 .
28 10 6 mm 4
I y
3 .
89 10 6 mm 4 42
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Following students are advised to meet the Instructor immediately
•
Div. I
•
Ankit Tamta (130104010)
•
Lehar Bhandari (130104033)
•
Div. III
•
Ananti Saroj (130121004)
•
Divya Kumar Meena (130121012)
•
Nikhil Anand (130121021)
•
Vikram Kaviya (130123042)
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