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An in-depth explanation of moments of inertia, including definitions, formulas, and the Parallel-Axis Theorem. It also covers the calculation of moments of inertia for common shapes and includes examples. This information is crucial for understanding beam and column behavior.
Typology: Study notes
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y
x x (^) eldx
dA = y⋅dx
θ pole
o
r
x x x x
2
2
We can define a single integral using a narrow strip:
for I (^) x,, strip is parallel to x for I (^) y, strip is parallel to y
2 2 2
J (^) o = Ix + I y
I (^) x = rx^2 A ⇒ A
rx = x radius of gyration in x
r (^) y = y radius of gyration in y
ro = o polar radius of gyration, and r (^) o^2 = rx^2 + r (^) y^2
axis through centroid at a distance d away from the other axis
axis to find moment of inertia about
y
dA
y′
d
The Parallel-Axis Theorem
y dA d ydA d dA
I y dA y-d dA 2 2
2 2
2
2 I (^) x = Icx + Ad y (text notation) or 2 I (^) x = Ix + Ady where I (^) cx ( or Ix ) is the moment of inertia about the centroid of the area about an x axis and dy is the y distance between the parallel axes
Similarly 2 I (^) y = Iy + Adx Moment of inertia about a y axis J J Ad^2 o =^ c + Polar moment of Inertia r^2 r^2 d^2 o =^ c + Polar radius of gyration r^2 = r^2 + d^2 Radius of gyration
Composite Areas:
I = ∑ I +∑ Ad^2 where I is the moment of inertia about the centroid of the component area d is the distance from the centroid of the component area to the centroid of the composite area (ie. dy = y ˆ^ - y )
Basic Steps
Example 1 (pg 257)
Find the moments of inertia ( x ˆ = 3.05”, yˆ = 1.05”).
Example 2 (pg 253)
Example 3 (pg 258)