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)