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

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