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This essay constitutes a sustained reading of "The Impossible" by Georges Bataille, focusing upon the idea of the "lost moment." Drawing upon sources such as Freud, Kierkegaard and Proust, this ess...
4. MOMENT OF INERTIA Moment of Inertia( Second moment 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 Iox= da1 y12 + da2 y22+ da3 y32+ -= ∑ da y2 Ioy = da1 x12 + da2 x22 + da3 x32+ ----
y
dA x
= ∑ da x2 y
Radius of Gyration Elemental area
A
r1
A
k
r2
k
r3
k
B
B
Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out altering the total moment of inertia. I
= da k 2 + da k 2
I
A k 2
Radius of Gyration Elemental area
A
r1
A
k
r2
k
r3
k
B
B
Radius of gyration is defined as a constant distance of all elemental areas which have been rearranged with out altering the total moment of inertia. I
= da k 2 + da k 2
I
A k 2
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 I zz or J or Ip. The moment of inertia of an area in xy plane w.r.to z. axis is I zz = Ip = J = ∫r 2dA = ∫(x2 + y2) dA = ∫x2dA + ∫y2dA = Ixx +Iyy
Y
x
r y O
Polar moment of Inertia (Perpendicular Axes 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.
Parallel Axis Theorem
dA y
x
* G
x
d A
B
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.
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
EXERCISE PROBLEMS
Q.2. Determine second moment of area about the centroidal horizontal and vertical axes. 300mm 300mm 200
200mm 900mm
[Ans: X = 99.7mm from A, Y = 265 mm
EXERCISE PROBLEMS
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: I = 45.54 x 106mm4, I = 24.15 x 106mm4
EXERCISE PROBLEMS
Q.4. Determine the horizontal and vertical centroidal M.I. Of the shaded portion of the figure.
60
X
20
60 [Ans: X
20
60
83.1mm
X
EXERCISE PROBLEMS
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
EXERCISE PROBLEMS
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 10 4mm4 Iyy = 2011.7 x 104mm4
160mm 2500m
Cross sectional area=6971mm
2
160mm
EXERCISE PROBLEMS
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 mm 2.
Properties of ISA Cross sectional area = 4400mm 2 Ixx = Iyy ;Cxx = Cyy =18.5mm 200mm
18.5mm 18.5mm
20mm
EXERCISE PROBLEMS
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 I xx = Iyy for the composite section. Properties of ISMC300 C/S Area =