Transcript Document

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.
y
dA
2
Ixx = ∫dA. y
x
Iyy = ∫dA. x2
y
x
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.
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
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.
T-5
Parallel Axis Theorem
dA
x0
y´
*G
y
x
x0
_
d
x
T-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
( since Ist moment of area about centroidal axis = 0)
_
Ix x = Ix x +Ad2
0 0
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.
T-8
2
Iyy = A.ryy
y
Ixx = A.rxx2
ryy = √ Iyy/A
A
And rxx = √ Ixx /A
ryy
y
x
rxx and ryy are called the radii of gyration
x
T-9
MOMENT OF INERTIA BY DIRECT INTEGRATION
M.I. about its horizontal centroidal axis :
RECTANGLE :
dAy2
+d/2
=-d/2∫
.
IXoXo = -d/2 ∫
+d/2
(b.dy)y2
= bd3/12
d/2
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
x0
y
d
G
x
x0
x
b
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
h
Ixx =  (b . (h-y)y2.dy)/h
0
= b[ h (y3/3) – y4/4 ]/h
= bh3/12
(h-y)
dy
x
x0
y
h/3
x0
x
b
T-11
(b) Moment of inertia about its centroidal axis:
_
Ixx = Ix x + Ad2
0 0
_
Ix x = Ixx – Ad2
0 0
= bh3/12 – bh/2 . (h/3)2 = bh3/36
T-12
3. CIRCULAR AREA:
Ix
x =
0 0
R 2
 dA . y2
d
=   (x.d.dr) r2Sin2
0 0
r
R 2
x0
=  r3.dr Sin2 d
=
0
r3
2
x
dr  {(1- Cos2)/2} d
R
=[r4/4]
0
2
[/2 – Sin2/4]
0
x0
R
0 0
R

y=rSin
0
= R4/4[ - 0] = R4/4
IXoXo =  R4/4 = D4/64
x
T-13
4. SEMI CIRCULAR AREA:
Ixx =  dA . y2
R 
=   (r.d.dr) r2Sin2
0 0
R

= r3.dr  Sin2
0
y0
d
0
R 
=  r3 dr (1- Cos2)/2) d
0 0

=[R4/4] [/2 – Sin2/4]
x0
4R/3
x
R
x
0
= R4/4[/2 - 0] = R4/8
x0
y0
T-14
About horizontal centroidal axis:
Ixx = Ix
x
0 0
Ix
x
0 0
+ A(d)
= Ixx – A(d)
=  R4/8
Ix
x
0 0
= 0.11R4
2
2
R2/2 . (4R/3)2
T-15
QUARTER CIRCLE:
y
Ixx = Iyy
y0
R /2
Ixx =   (r.d.dr). r2Sin2
0 0
R
x0
/2
= r3.dr  Sin2 d
0
R
4R/3π
0
x
/2
= r3 dr  (1- Cos2)/2) d
0
=[R4/4]
x0
0
/2
[/2 – (Sin2 )/4]
0
= R4 (/16 – 0) = R4/16
y
4R/3π
y0
x
T-16
Moment of inertia about Centroidal axis,
_
Ix x = Ixx - Ad2
0 0
= 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.
Sl.No
Y
1
2
x0
x
Figure
b
d
x0
x
d/2
Y
Xo
h
x0
x0
x
b
3
x0
y0
y0
x
y
y0
y0
x0
-y
0 0
I xx
bd3/12
-
bd3/3
-
bh3/36
-
bh3/12
-
R4/4
R4/4
-
-
0.11R4
R4/8
R4/8
-
0.055R4
0.055R4
R4/16
R4/16
4R/3π
x0
x
x0
4R/3π
Iy
y0
O
x0
-x
0 0
h/3
x
R
4
5
Ix
T-17
I yy
4R/3π
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
P-2
Solution:10
10
25
2
5
5
Ixx = Ixx1+ Ixx2 -Ixx3
30mm
3
1
20mm
15
X
X
5 5 5 5
20mm
20 * 203

 (20 * 20)102
12
20 * 303 1

 (20 * 30) * 302
36
2
10 *103
[
 10 *10 * 202 ]
12
Ixx  297.5 *103 mm 4
P-3
Q.2. Compute the M.I. about the base(bottom) for the area
given in fig.
100mm
30mm
25
mm
80mm
20mm
20mm
x
x
200mm
P-4
SOLUTION :100mm
5
30mm
80mm
25
2
3
4
1
x
20mm
20mm
x
200mm
P-5
Ix x =
3
3
2
200*20 /3+[25*100 /12+(25*100)70 ]
3
2
+2[87.5*20 /36+0.5*87.5*20*(26.67) ]
3
2
+[100*30 /12+100*30*135 ]
6
Ix x=71.05*10 mm
4
P-6
Q.3. Find M.I. about the horizontal centroidal axis for the area
fig. No.3, and also find the radius of gyration.
200mm 200mm
400mm
463.5mm
100
xo
250mm
250mm
xo
400mm
y=436.5mm
100mm
1100mm
P-7
Solution:200mm 200mm
400mm
3
xo
6
5
100mm
4
100
250mm
400mm
463.5mm
250mm
2
1
1100mm
xo
y=436.5mm
Solution to prob. No.03
P-8
∑A=1100*100+400*100+400*400+2(1/2*100*400)-π*502
2
=3,42,150 mm
∑AY=1100*100*50+100*400*300+400*400*700+[(1/2)*100*
400*633.3]*2 - π *502*700
3
=14,93,38,200mm
Y= ∑AY / ∑A=436.5mm
P-9
Moment of Inertia about horizontal centroidal
Axis:3
IXoXo =[1100*100 /12
2
3
+1100*100(386.5) ]+[100*400 /12
2
3
+(100*400)*(136.5) ]+[400*400 /12+400*400*(263.
2
3
2
5) ]+2[100*400 /36+(1/2*400*100)*(196.8) ]4
2
2
[π*(50) /4+π*50 *(263.5) ]
9
4
IXoXo =32.36*10 mm
rXoXo=√(IXoXo/A)=307.536mm.
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
C
 = Sin-1(4/5)
A
X
X
B
P-11
sin =4/5,  =53.13
Solution:-
o
= From geometry of fig ,
D
o
F
N
A
X
M
BK=ABsin(90-53.13 )
C
o
=100sin(90-53.13 )=60mm
 = Sin-1(4/5) ND=BK=60mm
K
L
E
B
X
FD=
o
60/sin53.13 = 75mm
AF=150-FD=75mm
o
FL=ME=75sin53.13 =60mm
AB
100
AE  FC 

 125mm
cos90    0.8
P-12
IXX=IDFC+IFCE+IFEA+IAEB
3
=125 (60) / 36+ (1/2)*125*60*(60+60/3)
3
+125(60) /36+(1/2)*125*60*40
2
3
+125*60 /36 +(1/2)*125*60 *20
3
+125*60 /36 +(1/2)*125*60*20
4
Ixx = 36,00,000 mm
2
2
2
P-13
Q.5. Find the M.I. of the shaded area shown in fig.,about AB.
80mm
40mm
A
B
40mm
40mm
40mm
P-14
Solution:-
IAB =IAB1+IAB2+IAB3
3
80mm
1
[80*80 /36
+(1/2)*80*80(80/3)2
4
+[(0.11*40 )+(1/2)π(40)2
40mm
A
B
4
4
3
40mm
2
40mm
2
(0.424r) ] –[π*20 /4]
40mm
4
IAB =429.3*10 mm
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
P-16
Solution:250mm
A
B
2
50mm
5
3
4
40mm
10mm
1
10mm
6
50mm
250mm
40mm
Y=73.03mm
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=6600mm
2
∑Ai Yi =
=250*10*5+2*40*10*30+40*40*15+40*10*255
+250*10*135
3
=4,82,000mm
P-18
Y= ∑AiYi / ∑A=482000/6600=73.03mm
Moment of Inertia about centroidal axis
=Sum of M.I. of individual rectangles
3
= 250*10 / 12+250*10*68.03
2
3
2
+ [10*40 /12 +40*10*(43.03) *2
3
2
2
+40*10 /12+40*10 (58.03) +10*250 /12+250*
2
3
10(73.05-135) +40*10 /12+40*10(73.05-255)
4
IXoXo =5,03,99395mm
2
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
P-20
solution:-
30mm
20mm
Xo
40mm
50mm
3
2
Xo
1
4
20mm
31.5mm
R=20
40mm
20mm
P-21
∑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-22
Q.8. Find the M.I. about top of section and about two
centroidal
150mm
axes.
10mm
150mm
10mm
P-23
solution:150mm
Yo
41.21mm
10mm
Xo
Xo
150mm
Yo
10mm
P-24
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)
2
4
*(80) =11296667mm
IXX = IXoXo + A(d)2 ,where A(d)2
IXoXo+(150*10+140*10)*(41.21)2=11296667mm3
4
IXoXo =6371701.10 mm
IYoYo =( 10*1503 /12) + (140*103/12)=2824166.7mm4
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
P-26
solution:10mm
A
Yo
B
50mm
Xo
Xo
F
E
Y=17.5mm
C
D
Yo
40mm
X=12.5mm
G
10mm
P-27
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
3
2
IXoXo = [10*50 /12+(50*10)(17.5-5) ]+
3
2
4
[30*10 /12+(30*10) (12.5) =181666.66mm
3
2
3
IYoYo=[(50*10 /12)+(50*10)(7.5) ]+[10*30 /12
2
4
+(30*10)*(12.5) ]=101666.66mm
rxx=√(181666.66/800) = 15.07 mm
ryy=√(101666.66/800)= 11.27 mm
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
P-29
solution:-
60 mm
100 mm
Xo
2
1
Xo
Y=40.3mm
100 mm
P-30
2
2
Y=[(100*100)50-(π/4)60 *74.56]/[(100*100)- (π/4)60 ]
=40.3mm
3
2
IXoXo= [(100*100 /12)+100*100*(9.7) ]4
2
2
[0.55*(60) +0.785*(60) *(34.56) ]
4
=83,75,788.74mm
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]
EP-2
Q.2. Determine second moment of area about the centroidal
300mm
horizontal and vertical axes.
300mm
200
200mm
900mm
[Ans: X = 99.7mm from A, Y = 265 mm
Ixx = 10.29 x 109mm4, Iyy = 16.97 x 109mm4]
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
60
140mm
20
100mm
[Ans: Ixx = 45.54 x 106mm4, Iyy = 24.15 x 106mm4
r = 62.66mm, r = 45.63mm]
EP-4
Q.4. Determine the horizontal and vertical centroidal M.I. Of
the shaded portion of the figure.
60
X
60
20
20
X
60
[Ans: X = 83.1mm
I = 2228.94 x 104mm4, I = 4789.61 x 104mm4]
EP-5
Q.5. Determine the spacing of the symmetrically placed
vertical blocks such that Ixx = Iyy for the shaded area.
200mm
400mm
200mm
d
200mm
200mm
600mm
[Ans: d/2 = 223.9mm d=447.8mm]
EP-6
Q.6. Find the horizontal and vertical centroidal moment of
inertia of the section shown in Fig. built up with R.S.J. (ISection) 250 x 250 and two plates 400 x 16 mm each attached
one to each.
Properties of I section are 160mm
Ixx = 7983.9 x 104mm4
2500mm
Iyy = 2011.7 x 104mm4
2
Cross sectional area=6971mm
160mm
4000mm
[Ans: I = 30.653 x 107mm4, I = 19.078 x 107mm4]
EP-7
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.
Properties of ISA
Cross sectional area = 4400mm2
Ixx = Iyy ;Cxx = Cyy =18.5mm
200mm
18.5mm
18.5mm
20mm
[Ans: I = 111.078 x 107mm4,
300mm
I = 39.574 x 107mm4]
EP-8
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.
Lacing
Properties of ISMC300
Y
C/S Area = 4564mm2
Ixx = 6362.6 x 104mm4
Iyy = 310.8 x 104mm4
23.6mm
X 380mm
X
Cyy = 23.6mm
[Ans: d = 183.1mm]
d
Y
EP-9
Q9. Determine horizontal and vertical centroidal M.I. for the
section shown in figure.
40mm
160mm
40mm
40mm
90mm
[Ans: I = 2870.43 x 104mm4, I = 521.64 x 104mm4]