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Learning Objectives:

Define centroid.
Define and compute centroid of a simple area.
Calculate centroid of composite shapes, with or without cavities, using formula for centroid of primitive shapes.

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:

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

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

C. $14$ cm

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

\[ R = \dfrac{66}{2} = 33\ cm\] \[ CG = \dfrac{4R}{3 \pi} = \dfrac{4 \times 33}{3 \pi} = 14\ cm\]

Correct Answer: C

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.

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

\begin{align*} \bar{x} &= \dfrac{\Sigma A_i x_i}{\Sigma A_i}\\ &= \dfrac{5400(-30) + 2000 (25) +1963.5 (21.22) - 3375(-35)}{5400 + 2000 + 1963.5 - 3375}\\ &= \dfrac{47790.5}{5988.5} = 7.98\ \mathrm{mm} \end{align*} \begin{align*} \bar{y}&= \dfrac{\Sigma A_i y_i}{\Sigma A_i}\\ &= \dfrac{5400(5) + 2000(-20) + 1963.5 (21.22) -3375(-10)}{5400 + 2000 + 1963.5 - 3375}\\ &= \dfrac{62415.5}{5988.5} = 10.42\ \mathrm{mm} \end{align*}

Correct Answer: A

Solved Example: 3435345

The centroid of a composite shape is calculated using:

A. The average of the centroids

B. The weighted average of the centroids based on area or mass

C. The geometric center of the shape

D. The centroid of the largest component

Correct Answer: B

Solved Example: 34535

Which of the following shapes has its centroid outside the shape?

A. Triangle

B. Rectangle

C. Ellipse

D. L-shape

Correct Answer: D

Learning Objectives:

Define area moment of inertia.
Understand importance of moment of inertia in rotational motion and bending.

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.

すじにくシチュー, CC0, via Wikimedia Commons

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?

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

Consider an elemental strip parallel to x-axis. Its area will be:

\[dA = (a-x) dy = (a-x) (2kxdx)\]

Its moment of inertia will be:

\[dI =dA \times y^2 =(a-x) (2kxdx) \times (kx^2)^2 = 2k^3 x^5 (a-x)dx\]

The limits are x=0 to x=a.

\begin{align*} I_x &= 2k^3 \int_{0}^{a} \left[ax^5 - x^6 \right]dx\\ &= 2k^3 \left[\dfrac{ax^6}{6} - \dfrac{x^7}{7} \right]_0 ^a\\ &= 2k^3 \left[\dfrac{a^7}{6} - \dfrac{a^7}{7}\right] = \dfrac{k^3 a^7}{21} \end{align*}

Using, $b = ka^2$ as it satisfies the condition y = kx$^2$ since the point (a,b) is on the curve.

\[I_x = \dfrac{k^3a^6.a}{21} = \dfrac{b^3.a}{21} = \dfrac{ab^3}{21}\]

Correct Answer: D

Solved Example: 345345

Which of the following shapes has the largest moment of inertia for the same mass and radius?

A. Solid sphere

B. Hollow sphere

C. Solid cylinder

D. Hollow cylinder

Correct Answer: D

Learning Objectives:

Calculate polar moment of inertia using perpendicular axis 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.

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:

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

Solved Example: 26-4-02

The moment of inertia of a thin uniform circular disc about one of its diameter is I. The moment of inertia about an axis perpendicular to the circular surface and passing through its centre is:

A. $\sqrt{2}I$

B. 2I

C. $\dfrac{1}{2}I$

D. $\dfrac{1}{\sqrt{2}}I$

Correct Answer: B

Solved Example: 343245

What does the Perpendicular Axis Theorem state?

A. The moment of inertia about any axis is the sum of the moments of inertia about the perpendicular axes.

B. The moment of inertia about a given axis is equal to the sum of the forces acting on an object.

C. The moment of inertia is independent of the axis chosen.

D. None of the above.

Correct Answer: A

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.

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

Learning Objectives:

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

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.

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

Solved Example: 26-3-04

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:

\[dI =dA \times r^2 =0.1(1600-x^2)dx \times x^2\] The limits are 0 to 40 (half parabola) and then multiply by 2. \begin{align*} I_y &= \int_{0}^{40} 0.1(1600-x^2)dx \times x^2 \times 2\\ &= \left[\dfrac{160x^3}{3}-0.1\dfrac{x^5}{5} \right]_0^{40} \times 2\\ &=2730666\ mm^4 \end{align*} Again, the limits are 0 to 40 (half parabola) and then multiply by 2. \begin{align*} A &= \int_{0}^{40} y dx \times 2\\ &= \int_{0}^{40} 0.1(1600-x^2) dx \times 2\\ &= \left[160x-0.1\dfrac{x^3}{3} \right]_0^{40} \times 2\\ &= 8533~mm^2 \end{align*}

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

\[k_y =\sqrt{\dfrac{I_y}{A}} =\sqrt{\dfrac{2730666}{8533}} =17.89\ mm\]

Correct Answer: B

Centroids and Moment of Inertia