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LECTURE+20+COE-3001-A

Increases as the reference axis is moved

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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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