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ME 101: Engineering Mechanics Rajib Kumar Bhattacharjya Department of Civil Engineering Indian Institute of Technology Guwahati

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ME 101: Engineering Mechanics Rajib Kumar Bhattacharjya Department of Civil Engineering Indian Institute of Technology Guwahati 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 ̓ = ͖ℎͧ 3 3 1 Moment of inertia about ͭ-axis ̓ = ͖ͧℎ 4 3 1 ɑ ͧ Moment of inertia about ͬ -axis ̓ ɒ = ͖ℎ 3 12 1 ͧ ɑ ̓ ɒ = ͖ ℎ Moment of inertia about ͭ -axis 4 12 1 Moment of inertia about ͮ-axis passing through C ̓ = ͖ℎ ͖ͦ + ℎͦ 12 1 Moment of inertia about ͬ-axis ̓ = ͖ℎͧ 3 12 1 ɑ ͧ Moment of inertia about ͬ -axis ̓ ɒ = ͖ℎ 3 36 1 Moment of inertia about ͬɑ-axis ̓ = ͦͨ 3 4 1 Moment of inertia about ͮ-axis passing through O ̓ = ͦͨ ͤ 2 1 Moment of inertia about ͬɑ-axis ̓ = ̓ = ͦͨ 3 4 8 1 Moment of inertia about ͮ-axis passing through O ̓ = ͦͨ ͤ 4 1 Moment of inertia about ͬɑ-axis ̓ = ̓ = ͦͨ 3 4 16 1 Moment of inertia about ͮ-axis passing through O ̓ = ͦͨ ͤ 8 1 Moment of inertia about ͬ-axis ̓ = ͕͖ͧ 3 4 1 Moment of inertia about ͭ-axis ̓ = ͕͖ͧ 4 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 67×10 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 7= . 02×10 + (12. 27×10 )(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 ̓ = ͖ℎͧ = × 80 × 60 ͧ = 5.76 × 10 ͪ ͡͡ ͨ 3 3 3 Consider area (2) ͨ 1 ͦ ͨ ͪ ͨ ̓ ɒ = = 30 = 0.1590 × 10 ͡͡ 3 4 4 16 ̓ŭ = 0.1590 × 10 ͪ − 30 ͦ × 12.73 ͦ = 0.0445 × 10 ͪ ͡͡ ͨ 3 4 ̓ = 0.0445 × 10 ͪ + 30 ͦ 60 − 12.73 = 1.624 × 10 ͪ ͡͡ ͨ 3 4 Consider area (3) 1 1 ̓ = ͖ℎͧ = × 40 × 30 ͧ = 0.09 × 10 ͪ ͡͡ ͨ 3 12 12 ͪ ͪ ͪ ̏ ̍ ̓3 = 5.76 × 10 − 1.624 × 10 − 0.09 × 10 = ̍.̉̎ × ̊̉ ÃÃ ͪ ̓3 4.05 × 10 1 1 ͟3 = = = ̌̍. ̉̉ÃÃ ̻ = 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. 1 1 ̓1 Ɣ ͖ͧ͘ Ɣ Ɛ 24 Ɛ 6ͧ Ɣ 432 ͡͡ ͨ 3 12 12 1 1 ̓2 Ɣ ͖ͧ͘ Ɣ Ɛ 8 Ɛ 48 ͧ Ɣ 73728 ͡͡ ͨ 3 12 12 1 1 ̓3 Ɣ ͖ͧ͘ Ɣ Ɛ 48 Ɛ 6ͧ Ɣ 864 ͡͡ ͨ 3 12 12 ͦ ͦ ͨ ̓13 Ɣ ̓1 3 ƍ ̻͜ Ɣ 432 ƍ 24 Ɛ 6 Ɛ 24 ƍ 3 Ɣ 105408 ͡͡ ͦ ͦ ͨ ̓33 Ɣ ̓33 ƍ ̻͜ Ɣ 864 ƍ 48 Ɛ 6 Ɛ 24 ƍ 3 Ɣ 210816 ͡͡ ͧ ͨ ̓3 Ɣ 105408 ƍ 73728 ƍ 210816 Ɣ 390 Ɛ 10 ͡͡ ̓ 390 Ɛ 10 ͧ ͟ Ɣ 3 Ɣ Ɣ ̋̊. ̒ ÃÃ 3 ̻ 816 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 + ysin 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 = ̓3 + ̓4 ̓3 − ̓4 ̓3 + ̓4 ̓3 − ̓4 ̓ ɒ = + ͗ͣͧ2 − ̓ ͧ͢͝2 ̓ ɒ − = ͗ͣͧ2 − ̓ ͧ͢͝2 3 2 2 34 3 2 2 34 ̓3 − ̓4 ̓ ɒ ɒ = ͧ͢͝2 + ̓ ͗ͣͧ2 3 4 2 34 Squaring both the equation and adding ͦ ͦ ͦ ̓3 + ̓4 ͦ ̓3 − ̓4 ̓3 − ̓4 ̓ ɒ − + ̓ ɒ ɒ = ͗ͣͧ2 − ̓ ͧ͢͝2 + ͧ͢͝2 + ̓ ͗ͣͧ2 3 2 3 4 2 34 2 34 ͦ ͦ īͮ Ĭ ͦ īͯ Ĭ ͦ īͯ Ĭ ͦ ͦ ̓ ɒ − + ̓ ɒ ɒ = ͗ͣͧ 2 − 2 ̓ ͗ͣͧ2ͧ͢͝2 + ̓ ͧ͢͝ 2 3 ͦ 3 4 ͦ ͦ 34 34 ͯ ͦ ͯ ͦ + ī Ĭ ͧ͢͝ ͦ2 + 2 ī Ĭ ̓ ͗ͣͧ2ͧ͢͝2 + ̓ ͧ͢͝ ͦ2 ͦ ͦ 34 34 ͦ ͦ ̓3 + ̓4 ͦ ̓3 − ̓4 ͦ ̓ ɒ − + ̓ ɒ ɒ = + ̓ 3 2 3 4 2 34 ͦ ͦ ̓3 + ̓4 ͦ ̓3 − ̓4 ͦ ̓ ɒ − + ̓ ɒ ɒ = + ̓ 3 2 3 4 2 34 ͦ ̓ − ̓ ͦ Defining ̓3 + ̓4 3 4 = ̓ And ͌ = + ̓34 2 1" 2 ͦ ͦ ͦ ̓3ɒ − ̓1" + ̓3ɒ4ɒ = ͌ Which is a equation of circle with center ̓1" , 0 and radius ͌ ̓34 ͌ 2 ̓ ̓1" 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.
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