Substute bending moment expression mmx

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Unformatted text preview: ng across the en?re span of the beam. Determine: 1.  equaBon of the deflecBon curve 2.  maximum deflecBon δmax at the midpoint 3.  angles of rotaBon ΘA and ΘB at the supports SOLUTION: 1.  First must obtain bending- moment funcBon M(x) in one of 2 ways ☞ free- body diagram and simple staBcs ☞ invoke derivaBve relaBonships: dV = dx 4/9/13 q V= dM dX M. Mello/Georgia Tech Aerospace 19 EXAMPLE 8- 1 (cont.) (a) Apply basic staBcs ☞ C x V M x/2 qx RA = X qL 2 Mc = 0 M + qx(x/2) M ( x) = 4/9/13 M. Mello/Georgia Tech Aerospace qL x=0 2 q x2 qLx + =0 2 2 20 EXAMPLE 8- 1 (cont.) (b) invoke derivaBve relaBonships ☞ dV dx = V ( x) = dM dx q q x + C1 obtain C1 c V (0) = RA = ) C1 = V ( x) = 4/9/13 =V M ( x) = Rx V ( ⇠ ) d⇠ + C 2 obtain C2 qL 2 ) C2 = 0 qL 2 qx + M (0) = 0 qL 2 M ( x) = q x2 2 + qLx 2 M. Mello/Georgia Tech Aerospace 21 EXAMPLE 8- 1 (cont.) 1.  equaBon of the deflecBon curve 00 INTEGRATE THE BENDING- MONENT EQUATION: EIv (x) = M (x) (8- 12a) Zx 3 2 0 EIv (x) = 0 EIv (x) = 0 M ( ⇠ ) d⇠ + C 1 Z x 0 EIv (x) = EIv (x) = q⇠2 qL⇠ + d⇠ + C 1 2 2 q x3 qLx2 + + C1 6 4 qx qLx + 6 4 q L3 24 slope of the beam: q (4x3 24EI 0 v ( x) = 6Lx2 + L3 ) Now, integrate a 2nd ?me: 0 q ( x4 24EI Invoke symmetry condi?on: v (L/2) = 0 v ( x) = q L3 qL3 0= + + C1 48 16 qL3 q L3 C1 = 48 16 Invoke boundary condi?on: v (0) = 0 C1 = ) C2 = 0 v ( x) = q L3 24 q ( x4 24EI 2Lx3 + L3 x) Beam deflec?on: v ( x) = 4/9/13 2Lx3 + L3 x + C2 ) M. Mello/Georgia Tech Aerospace qx ( x3 24EI 2Lx2 + L3 ) 22 EXAMPLE 8- 1 (cont.) 2.  maximum deflecBon δmax at the midpoint max v ( x) = qx ( x3 24EI 2Lx2 + L3 ) = |v (L/2)| v ( x) = q (L/2) L3 ( 24EI 8 v ( x) = q ( L4 ) 1 ( 24EI 16 11 +) 42 v ( x) = q ( L4 ) 1 ( 24EI 16 4 8 +) 16 16 v ( x) = 5qL4 = (24)(16)EI max max 5qL4 384EI 5qL4 = 384EI 5qL4 = 384EI L2 2L + L3 ) 4 4/9/13 Maximum deflec?on at center of beam M. Mello/Georgia Tech Aerospace 23 EXAMPLE 8- 1 (finish) 3.  angles of rotaBon ΘA and ΘB at the supports We must invoke small angle approxima?on: ✓ = 0 Recall we derived slope of the beam: v (x) = 0 At x=0: ✓A = v (0) = dv dx q (4x3 24EI 6Lx2 + L3 ) q L3 (note the nega?ve slope as an?cipated) 24EI qL3 Express angle ✓ A as posi?ve quan?ty: ✓A = 24EI qL3 Similarly, ✓B = v (L) = (note the posi?ve slope as an?cipated) 24EI 0 Note these angles are indeed small as can be verified by subs?tu?ng realis?c values for these variables (see Gere and Goodno) 4/9/13 M. Mello/Georgia Tech Aerospace 24 DEFLECTIONS BY INTEGRA...
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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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