Moments and Product of nertia
Contents ntroduction( 绪论 ) Moments of nertia of an Area( 平面图形的惯性矩 ) Moments of nertia of an Area b ntegration( 积分法求惯性矩 ) Polar Moments of nertia( 极惯性矩 ) Radius of Gration of an Area( 惯性半径 ) Parallel Ais Teorem( 平行移轴定理 ) Moments of nertia of Common Sapes of Areas( 常见平面图形的惯性矩 ) Product of nertia( 惯性积 ) Principal Aes and Principal Moments of nertia( 主惯性轴与主惯性矩 ) Mor s Circle for Moments of nertia( 惯性矩和惯性积莫尔圆 ) Principal Points( 主惯性点 )
ntroduction Previousl considered distributed forces wic were proportional to te area or volume over wic te act. - Te resultant was obtained b summing or integrating over te areas or volumes. - Te moment of te resultant about an ais was determined b computing te first moments of te areas or volumes about tat ais. Will now consider forces wic are proportional to te area or volume over wic te act but also var linearl wit distance from a given ais. - t will be sown tat te magnitude of te resultant depends on te first moments of te force distribution wit respect to te ais. - Te point of application of te resultant depends on te second moments of te distribution wit respect to te ais. Current capter will present metods for computing te moments and products of inertia for areas.
Moments of nertia of an Area Consider distributed forces F wose magnitudes are proportional to te elemental areas A on wic te act and also var linearl wit te distance of A from a given ais. Eample: Consider a beam subjected to pure bending. nternal forces var linearl wit distance from te neutral ais wic passes troug te section centroid. F ka R k da 0 da S first moment M k da da second moment Eample: Consider te net drostatic force on a submerged circular gate. F pa ga R g da M g da
Moments of nertia of an Area b ntegration Second moments or moments of inertia of an area wit respect to te and aes, da da Evaluation of te integrals is simplified b coosing da to be a tin strip parallel to one of te coordinate aes. For a rectangular area, da 0 bd b Te formula for rectangular areas ma also be applied to strips parallel to te aes, d d d da d 5
6 Polar Moments of nertia Te polar moments of inertia is an important parameter in problems involving torsion of clindrical safts and rotations of slabs. p r da Te polar moments of inertia is related to te rectangular moments of inertia, p r da da da da
7 Radius of Gration of an Area Consider area A wit moments of inertia. magine tat te area is concentrated in a tin strip parallel to te ais wit equivalent. i i i = i A i A radius of gration wit respect to te ais Similarl, i A i A i p i A i p p p p A i i i p
8 Sample Problem SOLUTON: A differential strip parallel to te ais is cosen for da. d da da l d Determine te moments of inertia of a triangle wit respect to its base. For similar triangles, l b l b ntegrating d from = 0 to =, da b 0 b d 0 da b d b 0 d b
9 Sample Problem SOLUTON: An annular differential area element is cosen, d p u da da u du d u u du u du p p r r p 0 0 r a) Determine te centroidal polar moments of inertia of a circular area b direct integration. b) Using te result of part a, determine te moments of inertia of a circular area wit respect to a diameter. From smmetr, =, r p r diameter
0 Parallel Ais Teorem Consider moments of inertia of an area A wit respect to te ais AA AA da Te ais BB passes troug te area centroid and is called a centroidal ais. AA da d da da d da d da Ad Ad AA BB For a group of parallel aes, te moment of inertia reaces te minimum value wen te reference ais is te centroid ais.
Parallel Ais Teorem Moments of inertia T of a circular area wit respect to a tangent to te circle, Ad r r r T 5 r Moments of inertia of a triangle wit respect to a centroidal ais, AA BB 6 BB AA b Ad Ad b b
Moments of nertia of Common Sapes of Areas Te moments of inertia of a composite area A about a given ais is obtained b adding te moments of inertia of te component areas A, A, A,..., wit respect to te same ais. p p p p p
Sample Problem SOLUTON: Compute te moments of inertia of te bounding rectangle and alf-circle wit respect to te ais. Determine te moments of inertia of te saded area wit respect to te ais. Te moments of inertia of te saded area is obtained b subtracting te moments of inertia of te alf-circle from te moments of inertia of te rectangle.
SOLUTON: Compute te moments of inertia of te bounding rectangle and alf-circle wit respect to te ais. Rectangle: 6 00 8. 0 b mm r 90 a 8. mm b 0 - a 8.8 mm A r.70 90 mm Half-circle: moments of inertia wit respect to AA, 6 r 90 5.760 mm AA 8 8 moments of inertia wit respect to, Aa 5.760.70 8. AA 7.00 mm 6 6 moments of inertia wit respect to, Ab 9.0 6 7.00 mm 6.70 8.8
5 Te moments of inertia of te saded area is obtained b subtracting te moments of inertia of te alf-circle from te moments of inertia of te rectangle. 8. 0 6 mm 9.0 6 mm 5.9 0 6 mm
6 Product of nertia Product of nertia: da Wen te ais, te ais, or bot are an ais of smmetr, te product of inertia is zero. Parallel ais teorem for products of inertia: A
7 Sample Problem SOLUTON: Determine te product of inertia using direct integration wit te parallel ais teorem on vertical differential area strips Appl te parallel ais teorem to evaluate te product of inertia wit respect to te centroidal aes. Determine te product of inertia of te rigt triangle (a) wit respect to te and aes and (b) wit respect to centroidal aes parallel to te and aes.
SOLUTON: Determine te product of inertia using direct integration wit te parallel ais teorem on vertical differential area strips b d b d da b el el ntegrating d from = 0 to = b, b b b el el b b d b b d b da d 0 0 0 8 b 8
9 Appl te parallel ais teorem to evaluate te product of inertia wit respect to te centroidal aes. b Wit te results from part a, A b b b b 7
Principal Aes and Principal Moments of nertia Te cange of aes ields cos sin sin cos cos sin Given da da We wis to determine moments and product of inertia wit respect to new aes and. da Te equations for and are te parametric equations for a circle, R ave ave R Note: cos sin cos sin Te equations for and lead to te same circle. 0
Principal Aes and Principal Moments of nertia At te points A and B, = 0 and is a maimum and minimum, respectivel. tan m Te equation for m defines two angles, 90 o apart wic correspond to te principal aes of te area about O. R ave ave R One metod to determine te principal moments of inertia of te area about O is to substitute tese m back into te equation for. Te advantage is tat we know wic of te two principal angles corresponds to eac principal moment of inertia. Alternativel, te principal moments of inertia ma be determined b ma, min ave R
Sample Problem SOLUTON: Compute te product of inertia wit respect to te aes b dividing te section into tree rectangles and appling te parallel ais teorem to eac. Determine te orientation of te principal aes and te principal moments of inertia. For te section sown, te moments of inertia wit respect to te and aes are = 0.8 in and = 6.97 in. Determine (a) te orientation of te principal aes of te section about O, and (b) te values of te principal moments of inertia about O.
SOLUTON: Compute te product of inertia wit respect to te aes b dividing te section into tree rectangles. Appl te parallel ais teorem to eac rectangle, A Note tat te product of inertia wit respect to centroidal aes parallel to te aes is zero for eac rectangle. Rectangle Area, in.5.5.5, in..5 0.5, in..75 0.75 A A,in.8 0.8 6.56 A 6.56 in
Determine te orientation of te principal aes and te principal moments of inertia. tan m m 75. and 55. 6.56 0.8 6.97.85 m 7.7 and m 7. 7 0.8 in 6.97 in 6.56 in ma,min 0.8 6.97 0.8 6.97 6.56 a b ma min 5.5 in.897 in
5 Mor s Circle for Moments of nertia Te moments and product of inertia for an area are plotted as sown and used to construct Mor s circle, ave R Mor s circle ma be used to grapicall or analticall determine te moments and product of inertia for an oter rectangular aes including te principal aes.
6 Sample Problem SOLUTON: Te moments and product of inertia wit respect to te and aes are = 7.06 mm, =.606 mm, and = -.50 6 mm. Using Mor s circle, determine (a) te principal aes about O, (b) te values of te principal moments about O, and (c) te values of te moments and product of inertia about te and aes Plot te points (, ) and (,- ). Construct Mor s circle based on te circle diameter between te points. Based on te circle, determine te orientation of te principal aes and te principal moments of inertia. Based on te circle, evaluate te moments and product of inertia wit respect to te aes.
7 SOLUTON: Plot te points (, ) and (,- ). Construct Mor s circle based on te circle diameter between te points. OC CD R ave.50.950 6 CD DX.7 0 mm 6 mm 6 mm 7.0.60 6 6.50 mm mm 6 mm Based on te circle, determine te orientation of te principal aes and te principal moments of inertia. tan m DX.097 m 7. 6 CD m. 8 ma OA ave R ma 8.60 6 mm min OB ave R min.90 6 mm
Based on te circle, evaluate te moments and product of inertia wit respect to te aes. Te points X and Y corresponding to te and aes are obtained b rotating CX and CY counterclockwise troug an angle (60 o ) = 0 o. Te angle tat CX forms wit te aes is f = 0 o - 7.6 o = 7. o. ' cos cos7. o OF OC CX f ave R 5.960 6 mm ' cos cos7. o OG OC CY f ave R.890 6 mm ' FX CY sinf Rsin 7. o OC R ave.7 0.950 6 mm 6 mm.80 6 mm 8
9 Principal Points Consider a pair of principal aes wit origin at a given point O. f tere eists a different pair of principal aes troug tat same point, ten ever pair of aes troug tat point is a set of principal aes and all moments of inertia are te same. cos sin sin cos A point so located tat ever ais troug te point is a principal ais, and ence te moments of inertia are te same for all aes troug te point, is called a principal point. n general, ever plane area as two principal points. Tese points lie equidistant from te centroid on te principal centroidal ais aving te larger principal moment of inertia.
0 Principal Points Appl te concepts described above to aes troug te centroid of an area. f an area as tree or more aes of smmetr, te centroid is a principal point and ever ais troug te centroid is a principal ais and as te same moment of inertia. Tese conditions are fulfilled for a circle, for all regular polgons (equilateral triangle, square, regular pentagon, regular eagon, and so on), and for man oter smmetric sapes. Wen te two principal centroidal moments of inertia are equal; ten te two principal points merge at te centroid, wic becomes te sole principal point.