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The concept of moment of inertia, which is the product of the elemental area and the square of the perpendicular distance between the centroid of the area and the axis of reference. It covers topics such as the radius of gyration, polar moment of inertia, and the parallel axis theorem. The document also provides formulas and examples for calculating the moment of inertia of various geometric shapes, including rectangles, circles, and semi-circles. Additionally, it includes numerical problems and solutions related to the calculation of moment of inertia for different areas. Overall, this document provides a comprehensive overview of the moment of inertia and its applications in engineering and physics.
Typology: Exercises
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Moment of Inertia( Second moment of area) 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
o
ox = da 1 y 1 2
= ∑ da x 2
Radius of Gyration k k k k k k Elemental area Elemental area r r 11 r r 22 r r 33 A A AA B B BB Radius of gyration is defined as a constant distance of all Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out elemental areas which have been rearranged with out altering the total moment of inertia. altering the total moment of inertia. I IABAB= da k 2
I IABAB = ∑ da k 2 I IABAB= A k 2 k=√I k=√IABAB/A/A
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. Polar moment of Inertia (Perpendicular Axes theorem)
Parallel Axis Theorem
G Moment of inertia of any area about an axis AB 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.
M.I. about its horizontal centroidal axis : Find the moment of inertia of rectangular area about centroidal horizontal axis by direct integration
3 / 2 / 2 2 2 / 2 / 2
D D D D x x
G
dy y D D/
Find the Moment of Inertia of rectangular area about its base (about the line AB) using Parallel Axis Theorem I AB
XX
A
G
x x
dy y D D/ =BD 3 /12+(BD)(D/2) 2 =BD 3 /12+BD 3 / =BD 3 /
xx = dA. y 2 R 2 = (x.d.dr) r 2 Sin 2 0 0 R 2 = r 3 .dr Sin 2 d 0 0 R 2 = r 3 dr {(1- Cos2)/2} d 0 R 0 2 =[r 4 /4] [/2 – Sin2/4] 0 0 = R 4 /4[ - 0] = R 4 / I XX
4 /4 = D 4 /
d y=rSin r Find the moment of inertia of Circular area about the centroidal horizontal axis
AB = dA. y 2 R = (r.d.dr) r 2 Sin 2 0 R 0 = r 3 .dr Sin 2 d 0 0 R = r 3 dr (1- Cos2)/2) d 0 0 =[R 4 /4] [/2 – Sin2/4] 0 = R 4 /4[/2 - 0] = R 4 /8 = (πD 4 /32)
0
0
R Find the moment of inertia of Semi-circular area about the Base & centroidal horizontal axis
AB
CD R / IAB = (r.d.dr). r 2 Sin 2 0 0 R / = r 3 .dr Sin 2 d 0 0 R / = r3 dr (1- Cos2)/2) d 0 0 / =[R 4 /4] [/2 – (Sin2 )/4] 0
4R/3π 4R/3π Find the moment of inertia of Quarter-circular area about the base & centroidal horizontal axis
Moment of inertia about Centroidal axis, I xx
AB
Find the moment of Inertia of the shaded area shown in fig.about its base. 20mm 5 5 5 5 5 25 15 10 10 5 20mm 30mm X X Q 8 Numerical Problems & Solutions
20mm 5 5 5 5 5 25 15 10 10 5 20mm 30mm
xx
xx
xx
xx
x x 2 3 2 3 2 3
X X 1 3 2 Solution:-
100mm 200mm 20mm 20mm 80mm 30mm
1 2 3 4 5
25
x x
3 /3+[25* 3 /12+(25100) 2 ] +2[87.5 3 /36+0.587.520(26.67) 2 ] +[100 3 /12+10030 2 ] I x x
6 mm 4
