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,
- Area: A=òòD dxdy.
- Mass: M=òòD r(x,y)dxdy.
- Moment about the x-axis: Mx=òòD yr(x,y)dxdy.
- Moment about y-axis: My=òòD xr(x,y)dxdy.
- Moment of inertia about the x-axis: Ixx=òòD y2r(x,y)dxdy.
- Moment of inertia about the y-axis: Iyy=òòD x2r(x,y)dxdy.
- Polar moment of inertia: IO=Ix2+y2 = òòD (x2+y2)r(x,y)dxdy=Ixx+Iyy.
- Product of inertia: Ixy=òòD xyr(x,y)dxdy.
The centre of mass (or centroid) is given by
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:
- Mass: M=òòD dxdy.
- Moment about the x-axis: Mx=òòD y dxdy.
- Moment about y-axis: My=òòD x dxdy.
- Moment of inertia about the x-axis: Ixx=òòD y2 dxdy.
- Moment of inertia about the y-axis: Iyy=òòD x2 dxdy.
- Polar moment of inertia: IO=Ixx+Iyy.
- Product of inertia: Ixy=òòD xy dxdy.
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=òòy2 dxdy=òò(yc+h)2dxdy = yc2òòdxdy+2ycòòhdxdy+òòh2dxdy
=ycA+2ycòòhdxdh+òòh2dxdh.
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., òòhdxdh = 0. Thus, we have
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Ixx=yc2A+Ixx or Ixx=Ixx- yc2A. |
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Similarly, we can drive:
- Ixh=Ixy-xcycA
- Ihh=Iyy-xc2A
- Ix2+h2=Ixx+Ihh.
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
Therefore, we can derive that
- Iuu=òv2 dA=Ixx cos2q+Ihh sin2q -2Ixh sinqcos q.
- Ivv=òu2 dA=Ixx sin2q+Ihh cos2q +2Ixh sinqcosq.
- Iuv=òuv dA=Ixx sinqcosq-Ihh sinqcosq+2Ixh(cos2q-sin2q).
- Iu2+v2=Ix2+h2.
The above equations may be simplified by using the trigonometric identities
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sin(2q)=2sinqcosq and cos(2q)=cos2q-sin2q, |
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in which case
- Iuu=[(Ixx+Iyy)/2]+[(Ixx-Iyy)/2]cos(2q) -Ixysin(2q) (1)
- Ivv=[(Ixx+Iyy)/2]-[(Ixx-Iyy)/2]cos(2q) +Ixysin(2q)
- Iuv=[(Ixx-Iyy)/2]sin(2q)+Ixycos(2q)
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,
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dIuu dq
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=-2 |
Ixx-Ihh 2
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sin2q -2Ixh cos2q = 0. |
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Consequently, at q = qp, we have
and
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Imax, Imin= |
Ixx+Ihh 2
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± |
æ ú
Ö
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æ ç
è
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Ixx-Ihh 2
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ö ÷
ø
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2
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+Ixh2 |
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. |
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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
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u=x cosqp+y sinqp and v=y cosqp+x sinqp. |
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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
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ó õ
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ó õ
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D
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æ ç
è
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dQ dx
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- |
dP dy
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ö ÷
ø
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dxdy = | ó (ç) õ
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C
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Pdx+Qdy |
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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 Î [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
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A= |
ó õ
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ó õ
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D
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dxdy= |
ó õ
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ó õ
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D
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æ ç
è
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dQ dx
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- |
dP dy
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ö ÷
ø
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dxdy=- | ó (ç) õ
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C
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ydx = - |
ó õ
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t1
t0
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y(t)x¢(t)dt. |
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As an example, we compute an area of a circle x(q)=Rcos(q)
and y(q)=Rsin(q):
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A=- |
ó õ
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2p
0
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y(q)x¢(q)dq = R2 |
ó õ
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2p
0
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(sinq)2dq = R2 |
æ ç
è
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q 2
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- |
1 4
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sin2q |
ö ÷
ø
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2p
0
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=R2p. |
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Similarly, we can derive
Mx=òòydxdy=-1/2(ò)y2dx=-1/2òt0t1 y2(t)x¢(t)dt (P=-1/2y2, Q=0)
My=òòxdxdy=1/2(ò)x2dy=1/2òt0t1x2(t)y¢(t)dt (P=0, Q=1/2 x2)
Ixx=òòy2dxdy=-1/3(ò)y3dx=-1/3òt0t1y3(t)x¢(t)dt (P=-1/3y3, Q=0)
Ixy=òòxydxdy=-1/2(ò)xy2dx=-1/2òt0t1y2(t)x(t)x¢(t)dt (P=-1/2xy2, Q=0)
Iyy=òòx2dxdy=1/3(ò)x3dy=1/3òt0t1x3(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:
To compute the above integration numerically, the well-known method is the
Gauss-Legendre integration rule which has the following form:
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ó õ
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b
a
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f(t)dt @ |
n å
i=0
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Aif(xi) |
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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
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| ó (ç) õ
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C
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P(x,y)dx+Q(x,y)dy= |
å
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ó õ
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Ci
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P(x,y)dx+Q(x,y)dy. |
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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
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| ó (ç) õ
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C
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P(x,y)dx+Q(x,y)dy= |
å
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ó õ
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xi+1
xi
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[P(x,y)+Q(x,y)y¢(x)]dx. |
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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
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y=y1+ |
y2-y1 x2-x1
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(x-x1)=y1+l(x-x1). |
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Referring to previous
sections, the area of lamina is given by
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A=- | ó (ç) õ
| ydx=- |
å
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ó õ
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x2
x1
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y dx=- |
1 2
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å
| (y1+y2) (x2-x1) |
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We note that (y1+y2)(x2-x1)/2 is the area of trapezoid. The moments of
lamina are given by
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Mx=- |
1 2
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| ó (ç) õ
| y2dx=- |
1 2
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å
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ó õ
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x2
x1
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[y1+l(x-x1)]2dx=- |
1 6
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å
| (x2-x1)(y12+y1y2+y22) |
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My= |
1 2
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| ó (ç) õ
| x2dy= |
1 2
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å
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ó õ
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y2
y1
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[x1+ |
1 l
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(y-y1)]2dy= |
1 6
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å
| (y2-y1)(x12+x1x2+x22) |
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The moments of inertia are given by
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Ixx=- |
1 3
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| ó (ç) õ
| y3dx=- |
1 3
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å
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ó õ
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x2
x1
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[y1 +l(x-x1)]3dx=- |
1 12
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å
| (x2-x1)(y13+y12y2+y1y22 +y23) |
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Iyy= |
1 3
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| ó (ç) õ
| x3dy= |
1 3
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å
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ó õ
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y2
y1
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[x1+ |
1 l
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(y-y1)]3dy= |
1 12
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å
| (y2-y1) (x13+x12x2+x1x22+x23) |
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Ixy=-1/2(ò)xy2dx=-1/2åòx1x2 x[y1+l(x-x1)]2dx
= -[1/(6l)]å{x[y1+l(x-x1)]3-òx1x2 [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
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y=y1 and x=x1+(x2-x1)t, t Î [0,1]. |
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Thus,
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Ixy=- |
1 2
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| ó (ç) õ
| xy2dx=- |
1 4
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y12(x22-x12). |
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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
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u(q)=acosq and v(q)=bsinq. |
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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
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æ ç
è
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ö ÷
ø
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= |
æ ç
è
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ö ÷
ø
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æ ç
è
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ö ÷
ø
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= |
æ ç
è
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ö ÷
ø
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. |
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In calculus, we have learned that
- (ò)cos2qdq = (ò)sin2qdq = p,
- (ò)cos3qdq = (ò)sin3qdq = 0,
- (ò)sin2qcos2qdq = p/4,
- (ò)cos4qdq = (ò)sin4qdq = 3p/4,
- (ò)cos3qsinqdq = (ò)sin3qcosqdq = 0.
Let as=asinf, ac=acosf, bs=bsinf, and bc=bcosf.
Then, x(q)=accosq-bssinq and y(q)=ascosq+bcsinq. Accordingly, we can derive
- Ixx=-1/3(ò)y3(q)dx(q) = [(p)/4](acbc+asbs)(as2+bc2),
- Ixy=-1/2(ò)y2(q)x(q)dx(q) = [(p)/4](acbc+asbs)(acas-bcbs),
- Iyy=1/3(ò)x3(q)dy(q) = [(p)/4](acbc+asbs)(ac2+bs2).
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:
- Convert B-spline curve into Bézier curves.
- With respect to each Bézier curve, compute signed area Ai, moments
(Mx and My) and moments of inertia (Ixx, Ixy, Iyy).
- Compute åAi, åMx, åMy, etc..
RETURN