Centroid, Area, and Moments of Inertia for Laminas

1  Definition

Definitions of an area, centroid, and moments of inertia for a lamina may be found in most text books of engineering mechanics. For completeness, we give a brief review of these definitions in this section.

Let D be a region consisting of the boundary of the lamina and its interior. Then,

The centre of mass (or centroid) is given by


xc= My
M
,       yc= Mx
M
.
In most engineering applications, the density of mass of lamina is constant so that r can be taken out of the integration sign. We may further assume that r = 1 by disregarding the constant factor. Therefore, we have the following simpler formulas:

In engineering mechanics, we often want to know the moments of inertia about the axes passing through the centroid. Let a user defined coordinate system be (O,x,y) and the centroid of a lamina be C=(xc,yc). Assume the new coordinate system is (C,x,h) with x- and h-axis parallel to the x- and y-axis. Then, we have

    Ixx=y2 dxdy=(yc+h)2dxdy = yc2dxdy+2ychdxdy+h2dxdy

          =ycA+2ychdxdh+h2dxdh.

It is noted that the third term represents the moment of inertia about the centroidal axis x. The second term is zero since the x- and h-axis pass through the centroid, i.e., hdxdh = 0. Thus, we have


Ixx=yc2A+Ixx   or    Ixx=Ixx- yc2A.
Similarly, we can drive:

It should be pointed out that Ixh=0 if a lamina is symmetric to either x- or h-axis.

2  Principal moments of inertia

Let Ixx, Ihh, Ix2+h2, and Ixh be moments of inertia defined in the coordinate system (C,x,h), where C is the centroid of lamina. We now want to compute moments of inertia about a new system (C,u,v) which is obtained by rotating (C,x,h) by q. It is noted that


u=x cosq+y sinq


v=h cosq-x sinq
Therefore, we can derive that

The above equations may be simplified by using the trigonometric identities


sin(2q)=2sinqcosq   and    cos(2q)=cos2q-sin2q,
in which case

Different q results in different moments of inertia. We now want to determine the qp such that Iuu and Ivv are maximum and minimum. This particular set of axes is called the principal axes, and the corresponding moments of inertia with respect to these axes are called the principal moments of inertia. The qp may be found by differentiating, for example, Iuu with respect to q and setting the result equal to zero. Thus,


dIuu
dq
=-2 Ixx-Ihh
2
 sin2q -2Ixh cos2q = 0.
Consequently, at q = qp, we have


tan2qp=- 2Ixh
Ixx-Ihh
(2)
and


Imax, Imin= Ixx+Ihh
2
  
 




Ixx-Ihh
2


2

 
+Ixh2
 
.
It is noted that equation (2) gives two solutions, q1 and q2, which are 90o apart. To find which one is correct, we may substitute q into equation (1) and check if Iuu matches Imax. If not, we should set q = q+p/2.

We now consider two principal axes. Let the x- and y-axis be defined respectively by two unit vectors x=(x1,x2) and y=(y1,y2). Then, the unit vectors that define the principal axes are


u=x cosqp+y sinqp   and    v=y cosqp+x sinqp.
Replacing q in (1), we can obtain Iuu and Ivv which correspond to Imin and Imax.

3  Green Theorem

We start our discuss by considering the Green's theorem. Theorem 1 [Green Theorem] Let C be a piecewise smooth simple closed curve and D be the region consisting of C and its interior. If P(x,y) and Q(x,y) are functions that are continuous and have continuous first partial derivatives throughout an open region R containing D, then






D 


dQ
dx
- dP
dy


dxdy =
()



C 
Pdx+Qdy

The sign of the right side integration is associated with the orientation of the curve. If we walk along the curve and see the region D on our left side, then the integration is positive. This requirement coincides with the definition of the ACIS about an orientation of a curve. Green's theorem can also be applied to a region D which contains holes, provided we integrated over entire boundary and always keep the region D to the left. This requirement again matches the definition of the ACIS. Hence, Green's theorem will be used to compute properties of lamina which have holes inside.

In CAD systems, curves are often given parametrically as r(t)=(x(t),y(t)) with t [t0,t1]. Let P(x,y)=-y and Q(x,y)=0. From Green theorem, a formula of computing an area for a lamina is given by


A=



D 
dxdy=



D 


dQ
dx
- dP
dy


dxdy=-
()



C 
ydx = -
t1

t0 
y(t)x(t)dt.
As an example, we compute an area of a circle x(q)=Rcos(q) and y(q)=Rsin(q):


A=-
2p

0 
y(q)x(q)dq = R2
2p

0 
(sinq)2dq = R2

q
2
- 1
4
sin2q

2p

0 
=R2p.

Similarly, we can derive

    Mx=ydxdy=-1/2()y2dx=-1/2t0t1 y2(t)x(t)dt     (P=-1/2y2, Q=0)

    My=xdxdy=1/2()x2dy=1/2t0t1x2(t)y(t)dt     (P=0, Q=1/2 x2)

    Ixx=y2dxdy=-1/3()y3dx=-1/3t0t1y3(t)x(t)dt     (P=-1/3y3, Q=0)

    Ixy=xydxdy=-1/2()xy2dx=-1/2t0t1y2(t)x(t)x(t)dt     (P=-1/2xy2, Q=0)

    Iyy=x2dxdy=1/3()x3dy=1/3t0t1x3(t)y(t)dt     (P=0, Q=1/3x3)

4  Gauss-Legendre integration

It is seen that computation of an area and moments is actually a requirement of computing the integration of the type:



b

a 
f(t)dt.
To compute the above integration numerically, the well-known method is the Gauss-Legendre integration rule which has the following form:



b

a 
f(t)dt @ n

i=0 
Aif(xi)
The above integration rule is exact if f(x) is a polynomial of degree not more than 2n+1. In our case, f(t) is either polynomial or rational function. If it is former case, f(t) is a polynomial and thus the integration rule is exact. If it is the latter case, f(t) can be approximated by a polynomial function since f(t) is an analytic function. Therefore, we can usually obtain accurate results very efficiently using the Gauss-Legendre integration rule.

5  Applying Green's theorem to a composite curve

It is noted that Green's theorem do not require C to be a single closed curve. Instead, C could be a piecewise closed curve. If C is made of piecewise curves Ci (Ci may be lines, arcs, and B-spline curves), then the Green's formula becomes



()



C 
P(x,y)dx+Q(x,y)dy=



Ci  
P(x,y)dx+Q(x,y)dy.
As an extreme case where Ci are all lines defined by points (xi,yi) and (xi+1,yi+1), then the Green formula is given by



()



C 
P(x,y)dx+Q(x,y)dy=

xi+1

xi 
[P(x,y)+Q(x,y)y(x)]dx.
We now consider area, moments, and moments of inertia. For the purpose of simplifying notation, let any two adjacent points be denoted by (x1,y1) and (x2,y2). Then, a line joining the two points is given by


y=y1+ y2-y1
x2-x1
(x-x1)=y1+l(x-x1).
Referring to previous sections, the area of lamina is given by


A=-
()

ydx=-

x2

x1 
y dx=- 1
2

(y1+y2) (x2-x1)
We note that (y1+y2)(x2-x1)/2 is the area of trapezoid. The moments of lamina are given by


Mx=- 1
2

()

y2dx=- 1
2


x2

x1 
[y1+l(x-x1)]2dx=- 1
6

(x2-x1)(y12+y1y2+y22)


My= 1
2

()

x2dy= 1
2


y2

y1 
[x1+ 1
l
(y-y1)]2dy= 1
6

(y2-y1)(x12+x1x2+x22)
The moments of inertia are given by


Ixx=- 1
3

()

y3dx=- 1
3


x2

x1 
[y1 +l(x-x1)]3dx=- 1
12

(x2-x1)(y13+y12y2+y1y22 +y23)


Iyy= 1
3

()

x3dy= 1
3


y2

y1 
[x1+ 1
l
(y-y1)]3dy= 1
12

(y2-y1) (x13+x12x2+x1x22+x23)

    Ixy=-1/2()xy2dx=-1/2x1x2 x[y1+l(x-x1)]2dx

          = -[1/(6l)]{x[y1+l(x-x1)]3-x1x2 [y1+l(x-x1)]3dx}

          = -[1/(6l)][x2y23-x1y13-1/4(x2-x1) (y23+y22y1+y2y12+y13)]

It is noted that if y1=y2, then [1/(l)]=. In this case, we represent the line parametrically as


y=y1    and    x=x1+(x2-x1)t,    t [0,1].
Thus,


Ixy=- 1
2

()

xy2dx=- 1
4
y12(x22-x12).

6  Ellipse: a special case

Let u and v denote the unit vectors of the major and minor axes of an ellipse, and a, b the length of major and minor axes respectively. Then, we may write an ellipse parametrically as


u(q)=acosq   and    v(q)=bsinq.
Let f be the angle measured anti-clock-wisely from the x-axis to the major axis u. Then, the x and y coordinates are given by




x(q)
y(q)


=

cosf
-sinf
sinf
cosf




u(q)
v(q)


=

acosfcosq-bsinfsinq
asinfcosq+bcosfsinq


.
In calculus, we have learned that

Let as=asinf, ac=acosf, bs=bsinf, and bc=bcosf. Then, x(q)=accosq-bssinq and y(q)=ascosq+bcsinq. Accordingly, we can derive

7  General organisation

In applications, a lamina may be defined by a curve made of lines, arcs, and B-spline curves. For efficient purpose, we may compute line/curve integrals of area and moments along lines and curves. Due to the additive property of areas and moments, we can obtain the total area and moments of lamina by simply add individuals together. To do so, it is however important to ensure that the orientation of the curve is same.

It is easy to implement line integrals of area and moments. In the subsequent sections, we discuss only the implementation of curve integrals of which a curve is represented in B-spline form. The procedures are summarised below:

  1. Convert B-spline curve into Bézier curves.
  2. With respect to each Bézier curve, compute signed area Ai, moments (Mx and My) and moments of inertia (Ixx, Ixy, Iyy).
  3. Compute Ai, Mx, My, etc..

RETURN