东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A07 Moments and Product of Inertia

Moments and Product of Inertia

Moments and Product of Inertia

Contents·Introduction（绪论）·MomentsofInertiaofanArea（平面图形的惯性矩）·Moments ofInertiaof anAreabyIntegration（积分法求惯性矩）·PolarMoments of Inertia（极惯性矩）·Radius ofGyrationof anArea（惯性半径）·ParallelAxisTheorem（平行移轴定理）·Moments of Inertia of Common Shapes of Areas（常见平面图形的惯性矩）·ProductofInertia（惯性积）·Principal Axes and Principal Moments of Inertia（主惯性轴与主惯性矩）·Mohr'sCircleforMomentsofInertia（惯性矩和惯性积莫尔圆）·PrincipalPoints（主惯性点）2

Contents • Introduction（绪论） • Moments of Inertia of an Area（平面图形的惯性矩） • Moments of Inertia of an Area by Integration（积分法求惯性矩） • Polar Moments of Inertia（极惯性矩） • Radius of Gyration of an Area（惯性半径） • Parallel Axis Theorem（平行移轴定理） • Moments of Inertia of Common Shapes of Areas（常见平面图形 的惯性矩） • Product of Inertia（惯性积） • Principal Axes and Principal Moments of Inertia（主惯性轴与主 惯性矩） • Mohr’s Circle for Moments of Inertia（惯性矩和惯性积莫尔圆） • Principal Points（主惯性点） 2

Lntroduction: Previously considered distributed forces which were proportional to thearea or volume over which they act-The resultant was obtained by summing or integrating over theareas or volumes.- The moment of the resultant about any axis was determined bycomputing the first moments of the areas or volumes about thataxis.: Will now consider forces which are proportional to the area or volumeover which they act but also vary linearly with distance from a given axis- It will be shown that the magnitude of the resultant depends on thefirst moments of the force distribution with respect tothe axis.- The point of application of the resultant depends on the secondmomentsofthedistributionwithrespecttotheaxis.: Current chapter will present methods for computing the moments andproducts of inertia for areas.3

Introduction • Previously considered distributed forces which were proportional to the area or volume over which they act. - The resultant was obtained by summing or integrating over the areas or volumes. - The moment of the resultant about any axis was determined by computing the first moments of the areas or volumes about that axis. • Will now consider forces which are proportional to the area or volume over which they act but also vary linearly with distance from a given axis. - It will be shown that the magnitude of the resultant depends on the first moments of the force distribution with respect to the axis. - The point of application of the resultant depends on the second moments of the distribution with respect to the axis. • Current chapter will present methods for computing the moments and products of inertia for areas. 3

Moments of Inertia of an Area. Consider distributed forces F whose magnitudes areproportional to the elemental areas A on which theyAAact and also vary linearly with the distance of AAF-kyAAfrom a given axis.: Example: Consider a beam subjected to pure bendingInternal forces vary linearly with distancefrom theneutral axis which passes through the section centroidAF = kyAAR=k[ydA=0 JydA=S, = first momentM=k[y’dA[y’dA= second momentExample:Considerthenethydrostaticforceonasubmerged circular gate△F = pAA= pgyAAAF=YyAAR= pgJydAM,=pgJy'dA4

Moments of Inertia of an Area • Consider distributed forces whose magnitudes are proportional to the elemental areas on which they act and also vary linearly with the distance of from a given axis. F   A A • Example: Consider a beam subjected to pure bending. Internal forces vary linearly with distance from the neutral axis which passes through the section centroid. 2 2 0 first moment second moment x F ky A R k y dA y dA S M k y dA y dA               • Example: Consider the net hydrostatic force on a submerged circular gate. 2 x F p A gy A R g y dA M g y dA             4

Moments of Inertia of an Area by Integration7dA =dxdy: Second moments or moments of inertia ofdxan area with respect to the x and y axes,IdyIx =[y?dAI,=[x?dAdl,=xedadlr=yedaydA=(a-x)dydA=ydx. Evaluation of the integrals is simplified bytchoosing dA to be a thin strip parallel tooneofthe coordinateaxes.ddly=x2dAdlr=y-dA·Forarectangulararea.Ix =[y2dA= y?bdy=1bh3dA-bdy0dy: The formula for rectangular areas may alsobe applied to strips parallel to the axes,dxdlx=-y'dxd,=reydxdl,=x2dA=x2ydx5

Moments of Inertia of an Area by Integration • Second moments or moments of inertia of an area with respect to the x and y axes, I x   y dA I y   x dA 2 2 • Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. • For a rectangular area, 3 3 1 0 2 2 I y dA y bdy bh h x      • The formula for rectangular areas may also be applied to strips parallel to the axes, dI y dx dI x dA x y dx x y 3 2 2 3 1    5

Polar Moments of Inertia. The polar moments of inertia is an importantparameter in problems involving torsion of1cylindrical shafts and rotations of slabs.dAr?dAx: The polar moments of inertia is related to therectangular moments of inertia,I, =JrdA=[(x? +y2)dA= [x’dA+[y'dA=I I6

Polar Moments of Inertia • The polar moments of inertia is an important parameter in problems involving torsion of cylindrical shafts and rotations of slabs. 2 p I r dA   • The polar moments of inertia is related to the rectangular moments of inertia,   2 2 2 2 2 p y x I r dA x y dA x dA y dA I I            6

Radius of Gyration of an AreaConsiderareaAwithmomentsofinertia Ir. Imagine that the area isconcentrated in a thin strip parallel tothe x axis with equivalent I1L=iAAi= radius of gyration with respectto the x axis.Similarly0XADA:2心pY

Radius of Gyration of an Area • Consider area A with moments of inertia Ix . Imagine that the area is concentrated in a thin strip parallel to the x axis with equivalent Ix . 2 x x x x I I i A i A   i x = radius of gyration with respect to the x axis • Similarly, 2 2 y y y y p p p p I I i A i A I I i A i A     2 2 2 p x y i i i   7 p i y i x i

Sample ProblemuSOLUTION:. A differential strip parallel to the x axis is chosen fordA.dl, =ydAdA=ldyForsimilartrianglesx1h-yh-y1-dAbhhhDeterminethemomentsofinertia ofatrianglewith respect. Integrating dl, from y= O to y = h,to its base.h-yI, =[y?dA={?h0bh3128

Sample Problem Determine the moments of inertia of a triangle with respect to its base. SOLUTION: • A differential strip parallel to the x axis is chosen for dA. dI y dA dA l dy x   2 • For similar triangles, dy h h y dA b h h y l b h h y b l       • Integrating dIx from y = 0 to y = h,   h h h x y y h h b hy y dy h b dy h h y I y dA y b 0 3 4 0 2 3 0 2 2 3 4                 12 3 bh I x  8

Sample ProblemySOLUTION:: An annular differential area element is chosendl,=u'dAdA=2元uduI,=[dl,=[u(2πudu)=2元[u'dux元2a)Determinethecentroidalpolar. From symmetry, I=Imoments of inertiaofacircular元area by direct integration=21I,=I +I, =212b) Using the result of part a.determinethemomentsofinertia元of a circular area with respect to adiameter4diameter.9

Sample Problem a) Determine the centroidal polar moments of inertia of a circular area by direct integration. b) Using the result of part a, determine the moments of inertia of a circular area with respect to a diameter. SOLUTION: • An annular differential area element is chosen,   2 2 3 0 0 2 2 2 p r r p p dI u dA dA u du I dI u u du u du            4 2 p I r    • From symmetry, Ix = Iy , 4 2 2 2 p x y x x I I I I r I      diamete 4 r 4 x I r I     9

ParallelAxis Theorem.Considermoments of inertiaIofanareaAwith respect to the axis AA'dAB'BCIA = y'dALA'.The axis BB'passes through the area centroidandiscalledacentroidalaxisIA =Jy'dA=f('+d) dAJ y"dA+ 2dfydA +d'[ dA= IA = IB + Ad? = + Ad? For a group of parallel axes, the moment of inertia reaches the minimum valuewhenthereferenceaxisisthecentroidaxis10

Parallel Axis Theorem • Consider moments of inertia I of an area A with respect to the axis AA’ 2 AA I y dA    • The axis BB’ passes through the area centroid and is called a centroidal axis.   2 2 2 2 AA I y dA y d dA y dA d y dA             2  d dA  2 2 AA BB      I I Ad I Ad   10 • For a group of parallel axes, the moment of inertia reaches the minimum value when the reference axis is the centroid axis