MOMENT OF INERTIA TOPICS, Exercises of Electrical and Electronics Engineering

Download MOMENT OF INERTIA TOPICS and more Exercises Electrical and Electronics Engineering in PDF only on Docsity! MOMENT OF INERTIA Moment of Inertia( Second moment of area)   The product of the elemental area and square of the perpendicular distance between the centroid of area and the axis of reference is the “Moment of Inertia” about the reference axis x y x y dA o Iox= da1 y1 2 + da2 y2 2+ da3 y3 2+ -- = ∑ da y2 Ioy = da1 x1 2 + da2 x2 2 + da3 x3 2+ ---- = ∑ da x2 Radius of Gyration k k k Elemental area r1 r2 r3 A A B B Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out altering the total moment of inertia. IAB= da k2 + da k2 + ----------- IAB = ∑ da k2 IAB= A k2 k=√IAB/A Parallel Axis Theorem d A B x x dA y *G Moment of inertia of any area about an axis AB is equal to the M.I. about parallel centroidal axis plus the product of the total area and square of the distance between the two axes. IAB =∑dA (d +y)2 = ∑dA (d2 + y2 + 2 × d × y) =∑dA. d2 +∑dA y2 + ∑ 2×d×dA y = ∑dA. d2 +∑dA y2 + (2×d).A ∑ y dA A In the above term (2×d).A is constant & ∑ y dA = 0 A IAB = Ixx + A.d2 MOMENT OF INERTIA BY DIRECT INTEGRATION M.I. about its horizontal centroidal axis : Find the moment of inertia of rectangular area about centroidal horizontal axis by direct integration   12 )( 3 2/ 2/ 2 2 2/ 2/ BD ydyB ydAI D D D D xx         G B x x dy yD D/2 Q 1 Ixx =  dA . y2 R 2 =   (x.d.dr) r2Sin2 0 0 R 2 =  r3.dr Sin2 d 0 0 R 2 = r3 dr  {(1- Cos2)/2} d 0 R 0 2 =[r4/4] [/2 – Sin2/4] 0 0 = R4/4[ - 0] = R4/4 IXX =  R4/4 = D4/64 BA xx R  d  y=rSin r Find the moment of inertia of Circular area about the centroidal horizontal axis  Q 5 IAB =  dA . y2 R  =   (r.d.dr) r2Sin2 0 R 0  = r3.dr  Sin2 d 0 0 R  =  r3 dr (1- Cos2)/2) d 0 0  =[R4/4] [/2 – Sin2/4] 0 = R4/4[/2 - 0] = R4/8 = (πD4/32) 4R/3 y0 y0 BA xx R Find the moment of inertia of Semi-circular area about the Base  & centroidal horizontal axis Q 6 using parallel axis theorem: IAB = Ixx + A(d)2 Ixx = IAB – A(d)2 =  R4/8 R2/2 . (4R/3)2 Ixx = 0.11R4 moment of inertia of Semi-circular area about the centriodal horizontal axis   Sl.No Figure I x0-x0 I y0-y0 I xx I yy 1 bd3/12 - bd3/3 - 2 bh3/36 - bh3/12 - 3 R4/4 R4/4 - - 4 0.11R4 R4/8 R4/8 - 5 0.055R4 0.055R4 R4/16 R4/16 b d x0 x x0 x d/2 b h xx x0 x0 h/3 x0x0 y0 y0 O R y0 y0 xx x0 x0 4R/3π x0 y y0 4R/3π 4R/3π Y Y Xo Find the moment of Inertia of the shaded area shown in fig.about its base. 20mm 5 5 5 5 5 25 15 1010 5 20mm 30mm X X Numerical Problems & SolutionsQ 8 20mm 5 5 5 5 5 25 15 1010 5 20mm 30mm Ixx = Ixx1+ Ixx2 - Ixx3 4mm310*297.5I ]20*10*10 12 10*10[ 30*)30*20( 2 1 36 30*20 10)20*20( 12 20*20 xx 2 3 2 3 2 3     X X 1 3 2 Solution:- Q 8 Ix x = 200*203/3+[25*1003/12+(25*100)702] +2[87.5*203/36+0.5*87.5*20*(26.67)2] +[100*303/12+100*30*1352] Ix x=71.05*106mm4 Q 9 Find M.I. about the horizontal centroidal axis for the area shown and also find the radius of gyration. xo 463.5mm 250mm250mm 100 100 200mm200mm 100mm 400mm 400mm 1100mm y=436.5mm xo Q 10 Solution:- xo 463.5mm 250mm250mm 100 100 200mm200mm 100mm 400mm 400mm 1100mm y=436.5mm xo 1 2 3 4 5 6 Q 10 Compute the M.I. of 100 mm x 150mm rectangular shown in fig.about x-x axis to which it is inclined at an angle of   = Sin-1(4/5) 15 0m m 100mm X X  = Sin-1(4/5)A B C D Q 11 15 0m m 100mm X X  = Sin-1(4/5)A B C D M N K L sin =4/5,  =53.13o = From geometry of fig , BK=ABsin(90-53.13o) =100sin(90-53.13o )=60mm ND=BK=60mm FD= 60/sin53.13o= 75mm AF=150-FD=75mm FL=ME=75sin53.13o=60mm   mmABFCAE 125 8.0 100 90cos     E F Solution:- Q 11 IXX=IDFC+IFCE+IFEA+IAEB =125 (60)3/ 36+ (1/2)*125*60*(60+60/3)2 +125(60)3 /36+(1/2)*125*60*402 +125*603 /36 +(1/2)*125*60 *202 +125*603/36 +(1/2)*125*60*202 Ixx = 36,00,000 mm4 Q 11 Calculate the moment of inertia of the built- up section shown in fig.about the centroidal axis parallel to AB. All members are 10mm thick. 250mm 250mm 50mm A B 50mm 10mm 10mm 50mm 50mm Q 13 250mm 250mm 40mm A B 50mm 10mm 10mm 1 2 3 4 5 6 50mm 40mm Solution:- Y=73.03mm Q 13 It is divided into six rectangles. Distance of centroidal X- axis from AB=Y=∑Ai Yi /∑A ∑A=2*250*10+40*4*10=6600mm2 ∑Ai Yi = =250*10*5+2*40*10*30+40*40*15+40*10*255 +250*10*135 =4,82,000mm3 Q 13 solution:- R=20 20mm 40mm 20mm 40mm 20mm 30mm 50mm 1 2 3 4 XoXo 31.5mm Q 14 ∑A=40*80+1/2*30*30+1/2*50*30-1/2*π*(20)2 =3772mm2 ∑AiXi =3200*40+450*2/3*30+750*(30+50/3) -1/2* π*202 *40 =146880mm3 ∑AiYi =3200*20+450*50+750*50-628*4*20/3 π= 118666.67mm3 Y =118666.67/3772= 31.5mm IXoXo = [80*403/12+(80*40)(11.5)2 ]+[30*303/36+ 1/2 *30*30(18.5)2 ]+ [50*303 /36 +1/2+50*30*(18.50)2]-[0.11*204 ) +π/2*(20)2 (31.5-0.424*20)] = 970.3*103mm4 Q 14 Find the M.I. about top of section and about two centroidal axes. 10mm 150mm 10mm 150mm Q 15 Find the M.I. about centroidal axes and radius of gyration for the area in given fig. 40mm 10mm 50mm 10mm Q 16 solution:- 40mm 10mm 50mm 10mm Xo Xo Yo Yo A B C D E G FY=17.5mm X=12.5mm Q 16 Centroid X=∑ax/∑a= [(50*10)5+(30*10)25]/800=12.5mm Y=∑ay/∑a= [(50*10)25+(30*10)5]/800=17.5mm Q 16 IXoXo = [10*503/12+(50*10)(17.5-5)2]+ [30*103/12+(30*10) (12.5)2 =181666.66mm4 IYoYo=[(50*103/12)+(50*10)(7.5)2]+[10*303/12 +(30*10)*(12.5)2]=101666.66mm4 rxx=√(181666.66/800) = 15.07 mm ryy=√(101666.66/800)= 11.27 mm Y=[(100*100)50-(π/4)602*74.56]/[(100*100)- (π/4)602] =40.3mm IXoXo= [(100*1003/12)+100*100*(9.7)2]- [0.55*(60)4+0.785*(60)2*(34.56)2 ] =83,75,788.74mm4 Q 17 EXERCISE PROBLEMS Q.1. Determine the moment of inertia about the centroidal axes. 100mm 20 30mm 30mm 30mm [Ans: Y = 27.69mm Ixx = 1.801 x 106mm4 Iyy = 1.855 x 106mm4] Q.2. Determine second moment of area about the centroidal horizontal and vertical axes. [Ans: X = 699.7mm from A, Y = 265 mm Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4] 200mm 200 300mm 300mm 900mm EXERCISE PROBLEMS