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LECTURE+20+COE-3001-A

Of the theory presented so far only

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Unformatted text preview:   The good news is … the flexure formula sWll works pre=y well! Normal stress calculated from the flexure formula is sWll fairly accurate even in the presence of shear forces and associated warping. •  The flexure formula can therefore be applied to beams subjected to nonuniform bending •  Must sWll avoid locaWons where the beam abruptly changes shape or where there may be sharp disconWnuiWes in loading such as near the supports or close to a concentrated load. 2/22/13 M. Mello/Georgia Tech Aerospace 5 MOMENTS OF INERTIA and FLEXURAL RIGIDITY y Z y Pure bending in the y- z plane Pure bending in the x- z plane ~ M M M Z x y ~ M z ~ M z • relevant moment of inertia: Ix • flexural rigidity: EIx 2/22/13 • relevant moment of inertia: Iy • flexural rigidity: EIy • Iy > Ix ! EIy > EIx M. Mello/Georgia Tech Aerospace 6 MOMENT OF INERTIA OF PLANE AREAS Ix = Iy = 2/22/13 Z Z y 2 dA (12- 9a) x2 dA (12- 9b) M. Mello/Georgia Tech Aerospace 7 Moment of InerWa: A common and most useful example ²་  MOMENT OF INERTIA OF A RECTANGULAR CROSS- SECTIONAL AREA (OF A BEAM) dx dAx Ix = dAy Z y 2 dAy dAy = bdy Ix = Z h/2 y 2 bdy h /2 bh3 Ix = 12 Iy = Z x2 dAx dAx = hdx Z b/2 Iy = x2 hdx b /2 hb3 Iy = 12 •  Note that if h>b (as the diagram suggests), then Ix > Iy •  Rectangular beam exhibits greater resistance to bending in the y- z plane (increased flexural rigidity) •  i.e., EIx > EIy 2/22/13 M. Mello/Georgia Tech Aerospace 8 Moment of InerWa (IBB) of the rectangular cross- secWonal with respect to BB dAy IBB = IBB = IBB Z Z y 2 dAy h y 2 bdy 0 bh3 = 3 (WITH RESPECT TO X- AXIS) ²་  In general, the moment of inerWa increases as the reference axis is moved parallel to itself and away from the centroid. 2/22/13 M. Mello/Georgia Tech Aerospace 9 Example 12- 3: Moment of inerWa of parabolic semisegment ✓ y = f ( x) = h 1 Area element: ✓ dA = ydx = h 1 2/22/13 2 ◆ x dx b2 x2 b2 ◆ DETERMINE: Ix and Iy , i.e., the moments of inerWa of the enWre area with respect to the x and y axes, respecWvely. Z x2 dA •  SOLVE for Iy first : Iy = ✓ ◆ Zb x2 2hb3 2 I= xh 1 dx = y b2 15 0 •  FOR Ix we can sWll integrate along x by taking advantage of the soluWon for the moment of a rectangular cross secWon (just determined on previous page). bh3 •  i.e., recall I = (with r...
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