Unformatted text preview: espect to x axis) x 3 • and so for the diﬀerenWal area element this translates to 1
y3
3
dx dIx = dxy =
3
3 M. Mello/Georgia Tech Aerospace 10 Example 12 3: Moment of inerWa of parabolic semisegment (complete Ix derivaWon) ✓ x2
b2 y = f ( x) = h 1
Z b Z ◆ b y3
Ix =
dIx =
dx
3
0
0
◆3
Z b 3✓
h
x2
dx
Ix =
1
3
b2
0
16bh3
Ix =
105 Moment of inerWa of rectangular diﬀerenWal area element 1
y3
3
dIx = dxy = dx
3
3
2/22/13 M. Mello/Georgia Tech Aerospace 11 MOMENT OF INERTIA OF A COMPOSITE AREA ²་ Here are some common shapes which are oGen encountered when analyzing stresses in beams Hollow rectangular beam Wide ﬂange secWon Channel secWon beam ²་ MOMENT OF INERTIA OF A COMPOSITE BEAM = SUM OF THE MOMENTS OF INERTIA OF ITS PARTS WITH RESPECT TO THE SAME AXIS ²་ IN THE CASE OF THE HOLLOW BEAM ABOVE: ²་ COMPUTE THE MOMENT OF INERTIA WITH RESPECT TO THE AXIS PASSING THROUGH THE CENTROID ²་ TAKE THE ALGEBRAIC SUM OF OF THE MOMENTS OF INERTIA OF THE OUTER AND INNER RECTANGLES ²་ USING RESULTS FROM THE PREVIOUS EXAMPLE WE HAVE: Ix =
2/22/13 outer
Ix inner
Ix bh3
Ix =
12 b 1 h3
1
12 M. Mello/Georgia Tech Aerospace Note: A similar approach may be applied for Iy 12 MOMENT OF INERTIA OF A COMPOSITE AREA ²་ For the wide ﬂange secWon it is oGen necessary to compute the moment of inerWa with respect to the XX or YY axis, depending upon how the beam is loaded. ²་ Although we certainly know how to compute the moment of inerWa of the 3 consWtuent rectangular secWons, what it is not immediately clear is how to incorporate the physical distance of the two outer secWons from the centroid locaWon. ²་ We have a similar problem in the case of the channel secWon… ²་ The determinaWon of moment of inerWa in these cases is achieved with the aid of a special theorem known as the parallel axis theorem. We’ll come back to these two cases… WIDE FLANGE SECTION 2/22/13 Channel secWon beam M. Mello/Georgia Tech Aerospace 13 PARALLEL AXIS THEOREM FOR MOMENTS OF INERTIA • xC,yC axes have origin at the centroid • Parallel set of x,y axes have origin at point O • Consider: 0 Z
Z
Z
Z
Ix = (y + d1 )2 dA = y 2 dA + 2d1 ydA + d2 dA
1 Ixc ARBITRARY SHAPE WITH CENTROID C A
First moment of the area with respect to...
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 Spring '09
 ZHU
 Second moment of area, Ibeam, M., Mello/Georgia

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