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Tutorial 6
G11-20, Engineering Mechanics - ME102
Courtesy: TMH
rraj@iitp.ac.in 2
Question 1
Using the parallel-axis theorem, determine the product of inertia of
the area shown with respect to the centroidal x and y axes.
rraj@iitp.ac.in 3
Solution
1 3
2
We have, 𝐼𝑥𝑦 = 𝐼𝑥𝑦 1
− 𝐼𝑥𝑦 2
- 𝐼𝑥𝑦 3
, where bar on the top denotes “about the
centroidal axes x and y
Symmetry implies, 𝐼𝑥𝑦 1
= 0
For the triangles, 𝐼𝑥𝑦 2
= 𝐼𝑥𝑦 3
, since the distribution of these masses about the set
of x and y axes is equal
And for each triangle, 𝐼𝑥𝑦 2/3
= 𝐼𝑥′𝑦′ 2/3
+ 𝑥2/3𝑦2/3𝐴2/3, where 𝐼𝑥′𝑦′ 2/3
= −
𝑏2ℎ2
72
(product of inertia about triangle's centroidal axes, solved in class), and 𝑥2/3 or 𝑦2/3
is distance of the centroid axis of the respective triangle (denoted by red and blue
dots) from the current set of x and y axes.
rraj@iitp.ac.in 4
Solution
1 3
2
For the triangles “2’ and “3”
rraj@iitp.ac.in 5
Solution
1 3
2
Therefore,
𝐼𝑥𝑦 = 𝐼𝑥𝑦 1
− 𝐼𝑥𝑦 2
- 𝐼𝑥𝑦 3
𝐼𝑥𝑦 = 0 − 𝐼𝑥′𝑦′
2
− 𝑥2𝑦2𝐴2 − 𝐼𝑥′𝑦′
3
− 𝑥3𝑦3𝐴3 (calculations in the previous slide)
𝐼𝑥𝑦 = − −
𝑏2ℎ2
72
− 11.52 × 106 − −
𝑏2ℎ2
72
− 11.52 × 106
rraj@iitp.ac.in 6
Question 2
Using Mohr’s circle, determine the moments of inertia and
the product of inertia of the area with respect to new
centroidal axes obtained by rotating the x and y axes
30°counterclockwise.
X’
Y’
𝐼𝑥 = 68.96 × 106
mm4
𝐼𝑦 = 132.48 × 106
mm4
𝐼𝑥𝑦 = −21.6 × 106
mm4
30°
rraj@iitp.ac.in 7
Solution
(68.96 × 106, −21.6 × 106)
(132.48 × 106
, +21.6 × 106
)
X
Y
X’
Y’
60°
30°
30° rotation counterclockwise in
physical coordinate system implies
60° counterclockwise rotation on
Mohr’s circle
𝐼𝑥/𝐼𝑦
𝐼𝑥𝑦
rraj@iitp.ac.in 8
where, Iave denotes x − ccordinate of the centre of the Mohr′
s circle
2𝜃𝑚
rraj@iitp.ac.in 9
2𝜃𝑚 𝛼
rraj@iitp.ac.in 10
Question 3: Solve and Submit
Using Mohr’s circle, determine for the area indicated the
orientation of the principal centroidal axes (w.r.t. x-y shown
below) and the corresponding values of the moments of
inertia.
𝐼𝑥 = 68.96 × 106 mm4
𝐼𝑦 = 132.48 × 106
mm4
𝐼𝑥𝑦 = −21.6 × 106 mm4
rraj@iitp.ac.in 11
Solution
(68.96 × 106, −21.6 × 106)
(132.48 × 106
, 21.6 × 106
)
X
Y
X’’
Y’’
Principal centroidal axes (X”,Y”) implies points on the x-axis of the Mohr’s
circle such that 𝐼𝑥𝑦=0.
Since 2𝜃𝑚 = −34.22, the principal axes are obtained by rotating the x-y
axes by 17.11° counterclockwise.
𝐼𝑥/𝐼𝑦
𝐼𝑥𝑦
2𝜃𝑚
3 M
rraj@iitp.ac.in 12
Solution
(68.96 × 106, −21.6 × 106)
(132.48 × 106
, 21.6 × 106
)
X
Y
X’’
Y’’
𝐼𝑥/𝐼𝑦
𝐼𝑥𝑦
2𝜃𝑚
𝐼𝑚𝑎𝑥 = 𝐼𝑦" = 139.1 × 106 mm4
𝐼𝑚𝑖𝑛 = 𝐼𝑥" = 62.3 × 106 mm4
3 M
4 M

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29 Solution Tutorial 6 Group no.11-20.pptx

  • 1. Tutorial 6 G11-20, Engineering Mechanics - ME102 Courtesy: TMH
  • 2. rraj@iitp.ac.in 2 Question 1 Using the parallel-axis theorem, determine the product of inertia of the area shown with respect to the centroidal x and y axes.
  • 3. rraj@iitp.ac.in 3 Solution 1 3 2 We have, 𝐼𝑥𝑦 = 𝐼𝑥𝑦 1 − 𝐼𝑥𝑦 2 - 𝐼𝑥𝑦 3 , where bar on the top denotes “about the centroidal axes x and y Symmetry implies, 𝐼𝑥𝑦 1 = 0 For the triangles, 𝐼𝑥𝑦 2 = 𝐼𝑥𝑦 3 , since the distribution of these masses about the set of x and y axes is equal And for each triangle, 𝐼𝑥𝑦 2/3 = 𝐼𝑥′𝑦′ 2/3 + 𝑥2/3𝑦2/3𝐴2/3, where 𝐼𝑥′𝑦′ 2/3 = − 𝑏2ℎ2 72 (product of inertia about triangle's centroidal axes, solved in class), and 𝑥2/3 or 𝑦2/3 is distance of the centroid axis of the respective triangle (denoted by red and blue dots) from the current set of x and y axes.
  • 4. rraj@iitp.ac.in 4 Solution 1 3 2 For the triangles “2’ and “3”
  • 5. rraj@iitp.ac.in 5 Solution 1 3 2 Therefore, 𝐼𝑥𝑦 = 𝐼𝑥𝑦 1 − 𝐼𝑥𝑦 2 - 𝐼𝑥𝑦 3 𝐼𝑥𝑦 = 0 − 𝐼𝑥′𝑦′ 2 − 𝑥2𝑦2𝐴2 − 𝐼𝑥′𝑦′ 3 − 𝑥3𝑦3𝐴3 (calculations in the previous slide) 𝐼𝑥𝑦 = − − 𝑏2ℎ2 72 − 11.52 × 106 − − 𝑏2ℎ2 72 − 11.52 × 106
  • 6. rraj@iitp.ac.in 6 Question 2 Using Mohr’s circle, determine the moments of inertia and the product of inertia of the area with respect to new centroidal axes obtained by rotating the x and y axes 30°counterclockwise. X’ Y’ 𝐼𝑥 = 68.96 × 106 mm4 𝐼𝑦 = 132.48 × 106 mm4 𝐼𝑥𝑦 = −21.6 × 106 mm4 30°
  • 7. rraj@iitp.ac.in 7 Solution (68.96 × 106, −21.6 × 106) (132.48 × 106 , +21.6 × 106 ) X Y X’ Y’ 60° 30° 30° rotation counterclockwise in physical coordinate system implies 60° counterclockwise rotation on Mohr’s circle 𝐼𝑥/𝐼𝑦 𝐼𝑥𝑦
  • 8. rraj@iitp.ac.in 8 where, Iave denotes x − ccordinate of the centre of the Mohr′ s circle 2𝜃𝑚
  • 10. rraj@iitp.ac.in 10 Question 3: Solve and Submit Using Mohr’s circle, determine for the area indicated the orientation of the principal centroidal axes (w.r.t. x-y shown below) and the corresponding values of the moments of inertia. 𝐼𝑥 = 68.96 × 106 mm4 𝐼𝑦 = 132.48 × 106 mm4 𝐼𝑥𝑦 = −21.6 × 106 mm4
  • 11. rraj@iitp.ac.in 11 Solution (68.96 × 106, −21.6 × 106) (132.48 × 106 , 21.6 × 106 ) X Y X’’ Y’’ Principal centroidal axes (X”,Y”) implies points on the x-axis of the Mohr’s circle such that 𝐼𝑥𝑦=0. Since 2𝜃𝑚 = −34.22, the principal axes are obtained by rotating the x-y axes by 17.11° counterclockwise. 𝐼𝑥/𝐼𝑦 𝐼𝑥𝑦 2𝜃𝑚 3 M
  • 12. rraj@iitp.ac.in 12 Solution (68.96 × 106, −21.6 × 106) (132.48 × 106 , 21.6 × 106 ) X Y X’’ Y’’ 𝐼𝑥/𝐼𝑦 𝐼𝑥𝑦 2𝜃𝑚 𝐼𝑚𝑎𝑥 = 𝐼𝑦" = 139.1 × 106 mm4 𝐼𝑚𝑖𝑛 = 𝐼𝑥" = 62.3 × 106 mm4 3 M 4 M