INTRODUCTION
When I studied Mechanics of Materials, I thought ' This is the most realistic Study I have ever done '. I guess that's why Elon Musk answered at interview the question of What is most important Study in Engineering?
Mechanics of Materials let me know how much the Force is applied and different depends on its direction. For instance, Normal & Shear Forces, Stress, Torque, Deformation and so on.
So in this post, I'm going to introduce the fundamental concept of Mechanics of Materials, ' Moment of Inertia '. But you may refute that there're lots of blogs introduce it, why should I read this?
My answer is 'No'. There's no reason you must read this. But the difference is Not only get the result of each cases, also PROVE it !
Actually, I finished writing this post and came back for intros, so If you just need values of Moment of Inertia, it is highlighted with Blue color. If not, Let's study what Moment of Inertia is, together.
WHAT IS THE MOMENT OF INERTIA ?
The word ' Inertia ' is the Property that sustain current motion state (There's detailed description for 'Inertia' in IMU post https://loookup.tistory.com/4). Then we can guess the meaning of ' Moment of Inertia '.
Moment of Inertia is magnitude of property that an object who rotates on the rotational axis to sustain rotation. In other words, It is a quantity of how much it sensitively reacts to external Torques.
So if an object which has large Moment of Inertia, it's not only difficult to rotate but also difficult to stop when it's rotating. And it depends on Mass, Shape, Axis, Direction and so on.
It's denoted by ' I ' and expressed in Kg*m^2 in SI unit. In formula, I = L / ω ( L : Angular Momentum, ω : Angular Velocity). So If there're no external forces, Angular Velocity will be small when Moment of Inertia is large.
In Point-Mass, Moment of Inertia is defined as I = (Mass of Particle) * (Distance between Particle and Rotation-Axis)^2 = mr^2.
In System consists of particles, Moment of Inertia is defined as I = summation of (each particle's mass) * (Distance between Particle and Rotation-Axis)^2 = ∑ (mi)(ri)^2.
Let's find out Moment of Inertia of Areas also known as ' Second Moment of Area ' in 2-dimension. It is frequently used in engineering for structural design of beams and columns with cross-sectional areas.
But, it's obvious that ' First Moment of Area ' must exist if 'second' does. So let's find out both at once.
FIRST MOMENT OF AREA
' First moment of Area ' is a measure of the spatial distribution of a shape in relation to an axis. (From Wikipedia). It's not intuitive concept, so let's check the figure below.
G : Centroid
A : Area of plane figure
dA : Differential Area element
x̄, ȳ : Distance between Axis and Centroid
What is the Centroid ? Is it different from the Center of Mass ?
You might familiar with word ' Center of Mass ', but what is ' Centroid ' ? These are quite similar concept but the difference is What it depends on.
Centroid is geomatrical center of an object and it's not differ no matter what it's properties, but its shape. But Center of Mass depends on its density. If an object has uniform density, its Centroid and Center of Mass would be same, but if it's not, there're not same.
SECOND MOMENT OF AREA
In Rectangle
Then let's find out the Centroid and First,Second moment of Area in different Shape.
First case is Rectangle.
In this figure we suppose that this has uniform density, so Centroid and Center of Mass is same. We already knew where is the Center of Mass in Rectangle : ȳ = h/2, x̄ = b/2. And next, we can figure First Moment of Area for X,Y axis Qx, Qy.
First Moment of Area for X-axis Qx = y1A1 + y2A2 + ... = ∑Aiyi = ∫ y dA = ȳA.
First Moment of Area for Y-axis Qy = x1A1 + x2A2 + ... = ∑Aixi = ∫ x dA = x̄A.
Since A = bh, Qx is Qx = h/2 * bh = bh^2 /2, or by integral ∫ y dA = ∫∫ y dxdy = ∫ b*y dy = [by^2/2](0~h) = bh^2 /2.
And Qy is Qy = b/2 * bh = hb^2 /2, or by integral ∫ y dA =∫∫ y dxdy =∫ h*x dx = [hx^2/2](0~b) = hb^2 /2.
And next is Moment of Inertia, the Second Moment of Area. Not like first, we must consider rotational axis is on the Centroid or not. If it's on the Centroid,
Second Moment of Area for X-axis Ix = ∫ y^2 dA.
Second Moment of Area for Y-axis Iy = ∫ x^2 dA.
If it's not on the Centroid, Ix' = Ix + Ady^2 and Iy' = Iy + Adx^2 calls ' Parallel Axis Theorem '. A is the Area of an object and dx,dy is the Distance between rotational axis and parallel axis on the centroid.
Following the Definition, figure above's coordinate axis is not placed on the Centroid, so we move coordinate axis to the Centroid. Then domain X : -b/2 ~ b/2, and codomain Y : -h/2 ~ h/2.
Ix = ∫ y^2 dA = ∫∫ y^2 dxdy = ∫ [xy^2](-b/2~b/2) dy = ∫ by^2 dy = [by^3/3](-h/2~h/2) = bh^3/12.
Iy = ∫ x^2 dA =∫∫ x^2 dxdy =∫ [x^3/3](-b/2~b/2)dx =∫ b^3/12 dy = [yb^3/12](-h/2~h/2) = hb^3/12.
Go back to original figure, there's a distance between coordinate axis and centroid dy = h/2, dx = b/2.
Ix' = Ix + Ady^2 = bh^3/12 + bh*(h/2)^2 = hb^3/3.
Iy' = Iy + Adx^2 = hb^3/12 + bh*(b/2)^2 = bh^3/3.
So we got Moment of Inertia in Rectangle.
In Triangle
Second case is Triangle.
The Center of Mass in Triangle is the intersection with connected line between each vertex and middle point of opposite line. So we find out height of Centroid is height/3 : ȳ = h/3, but how can we get the width part?
Figure above is the common form of Triangle. You can always lay X-axis parallel to the base line of Triangle. Before we start, let's divide it into two right triangles because it is easy to calculate. Mechanics of Materials always push you to divide whole form into your taste.
Not like rectangle, 'A' should consider carefully.
In left side of triangle, A = ∫ dA = ∫ (h/a)x dx = h/a * [x^2/2](0~a) = ha/2.
And First Moment of Area for Y-axis Qy1 = ∫ x dA = ∫ (h/a)x^2 dx = h/a * [x^3/3](0~a) = ha^2/3.
Next in right side of triangle A = ∫ dA = ∫ (h/(b-a))(b - x) dx = h/(b-a) * [bx - x^2/2](a~b) =(b-a)*h/2.
And First Moment of Area for Y-axis Qy2 = ∫ x dA = ∫ (h/(b-a))(b - x)x dx
= h/(b-a) * [bx^2/2 - x^3/3](a~b)
= h/(b-a) * (b^3 -3a^2b + 2a^3)/6
= h/(b-a) * (b+2a) * (b-a)^2 / 6
= h(b-a)(b+2a)/6.
So Qy = Qy1 + Qy2 = ha^2/3 + h(b-a)(b+2a)/6 = h(b^2 + ab)/6 = hb(b+a)/6 and A = bh/2.
Then by equation above, Qy = x̄A can modifiy into x̄ = Qy/A = (a+b)/3.
Consequently, we get centroid x̄ = (a+b)/3 and ȳ = h/3, and also First Moment of Area for Y-axis Qy in triangle. (Qx is simple. Try it :)
Another Case in Triangle
And same with rectangle case, we move coordinate axis to the Centroid of triangle and find out Moment of Inertia. At previous process, we recognized all triangle form can divide with right triangles, so this time we assume right triangle form.
In this case be careful about Y-axis is placed in 1/3 of height.
Ix = ∫ y^2 dA = 2*∫ (b/2h)*(2h/3-y)*y^2 dy
= 2*(b/2h)*[2h/9*y^3 - y^4/4](-h/3~2h/3)
= (b/h)*(16/243-4/81+2/243+1/324)*h^4
= bh^3/36.
Wait, I intuitively expect Ix in triangle would be 1/2 compare with Ix in rectangle, but result is 1/3.
And next, we're going to get Iy with figure below.
At Parallel Axis Theorem, we knew Iy' equals Iy + Adx^2. In that equation, dx means perpendicular distance between Y-axis and Centroid so according to figure above, dx = 0.
Iy = ∫ x^2 dA = 2*∫ (2h/b)(b/2 -x)*x^2 dx
= (4h/b)*[(b/6)x^3 -x^4/4](0~b/2)
= (4h/b)*(b^4/48 - b^4/64)
= hb^3/48.
Not like Rectangle, Second Moment of Areas for X & Y-axis in Triangle are different. Be careful !
So we got Moment of Inertia in Triangle. If you figure out better idea, apply it.
In Half / Full Circle
Final case is Half / Full Circle !
First, Let's find out where Centroid is.
It is obvious that x̄ is placed at the center of half circle in this figure(x̄=0), so we're going to calculate ȳ only.
According to above equation, Qx = ∫ y dA = ȳA. Since this Circle's equation is x^2 + y^2 = r^2, we can modify it into fuction f(y) form :
y = sqrt(r^2 - x^2) (0 <= x <= r)
y = -sqrt(r^2 - x^2) (-r <= x <= 0)
At first, we're going to deal with quarter of circle at The First Quadrant (X>0 & Y>0), so
A = ∫ dA = ∫ y dx = ∫ sqrt( r^2 - x^2 ) dx(0~r)
-> Substitute x to rcosθ , then dx = -rsinθ dθ and The domain should change either x : 0~r >> θ : -π/2~0 or -θ : 0~π/2
= ∫ sqrt(r^2 - cos^2θ*r^2)*rsinθ dθ (0~π/2)
= ∫ r^2*sinθ^2 dθ
= ∫ r^2*(1-cos2θ)/2 dθ
= (r^2/2)*[θ - sin2θ/2] (0~π/2)
= (r^2/2)*π/2 = π*r^2/4.
It could be easier to get result with 'Circle's area is π*r^2 and its quarter is π*r^2/4', but Mathematics starts with Calculating more complex and fundamental ways. As I mentioned, ∫ dA = ∫∫ dxdy, So
Qx = ∫ y dA = ∫∫ y dydx = ∫ y^2/2 dx
= (1/2)*∫ r^2 - x^2 dx
= (1/2)*[ r^2*x - x^3/3 ] (0~r)
= (1/2)*(r^3 - r^3/3) = r^3/3.
Consequently, Y-Coordinate of Centroid at of Half / Quarter Circle ȳ = Qx/A = ∫ y dA/∫ dA = (r^3/3)/(π*r^2/4) = (4/3pi)r.
Although we calculated quarter case, just double A and Qx is the value of half case and the outcome will still be same.
And next step is Second Moment of Area for X & Y axis of Full Circle.
Ix/4 = ∫ y^2 dA = ∫∫ y^2 dydx = ∫ y^3/3 dx
= (1/3)*∫ (r^2 - x^2)^(3/2) dx
-> Substitute x to rcosθ, Check previous process.
= (1/3)*∫ (rsinθ)^3 * rsinθ dθ (0~π/2)
= r^4/3 * ∫ sinθ^4 dθ
= r^4/3 * ∫ ((1-cos2θ)/2)^2 dθ
= r^4/3 * ∫ (1 - 2cos2θ + cos2θ^2)/4 dθ
= r^4/12 * ∫ 1 - 2cos2θ + (cos4θ+1)/2 dθ
= r^4/24 * ∫ 3 - 4cos2θ +cos4θ dθ
= r^4/24 * [3θ -2sin2θ + sin4θ/4] (0~π/2)
= r^4/24 * 3/2 * π = π*r^4/16
>> Ix = π*r^4/4 = π*D^4/64 (D : Diameter)
Since Moment of Inertia depends on its shape, both Ix and Iy are definitely same.
Iy = Ix = π*r^4/4 = π*D^4/64
You can also get results using Polar Coordinate, like x = rcosθ , y = rsinθ, It's up to you.
HOW ABOUT THE POLAR COORDINATE ?
Eventually, we figured out 3 Concepts which are Centroid, First and Second Moment of Area (Moment of Inertia). These are defined in Orthogonal Coordinate(2D), but there's another familiar coordinate, Polar Coordinate. How about it?
In Polar Coordinate also has First and Second polar Moment of Area, it is quantity used to describe resistance to Torsional deformation. I hope there're chances to introduce Torsional deformation. Mechanics of Materials has lots of Interesting Theory.
In Polar Coordinate, it needs rotational axis but it's not belong to 2-Dimension. Its rotational axis is parallel to Z-axis. So Second polar Moment of Area Ip is also denoted in Iz.
Ip = Iz = Ix + Iy
EXAMPLE
Finally, we have learned Concept of Moment of Inertia in different cases. But in real problems, there isn't one shape of Structures. Although they're consist of different shapes, If you can divide with proper shapes, these are not hard problems.
In Mechanics of Materials, Capital letter 'I' shape is most often used for calculating of design. So I'm going to finish with one example
Three rectangles have same shape and uniform density. Use Ix' = Ix + Ad^2 and calculates Second Moment of Inertia for reference axis.
Thanks for your reading and I hope you got what you wanted !
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