Area Moment of Inertia

Author

Eric W. Weisstein
July 18, 2014

This notebook downloaded from http://mathworld.wolfram.com/notebooks/Laminae/AreaMomentofInertia.nb.

For more information, see Eric's MathWorld entry http://mathworld.wolfram.com/AreaMomentofInertia.html.

©2014 Wolfram Research, Inc. except for portions noted otherwise

Code

AreaInertiaTensor[xy__]:=Module[{x,y,i,j},
Table[{xy}.{xy}KroneckerDelta[i,j]-{xy}[[i]]{xy}[[j]],{i,2},{j,2}]
]

Tensor

AreaInertiaTensor[x,y]//TraditionalForm

Examples

Annulus about centroid

Integrate[AreaInertiaTensor[x,y]Boole[b^2<x^2+y^2<a^2],{x,-a,a},{y,-a,a},Assumptions->0<b<a]

FullSimplify[%,0<b<a]//TraditionalForm

Disk about Centroid

Integrate[AreaInertiaTensor[x,y]Boole[x^2+y^2<a^2],{x,-a,a},{y,-a,a},Assumptions->a>0]//TraditionalForm

Ellipse about Centroid

Integrate[AreaInertiaTensor[x,y]Boole[((x^2)/(a^2))+((y^2)/(b^2))<1],{x,-a,a},{y,-b,b},Assumptions->a>0&&b>0]//TraditionalForm

Half-Disk about Axis

Integrate[AreaInertiaTensor[x,y]Boole[x^2+y^2<a^2],{x,-a,a},{y,0,a},Assumptions->a>0]//TraditionalForm

Hexagon about a diameter

RegionPlot[(1/(2(Sqrt[3])))(2Abs[y]+Abs[y-x(Sqrt[3])]+Abs[y+x(Sqrt[3])])<1,{x,-1,1},{y,-1,1}]

Integrate[AreaInertiaTensor[x,y]Boole[(1/(2(Sqrt[3])))(2Abs[y]+Abs[y-x(Sqrt[3])]+Abs[y+x(Sqrt[3])])<a],{x,-∞,∞},{y,-∞,∞},Assumptions->a>0]//TraditionalForm

ineq=Inequalities[RegularPolygon[6,a Csc[π/6]/2],{x,y}]//FullSimplify[#,Assumptions->a>0]&

RegionPlot[ineq/.a->1,{x,-1,1},{y,-1,1}]

(i=FullSimplify[Integrate[AreaInertiaTensor[x,y]Boole[ineq],{x,-∞,∞},{y,-∞,∞},Assumptions->a>0]])//Timing

5 Sqrt[3]/16//N

Octagon about a diameter

MaxValue[{x,(1/4) (-1+(Sqrt[2]))(2(Abs[x]+Abs[y])(Sqrt[2])(Abs[x-y]+Abs[x+y]))<1},{x,y}]

R==(1/2)(Sqrt[4+2(Sqrt[2])])s

FullSimplify[((%)/((1/2)(Sqrt[4+2(Sqrt[2])])))]

RegionPlot[(1/4) ((Sqrt[2])-1)(2(Abs[x]+Abs[y])(Sqrt[2])(Abs[x-y]+Abs[x+y]))<1,{x,-2,2},{y,-2,2}]

Integrate[AreaInertiaTensor[x,y]Boole[(1/4) ((Sqrt[2])-1)(2(Abs[x]+Abs[y])(Sqrt[2])(Abs[x-y]+Abs[x+y]))<a],{x,-∞,∞},{y,-∞,∞},Assumptions->a>0]//FullSimplify//TraditionalForm

Pentagon about a center-vertex axis

ineq=Inequalities[RegularPolygon[5,a Csc[π/5]/2],{x,y}]//FullSimplify[#,Assumptions->a>0]&

RegionPlot[ineq/.a->1,{x,-1,1},{y,-1,1}]

(i=FullSimplify[Integrate[AreaInertiaTensor[x,y]Boole[ineq],{x,-∞,∞},{y,-∞,∞},Assumptions->a>0]])//Timing

1/96 Sqrt[265+118 Sqrt[5]]//N

Quarter-Disk about Axis

Integrate[AreaInertiaTensor[x,y]Boole[x^2+y^2<a^2],{x,0,a},{y,0,a},Assumptions->a>0]//TraditionalForm

Rectangle about Centroid

Integrate[AreaInertiaTensor[x,y],{x,-a/2,a/2},{y,-b/2,b/2}]//TraditionalForm

%/.b->a

Regular Polygon

FullSimplify[With[{A=(1/4)n (a^2)Cot[(π/n)],y1=(1/2)a Csc[(π/n)],y2=(1/2)a Cot[(π/n)]},{{A(((6(y1^2)-(a^2)))/24),0},{0,A(((12(y2^2)+(a^2)))/48)}}],n>2&&n∈Integers]

{{(1/192) (a^4) n (2+Cos[((2 π)/n)]) Cot[(π/n)] ((Csc[(π/n)])^2),0},{0,(1/192) (a^4) n Cot[(π/n)] (1+3 ((Cot[(π/n)])^2))}}//TraditionalForm

Column[(i=Join[{0,0},Table[{{(1/192) (a^4) n (2+Cos[((2 π)/n)]) Cot[(π/n)] ((Csc[(π/n)])^2),0},{0,(1/192) (a^4) n Cot[(π/n)] (1+3 ((Cot[(π/n)])^2))}},{n,3,10}]]//FullSimplify)]//TraditionalForm

Heptagon

i[[7]]

FullSimplify[With[{n=7},(Sqrt[((Tr[i[[n]],List])/((1/4)n (a^2)Cot[(π/n)]))])],a>0]

Nonagon

i[[9]]

FullSimplify[With[{n=9},(Sqrt[((Tr[i[[n]],List])/((1/4)n (a^2)Cot[(π/n)]))])],a>0]

Square about Centroid

Integrate[AreaInertiaTensor[x,y],{x,-a/2,a/2},{y,-a/2,a/2}]//TraditionalForm

Triangle about Axis

(y-y1)==(((y2-y1))/((x2-x1)))(x-x1)

With[{b=1,h=.5,a=.2},RegionPlot[(0<x<a&&y>0&&(y-h)<h/a(x-a))||(a<x<b&&y>0&&(y-h)<((-h)/(b-a))(x-a)),{x,-.01,1.1},{y,-.01,.6},MaxRecursion->8]]

FullSimplify[Integrate[Boole[(0<x<a&&y>0&&(y-h)<h/a(x-a))||(a<x<b&&y>0&&(y-h)<((-h)/(b-a))(x-a))],{x,-∞,∞},{y,-∞,∞},Assumptions->0<=a<=b&&h>0],0<=a<=b&&h>0]

FullSimplify[(1/((1/2)b h))Integrate[{x,y}Boole[(0<x<a&&y>0&&(y-h)<h/a(x-a))||(a<x<b&&y>0&&(y-h)<((-h)/(b-a))(x-a))],{x,-∞,∞},{y,-∞,∞},Assumptions->0<=a<=b&&h>0],0<=a<=b&&h>0]

FullSimplify[%]

Integrate[AreaInertiaTensor[x,y]Boole[(0<x<a&&y>0&&(y-h)<h/a(x-a))||(a<x<b&&y>0&&(y-h)<((-h)/(b-a))(x-a))],{x,-∞,∞},{y,-∞,∞},Assumptions->0<=a<=b&&h>0]

i=FullSimplify[%,0<=a<=b&&h>0]

FullSimplify[i/.a->0,b>0]

FullSimplify[i/.a->b,b>0]