Engineering Mechanics MEI 101
Distributed Forces
Dr. Antarip Poddar
Department of Mechanical Engineering
IIT (ISM) Dhanbad
Email: antarip@iitism.ac.in
Center of Gravity of 2D Body
A distributed system of forces can be replaced by a
single resultant force acting at a specific point on an object.
The specific point is called the object’s center of gravity.
Center of
gravity of a
plate
Forces
acting in
3D
𝑀𝑦 :
𝑥𝑊
𝑀𝑥 :
𝑦𝑊 =
=
𝑥Δ𝑊 =
𝑦Δ𝑊 =
𝑥𝑑𝑊
𝑦𝑑𝑊
Center of Gravity of 2D Body
Center of gravity of a wire
The center of gravity of a wire is the point where the
resultant weight of the wire acts. The center of gravity
may not actually be located on the wire.
Centroids of Areas and Lines
Centroid of an area:
𝑥𝑊 =
𝑥𝑑𝑊
𝑦𝑊 =
𝑦𝑑𝑊
𝑊 =𝛾×𝑡×𝐴
𝑥𝐴 =
𝑥𝑑𝐴 = 𝑄𝑦
𝑑𝑊 = 𝛾 × 𝑡 × 𝑑𝐴
Assume:
homogeneous plate of
uniform thickness
= first moment with respect to 𝑦
𝑦𝐴 =
𝑦𝑑𝐴 = 𝑄𝑥
= first moment with respect to 𝑥
Centroids of Areas and Lines
Centroid of a line
Assume:
homogeneous wire of
uniform cross section
An area is symmetric with respect to an axis BB’ if
for every point P there exists a point P’ such that PP’
is perpendicular to BB’ and is divided into two equal
parts by BB’.
The first moment of an area with respect to a
line of symmetry is zero.
𝑥𝐴 =
𝑥𝑑𝐴 = 𝑄𝑦
= first moment with respect to 𝑦
• If an area possesses an axis of symmetry, its
centroid lies on that axis.
If an area possesses two lines of symmetry?
its centroid lies at their intersection.
• An area is symmetric with respect to a center O
if for every element dA at (x,y) there exists an
area dA’ of equal area at (-x,-y).
𝑥𝐴 =
𝑥𝑑𝐴 = 𝑄𝑦 = first moment with respect to 𝑦
𝑦𝐴 =
𝑦𝑑𝐴 = 𝑄𝑥 = first moment with respect to 𝑥
Here we have
It follows:
The centroid of the area coincides with the
center of symmetry.
Centroids of Common Shapes
Composite Plates/Areas
In many instances, we can divide a flat plate into
rectangles, triangles, or the other common shapes
𝑜𝑟, 𝑋
We can determine the abscissa 𝑋 of the plate’s center of
gravity G from the abscissas 𝑥1 , 𝑥2 , . . . , 𝑥𝑛 of the centers
of gravity of the various parts.
Equate the moment of the weight of the whole plate about
the y-axis to the sum of the moments of the weights of the
various parts about the same axis:
𝑊=
𝑥𝑊
Centroid of Areas: Example 1
For the plane area shown, determine
the
(a) first moments with respect to the x
and y axes and
(b) the location of the centroid.
Strategy: Break up the given area into simple components,
find the centroid of each component, and
then find the overall first moments and centroid.
Centroid of Areas: Example 1
𝑄𝑥 = +506.2 × 103 mm3
𝑄𝑦 = +757.7 × 103 mm3
𝑋=
𝑥𝐴 +757.7 × 103 mm3
=
𝐴
13.828 × 103 mm2
𝑋 = 54.8 mm
𝑌=
𝑦𝐴 +506.2 × 103 mm3
=
𝐴
13.828 × 103 mm2
𝑌 = 36.6 mm
Determination of Centroids by Integration
For an area bounded by analytical curves
(i.e., curves defined by algebraic equations)
If the element of area dA is a small rectangle of sides
dx and dy, evaluating each of these integrals requires
a double integration with respect to x and y.
Double integration to find the first moment may be
avoided by defining dA as a thin rectangle or strip.
𝑥𝑒𝑙 , 𝑦𝑒𝑙 : coordinates of the centroid of the element dA
Vertical rectangular strip
𝑥𝐴 =
𝑥𝑒𝑙 𝑑𝐴 =
𝑥 𝑦𝑑𝑥
𝑦𝐴 =
𝑦𝑒𝑙 𝑑𝐴 =
𝑦
2
𝑦𝑑𝑥
Determination of Centroids by Integration
Horizontal rectangular strip
Choose the coordinate
system which best matches
the boundaries of the figure.
Triangular
sector
Polar coordinates
A double integration is also necessary if we use
polar coordinates for which
dA is a small element with sides r and r dθ
𝑥𝐴 =
=
𝑥𝑒𝑙 𝑑𝐴
2𝑟
1
cos 𝜃 𝑟 2 𝑑𝜃
3
2
𝑦𝐴 =
=
𝑦𝑒𝑙 𝑑𝐴
2𝑟
1
sin 𝜃 𝑟 2 𝑑𝜃
3
2
Centroid by Integration: Example 2
SOLUTION:
• Determine the constant k.
𝑦 = 𝑘𝑥 2
𝑘𝑎2
𝑏=
𝑏
⇒ 𝑘= 2
𝑎
𝑏 2
𝑦 = 2𝑥
𝑎
• Evaluate the total area.
Determine by direct integration
the location of the centroid of a
parabolic spandrel.
𝐴=
𝑎
=
0
=
𝑑𝐴 =
𝑦𝑑𝑥
𝑏 2
𝑏 𝑥3
𝑥 𝑑𝑥 = 2
2
𝑎
𝑎 3
𝑎𝑏
3
𝑎
0
Centroid by Integration: Example 2
First moments:
𝑎
𝑄𝑦 =
𝑥𝑒𝑙 𝑑𝐴 =
𝑥𝑦𝑑𝑥 =
0
𝑄𝑥 =
𝑦𝑒𝑙 𝑑𝐴 =
𝑦
𝑦𝑑𝑥 =
2
𝑏 2
𝑏 𝑥4
𝑥 2 𝑥 𝑑𝑥 = 2
𝑎
𝑎 4
𝑎
0
1 𝑏 2
𝑥
2 𝑎2
• Centroid coordinates:
xA  Q y
ab a 2 b
x

3
4
3
x a
4
yA  Q x
ab ab 2
y

3
10
y
3
b
10
2
𝑎
0
𝑎2 𝑏
=
4
𝑏2 𝑥 5
𝑑𝑥 =
2𝑎4 5
𝑎
0
𝑎𝑏 2
=
10
Moment of Inertia
Whenever a distributed load acts perpendicular to an area and
its intensity varies linearly, the calculation of the moment of
the loading about an axis will involve an integral of the form
𝑦 2 𝑑𝐴
“second moment” of the area
about an axis (the x axis), or
moment of inertia of the area.
Example: Consider the plate, submerged in a fluid
and subjected to the pressure p.
This pressure varies linearly with depth, such that 𝑝 = 𝛾𝑦
∴ Force acting on differential area dA of the plate:
𝑑𝐹 = 𝑝 𝑑𝐴 = 𝛾𝑦 𝑑𝐴
The moment of this force about the x axis:
𝑑𝑀 = 𝑦 𝑑𝐹 = 𝛾𝑦 2 𝑑𝐴
Integrating dM over the entire area of the plate: 𝑀 = 𝛾
This integral has no physical meaning, it often arises in formulas used in
fluid mechanics, mechanics of materials, structural mechanics, and design.
𝑦 2 𝑑𝐴
Polar Moment of Inertia
Rectangular moments of inertia,
• The polar moment of inertia is an important parameter in
problems involving torsion of cylindrical shafts and rotations of
slabs.
𝑟 2 𝑑𝐴
𝐽0 =
• The polar moment of inertia is related to
rectangular moments of inertia,
𝐽0 =
=
=
𝑟 2 𝑑𝐴
𝑥 2 + 𝑦 2 𝑑𝐴
𝑥 2 𝑑𝐴 +
= 𝐼𝑦 + 𝐼𝑥
𝑦 2 𝑑𝐴
It is seen that 𝐼𝑥 , 𝐼𝑦 , 𝐽0 will
always be positive since they
involve the product of
distance squared and area.
Radius of Gyration of an Area
Consider area A with moment of inertia Ix.
Imagine that the same total area A is concentrated in a thin
strip parallel to the x axis with equivalent Ix.
𝐼𝑥 = 𝑘𝑥2 𝐴
𝑘𝑥 =
kx = radius of gyration with respect
to the x axis
𝐼𝑥
𝐴
Similarly,
𝐼𝑦 =
𝑘𝑦2 𝐴
𝐽𝑂 =
𝑘𝑦 =
𝑘𝑂2 𝐴
𝐼𝑦
𝐴
𝑘𝑂 =
𝐽𝑂
𝐴
𝑘𝑂2 = 𝑘𝑥2 + 𝑘𝑦2
Moment of Inertia: Example
SOLUTION:
• A differential strip parallel to the x axis is chosen for dA.
𝑑𝐼𝑥 = 𝑦 2 𝑑𝐴
𝑑𝐴 = 𝑙𝑑𝑦
• For similar triangles,
𝑙 ℎ−𝑦
=
𝑏
ℎ
ℎ−𝑦
𝑙=𝑏
ℎ
ℎ−𝑦
𝑑𝐴 = 𝑏
𝑑𝑦
ℎ
• Integrating dIx from y = 0 to y = h,
Determine the moment of
inertia of a triangle with
respect to its base.
𝑦 2 𝑑𝐴
𝐼𝑥 =
ℎ
𝑦2𝑏
=
0
ℎ−𝑦
𝑑𝑦
ℎ
𝑏ℎ3
𝐼𝑥 =
12
Moment of Inertia: Example
Why not a differential strip perpendicular to the base?
A
𝑙1
𝑙2
B
C
D
𝑑𝐼𝑥 = 𝑦 2 𝑑𝐴
𝑑𝐼𝑦 = 𝑥 2 𝑑𝐴
𝑑𝐴 = 𝑙𝑑𝑦
𝑑𝐴 = 𝑙 𝑑x
𝑙 is step function. Thus, two integrals will be required.
𝐼𝑦 =
𝑥 2 𝑙 𝑑𝑥 =
𝐷 2
𝑥 𝑙1
𝐵
𝑑𝑥+
𝐶 2
𝑥 𝑙2
𝐷
𝑑𝑥
Example
SOLUTION:
• An annular differential area element is chosen,
𝑑𝐽𝑂 = 𝑢2 𝑑𝐴
𝐽𝑂 =
𝑟
𝑑𝐴 = 2𝜋𝑢𝑑𝑢
𝑢2 2𝜋𝑢𝑑𝑢
𝑑𝐽𝑂 =
0
𝜋 4
𝐽𝑂 = 𝑟
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.
• From symmetry, Ix = Iy,
JO  I x  I y  2I x

2
r 4  2I x
I diameter  I x 

4
r4
Parallel Axis Theorem
0
The second integral is zero since the x’
axis passes through the area’s centroid C
The moment of inertia of an area with respect
to an axis 𝑥 can be determined from its
moment of inertia with respect to the
centroidal axis 𝑥′ by a calculation involving
the distance 𝑑𝑦 between the axes.
Similarly,
Moment 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.
Moment of Inertia of Composite Areas: Example
SOLUTION:
1 3 1
Rectangle: 𝐼𝑥 = 𝑏ℎ = 240 120
3
3
6
= 138.2 × 10 mm4
Determine the moment of
inertia of the shaded area
with respect to the x axis.
From chart:
Moment of Inertia of Composite Areas: Example
4𝑟
4 90
𝑎=
=
= 38.2 mm
3𝜋
3𝜋
b = 120−a = 81.8 mm
1
2
1
2
𝐴 = 𝜋𝑟 2 = 𝜋 90
Moment of inertia with respect to AA’,
Use parallel axis theorem to get the
MOI about centroidal axis x’:
Moment of inertia with respect to x,
1
1
𝐼𝐴𝐴′ = 𝜋𝑟 4 = 𝜋 90
8
8
4
2
mm2
= 25.76 × 106 mm4
𝐼𝑥′ = 𝐼𝐴𝐴′ − 𝐴𝑎2 = 25.76 × 106 12.72 × 103 = 7.20 × 106 mm4
𝐼𝑥 = 𝐼𝑥′ + 𝐴𝑏 2 = 7.20 × 106 + 12.72 × 103 81.8
= 92.3 × 106 mm4
2
Moment of Inertia of Composite Areas: Example
• The moment of inertia of the shaded area is obtained by subtracting the moment of
inertia of the semicircle from the moment of inertia of the rectangle.
𝐼𝑥
=
138.2 × 106 mm4
−
𝐼𝑥 = 45.9 × 106 mm4
92.3 × 106 mm4
Product of Inertia
The moments of inertia of an area can have different values depending
on what axes we use to calculate them.
Important to determine the maximum and minimum values of the
moments of inertia, which means finding the particular orientation of
axes that produce these values.
The first step in calculating moments of inertia with regard to rotated
axes is to determine a new kind of second moment, called the product
of inertia.
Product of Inertia with
respect to x and y axis:
Product of Inertia
Since x or y may be negative, the product of inertia
may either be positive, negative, or zero, depending on
the location and orientation of the coordinate axes.
𝑑𝐼𝑥 = 𝑥𝑦 𝑑𝐴
When the x axis, the y axis, or both are an axis
of symmetry, the product of inertia is zero.
𝑑𝐼𝑥 = −𝑥𝑦 𝑑𝐴
Parallel axis theorem for products of inertia
The second and third integrals are zero since the moments
of the area are taken about the centroidal axis.
Principal Axes and Principal Moments of Inertia
Given
𝐼𝑥 =
𝑦 2 𝑑𝐴
𝐼𝑦 =
𝑥 2 𝑑𝐴
𝐼𝑥𝑦 =
𝑥𝑦𝑑𝐴
we wish to determine moments and product of
inertia with respect to new axes x’ and y’, rotating
axes about the origin through angle θ.
′
𝑥 = 𝑥 cos 𝜃 + 𝑦 sin 𝜃
𝑦 ′ = 𝑦 cos 𝜃 − 𝑥 sin 𝜃
Learn about
coordinate transformation
matrix/ Rotation matrix
Principal Axes and Principal Moments of Inertia
𝐼𝑥 ′ =
𝐼
𝑦′
=
𝐼𝑥 +𝐼𝑦
2
𝐼𝑥 +𝐼𝑦
𝐼𝑥 ′ 𝑦 ′ =
2
+
−
𝐼𝑥 −𝐼𝑦
2
𝐼𝑥 −𝐼𝑦
2
𝐼𝑥 −𝐼𝑦
2
cos 2 𝜃 − 𝐼𝑥𝑦 sin 2 𝜃
(i)
cos 2 𝜃 + 𝐼𝑥𝑦 sin 2 𝜃
(ii)
sin 2 𝜃 + 𝐼𝑥𝑦 cos 2 𝜃
(iii)
The equations for Ix’ and Ix’y’ are the
parametric equations for a circle,
i.e. for any given value of the parameter
θ, all of the points will lie on a circle.
MOHR’S CIRCLE
𝐼𝑥′ − 𝐼𝑎𝑣𝑒
Eliminate θ from
(i) and (iii):
𝐼𝑎𝑣𝑒
𝐼𝑥 + 𝐼𝑦
=
2
2
+ 𝐼𝑥2′ 𝑦′ = 𝑅2
𝑅=
𝐼𝑥 − 𝐼𝑦
2
2
2
+ 𝐼𝑥𝑦
Principal Axes and Principal Moments of Inertia
Eqs. (ii) and (iii) are parametric equations of the same circle.
Because of the symmetry of the circle about the horizontal axis,
we would obtain the same result if we plot a point N of
coordinates Iy’ and -Ix’y’
Principal Axes and Principle Moments of Inertia
• At the points A and B, Ix’y’ = 0 and Ix’ is a maximum
and minimum, respectively.
I max, min  I ave  R
2θm +
Ix’y’ = 0
180o
2θm
𝐼𝑎𝑣𝑒
2
2
sin 2 𝜃 + 𝐼𝑥𝑦 cos 2 𝜃 = 0
2𝐼𝑥𝑦
tan 2 𝜃𝑚 = −
𝐼𝑥 − 𝐼𝑦
N
𝐼𝑥′ − 𝐼𝑎𝑣𝑒
∴
𝐼𝑥 −𝐼𝑦
• The equation for 2θm defines two angles, 180o apart
which correspond to the principal axes of the area
about O. or,
+ 𝐼𝑥2′ 𝑦′ = 𝑅2
𝐼𝑥 + 𝐼𝑦
=
2
𝑅=
𝐼𝑥 − 𝐼𝑦
2
• θm : two angles 90o apart.
2
2
+ 𝐼𝑥𝑦
• Imax and Imin are the principal moments of inertia of the
area about O.
Principal Moment of Inertia: Example
For the section shown, the moments of
inertia with respect to the x and y axes are Ix
= 4.32 x 10-6 m4 and Iy = 2.901 x 10-6 m4.
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.
Note: Product of inertia is not directly asked.
But it is required for mag. & orientation of principle MOI.
Principal Moment of Inertia: Example
SOLUTION:
Add three rectangles.
Apply the parallel axis theorem to each rectangle,
𝐼𝑥𝑦 =
𝐼𝑥′ 𝑦′ + 𝑥 𝑦𝐴
Note that the product of inertia with respect to centroidal
axes parallel to the xy axes is zero for each rectangle.
𝐼𝑥′ 𝑦′ = 0 (Recall axis of symmetry)
Rectangle Area, cm2
I
9.652
II
9.4742
III
9.652
𝑥, cm 𝑦, cm
−3.2 +4.4
0
0
+3.2 −4.4
xy𝐴,cm4
−135.9
0
−135.9
𝑥 𝑦𝐴 = −271.8
𝐼𝑥𝑦 =
𝑥 𝑦𝐴 = −2.718 x 10−6 m4
Principal Moment of Inertia: Example
Determine the orientation of the principal axes
and the principal moments of inertia.
2𝐼𝑥𝑦
−2 × 2.718 × 10−6
tan 2𝜃𝑚 = −
=−
= 3.83
𝐼𝑥 − 𝐼𝑦
4.32 − 2.901 × 10−6
𝜃𝑚 = 37.7° and 𝜃𝑚 = 127.7°
𝐼𝑚𝑎𝑥,𝑚𝑖𝑛
𝐼𝑥 + 𝐼𝑦
=
±
2
𝐼𝑥 − 𝐼𝑦
2
Given:
Ix = 4.32 x 10-6 m4
Iy = 2.901 x 10-6 m4
𝐼𝑚𝑎𝑥 = 6.4195 x 10−6 m4
𝐼𝑚𝑖𝑛 = 0.8015 x 10−6 m4
2
2
+ 𝐼𝑥𝑦
Mohr’s Circle for Moments and Products of Inertia
The moments and product of inertia for an area
are plotted as shown and used to construct
Mohr’s circle,
𝐼𝑎𝑣𝑒
𝑅=
𝐼𝑥 + 𝐼𝑦
=
2
𝐼𝑥 − 𝐼𝑦
2
+ 𝐼𝑥𝑦
2
• Mohr’s circle may be used to graphically or analytically
determine the moments and product of inertia for any other
rectangular axes including the principal axes and principal
moments and products of inertia.
Product of Inertia: Example
SOLUTION:
𝑦 =ℎ 1−
𝑥
𝑏
𝑥
𝑑𝑥
𝑏
1
1
𝑥
𝑦𝑒𝑙 = 𝑦 = ℎ 1 −
2
2
𝑏
𝑑𝐴 = 𝑦𝑑𝑥 = ℎ 1 −
𝑥𝑒𝑙 = 𝑥
b
I xy   dI xy   xel yel dA   x
0
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.
b

1
2
2
x
h 1   dx
 b
2
2
b
2

 x 2 x3 x 4 
x
x
x
2 
dx  h   
h  

2
 2 b 2b 2 
4
3
b
8b  0
0


𝐼𝑥𝑦
2
1 2 2
=
𝑏 ℎ
24
3
Product of Inertia: Example
• Apply the parallel axis theorem to evaluate the product of inertia with
respect to the centroidal axes.
1
𝑥= 𝑏
3
1
𝑦= ℎ
3
With the results from part (a),
𝐼𝑥𝑦 = 𝐼𝑥″𝑦″ + 𝑥 𝑦 𝐴
𝐼𝑥 ″ 𝑦 ″
𝐼𝑥 ″ 𝑦 ″
1 2 2
1
=
𝑏 ℎ −
𝑏
24
3
1 2 2
=− 𝑏 ℎ
72
1
ℎ
3
1
𝑏ℎ
2
Mohr’s Circle: Example
SOLUTION:
• Plot the points (Ix , Ixy) and (Iy ,-Ixy). Construct Mohr’s circle based
on the circle diameter between the points.
OC  I ave  12 I x  I y   4.925  106 mm 4
CD  12 I x  I y   2.315  106 mm 4
R
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
CD 2  DX 2
 3.437  106 mm 4
• Based on the circle, determine the
orientation of the principal axes and the
principal moments of inertia.
tan 2 m 
DX
 1.097 2 m  47.6
CD
 m  23.8
Mohr’s Circle: Example
• Based on the circle, evaluate the moments and product of inertia with
respect to the x’y’ axes.
𝐼𝑚𝑎𝑥 = 8.33 × 106 mm4
𝐼𝑚𝑖𝑛 = 1.47 × 106 mm4
Points X’ and Y’ corresponding to the x’ and y’ axes are obtained by
rotating CX and CY counterclockwise through an angle 2θ = 2(60o) = 120o.
The angle that CX’ forms with the x’ axes is ϕ = 120o - 47.6o = 72.4o.
𝐼𝑥′ = 𝑂𝐹 = 𝑂𝐶 + 𝐶𝑋 ′ cos 𝜙 = 𝐼𝑎𝑣𝑒 + 𝑅 cos 7 2. 4𝑜
𝐼𝑥′ = 5.96 × 106 mm4
𝐼𝑦′ = 𝑂𝐺 = 𝑂𝐶 − 𝐶𝑌 ′ cos 𝜙 = 𝐼𝑎𝑣𝑒 − 𝑅 cos 7 2. 4𝑜
𝐼𝑦′ = 3.89 × 106 mm4
OC  I ave  4.925  106 mm 4
R  3.437  106 mm 4
𝐼𝑥′ 𝑦′ = 𝐹𝑋 ′ = 𝐶𝑌 ′ sin 𝜙 = 𝑅 sin 7 2. 4𝑜
𝐼𝑥′ 𝑦′ = 3.28 × 106 mm4