ME 0: Engineering Mechanics Rajib Kumar Bhattacharja Department of Civil Engineering ndian nstitute of Technolog Guwahati M Block : Room No 005 : Tel: 8 www.iitg.ernet.in/rkbc
Area Moments of nertia Parallel Ais Theorem Consider moment of inertia of an area A with respect to the ais AA da The ais BB passes through the area centroid and is called a centroidal ais. Parallel Ais theorem: M @ an ais M @ centroidal ais Ad The two aes should be parallel to each other. da da d ( d ) da da d Second term 0 since centroid lies on BB ( da c A, and c 0 da Ad Parallel Ais theorem
Area Moments of nertia Parallel Ais Theorem Moment of inertia T of a circular area with respect to a tangent to the circle, T 5 π r Ad π r ( π r ) Moment of inertia of a triangle with respect to a centroidal ais, AA BB BB AA 36 bh Ad Ad 3 bh 3 r bh ( h) The moment of inertia of a composite area A about a given ais is obtained b adding the moments of inertia of the component areas A, A, A 3,..., with respect to the same ais. 3
Area Moments of nertia: Standard Ms Moment of inertia about -ais Moment of inertia about -ais Answer 3 h 3 h Moment of inertia about -ais Moment of inertia about -ais h h Moment of inertia about -ais passing through C h h
Moment of inertia about -ais h Moment of inertia about -ais 36 h
Moment of inertia about -ais Moment of inertia about -ais passing through O
Moment of inertia about -ais 8 Moment of inertia about -ais passing through O
Moment of inertia about -ais 6 Moment of inertia about -ais passing through O 8
Moment of inertia about -ais Moment of inertia about -ais Moment of inertia about -ais passing through O
Area Moments of nertia Eample: Determine the moment of inertia of the shaded area with respect to the ais. SOLUTON: Compute the moments of inertia of the bounding rectangle and half-circle with respect to the ais. The moment of inertia of the shaded area is obtained b subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle.
Area Moments of nertia Eample: Solution ( )( 90) r a 38. mm 3π 3π b 0 - a 8.8 mm A πr π.7 0 3 ( 90) mm SOLUTON: Compute the moments of inertia of the bounding rectangle and half-circle with respect to the ais. Rectangle: 3 3 0 6 bh 0 0 38. mm 3 3 ( )( ) Half-circle: moment of inertia with respect to AA, 6 πr 90 5.76 0 mm AA π 8 8 ( ) Moment of inertia with respect to, AA Aa 7.0 0 6 mm 6 3 ( 5.76 0 ) (.7 0 )( 38.) moment of inertia with respect to, Ab 9.3 0 6 7.0 0 mm 6 3 (.7 0 )( 8.8)
Area Moments of nertia Eample: Solution The moment of inertia of the shaded area is obtained b subtracting the moment of inertia of the half-circle from the moment of inertia of the rectangle. 38. 0 6 mm 9.3 0 6 mm 5.9 0 6 mm
Consider area () 3 h 3 80 60 5.76 0 Consider area () 6 30 0.590 0 0.590 0 30.73 0.05 0 0.05 0 30 60.73.6 0 Consider area (3) h 0 30 0.09 0 5.76 0.6 0 0.09 0. 60 80 30 0 30390.05 0 390.
Determine the moment of inertia and the radius of gration of the area shown in the fig. 6 3 88 7378 3 86 86 36 3 0508 3 3 8686 3 086 050873780863900 3900 86.
Area Moments of nertia Products of nertia: for problems involving unsmmetrical cross-sections and in calculation of M about rotated aes. t ma be ve, -ve, or zero Product of nertia of area A w.r.t. - aes: da and are the coordinates of the element of area da When the ais, the ais, or both are an ais of smmetr, the product of inertia is zero. Parallel ais theorem for products of inertia: A - Quadrants -
Area Moments of nertia Rotation of Aes Product of inertia is useful in calculating M @ inclined aes. Determination of aes about which the M is a maimum and a minimum da Moments and product of inertia w.r.t. new aes and? Note: cosθ sinθ da cosθ sinθ da ' ' ' ' ' ' da da ' ' da ( cosθ sinθ ) da ( cosθ sinθ ) da ( cosθ sinθ )( cosθ sinθ )da cos θ cos θ sin θ cos θ sinθ cosθ / sin θ cos θ sin θ cos θ ' cos θ cos θ sin θ cos θ sin θ sin θ
Area Moments of nertia Rotation of Aes cosθ cosθ ' sinθ cosθ sinθ sinθ Adding first two eqns: z The Polar M @ O Angle which makes and either ma or min can be found b setting the derivative of either or w.r.t. θ equal to zero: ( ) 0 d sinθ cosθ dθ Denoting this critical angle b α ' tanα two values of α which differ b π since tanα tan(απ) two solutions for α will differ b π/ one value of α will define the ais of maimum M and the other defines the ais of minimum M These two rectangular aes are called the principal aes of inertia
Area Moments of nertia Rotation of Aes θ θ θ θ θ θ cos sin sin cos sin cos ' cos sin tan α α α Substituting in the third eqn for critical value of θ: 0 Product of nertia is zero for the Principal Aes of inertia Substituting sinα and cosα in first two eqns for Principal Moments of nertia: ( ) ( ) 0 @ min ma α
Squaring both the equation and adding
Defining And Which is a equation of circle with center,0 and radius
Area Moments of nertia Mohr s Circle of nertia: Following relations can be represented graphicall b a diagram called Mohr s Circle For given values of,, &, corresponding values of,, & ma be determined from the diagram for an desired angle θ. θ θ θ θ θ θ cos sin sin cos sin cos ' tan α ( ) ( ) 0 @ min ma α ( ) ave ave R R At the points A and B, 0 and is a maimum and minimum, respectivel. R ave ± ma,min
Area Moments of nertia Mohr s Circle of nertia: Construction tanα cosθ cosθ ' sinθ cosθ ma min @ α 0 ( ) ( ) sinθ sinθ Choose horz ais M Choose vert ais P Point A known {, } Point B known {, - } Circle with dia AB Angle α for Area Angle α to horz (same sense) ma, min Angle to θ Angle OA to OC θ Same sense Point C, Point D
Area Moments of nertia Eample: Product of nertia SOLUTON: Determine the product of inertia using direct integration with the parallel ais theorem on vertical differential area strips Appl the parallel ais theorem to evaluate the product of inertia with respect to the centroidal aes. Determine the product of inertia of the right triangle (a) with respect to the and aes and (b) with respect to centroidal aes parallel to the and aes.
Area Moments of nertia Eamples SOLUTON: Determine the product of inertia using direct integration with the parallel ais theorem on vertical differential area strips b h d b h d da b h el el ntegrating d from 0 to b, ( ) b b b el el b b h d b b h d b h da d 0 3 0 3 0 8 3 h b
Area Moments of nertia Eamples SOLUTON Appl the parallel ais theorem to evaluate the product of inertia with respect to the centroidal aes. b 3 3 With the results from part a, b h h A ( b)( h)( bh) 3 3 b h 7
Area Moments of nertia Eample: Mohr s Circle of nertia SOLUTON: Plot the points (, ) and (,- ). Construct Mohr s circle based on the circle diameter between the points. The moments and product of inertia with respect to the and aes are 7.06 mm,.606 mm, and -.50 6 mm. Using Mohr s circle, determine (a) the principal aes about O, (b) the values of the principal moments about O, and (c) the values of the moments and product of inertia about the and aes Based on the circle, determine the orientation of the principal aes and the principal moments of inertia. Based on the circle, evaluate the moments and product of inertia with respect to the aes.
Area Moments of nertia Eample: Mohr s Circle of nertia SOLUTON: Plot the points (, ) and (,- ). Construct Mohr s circle based on the circle diameter between the points. OC CD R ( ) ( ) ave.95 0.35 0 6 ( CD) ( DX ) 3.37 0 mm 6 mm 6 mm 7. 0.6 0 6 6.5 0 mm mm 6 mm Based on the circle, determine the orientation of the principal aes and the principal moments of inertia. DX tan θ m.097 θ m 7. 6 θ m 3. 8 CD
Area Moments of nertia Eample: Mohr s Circle of nertia OC ave R 3.37 0.95 0 6 mm 6 mm Based on the circle, evaluate the moments and product of inertia with respect to the aes. The points X and Y corresponding to the and aes are obtained b rotating CX and CY counterclockwise through an angle θ (60 o ) 0 o. The angle that CX forms with the horz is φ 0 o - 7.6 o 7. o. ϕ Rcos7. ' OF OC CX cos ave OG OC CY cosϕ Rcos7. ' 5.96 0 ave 6 mm o o 3.89 0 6 mm R sin 7. ' FX CY sinϕ o 3.8 0 6 mm
Determine the product of the inertia of the shaded area shown below about the - aes.
Solution: Parallel ais theorem: Ī d d A Both areas () and () are smmetric @ their centroidal Ais Ī 0 for both area. Therefore, for Area (): d d A 0 0 70 70 3.7 0 6 mm Similarl, for Area (): d d A 60 0 30 30.68 0 6 mm Total 8.0 0 6 mm
<, () >, () >, (-) <, (-)
Area Moments of nertia Mass Moments of nertia (): mportant in Rigid Bod Dnamics - is a measure of distribution of mass of a rigid bod w.r.t. the ais in question (constant propert for that ais). - Units are (mass)(length) kg.m Consider a three dimensional bod of mass m Mass moment of inertia of this bod about ais O-O: r dm ntegration is over the entire bod. r perpendicular distance of the mass element dm from the ais O-O
Area Moments of nertia Moments of nertia of Thin Plates For a thin plate of uniform thickness t and homogeneous material of densit ρ, the mass moment of inertia with respect to ais AA contained in the plate is AA r dm ρt r ρt AA, area da Similarl, for perpendicular ais BB which is also contained in the plate, BB ρt BB, area For the ais CC which is perpendicular to the plate, CC AA BB ( ) ρt JC, area ρt AA, area BB, area
Area Moments of nertia Moments of nertia of Thin Plates For the principal centroidal aes on a rectangular plate, 3 ( ) a b 3 ( ) ab m( a ) A A ρt AA, area ρt ma B B ρt BB, area ρt mb C C AA, mass BB, mass b For centroidal aes on a circular plate, ( r ) A A BB ρt AA area ρt π, mr
Area Moments of nertia Moments of nertia of a 3D Bod b ntegration Moment of inertia of a homogeneous bod is obtained from double or triple integrations of the form ρ r dv For bodies with two planes of smmetr, the moment of inertia ma be obtained from a single integration b choosing thin slabs perpendicular to the planes of smmetr for dm. The moment of inertia with respect to a particular ais for a composite bod ma be obtained b adding the moments of inertia with respect to the same ais of the components.
Q. No. Determine the moment of inertia of a slender rod of length L and mass m with respect to an ais 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 ais Solution
Area Moments of nertia M of some common geometric shapes
Q. Determine the angleαwhich locates the principal aes of inertia through point O for the rectangular area (Figure 5). Construct the Mohr s circle of inertia and specif the corresponding values of ma and min. b b O α
() () ( )() With this data, plot the Mohr s circle, and using trigonometr calculate the angle α α tan - (b /b ) 5 o Therefore, α.5 o (clockwise w.r.t. ) From Mohr s Circle: ma 3.08 min 0.5
Q. No. Determine the moments and product of inertia of the area of the square with respect to the '- ' aes. ; 0 ( ) Withθ30 o, using the equation of moment of inertia about an inclined aes we get, ' 0 60 ( ) 0.68 ' 0 60 ( ) 0.598 ' 0( ) ( )0.50 Alternativel, Mohr s Circle ma be used to determine the three quantities.