moment of inertia moment of inertia: the product of the elemental area and square of the...
TRANSCRIPT
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MOMENT OF INERTIAMoment 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
x
y
x
ydA
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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.
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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
O
y
x
r
z
x
Y
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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.
PERPENDICULAR AXIS THEOREMT-4
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Parallel Axis Theorem
y_d
x x
x0 x0
dA
y´*G
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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
( since Ist moment of area about centroidal axis = 0) _Ix x = Ix
0 x
0 +Ad2
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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.
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Iyy = A.ryy2
Ixx = A.rxx2
ryy = √ Iyy/A
And rxx = √ Ixx /A
y
y
A
rxx
A
x x
rxx and ryy are called the radii of gyration
ryy
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MOMENT OF INERTIA BY DIRECT INTEGRATION
d
dy
x0
xx
d/2
x0
y
b
M.I. about its horizontal centroidal axis :
G.
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
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h
x0
xh/3
x0
b
xy
(h-y)
dy
(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
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(b) Moment of inertia about its centroidal axis: _ Ixx = Ix
0x
0 + Ad2
_ Ix
0x
0 = Ixx – Ad2
= bh3/12 – bh/2 . (h/3)2 = bh3/36
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Ix0
x0 = 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
xx
x0x0
R
d
y=rSin
3. CIRCULAR AREA:
r
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Ixx = dA . y2
R
= (r.d.dr) r2Sin2
0
R0
= 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
4R/3
y0
y0
xxx0
x0
4. SEMI CIRCULAR AREA:
R
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About horizontal centroidal axis: Ixx = Ix
0x
0 + A(d)2
Ix0
x0
= Ixx – A(d)2
= R4/8 R2/2 . (4R/3)2
Ix0
x0
= 0.11R4
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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
x x
x0 x0
y
y y0
y0
4R/3π
4R/3π
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Moment of inertia about Centroidal axis, _ Ix
0x
0 = 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.
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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
dx0
x
x0
xd/2
b
h
xx
x0x0
h/3
x0x0
y0
y0
O
R
y0
y0xxx0
x0
4R/3π
x0
y y0
4R/3π
4R/3π
Y
Y Xo
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Q.1. Find the moment of Inertia of the shaded area shown in fig.about its base.
20mm5 5
5
5 5
25
15
1010
5
20mm
30mm
Problems on Moment of Inertia
X X
P-1
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20mm5 5
5
5 5
25
15
1010
5
20mm
30mm
Ixx = Ixx1+ Ixx2 -Ixx3
4mm310*297.5I
]20*10*1012
10*10[
30*)30*20(2
1
36
30*20
10)20*20(12
20*20
xx
23
23
23
X X
1
3
2
Solution:-
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Q.2. Compute the M.I. about the base(bottom) for the area given in fig.
100mm
200mm
20mm20mm
80mm
30mm
x x
25mm
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100mm
200mm
20mm20mm
80mm
30mm
x x1
2
3 4
5
SOLUTION :-
25
P-4
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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
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Q.3. Find M.I. about the horizontal centroidal axis for the area fig. No.3, and also find the radius of gyration.
xo
463.5mm
250mm250mm
100
100
200mm200mm
100mm
400mm
400mm
1100mm
y=436.5mm
xo
P-6
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Solution:-
xo
463.5mm
250mm250mm
100
100
200mm200mm
100mm
400mm
400mm
1100mm
y=436.5mm
xo
1
2
3 45
6
P-7
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Solution to prob. No.03
∑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
P-8
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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
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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)
150m
m
100mm
X X
= Sin-1(4/5)A
B
C
D
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150m
m
100mm
X X
= Sin-1(4/5)A
B
C
D
MN
KL
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
mmAB
FCAE 1258.0
100
90cos
E
F
Solution:-
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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
P-12
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Q.5. Find the M.I. of the shaded area shown in fig.,about AB.
40mm40mm
40mmA B
40mm
80mm
P-13
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40mm40mm
40mmA B
40mm
80mm
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*10 44mmmm44
Solution:-
1
2
3
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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
250mm
50mm
A B
50mm
10mm
10mm
50mm50mm
P-15
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250mm
250mm
40mm
A B
50mm
10mm
10mm
1
2
3 45
6
50mm
40mm
Solution:-
Y=73.03mm
P-16
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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
P-17
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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
P-18
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Q.7. Find the second moment of the shaded area shown in fig.about its centroidal x-axis.
R=20
20mm 40mm 20mm
40mm
20mm
30mm 50mm
P-19
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solution:-
R=20
20mm 40mm 20mm
40mm
20mm
30mm 50mm
1
2 3
4
XoXo
31.5mm
P-20
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∑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
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Q.8. Find the M.I. about top of section and about two centroidal axes.
10mm
150mm
10mm
150mm
P-22
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solution:-
10mm
150mm
10mm
150mm
Yo
Yo
Xo Xo
41.21mm
P-23
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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
IXoXo+(150*10+140*10)*(41.21)2=11296667mm3
P-24
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Q.9. Find the M.I. about centroidal axes and radius of gyration for the area in given fig.
40mm
10mm
50mm
10mm
P-25
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solution:-
40mm
10mm
50mm
10mm
Xo Xo
Yo
Yo
A B
C D
E
G
FY=17.5mm
X=12.5mm
P-26
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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
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Q. 10. Determine he moment of inertia about the horizontal centroidal axes for the area in fig.
100 mm
100 mm
60 mm
P-28
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solution:-
100 mm
100 mm
60 mm
Y=40.3mm
Xo Xo1 2
P-29
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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
P-30
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EXERCISE PROBLEMS ON M.I.
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]
EP-1
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Q.2. Determine second moment of area about the centroidal horizontal and vertical axes.
[Ans: X = 99.7mm from A, Y = 265 mm
Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]
200mm
200
300mm
300mm
900mm
EP-2
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Q.3. Determine M.I. Of the built up section about the horizontal and vertical centroidal axes and the radii of gyration.
[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4
rxx = 62.66mm, ryy = 45.63mm]
60
100mm
200mm
140mm
20
20
EP-3
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Q.4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure.
[Ans: X = 83.1mm
Ixx = 2228.94 x 104mm4, Iyy = 4789.61 x 104mm4]
60
60 60
X X2020
EP-4
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Q.5. Determine the spacing of the symmetrically placed vertical blocks such that Ixx = Iyy for the shaded area.
[Ans: d/2 = 223.9mm d=447.8mm]
200mm
600mm
d
400mm
200mm
200mm
200mm
EP-5
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Q.6. Find the horizontal and vertical centroidal moment of inertia of the section shown in Fig. built up with R.S.J. (I-Section) 250 x 250 and two plates 400 x 16 mm each attached one to each.
Properties of I section are
Ixx = 7983.9 x 104mm4
Iyy = 2011.7 x 104mm4
[Ans: Ixx = 30.653 x 107mm4, Iyy = 19.078 x 107mm4]
4000mm
2500mm
160mm
160mm
Cross sectional area=6971mm2
EP-6
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Q.7. Find the horizontal and vertical centroidal moment of inertia of built up section shown in Figure. The section consists of 4 symmetrically placed ISA 60 x 60 with two plates 300 x 20 mm2.
[Ans: Ixx = 111.078 x 107mm4, Iyy = 39.574 x 107mm4]300mm
Properties of ISA
Cross sectional area = 4400mm2
Ixx = Iyy ;Cxx = Cyy =18.5mm
18.5mm
18.5mm
20mm
200mm
EP-7
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Q.8. The R.S. Channel section ISAIC 300 are placed back to back with required to keep them in place. Determine the clear distance d between them so that Ixx = Iyy for the composite section.
[Ans: d = 183.1mm]
Properties of ISMC300
C/S Area = 4564mm2
Ixx = 6362.6 x 104mm4
Iyy = 310.8 x 104mm4
Cyy = 23.6mm
X X
Y
Y
Lacing
d
380mm
23.6mm
EP-8
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Q9. Determine horizontal and vertical centroidal M.I. for the section shown in figure.
[Ans: Ixx = 2870.43 x 104mm4, Iyy = 521.64 x 104mm4]90mm
160mm
40mm
40mm
40mm
EP-9