Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
(1) Considering the Fig. Prob. 01, show that the moment of inertia of the rectangular area
about the -axis
axis through one end may be used for its polar moment of inertia about point
where is considered small as compared with . What is the percentage error where
/ 1/10.
Solu.
Rectangle moment of inertia w.r.t. , Rectangle moment of inertia w.r.t. , Polar moment of inertia at , Evaluating error percentage, % ⇒/
1
4
⇒/
4
"# $"%
"%
& 100 '
& 100 /
1
4
)
'
'
$ ()
,
.
-
(*( +
1 1 4 & 100
& 100
1
0.01
& 100 /0.249%
4 1 1 & 0.01
4
1
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
(2) Determine the moment of inertia of the shaded area (shown in Fig. Prob. 02) about the and -axes.
axes. Use the differential element for both the calculations.
calculations
Solu.
Differential area, 4 / 4 /
Rectangle moment of inertia w.r.t. , 4
(
( 56 4
(
(
1 9
1 1
⇒ 8 / ; * / . 3 4 7 6
3 4 7
28
(
(
Rectangle moment of inertia w.r.t. , 56 4 56 /
= (
?
(
(
4 56 /
1 1
⇒ 8 / ; * / . 4 5 6
4 5
20
(
/ 56 4 56 / ) 4
(
(
4
2
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
(3) Determine the rectangular moments of inertia of the shaded area about the - and -axes
and the polar radius of gyration about point .
Solu.
Slant height in terms of , A
/
Differential area, 4 24
4 2 A
A
4 4
A
A
A
-
A
Rectangle moment of inertia w.r.t. , 5( 4 5( 4 F H ⇒ A
G / 4G
A
A
(
2 4 5( 4
A
A Rectangle moment of inertia w.r.t. , 5( 4
5(
A
A
1 1 G / 8
;
⇒ 12 G 4 ( 48 G
Polar moment of inertia, @ Total area, A
A
/ G /
(
A
Radius of gyration, B C D C 1 "
E
1 (
A
A
A
G / G / 3
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
(4) Determine the product of inertia about - axes of the circular area with three equal
square holes.
Solu.
Shape
Circle
Rectangle 1
Rectangle 2
Rectangle 3
Area (
), in.2
I10
10
/3 & 3
/3 & 3
/3 & 3
̅ , in.
0
3.5
/3.5
/3.5
T, in.
0
/3.5
/3.5
3.5
Total, ∑̅ T
̅ T, in.4
0
110.25
/110.25
110.25
110.25
(5) The maximum and minimum moments of inertia of the shaded area are 12 & 10J mm4
and 2 & 10J mm4, respectively, about axes passing through the centroid ^, and the
product of inertia with respect to the - axes has a magnitude of 4 & 10J mm4. Use the
proper sign for the product of inertia and calculate and the angle _ measured
counterclockwise from the -axis
axis to the axis of maximum moment of inertia.
Solu.
From figure we can easily depict that the sign of is negative (/).
). Also, from the equation of
principal moment of inertia, K( KVW 12 2 & 10J 14 & 10J mm4
Similarly, / CL2
K( / X
YZ / 4
MN2 & 12 / 14O & 10 / 4 & /4 & 10 √36 & 10 6 & 10J mm4
From above two expressions, we get
10 & 10J mm4 and 4 & 10J mm4
Principal axes,
tan 2] /
2
2 & /4 & 10J 4
/ /6 & 10J
3
4
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
2] 53.13° ⇒ ] 26.6°
(6) Derive the expression for the moment of inertia of the trapezoidal area about the -axis
through its base.
Solu.
Distort to a rectangle and a triangle without altering -distribution of area.
Rectangle G and Triangle /
For trapezoid, A)
/
$(
(A)
$(A)
3G
5
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
(7) Calculate the polar radius of gyration about point of the area shown.
shown Note that the
widths of the elements are small compared with their lengths.
lengths.
Solu.
For Part 1 and Part 2:
Area, , 10 & in2
Polar moment of inertia, , , , 10 & & 10 6)
in4
Part 3:
b6
Area, a
& c
b6
Polar moment of inertia, , a
d
c 10 b
d
& 10 in4
Combined:
b
Area, 2 & 10 & 10 & 26.78 in2
2 & 6)
b
d
d
10 1.68 & 10 in4
6
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
Radius of gyration, B C % C
"
E
.Jd&6)
J.9d
7.92 in
(8) A thin plate of mass e has the trapezoidal shape shown in the Fig. Prob. 08
08. Determine
the mass moment of inertia of the plate with respect to ()
( the -axis
axis and ()
( the -axis.
Solu.
()
Total area of the trapezoidal shape, 2 & & 2 & 3
Density of the material (as it has uniform thickness) f Rectangular moment of inertia, ,ghij. ,klV.
K
(
Moment of inertia of the sub-shape
shape rectangle, ,ghij. 2 Moment of inertia of the sub-shape
shape triangle, ,klV. =
2 Thus, the area moment of iner
inertia, Mass moment of inertia & f
J
J
J
5 e
5
& e
6
3
18
()
Polar area moment of inertia, ,ghij. ,klV.
Rectangle,
7
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
2
1
10 ,ghij. ,ghij. @,ghij. 2 3
3
3
Triangle, at the centroid of triangle ′
m,klV. m,klV
klV. @m,klV. 1
1
5 2 2 36
36
18
Distance between the centroid of the triangle sub
sub-shape and origin,
2
65
4 p / 0 *2 / 0. p 3
3
9
,klV. m,klV. 4 klV. Total polar area moment of inertia, Polar mass moment of inertia J=
J
6
K
135 15 5 65 18
18
9
2
=
J=
J=
J
& ( d e 3.61e
(9) Determine by direct integration of the mass moment of inertia with respect to -axis of
the pyramid shown assuming that it has a uniform density and a mass e.
y
b
a
h
x
z
Solu.
Volume of pyramid, n (
A
(A
8
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
K
Density of the material, f (A
Differential mass moment of inertia of the pyramid w.r.t. its centroid axis parallel to -axis is
4
1 q 4e
12
and q defines the variation of side of any differential rectangle area of thickness 4, both
varies w.r.t. as 2 1 / A and 2 1 / A , respectively. Similarly, differential mass 4e fq4 4f 1 / 4 .Hence, substituting the values back,
A
⇒ 4
4 1 / 4f4
G
12
⇒ fG e
15
5
A
4f ⇒ 5 4
r 1 / 4
G
3
6
(10) Shown is the cross section of ideal roller. Determine its mass moment of inertia and its
radius of gyration with respect to the axis m . (The density of bronze is 8580 kg/m3; of
aluminum, 2770 kg/m3; and of neoprene, 1250 kg/m3.)
Solu.
Cylinder's mass moment of inertia is e 9
Indian Institute of Technology, Guwahati
ME 101 – Engineering Mechanics
Assignment
___________________________________________________________________________
Mass, e I Gf
Bronze:
XeV V / es s Y 8.648 & 10$d kg m2
btA
XV / s Y b&d=d6&6.6u=
0.0045 / 0.003
003 Mass, e IGfXV / s Y I & 8580 & 0.01950.0045 / 0.003 5.913 & 10$ kg
Aluminum:
b&996&6.6J=
0.006 / 0.0045 6.360 & 10$d kg m2
Mass, e I & 2770 & 0.0165
01650.006 / 0.0045 2.261 & 10$ kg
Neoprene:
b&=6&6.6J=
0.0135 / 0.006 1.034 & 10$J kg m2
Mass, e I & 1250 & 0.0165
01650.0135 / 0.006 9.476 & 10$ kg
Combined:
8.648 & 10$d 6.360 & 10$d 1.034 & 10$J
1.184 & 10$J kg m2
Mass, e 5.913 & 10$ 2.261
2
& 10$ 9.476 & 10$ 1.765 & 10$ kg
Radius of gyration, 4 C C
"
K
.d&6v?
.9J=&6v
8.191 & 10$ m.
10