ME 101: Engineering Mechanics Rajib Kumar Bhattacharjya Department - PowerPoint PPT Presentation

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’ 2 � = I y dA • The axis BB’ passes through the area centroid and is called a centroidal axis. 2 � 2 � ( ) ′ = = + I y dA y d dA 2 2 � � � = ′ + ′ + y dA 2 d y dA d dA Parallel Axis theorem : MI @ any axis = • Second term = 0 since centroid lies on BB’ MI @ centroidal axis + Ad 2 ( � y’dA = y c A , and y c = 0 2 The two axes should be = + Parallel Axis theorem I I Ad parallel to each other.

Area Moments of Inertia Parallel Axis Theorem • Moment of inertia I T of a circular area with respect to a tangent to the circle, ( ) 2 4 2 2 1 I T = I + Ad = π r + π r r 4 4 5 = π r 4 • Moment of inertia of a triangle with respect to a centroidal axis, 2 = + I I Ad ′ ′ A A B B ( ) 2 2 3 1 1 1 = − = − I I Ad bh bh h ′ ′ B B A A 12 2 3 3 1 = bh 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 A 1 , A 2 , A 3 , ... , with respect to the same axis.

Area Moments of Inertia: Standard MIs Answer � � = 1 3 �ℎ � Moment of inertia about � -axis � � = 1 3 � � ℎ Moment of inertia about � -axis � � � = 1 12 �ℎ � Moment of inertia about � � -axis � � � = 1 12 � � ℎ Moment of inertia about � � -axis Moment of inertia about � -axis passing through C � � = 1 12 �ℎ � � + ℎ �

� � = 1 12 �ℎ � Moment of inertia about � -axis � � � = 1 36 �ℎ � Moment of inertia about � � -axis

� � = 1 Moment of inertia about � � -axis 4 �� � Moment of inertia about � -axis passing through O � � = 1 2 �� �

� � = � � = 1 Moment of inertia about � � -axis 8 �� � Moment of inertia about � -axis passing through O � � = 1 4 �� �

� � = � � = 1 Moment of inertia about � � -axis 16 �� � Moment of inertia about � -axis passing through O � � = 1 8 �� �

� � = 1 4 ��� � Moment of inertia about � -axis � � = 1 4 �� � � Moment of inertia about � -axis Moment of inertia about � -axis passing through O � � = 1 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: ( )( ) 3 3 6 4 1 1 = = = × I x bh 240 120 138 . 2 10 mm 3 3 Half-circle: ( )( ) 4 r 4 90 = = = moment of inertia with respect to AA’, a 38 . 2 mm π π 3 3 4 4 ( ) 6 4 1 1 = π r = π = × I A 90 25 . 76 10 mm ′ A 8 8 = = b 120 - a 81.8 mm Moment of inertia with respect to x’, ( ) 2 2 1 1 = π = π A r 90 2 2 ( ) ( ) ( ) 2 2 6 3 = − = × − × I I Aa 25 . 76 10 12 . 72 10 38 . 2 3 2 = 12 . 72 × 10 mm ′ ′ x A A 6 4 = × 7 . 20 10 mm moment of inertia with respect to x , ( ) ( ) 2 2 6 3 = + = × + × I I Ab 7 . 20 10 12 . 72 10 81 . 8 ′ x x 6 4 = 92 . 3 × 10 mm

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 mm 4 − 6 mm 4 I 138 . 2 × 10 92 . 3 × 10 x 6 mm 4 = × I 45 . 9 10 x

Consider area (1) � � = 1 3 �ℎ � = 1 3 × 80 × 60 � = 5.76 × 10 "" � Consider area (2) �� � � � � = 1 = � 16 30 � = 0.1590 × 10 "" � 4 4 $ = 0.1590 × 10 − � 4 30 � × 12.73 � = 0.0445 × 10 "" � � � � � = 0.0445 × 10 + � 4 30 � 60 − 12.73 = 1.624 × 10 "" � Consider area (3) � � = 1 12 �ℎ � = 1 12 × 40 × 30 � = 0.09 × 10 "" � � � = 5.76 × 10 − 1.624 × 10 − 0.09 × 10 = &. '( × )' * ++ & 4.05 × 10 � � - � = , = = .&. ''++ , = 60 × 80 − 1 4 � 30 � − 1 3490 2 40 × 30 = 3490"" �

Determine the moment of inertia and the radius of gyration of the area shown in the fig. �� � � � �� �/ � � � �� � �� � � � � ���!"" � �� � � � �� �/ � � � �� � � � �� � � �����!"" � �� � � � �� �/ � � � �� � �� � � � � ���!"" � �� � � �� � � ,� � � ��� � �� � � � �� � � � � ������ "" � �� � � �� � � ,� � � ��� � �� � � � �� � � � � ������ "" � � � � ������ � ����� � ������ � �#� � �� � "" � �#� � �� � � � - � � , � � 0)� 1!++ ���

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 I x y A xy xy - I xy + I xy Quadrants + I xy - I xy

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 = = θ − θ I y ' dA y cos x sin dA x ' � � ( ) 2 2 = = θ + θ I x ' dA x cos y sin dA y ' � � ( )( ) dA = = θ + θ θ − θ I x ' y ' dA x cos y sin y cos x sin x ' y ' − θ + θ 1 cos 2 1 cos 2 2 2 θ = θ = sin cos 2 2 2 2 θ θ = θ θ − θ = θ sin cos 1 / 2 sin 2 cos sin cos 2 � 2 � 2 I + I I − I = = I y dA I x dA x y x y x y = + θ − θ I cos 2 I sin 2 ′ x xy � 2 2 = I xy dA xy + − I I I I Moments and product of inertia x y x y = − θ + θ I cos 2 I sin 2 ′ w.r.t. new axes x’ and y’ ? y xy 2 2 Note: ′ − = θ + θ I I x x cos y sin x y = sin 2 θ + cos 2 θ I I ′ x y ' xy ′ y = y cos θ − x sin θ 2

Area Moments of Inertia Adding first two eqns: Rotation of Axes I x’ + I y’ = I x + I y = I z � The Polar MI @ O Angle which makes I x’ and I y’ either max or min can be found by setting the derivative of either I x’ or I y’ w.r.t. � equal to zero: �� ( ) � � � � � � � � ��� � � � ��� � � + − = − θ − θ = � � � � � � �� � ��� � � ��� � � = + θ − θ θ � �� ′ � � Denoting this critical angle by � � � � � + − � � � � � ��� � � ��� � = − θ + θ � �� ′ � � � � �� ��� α � � � = − � � � ��� � � ��� � � � = θ + θ − � � � �� ′ � � � � two values of 2 � which differ by � since tan2 � = tan(2 � + � ) � two solutions for � will differ by � /2 � one value of � 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 �� �� � ��� � ��� � ��� � α = α = α � � � � − − � � � � Substituting in the third eqn for critical value of 2 � : I x’y’ = 0 � Product of Inertia I x’y’ is zero for the � � � � + − Principal Axes of inertia � � � � � ��� � � ��� � = + θ − θ � �� ′ � � Substituting sin2 � and cos2 � in first two eqns � � � � + − � � � � � ��� � � ��� � = − θ + θ for Principal Moments of Inertia : � �� ′ � � � � − � � � � � ��� � � ��� � + � = θ + θ � � � �� � � ( ) ′ � � � � � � � � = + − + ��� � � �� � � � � + � � � ( ) � � � � � � � = − − + ��� � � �� � � � � = �� � α