LECTURE 21 COE-3001-A

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Unformatted text preview: force diagram dV dx •  = q ( x) dM dx = V ( x) Note abrupt jump of magnitude 10.8kN at B (equal to magnitude of reacTon force RB) Next, construct the bending- moment diagram: 2/22/13 M. Mello/Georgia Tech Aerospace 16 EXAMPLE 5- 4 (cont.) I.  NEUTRAL AXIS OF THE CROSS SECTION: 1.  z- axis defined as neutral axis of cross SecLon (passes through centroid) 2.  FIND CENTROID COORDINATES avg.thickness of sloping flanges ~ M web thickness Channel- shaped cross- secTon of beam ABC 3.  DEFINE 3 RECTANGLES (A1 , A2 , A3) 4.  ESTABLISH REFERENCE Z AXIS (Z- Z AXIS) 5.  Determine distances y1 , y2 , y3 rom the Z- Z Axis to centroids of A1 , A2 , A3, respecLvely. 2/22/13 M. Mello/Georgia Tech Aerospace 17 EXAMPLE 5- 4 (cont.) Dist. centroid Area Area 1 y1 = t/2 = 12mm/2 = 6mm A1 = (b- 2t)t = (300mm- 2(12mm))12mm = 3312mm2 Area 2 y2 = h/2 = 80mm/2 = 40mm A2 = ht = (80mm)(12mm) = 960mm2 Area 3 y3 = h/2 = 80mm/2 = 40mm A3 = ht = (80mm)(12mm) = 960mm2 Channel- shaped cross- secTon of beam ABC avg.thickness of sloping flanges ~ M web thickness 2/22/13 M. Mello/Georgia Tech Aerospace 18 EXAMPLE 5- 4 (cont.) 6.  Calculate the centroid of the channel- shaped cross secLon c1 = c1 = P y P i Ai Ai (6mm)(3312mm2 )+2(40mm)(960mm2 3312mm2 +2(960mm2 ) c1 = 18.48mm c2 = h c1 = 80mm 18.48mm = 61.52mm c2 = h c1 = 80mm 18.48mm Channel- shaped cross- secTon of beam ABC c2 = 61.52mm ) neutral axis position determined 2/22/13 M. Mello/Georgia Tech Aerospace 19 EXAMPLE 5- 4 (cont.) II.  Calculate the moment of inerLa of the channel- shaped cross secLon (apply the parallel- axis theorem) Area 1: (Iz )1 = (Ic )1 + A1 d2 1 1 3 2 (Iz )1 = 12 (b 2t)(t ) + (A1 )(c1 t/2)d1 (I ) = 555600mm4 z1 Similarly for A2, A3: (Iz )2 = (Iz )3 = 956600mm4 I = (I ) + (I ) + (I ) = 2.469 ⇥ 106mm4 z 2/22/13 z1 z2 Channel- shaped cross- secTon of beam ABC z3 M. Mello/Georgia Tech Aerospace 20 EXAMPLE 5- 4 (cont.) Channel- shaped cross- secTon of beam ABC III.  SECTION MODULI: S1 = IZ c1 = 133600mm3 S2 = IZ c2 = 40100mm3 IV.  MAXIMUM STRESSES: Recall maximum bending- moments x Mpos = 2.025kN · m Mneg = 2/22/13 3.6kN · m M. Mello/Georgia Tech Aerospace 21 EXAMPLE 5- 4 (finish) ²་  AT CROSS SECTION OF MAX. POSITIVE BENDING (x=1.125m): 2 = largest tensile stress (on bottom) Mpos T = 2 = S2 2.025kN T = 2 = 40100mm3 = 50.5M P a 1 = largest tensile compressive stress (on top) C = 1 = Mpos S1 2.025kN 15.2M P a C= 1= 133600 = ²་  AT CROSS SECTION OF MAX. NEGATIVE BENDING (x=3.0m): Mneg 3.6kN T = 1 = S1 = 133600mm3 = 26.9M P a C = 2 2/22/13 = Mneg S2 = 3.6kN 40100mm3 = 89.8M P a M. Mello/Georgia Tech Aerospace 22...
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This note was uploaded on 09/19/2013 for the course CEE 3001 taught by Professor Zhu during the Spring '09 term at Georgia Institute of Technology.

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