Derive the equation of the deflection curve and then determine the maximum

# Derive the equation of the deflection curve and then

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Derive the equation of the deflection curve, and then determine the maximum deflection max . Use the fourth- order differential equation of the deflection curve (the load equation). Solution 9.4-7 Simple beam (parabolic load) A y x 4 q 0 x L 2 B q = ( L x ) L L OAD EQUATION (E Q . 9-12 c) B . C . 1 C 2 0 B . C . 2 EIv ¿ q 0 30 L 2 ( 5 L 3 x 2 5 L x 4 2 x 5 ) C 3 C 1 q 0 L 3 EIv ( L ) 0 EIv (0) 0 EIv M EIv q 0 3 L 2 (2 Lx 3 x 4 ) C 1 x C 2 EIv 2 q 0 3 L 2 (3 Lx 2 2 x 3 ) C 1 EIv –– q 4 q 0 x L 2 ( L x ) 4 q 0 L 2 ( Lx x 2 ) B . C . 3 (Symmetry) B . C . 4 v (0) 0 C 4 0 max v ¢ L 2 61 q 0 L 4 5760 EI v q 0 x 90 L 2 EI (3 L 5 5 L 3 x 2 3 Lx 4 x 5 ) EIv q 0 30 L 2 ¢ L 5 x 5 L 3 x 3 3 L x 5 x 6 3 C 4 C 3 q 0 L 3 30 v ¿ ¢ L 2 0
SECTION 9.4 Deflections by Integration of the Shear Force and Load Equations 563 Problem 9.4-8 Derive the equation of the deflection curve for a simple beam AB carrying a triangularly distributed load of maximum intensity q 0 (see figure). Also, determine the maximum deflection max of the beam. Use the fourth-order differential equation of the deflection curve (the load equation). Solution 9.4-8 Simple beam (triangular load) A y q 0 x B L L OAD EQUATION (E Q . 9-12 c) B . C . 1 C 2 0 B . C . 2 B . C . 3 v (0) 0 C 4 0 EIv q 0 x 5 120 L q 0 Lx 3 36 C 3 x C 4 EIv ¿ q 0 x 4 24 L q 0 L x 2 12 C 3 C 1 q 0 L 6 EIv ( L ) 0 EIv (0) 0 EIv M EIv q 0 x 3 6 L C 1 x C 2 EIv q 0 x 2 2 L C 1 EIv –– q q 0 x L B . C . 4 v ( L ) 0 M AXIMUM DEFLECTION Set v 0 and solve for x : x 1 0.51933 L (These results agree with Case 11, Table G-2.) 0.006522 q 0 L 4 EI max v ( x 1 ) q 0 L 4 225 EI ¢ 5 3 2 3 A 8 15 1 2 x 2 1 L 2 ¢ 1 A 8 15 v ¿ q 0 360 LEI (7 L 4 30 L 2 x 2 15 x 4 ) v q 0 x 360 LEI (7 L 4 10 L 2 x 2 3 x 4 ) C 3 7 q 0 L 3 360 Problem 9.4-9 Derive the equations of the deflection curve for an overhanging beam ABC subjected to a uniform load of intensity q acting on the overhang (see figure). Also, obtain formulas for the deflection C and angle of rotation C at the end of the overhang. Use the fourth-order differential equation of the deflection curve (the load equation). Solution 9.4-9 Beam with an overhang A C B q y x L 2 L L OAD EQUATION (E Q . 9-12 c) (0 x L ) (0 x L ) (0 x L ) B . C . 1 C 2 0 B . C . 2 ¢ L x 3 L 2 EIv qx 2 2 3 qLx 2 C 4 C 3 3 qL 2 EIv ¢ 3 L 2 0 EIv V ¢ L x 3 L 2 EIv qx C 3 ¢ L x 3 L 2 EIv –– q EIv (0) 0 EIv M EIv C 1 x C 2 EIv C 1 EIv –– q 0 B . C . 3 B . C . 4 at x L (0 x L ) ¢ L x 3 L 2 EIv ¿ qx 3 6 3 qLx 2 4 9 qL 2 x 8 C 6 EIv ¿ qLx 2 16 C 5 C 1 qL 8 C 1 L qL 2 2 3 qL 2 2 9 qL 2 8 EI ( v ) Left EI ( v ) Right C 4 9 qL 2 8 EIv ¢ 3 L 2 0 EIv M (Continued)
564 CHAPTER 9 Deflections of Beams B . C . 5 at x L (a) (0 x L ) B . C . 6 v (0) 0 C 7 0 B . C . 7 v ( L ) 0 for 0 x L From Eq.(a): ¢ L x 3 L 2 EIv qx 4 24 3 qLx 3 12 9 qL 2 x 2 16 qL 3 x 2 C 8 C 6 qL 3 2 C 5 qL 3 48 EIv qLx 3 48 C 5 x C 7 C 6 C 5 23 qL 3 48 ( v ¿ ) Left ( v ¿ ) Right B .
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