724_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

724_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 09Ch09.qxd 9/27/08 718 1:31 PM Page 718 CHAPTER 9 Deflections of Beams Solution 9.3-14 Cantilever beam (partial uniform load) BENDING-MOMENT EQUATION (EQ. 9-12a) EI – q (a 2 M q2 (a 2 x)2 B.C. 2ax + x2) 3 n(0) 0 C3 0 3 EI qa x + C4 (a … x … L) 6 C2x + C4 (0 … x … a) q2 aa x 2 EI ¿ B.C. 1 n (0) EI ¿¿ 0 M EI ¿ C1 C2 B.C. ax2 + 0 4 (n)Left (0 … x … a) (a x (v¿)Right at x q a2x2 a 22 a ‹ C2 ax3 x4 + b + C3 3 12 qa3 6 a) (L) L) a qa3 (4L 24EI dB ‹ C4 4ax + x2) (0 … x … a) qa3 (4x 24EI L) x (n)Right at x qx2 (6a2 24EI 0 (a 2 (v¿)Left EI x3 b + C1 3 B.C. (a … x … L) ; ; a) (0 … x … a) y q0 MA beam AB supporting a distributed load of peak intensity q0 acting over one-half of the length (see figure). Also, obtain formulas for the deflections dB and dC at points B and C, respectively. (Note: Use the second-order differential equation of the deflection curve.) x L /2 A C L /2 RA Solution 9.3-15 BENDING-MOMENT EQUATION (x) EI – M(x) EI ¿ q0 L x2 8 3 q0 L x 24 EI B.C. n (0) B.C. n(0) L ¿a b 2 0 0 q0L2 x + C1 6 22 q0 L x + C1 x + C2 12 C1 C2 5 q0 L3 96 EI 0 0 For q0 L 3 (x 24 EI 2 L x2) ; dC L For 0 … x … 2 q0 L2 6 ; (These results agree with Case 2, Table G-1.) Problem 9.3-15 Derive the equations of the deflection curve for a cantilever q0 L x 4 qa4 24 L ab 2 q0 L4 64 EI ; L …x…L 2 EI – M(x) q0 L x 4 q0L2 6 ax L2 b 2 (L x) d a x q0 (L L 1 cq 20 L 22 b 23 x) 2 q0 L B ...
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