ME 101: Engineering Mechanics Rajib Kumar Bhattacharjya Department of Civil Engineering Indian Institute of Technology Guwahati
M Block : Room No 005 : Tel: 2428 www.iitg.ernet.in/rkbc Area Moments of Inertia Parallel Axis Theorem • Consider moment of inertia I of an area A with respect to the axis AA’ I = ∫ y2dA
• The axis BB’ passes through the area centroid and is called a centroidal axis.
I = ∫ y2dA = ∫(y′ + d )2 dA = y′2dA + 2d y′dA + d 2 dA Parallel Axis theorem : ∫ ∫ ∫ MI @ any axis = • Second term = 0 since centroid lies on BB’ MI @ centroidal axis + Ad 2 (∫y’dA = ycA, and yc = 0 The two axes should be 2 Parallel Axis theorem parallel to each other. I = I + Ad Area Moments of Inertia Parallel Axis Theorem
• Moment of inertia IT of a circular area with respect to a tangent to the circle, I = I + Ad 2 = 1 π r 4 + (π r 2 )r 2 T 4 = 5 π r 4 4
• Moment of inertia of a triangle with respect to a centroidal axis, 2 I AA′ = I BB′ + Ad 2 I = I − Ad 2 = 1 bh 3 − 1 bh ()1 h BB′ AA′ 12 2 3 = 1 bh 3 36 • The moment of inertia of a composite area A about a given axis is obtained by adding the
moments of inertia of the component areas A1, A2, A3, ... , with respect to the same axis. Area Moments of Inertia: Standard MIs Answer 1 Moment of inertia about -axis = ℎ 3 1 Moment of inertia about -axis = ℎ 3 1 Moment of inertia about -axis = ℎ 12
1 = ℎ Moment of inertia about -axis 12
1 Moment of inertia about -axis passing through C = ℎ + ℎ 12 1 Moment of inertia about -axis = ℎ 12
1 Moment of inertia about -axis = ℎ 36 1 Moment of inertia about -axis = 4
1 Moment of inertia about -axis passing through O = 2 1 Moment of inertia about -axis = = 8
1 Moment of inertia about -axis passing through O = 4 1 Moment of inertia about -axis = = 16
1 Moment of inertia about -axis passing through O = 8 1 Moment of inertia about -axis = 4
1 Moment of inertia about -axis = 4
1 Moment of inertia about -axis passing through O = + 4 Area Moments of Inertia Example: Determine the moment of inertia of the shaded area with respect to the x axis.
SOLUTION: • Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. • The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. Area Moments of Inertia
Example: Solution SOLUTION: • Compute the moments of inertia of the bounding rectangle and half-circle with respect to the x axis. Rectangle: 1 3 1 3 6 4 I x = 3 bh = 3 (240 )(120 ) = 138 2. ×10 mm Half-circle: 4r (4)(90 ) a = = = 38 2. mm moment of inertia with respect to AA’, 3π 3π I = 1πr4 = 1π (90 )4 = 25 76. ×10 6 mm 4 b = 120 - a = 81.8 mm AA′ 8 8
1 2 1 2 Moment of inertia with respect to x’, A = 2 πr = 2 π ()90 2 6 3 2 3 2 I I Aa = 12 72. ×10 mm x′ = AA′ − = (25 76. ×10 )− (12 72. ×10 )(38 2. ) = 7.20 ×10 6 mm 4 moment of inertia with respect to x, 2 6 3 2 I x = I x′ + Ab = 20.7 ×10 + (12 72. ×10 )(81 8. ) = 92 .3×10 6 mm 4 Area Moments of Inertia
Example: Solution
• The moment of inertia of the shaded area is obtained by subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.
6 4 6 4 Ix = 138 .2×10 mm − 92 .3×10 mm
6 4 Ix = 45 9. ×10 mm Consider area (1) 1 1 = ℎ = × 80 × 60 = 5.76 × 10 3 3 Consider area (2)
1 = = 30 = 0.1590 × 10 4 4 16 = 0.1590 × 10 − 30 × 12.73 = 0.0445 × 10 4 = 0.0445 × 10 + 30 60 − 12.73 = 1.624 × 10 4 Consider area (3) 1 1 = ℎ = × 40 × 30 = 0.09 × 10 12 12 = 5.76 × 10 − 1.624 × 10 − 0.09 × 10 = . × 4.05 × 10 1 1 = = = . = 60 × 80 − 30 − 40 × 30 = 3490 3490 4 2 Determine the moment of inertia and the radius of gyration of the area shown in the fig.
Area Moments of Inertia
Products of Inertia: for problems involving unsymmetrical cross-sections and in calculation of MI about rotated axes. It may be +ve, -ve, or zero • Product of Inertia of area A w.r.t. x-y axes : I xy = ∫ xy dA x and y are the coordinates of the element of area dA=xy
• When the x axis, the y axis, or both are an axis of symmetry, the product of inertia is zero. • Parallel axis theorem for products of inertia:
I xy = I xy + yx A - Ixy + Ixy
Quadrants
+ Ixy - Ixy Area Moments of Inertia Rotation of Axes Product of inertia is useful in calculating MI @ inclined axes. Determination of axes about which the MI is a maximum and a minimum
2 2 I x' = ∫ y' dA = ∫(y cos θ − xsin θ ) dA 2 2 I y' = ∫ x' dA = ∫()x cos θ + y sin θ dA
I x'y' = ∫ x' y'dA = ∫(x cos θ + y sin θ y cos )(θ − xsin θ dA ) 1− cos 2θ 1+ cos 2θ sin 2 θ = cos 2 θ = 2 2 sin θ cos θ = 2/1 sin 2θ cos 2 θ − sin 2 θ = cos 2θ 2 2 I x = ∫ y dA I y = ∫ x dA I x + I y I x − I y I x′ = + cos 2θ − I xy sin 2θ I xy = ∫ xy dA 2 2 Moments and product of inertia I + I I − I I = x y − x y cos 2θ + I sin 2θ w.r.t. new axes x’ and y’ ? y′ 2 2 xy Note: x′ = x cos θ + y sin θ I − I I = x y sin 2θ + I cos 2θ y′ = y cos θ − xsin θ x′y' 2 xy Area Moments of Inertia Rotation of Axes Adding first two eqns: Ix’ + Iy’ = Ix + Iy = Iz The Polar MI @ O
Angle which makes Ix’ and Iy’ either max or min can be found by setting the derivative
of either Ix’ or Iy’ w.r.t. θ equal to zero:
dIx' Ix + Iy Ix − Iy = (I − I )sin 2θ − 2I cos 2θ = 0 I = + cos2θ − I sin 2θ y x xy x′ 2 2 xy dθ
Ix + Iy Ix − Iy Denoting this critical angle by α I = − cos 2θ + I sin 2θ y′ 2 2 xy 2 Ixy Ix − Iy tan 2α = I ' = sin 2θ + I cos 2θ x′y 2 xy Iy − Ix two values of 2 α which differ by π since tan2 α = tan(2α+π) two solutions for α will differ by π/2 one value of α will define the axis of maximum MI and the other defines the axis of minimum MI These two rectangular axes are called the principal axes of inertia Area Moments of Inertia Rotation of Axes 2I 2I tan 2α = xy ⇒ sin 2α = cos 2α xy Iy − Ix Iy − Ix Substituting in the third eqn for critical value
of 2 θ: Ix’y’ = 0
Product of Inertia Ix’y’ is zero for the Ix + Iy Ix − Iy Principal Axes of inertia I = + cos2θ − I sin 2θ x′ 2 2 xy Substituting sin2 α and cos2 α in first two eqns Ix + Iy Ix − Iy I = − cos 2θ + I sin 2θ y′ 2 2 xy for Principal Moments of Inertia :
Ix − Iy I ' = sin 2θ + I cos 2θ x′y 2 xy Ix + Iy 1 2 2 Imax = + (I − I ) + 4I 2 2 x y xy
Ix + Iy 1 2 2 Imin = − ()I − I + 4I 2 2 x y xy 0 Ixy @α = + − + − = + 2 − 2 − = 2 − 2 2 2 2 2
− = 2 + 2 2 Squaring both the equation and adding
+ − − − + = 2 − 2 + 2 + 2 2 2 2
− + = 2 − 2 2 2 + 2
+ 2 + 2 2 2 + 2
+ − − + = + 2 2 + − − + = + 2 2
− Defining + = And = + 2 2 − + = Which is a equation of circle with center , 0 and radius
2 Area Moments of Inertia Mohr’s Circle of Inertia: Following relations can be represented graphically by a diagram called Mohr’s Circle
For given values of Ix, Iy, & Ixy , corresponding values of Ix’ , Iy’ , & Ix’y’ may be determined from the diagram for any desired angle θ. 2 Ix + Iy Ix − Iy I I = + cos 2θ − I sin 2θ tan 2 xy I x′ 2 2 xy α = xy Iy − Ix Ix + Iy Ix − Iy I = − cos 2θ + I sin 2θ y′ 2 2 xy
Ix − Iy I ' = sin 2θ + I cos 2θ x′y 2 xy
Ix + Iy 1 2 2 Imax = + (I − I ) + 4I I 2 2 x y xy
Ix + Iy 1 2 2 Imin = − ()I − I + 4I 2 2 x y xy 0 Ixy@α = 2 2 2 (Ix′ − Iave ) + Ix y′′ = R • At the points A and B, I x’y’ = 0 and Ix’ is 2 a maximum and minimum, respectively. I + I I − I x y x y 2 Iave = R = + Ixy Imax, min = Iave ± R 2 2 Ix + Iy Ix − Iy I = + cos 2θ − I sin 2θ Area Moments of Inertia x′ 2 2 xy Ix + Iy Ix − Iy I = − cos 2θ + I sin 2θ 2I y′ 2 2 xy Construction tan 2α = xy Mohr’s Circle of Inertia: Iy − Ix Ix − Iy I ' = sin 2θ + I cos 2θ x′y 2 xy
Ix + Iy 1 2 2 Imax = + (I − I ) + 4I 2 2 x y xy
Ix + Iy 1 2 2 Imin = − ()I − I + 4I 2 2 x y xy 0 Ixy@α = Choose horz axis MI Choose vert axis PI
Point A – known { Ix, Ixy } Point B – known { Iy, -Ixy } Circle with dia AB Angle α for Area Angle 2 α to horz (same
sense) Imax , Imin Angle x to x’ = θ Angle OA to OC = 2 θ Same sense
Point C Ix’ , Ix’y’ Point D Iy’ Area Moments of Inertia Example: Product of Inertia SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips • Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes.
Determine the product of inertia of the right triangle (a) with respect to the x and y axes and (b) with respect to centroidal axes parallel to the x and y axes. Area Moments of Inertia
Examples SOLUTION: • Determine the product of inertia using direct integration with the parallel axis theorem on vertical differential area strips x x y = h1− dA = y dx = h1− dx b b x x = x y = 1 y = 1 h1− el el 2 2 b
Integrating dI x from x = 0 to x = b, b x 2 I dI x y dA x()1 h2 1 dx xy = ∫ xy = ∫ el el = ∫ 2 − 0 b 2 b b x x2 x3 x2 x3 x4 = h2 ∫ − + dx =h − + 2 b 2 4 3b 2 0 2b 8b 0 I = 1 b2h2 xy 24 Area Moments of Inertia
Examples SOLUTION
• Apply the parallel axis theorem to evaluate the product of inertia with respect to the centroidal axes. x = 1 b y = 1 h 3 3
With the results from part a,
I xy = Ix y ′′′′ + yx A I = 1 b2h2 − ()1 b ()1 h (1 bh ) x y ′′′′ 24 3 3 2
I = − 1 b2h2 x y ′′′′ 72 Area Moments of Inertia Example: Mohr’s Circle of Inertia SOLUTION:
• Plot the points ( Ix , I xy ) and ( Iy ,-Ixy ). Construct Mohr’s circle based on the circle diameter between the points. • Based on the circle, determine the orientation of the principal axes and the The moments and product of inertia principal moments of inertia.
with respect to the x and y axes are Ix = • Based on the circle, evaluate the 4 4 7.24x106 mm , Iy = 2.61x106 mm , and moments and product of inertia with 6 4 Ixy = -2.54x10 mm . respect to the x’y’ axes. Using Mohr’s circle, determine (a) the principal axes about O, (b) the values of the principal moments about O, and (c) the values of the moments and product of inertia about the x’ and y’ axes Area Moments of Inertia Example: Mohr’s Circle of Inertia SOLUTION:
• Plot the points ( Ix , I xy ) and ( Iy ,-Ixy ). Construct Mohr’s circle based on the circle diameter between the points. OC = I = 1 (I + I )= .4 925 ×10 6 mm 4 ave 2 x y CD = 1 ()I − I = .2 315 ×10 6 mm 4 2 x y R = ()()CD 2 + DX 2 = .3 437 ×10 6 mm 4
6 4 Ix = 24.7 ×10 mm • Based on the circle, determine the orientation of the 6 4 principal axes and the principal moments of inertia. I y = 61.2 ×10 mm DX 6 4 tan 2θm = = .1 097 2θm = 47 6. ° θm = 23 8. ° Ixy = − 54.2 ×10 mm CD 6 4 Area Moments of Inertia OC = Iave = .4 925 ×10 mm 6 4 Example: Mohr’s Circle of Inertia R = 3.437 ×10 mm • Based on the circle, evaluate the moments and product of inertia with respect to the x’y’ axes. The points X’ and Y’ corresponding to the x’ and y’ axes are obtained by rotating CX and CY counterclockwise through an angle θ= 2(60 o) = 120 o. The angle that CX’ forms with the horz is φ = 120 o - 47.6 o = 72.4 o.
o Ix' = OF = OC + CX ′cos ϕ = Iave + Rcos 72 4. 6 4 Ix′ = 96.5 ×10 mm o I y' = OG = OC − CY′cos ϕ = Iave − Rcos 72 4.
6 4 I y′ = 89.3 ×10 mm o Ix′y' = FX ′ = CY′sin ϕ = Rsin 72 4.
6 4 Ix y′′ = 28.3 ×10 mm Determine the product of the inertia of the shaded area shown below about the x-y axes. Solution :
Parallel axis theorem: Ixy = Īxy +dx dy A Both areas (1) and (2) are symmetric @ their centroidal
Axis Īxy = 0 for both area.
Therefore, for Area (1): Ixy1 = dx1 dy1 A1 6 4 Ixy 1 = 20 ×140 ×70 ×70 = 13.72 ×10 mm
Similarly, for Area (2): Ixy2 = dx2 dy2 A2 6 4 Ixy 2 = 60 ×20 ×130 ×30 = 4.68 ×10 mm
6 4 Total Ixy = Ixy1 + Ixy2 = 18.40 ×10 mm Ix < Iy , Ixy (+)
Ix >Iy , Ixy (+)
Ix > Iy , Ixy (-)
Ix < Iy , Ixy (-) Area Moments of Inertia Mass Moments of Inertia ( I): Important in Rigid Body Dynamics - I is a measure of distribution of mass of a rigid body w.r.t. the axis in question (constant property for that axis). - Units are (mass)(length) 2 kg.m 2
Consider a three dimensional body of mass m Mass moment of inertia of this body about axis O-O: I = ∫ r 2 dm
Integration is over the entire body.
r = perpendicular distance of the mass element dm from the axis O-O Area Moments of Inertia Moments of Inertia of Thin Plates • For a thin plate of uniform thickness t and homogeneous material of density ρ, the mass moment of inertia with respect to axis AA’ contained in the plate is 2 2 I AA′ = ∫ r dm = ρt∫ r dA
= ρ It AA′,area
• Similarly, for perpendicular axis BB’ which is also contained in the plate,
I BB′ = ρ It BB′,area
• For the axis CC’ which is perpendicular to the plate,
ICC′ = ρt JC,area = ρt (I AA′,area + I BB′,area )
= I AA′ + I BB′ Area Moments of Inertia Moments of Inertia of Thin Plates
• For the principal centroidal axes on a rectangular plate,
I = ρ It = ρt (1 a3b)= 1 ma 2 AA′ AA′,area 12 12
I = ρ It = ρt (1 ab 3 )= 1 mb 2 BB′ BB′,area 12 12
I = I + I = 1 m(a2 + b2 ) CC′ AA′,mass BB′,mass 12
• For centroidal axes on a circular plate,
I = I = ρ It = ρt (1 π r 4 )= 1 mr 2 AA′ BB′ AA′,area 4 4 Area Moments of Inertia Moments of Inertia of a 3D Body by Integration • Moment of inertia of a homogeneous body is obtained from double or triple integrations of the form I = ρ∫ r 2dV
• For bodies with two planes of symmetry, the moment of inertia may be obtained from a single integration by choosing thin slabs perpendicular to the planes of symmetry for dm.
• The moment of inertia with respect to a particular axis for a composite body may be obtained by adding the moments of inertia with respect to the same axis of the components. Q. No. Determine the moment of inertia of a slender rod of length L and mass m with respect to an axis which is perpendicular to the rod and passes through one end of the rod.
Solution Q. No. For the homogeneous rectangular prism shown, determine the moment of inertia with respect to z axis
Solution Area Moments of Inertia MI of some common geometric shapes Q. Determine the angle α which locates the principal axes of inertia through point O for the rectangular area (Figure 5). Construct the Mohr’s circle of inertia and specify the corresponding values of Imax and Imin .
y b
y’ 2b
x’ α x O I = ( ) x I = ( ) y I = ( )( ) xy
With this data, plot the Mohr’s circle, and using trigonometry calculate the angle 2 α 2 α =tan -1(b4/b4) = 45 o Therefore, α = 22.5 o (clockwise w.r.t. x)
From Mohr’s Circle: I = + 3.08 max I = − 0. 5 min Q. No. Determine the moments and product of inertia of the area of the square with respect to the x'- y' axes.
I = = ; I = x y I = 0 + ( × × ) = xy With θ = 30 o, using the equation of moment of inertia about any inclined axes
+ − = + − + − = − + − = + we get, I = + 0 − 60° = ( − ) = 0.1168 Alternatively, Mohr’s Circle may be x' used to determine the three I = + 0 + 60° = ( + ) = 0.5498 y' quantities. I = 0 + ( × ) = ( × ) = 0.1 50 xy'