Learning Objectives:

• Define Center of Gravity.

• Define and compute centroid of a simple area.

Centroid or geometric center of a plane figure is the arithmetic mean position of all the points of an object or figure. It is the point at which entire area can be assumed to be concentrated. You can balance the figure on centroid point because all points are supposed to symmetrically distributed about centroid. Centroid of basic shapes can be calculated by referring tables given on the next page. e.g. Centroid of a circle is at its center and centroid of a rectangle of sides b and d is at b/2, d/2.

Solved Example: 26-1-01

Centroid of a right-angled triangle lies on a vertical line, that lies:

A. Two-thirds of base from lighter side

B. Two-thirds of base from heavier side

C. Two-fifth of base from lighter side

D. Three-fifth of base from lighter side

Centroid of a right-angled triangle is one third from its heavier side, or two-thirds from its lighter side.

Correct Answer: A

Solved Example: 26-1-02

The CG of a semicircular plate of 66 cm diameter, from its base, is: (ISRO SAC Tech Asst Jul 2018)

A. $\dfrac{8}{33}$ cm

B. $\dfrac{1}{14}$ cm

C. $14$ cm

D. $\dfrac{63}{8}$ cm

Correct Answer: C

Learning Objectives:

• Calculate centroid of composite shapes, with or without cavities, using formula for centroid of primitive shapes.

Centroid of a composite area is found out by breaking that shape into series of smaller shapes for which we can locate centroid locations.

\[\bar{x} = \dfrac{\Sigma A_i x_i}{\Sigma A_i}, \ \bar{y} = \dfrac{\Sigma A_i y_i}{\Sigma A_i}\] In the above formula, if a shape is added (let’s say by welding) then the terms related to it will be positive. If a shape is removed (let’s say, a hole, cut or cavity) then the terms related to it will be negative.

Solved Example: 26-2-01

The centre of gravity of a triangle is at the point where three:

A. Perpendicular bisectors of the sides of the triangle meet

B. Bisectors of the angle of the triangle meet

C. Medians of the triangle meet

D. None of these.

A centroid of a triangle is the point where the three medians of the triangle meet. A median of a triangle is a line segment from one vertex to the mid point on the opposite side of the triangle.

Correct Answer: C

Solved Example: 26-2-02

If an area has one line of symmetry the centroid will:

A. Lie somewhere along the line symmetry

B. Lie anywhere on the area

C. Lie in the midpoint of the line of symmetry

D. Not lie on the line of symmetry

The centroid of a 2D surface is a point that corresponds to the center of gravity of a very thin homogeneous plate of the same area and shape. If the area (or section or body) has one line of symmetry, the centroid will lie somewhere along the line of symmetry.

Correct Answer: A

Solved Example: 26-2-03

Calculate centroid of the figure shown.

A. $\bar{x}$ = 6.98 cm , $\bar{y}$ = 12.73 cm

B. $\bar{x}$ = 12.73 cm , $\bar{y}$ = 6.98 cm

C. $\bar{x}$ = 5.29 cm , $\bar{y}$ = 16.22 cm

D. $\bar{x}$ = 16.22 cm , $\bar{y}$ = 5.29 cm

• For quarter circle Area = 314.16 \(cm^2\)
  
  x = \(\dfrac{4R}{3\pi}\) = 8.4883 cm
  y = \(\dfrac{4R}{3\pi}\) = 8.4883 cm
  

• For semi circle Area = 157.08 \(cm^2\)
  
  x = \(\dfrac{20}{2}\) = 10 cm
  y = \(\dfrac{4R}{3\pi}\) = 4.2441 cm
  

\[\bar{x} = \dfrac{\Sigma A_x}{\Sigma A} = \dfrac{1095.87}{157.08} = 6.98\ cm\] \[\bar{y} = \dfrac{\Sigma A_y}{\Sigma A} = \dfrac{2000}{157.08} = 12.73\ cm\]

Correct Answer: A

Solved Example: 26-2-04

Considering bottom left corner point as origin, what is the y coordinate of the centroid of the object shown?

A. 1.85 cm

B. 1.67 cm

C. 1.33 cm

D. 1.15 cm

Let part 1 be the rectangle and part 2 be the semicircle. \[A_1 = 6 \times 3 = 18\ cm^2\] \[y_1 = 1.5\] \[A_2 = \dfrac{(\pi)(4)}{2} = 6.28\ cm^2\] \begin{equation*} \begin{split} y_2 & = 3 - \dfrac{4R}{3\pi} \\ & = 3 - \dfrac{8}{3 \pi} \\ & = 2.1511\ cm \end{split} \end{equation*} \begin{equation*} \begin{split} \bar{y} & = \dfrac{A_1 y_1 - A_2 y_2}{A_1 -A_2} \\ & = \dfrac{18 \times 1.5 - 6.28 \times 2.1511}{18 - 6.28} \\ & = \dfrac{27 - 13.509}{11.72} \\ & = 1.15\ cm \end{split} \end{equation*}

Correct Answer: D

Solved Example: 26-2-05

For the composite lamina shown in red in the figure, determine the coordinates of its centroid. (CBCGS Engineering Mechanics- Dec 2018)

A. $\bar{x}$ = 7.98mm, $\bar{y}$ = 10.42mm

B. $\bar{x}$ = 5.48mm, $\bar{y}$ = 9.11mm

C. $\bar{x}$ = 4.33mm, $\bar{y}$ = -2.82mm

D. $\bar{x}$ = 1.44mm, $\bar{y}$ = -3.50mm

Part I = Large Rectangle
Area = 90 $\times$ 60 = 5400 mm$^2$
$\bar{x}$ = -30, $\bar{y}$ = 5

Part II = Small Rectangle
Area = 50 $\times$ 40 = 2000 mm$^2$
$\bar{x}$ = 25, $\bar{y}$ = -20

Part III = Quarter Circle
Area = $\dfrac{\pi}{4}$ $\times$ 50$^2$ = 1963.5 mm$^2$
$\bar{x}$ = $\dfrac{4R}{3\pi}$=21.22
$\bar{y}$ = $\dfrac{4R}{3\pi}$=21.22

Part IV = Triangle (Removed part)
Area = $\dfrac{1}{2}$ $\times$ 75 $\times$ 90 = 3375 mm$^2$
$\bar{x}$ =-35, $\bar{y}$ =-10

Correct Answer: A

Learning Objectives:

• Define moment of inertia.

• Understand importance of moment of inertia in rotational motion and bending.

• Understand terms such as polar moment of inertia, radius of gyration.

Area Moment of Inertia is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis.
The area moment of inertia indicates the resistance of a beam to bending and deflection, around an axis that lies in the cross-sectional plane.
Beams with higher area moments of inertia are used in construction because they offer more resistance to rotation.
Polar Moment of Inertia of an area about a point is equal to the sum of moments of inertia of the area about any two perpendicular axes in the area and passing through the same point.
The polar moment of inertia is a measure of an object’s ability to resist torsion as a function of its shape.

\[J = I_x+ I_y\]

Radius of Gyration is the distance from a reference axis at which all of the area can be considered to be concentrated to produce the moment of inertia.

\[r_{x}=\sqrt{\dfrac{I_{x}}{A}}\]

where,
\(r_{x}\) = radius of gyration,
\(I_{x}\) = Moment of Inertia about x axis
A = Area of cross section

Solved Example: 26-3-01

The SI unit for polar moment of inertia is:

A. $kg.m^2$

B. $kg.m^4$

C. $m^4$

D. $m^2$

$J = \int \int r^2 dA$
The SI unit for polar moment of inertia, like the area moment of inertia, is meters to the fourth power (m$^4$), and inches to the fourth power (in$^4$) in imperial units.

Correct Answer: C

Solved Example: 26-3-02

Determine the radius of gyration k$_y$ of the area between parabola and x-axis.

A. 76.5 mm

B. 17.89 mm

C. 78.6 mm

D. 28.3 mm

Consider an elemental strip parallel to y-axis.

Its area will be: $dA = y dx = 0.1(1600-x^2) dx$

Its moment of inertia will be:

substitute the values of I,A in k$_y$ we get

Correct Answer: B

Solved Example: 26-3-03

The second moment of area is an important value which is used to ____________. It can also be called moment of inertia.

A. Determine the state of stress in a section

B. Calculate the resistance to buckling

C. Determine the amount of deflection in a beam

D. All of the above.

In structural engineering, the second moment of area of a beam is an important property used in the calculation of the beam's deflection and the calculation of stress caused by a moment applied to the beam.

Correct Answer: D

Solved Example: 26-3-04

Second moment of area is the product of:

A. Area and square of the distance from the reference axis

B. Area and distance from the reference axis

C. Square of the area and distance from the reference axis

D. Square of the area and square of the distance from the reference axis

For a section component having an axis of symmetry that is parallel to either of the section reference axes, the product second moment of area is the product of the coordinates of its centroid multiplied by its area.

Correct Answer: A

Solved Example: 26-3-05

In a square plate, a circular hole is drilled at the center. Which of the following will most likely happen? (Prof. Prashant More's Strength of Materials Exam- Dec 2010)

A. Its Center of Gravity (CG) will remain unchanged and Moment of Inertia (MI) will decrease.

B. Its Center of Gravity (CG) and Moment of Inertia (MI) both will remain unchanged as long as drilled hole is at the center of the plate.

C. Its Center of Gravity (CG) will remain unchanged and Moment of Inertia (MI) will increase.

D. Its Center of Gravity (CG) will shift and Moment of Inertia (MI) will increase.

In a square plate, when a circular hole is drilled at the center, the mass is removed symmetrically about the existing center of gravity, so its CG will not change.
However, the MI will decrease to reflect the reduction in mass because of removal of material.

Correct Answer: A

Solved Example: 26-3-06

The moment of inertia of the area under the curve y=kx$^2$ shown in the following figure about x-axis is:

A. $\dfrac{ab^3}{4}$

B. $\dfrac{ab^3}{9}$

C. $\dfrac{ab^3}{16}$

D. $\dfrac{ab^3}{21}$

Correct Answer: D

Learning Objectives:

• Calculate polar moment of inertia using perpendicular axes theorem.

The moment of inertia of an area about an axis perpendicular to its plane at any point, is equal to the sum of moment of inertia about any two mutually perpendicular axes through the same point and lying in the plane of the area.

\[I_{z}= I_{x} + I_{y}\] where,
\(I_{x}\), \(I_{y}\) and \(I_{z}\) represent Moment of Inertia about x, y and z axis respectively.

Solved Example: 26-4-01

The polar moment of inertia of a hollow shaft of outer diameter (D) and inner diameter (d) is: (UPRVUNL JE ME Oct 2021- Shift II)

A. $\dfrac{\pi}{16}(D^3-d^3)$

B. $\dfrac{\pi}{16}(D^4-d^4)$

C. $\dfrac{\pi}{32}(D^4-d^4)$

D. $\dfrac{\pi}{64}(D^4-d^4)$

The polar moment of inertia is a shaft or beam's resistance to being distorted by torsion, as a function of its shape. The solid shaft will have polar moment of inertia = $\dfrac{\pi}{32}(D^4)$ and the hollow shaft will have polar moment of inertia = $\dfrac{\pi}{32}(D^4-d^4)$.

Correct Answer: C

Learning Objectives:

• To understand the parallel-axis theorem and its applications.

• Use the parallel axis theorem to calculate moment of inertia of a body or system about various axes.

The moment of inertia of an area about any axis is defined as the moment of inertia of the area about a parallel centroidal axis plus a term equal to the area multiplied by the square of the perpendicular distance d from the centroidal axis to the axis in question.

\[I'_{x} = I_{x_{c}}+ d_{x}^{2}A\]

where,
\(I'_{x}\) = new moment of inertia about non-centroidal axis
\(I_{x_{c}}\) = moment of inertia about centroidal axis
\(d_{x}\)= distance between centroidal and non-centroidal axes
A = Area of cross section

Solved Example: 26-5-01

The parallel axis theorem gives you:

A. Location of the axes to get minimum moment of inertia

B. Polar moment of Inertia

C. Modified moment of inertia in case axis of rotation does not match with the centroidal axis.

D. Location of centroidal axes in case one of the principal moments of inertia are missing

The Parallel Axis Theorem provides a convenient way to calculate the moment of inertia for an object about an different parallel axes.
For example, the moments of inertia of a thin rod about its center is different from the moment of inertial about its end. The parallel axis theorem states that the moment of inertia about an axis parallel to the axis passing through the center of mass, and separated by a distance is: \[I = I_c+ A \cdot d^2\]

Correct Answer: C

Solved Example: 26-5-02

Moment of inertia of an area about an axis is equal to the sum of moment of inertia about an axis passing through the centroid parallel to the given axis and :

A. Area and square of the distance between two parallel axes

B. Area and distance between two parallel axes

C. Square of the area and distance between two parallel axes

D. Square of the area and square of the distance between two parallel axes

\[I = I_c+ A \cdot d^2\] So the extra added term is Ad$^2$. So the extra term is product of area and square of the distance between two parallel axes.

Correct Answer: A

Solved Example: 26-5-03

Moment of Inertia of disc about an axis which is tangent and parallel to its plane is I. Then the moment of inertia of disc about a tangent, but perpendicular to its plane will be:

A. $\dfrac{3I}{4}$

B. $\dfrac{3I}{2}$

C. $\dfrac{5I}{6}$

D. $\dfrac{6I}{5}$

MI of disc about tangent in a place \[= \dfrac{5}{4}MR^2 = I MR^2 = \dfrac{4}{5}I\] MI of disc about tangent perpendicular to plane \[I' = \dfrac{3}{2}MR^2\] \[I' = \dfrac{3}{2}(\dfrac{4}{5}I)R^2 = \dfrac{6}{5}I\]

Correct Answer: D

Solved Example: 26-5-04

A vertical pole of height 30m rotates and falls on ground without slipping. On hitting ground, the velocity, in m/s, at the top of the pole is: (take g = 10 $m/s^2$)

A. 3

B. 10

C. 30

D. 15

Potential Energy of center of gravity before falling, \[mgh = mg\dfrac{L}{2}\] Rotational Kinetic Energy after falling, \[KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2} \dfrac{mL^2}{3} \omega^2 \] Equating these two, \[\omega = 1 \ rad/s\] Velocity of tip of the rod, \[v = r .\omega = 30 . 1 = 30\ m/s\]

Correct Answer: C