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MOMENT OF INERTIA

T-1. MOMENT OF INERTIA. Moment of Inertia: 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. I xx = ∫dA. y 2 I yy = ∫dA. x 2. y. dA. x. y. x. T-2.

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MOMENT OF INERTIA

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  1. T-1 MOMENT OF INERTIA Moment of Inertia: 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. Ixx = ∫dA. y2 Iyy = ∫dA. x2 y dA x y x

  2. T-2 It is also called second moment of area because first moment of elemental area is dA.y and dA.x; and if it is again multiplied by the distance,we get second moment of elemental area as (dA.y)y and (dA.x)x.

  3. T-3 Polar moment of Inertia (Perpendicular Axes theorem) The moment of inertia of an area about an axis perpendicular to the plane of the area is called “Polar Moment of Inertia” and it is denoted by symbol Izz or J or Ip. The moment of inertia of an area in xy plane w.r.to z. axis is Izz = Ip = J = ∫r2dA = ∫(x2 + y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy Y x r y O x z

  4. T-4 PERPENDICULAR AXIS THEOREM  Hence polar M.I. for an area w.r.t. an axis perpendicular to its plane of area is equal to the sum of the M.I. about any two mutually perpendicular axes in its plane, passing through the point of intersection of the polar axis and the area.

  5. Parallel Axis Theorem T-5 dA y´ x0 x0 * G _ d y x x

  6. Ixx = ∫dA. y2 _ = ∫dA (d +y')2 _ _ = ∫dA (d2+ y'2 + 2dy') _ = ∫dA. d2 + ∫dAy΄2 + ∫ 2d.dAy' _ d2 ∫dA = A.(d)2 ∫dA. y'2 = Ix0x0 _ 2d ∫ dAy’ = 0 T-6 • ( since Ist moment of area about centroidal axis = 0) • _ • Ix x = Ix0 x0 +Ad2

  7. T-7 Hence, moment of inertia of any area about an axis xx 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. Radius of Gyration It is the perpendicular distance at which the whole area may be assumed to be concentrated, yielding the same second moment of the area above the axis under consideration.

  8. T-8 y • Iyy = A.ryy2 • Ixx = A.rxx2 • ryy = √Iyy/A Andrxx = √Ixx /A A A rxx ryy y x x rxx and ryy are called the radii of gyration

  9. T-9 MOMENT OF INERTIA BY DIRECT INTEGRATION M.I. about its horizontal centroidal axis : RECTANGLE :   IXoXo = -d/2∫ +d/2 dAy2 =-d/2∫+d/2 (b.dy)y2 = bd3/12 About its base IXX=IXoXo +A(d)2 Where d = d/2, the distance between axes xx and xoxo =bd3/12+(bd)(d/2)2 =bd3/12+bd3/4=bd3/3 . dy d/2 y d x0 x0 G x x b

  10. T-10 (2) TRIANGLE :  (a) M.I. about its base :   Ixx =  dA.y2 =  (x.dy)y2 From similar triangles b/h = x/(h-y)  x = b . (h-y)/h h Ixx =  (b . (h-y)y2.dy)/h 0 = b[ h (y3/3) – y4/4 ]/h = bh3/12 (h-y) dy h x x0 y x0 h/3 x b

  11. T-11 (b) Moment of inertia about its centroidal axis: _ Ixx = Ix0x0 + Ad2 _ Ix0x0 = Ixx – Ad2 = bh3/12 – bh/2 . (h/3)2 = bh3/36

  12. T-12 3. CIRCULAR AREA: Ix0x0 =  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 IXoXo =  R4/4 = D4/64 d r y=rSin  x0 x0 R x x

  13. T-13 4. SEMI CIRCULAR AREA: Ixx =  dA . y2 R  =   (r.d.dr) r2Sin2 0R0  = 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 y0 R x0 x0 4R/3 x x y0

  14. T-14 About horizontal centroidal axis: Ixx = Ix0x0 + A(d)2 Ix0x0= Ixx – A(d)2 =  R4/8 R2/2 . (4R/3)2 Ix0x0= 0.11R4

  15. T-15 QUARTER CIRCLE: Ixx = Iyy R /2 Ixx =   (r.d.dr). r2Sin2 0 0 R /2 = r3.dr  Sin2 d 0 0 R /2 = r3 dr  (1- Cos2)/2) d 0 0 /2 =[R4/4] [/2 – (Sin2 )/4] 0 = R4 (/16 – 0) = R4/16 y0 y x0 x0 4R/3π x x y y0 4R/3π

  16. T-16 Moment of inertia about Centroidal axis, _ Ix0x0= Ixx - Ad2 = R4/16 - R2. (0. 424R)2 = 0.055R4 The following table indicates the final values of M.I. about X and Y axes for different geometrical figures.

  17. T-17 b Y d x0 x0 d/2 x x Y Xo h x0 x0 h/3 x x b y0 R x0 x0 O y0 4R/3π y0 x0 x0 x x y0 y y0 x0 4R/3π 4R/3π

  18. P-1 Problems on Moment of Inertia Q.1. Find the moment of Inertia of the shaded area shown in fig.about its base. 10 10 25 30mm 5 5 20mm 15 X X 5 5 5 5 20mm

  19. P-2 Solution:- Ixx = Ixx1+ Ixx2 -Ixx3 10 10 25 30mm 2 5 3 5 20mm 1 15 X X 5 5 5 5 20mm

  20. P-3 Q.2.Compute the M.I. about the base(bottom) for the area given in fig. 100mm 30mm 25mm 80mm 20mm 20mm x x 200mm

  21. P-4 SOLUTION:- 100mm 5 30mm 80mm 25 2 20mm 3 4 20mm 1 x x 200mm

  22. 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 P-5

  23. P-6 200mm 200mm 100 400mm 463.5mm 100 250mm 250mm xo 400mm y=436.5mm 100mm 1100mm Q.3. Find M.I. about the horizontal centroidal axis for the area fig. No.3, and also find the radius of gyration. xo

  24. P-7 200mm 200mm 100 400mm 463.5mm 100 250mm 250mm xo 400mm y=436.5mm 100mm 1100mm Solution:- 6 3 4 5 xo 2 1

  25. P-8 Solution to prob. No.03 149338200mm ∑A=1100*100+400*100+400*400+2(1/2*100*400)-π*502 =3,42,150 mm2 ∑AY=1100*100*50+100*400*300+400*400*700+[(1/2)*100* 400*633.3]*2 - π *502*700 =14,93,38,200mm3 Y= ∑AY / ∑A=436.5mm

  26. Moment of Inertia about horizontal centroidal Axis:- IXoXo =[1100*1003/12 +1100*100(386.5)2]+[100*4003/12 +(100*400)*(136.5)2]+[400*4003/12+400*400*(263.5)2]+2[100*4003/36+(1/2*400*100)*(196.8)2]-[π*(50)4/4+π*502*(263.5)2] IXoXo =32.36*109mm4 rXoXo=√(IXoXo/A)=307.536mm. P-9

  27. P-10 Q. 4. 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) D 150mm C  = Sin-1(4/5) A X X B 100mm

  28. P-11 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 Solution:- D M 150mm C F N  = Sin-1(4/5) A K X X L E B 100mm

  29. P-12 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

  30. P-13 Q.5. Find the M.I. of the shaded area shown in fig.,about AB. 80mm 40mm A B 40mm 40mm 40mm

  31. P-14 Solution:- IAB =IAB1+IAB2+IAB3 [80*803/36 +(1/2)*80*80(80/3)2 +[(0.11*404)+(1/2)π(40)2 (0.424r)2] –[π*204/4] IAB =429.3*104mm4 80mm 1 40mm A B 3 40mm 2 40mm 40mm

  32. P-15 Q.6. 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 A B 50mm 50mm 50mm 10mm 250mm 10mm 50mm

  33. P-16 Solution:- 250mm A B 2 50mm Y=73.03mm 5 40mm 3 4 40mm 10mm 1 250mm 10mm 6 50mm

  34. P-17 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

  35. P-18 Y= ∑AiYi / ∑A=482000/6600=73.03mm Moment of Inertia about centroidal axis =Sum of M.I. of individual rectangles = 250*103/12+250*10*68.032 + [10*403/12 +40*10*(43.03)2 *2 +40*103/12+40*10 (58.03)2 +10*2502 /12+250* 10(73.05-135)2 +40*103 /12+40*10(73.05-255)2 IXoXo =5,03,99395mm4

  36. P-19 Q.7. Find the second moment of the shaded area shown in fig.about its centroidal x-axis. 30mm 50mm 20mm 40mm R=20 20mm 40mm 20mm

  37. P-20 solution:- 30mm 50mm 20mm 3 2 Xo Xo 40mm 1 31.5mm R=20 4 20mm 40mm 20mm

  38. ∑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 P-21

  39. P-22 Q.8. Find the M.I. about top of section and about two centroidal axes. 150mm 10mm 150mm 10mm

  40. P-23 solution:- 150mm Yo 10mm 41.21mm Xo Xo 150mm Yo 10mm

  41. It is symmetrical about Y axis, X=0 Y=∑AY/∑A =[ (10*150*5) +(10*140*80)]/[(10*150)+(10*140) =41.21mm from top IXX= 150*103 /12 + ( 150*10)*52 +10*1403 /12+(10*140) *(80)2=11296667mm4 IXX = IXoXo + A(d)2 ,where A(d)2 IXoXo =6371701.10 mm4 IYoYo =( 10*1503 /12) + (140*103/12)=2824166.7mm4 P-24 IXoXo+(150*10+140*10)*(41.21)2=11296667mm3

  42. P-25 Q.9. Find the M.I. about centroidal axes and radius of gyration for the area in given fig. 10mm 50mm 10mm 40mm

  43. P-26 solution:- 10mm Yo A B 50mm Xo Xo E F Y=17.5mm 10mm C D G Yo 40mm X=12.5mm

  44. 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 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 P-27

  45. P-28 Q. 10. Determine he moment of inertia about the horizontal centroidal axes for the area in fig. 60 mm 100 mm 100 mm

  46. P-29 solution:- 60 mm 100 mm 2 1 Xo Xo Y=40.3mm 100 mm

  47. P-30 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

  48. EP-1 EXERCISE PROBLEMS ON M.I. Q.1. Determine the moment of inertia about the centroidal axes. 30mm 30mm 20 30mm 100mm [Ans: Y = 27.69mm Ixx = 1.801 x 106mm4 Iyy = 1.855 x 106mm4]

  49. EP-2 Q.2. Determine second moment of area about the centroidal horizontal and vertical axes. 300mm 300mm 200 200mm 900mm [Ans: X = 99.7mm from A, Y = 265 mm Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]

  50. EP-3 Q.3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration. 200mm 20 140mm 60 20 100mm [Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4 rxx = 62.66mm, ryy = 45.63mm]

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