808_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 808_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 10Ch10.qxd 802 9/27/08 7:29 AM Page 802 CHAPTER 10 Statically Indeterminate Beams Solution 10.3-7 q0 cos a (a) Loading q REACTIONS px b L RA 24 V(0) p4 DIFFERENTIAL EQUATION EI –– q q0 cos a px b L EI –¿ q0 L px sin a b + C1 p L (1) (2) L2 px q0 a b cos a b + C1 x + C2 p L EI – M EI ¿ (3) (4) L4 px x3 q0 a b cos a b + C1 p L 6 EI B.C. B.C. 2. (0) B.C. 3. B.C. 0 4. (L) œ 0 (L) 0 0 ‹ C3 ‹ C1 ‹ C1 (5) 12 p a L + C2 2 p2 b q0 L2 12 1 p2 ; b q0 L2 ; 24 L4 px x3 q0 a b cos a b + 4 q0 L p L 6 p 12 p 4 v L4 b p q0 L2 4 1 p4EI c x2 L4 + q0 a b , or p 2 q0 L4 cos a px b + 4q0 Lx 3 L 6q0 L2x 2 + q0 L4 d 0 L3 L L2 + a C1 b 6 22 2q0 a 1 4 p4 0 L q0 a b p ‹ C4 a (counter-clockwise) EIv x + C3x + C4 2 1. ¿ (0) ; q0 L DEFLECTION CURVE (EQ. 5) 2 + C2 p4 (counter-clockwise) MB x2 + C2 x + C3 2 24 V(L) From equilibrium MA L3 px q0 a b sin a b p L + C1 RB ; q0 L q0 sin px/L (b) Loading q (6) ; FROM SYMMETRY: RA RB MA MB DIFFERENTIAL EQUATIONS SOLVE EQS. (6): C1 C2 24 p4 q0 L 12 4 p q0 L2 SHEAR FORCE (EQ. 2) V q0 L 24 px sin a b + 4 q0 L p L p Elv –– EI ¿– V EI – M EI ¿ q0 sin px/L q q0 L px + C1 cos p L q0 L2 px sin + C1x + C2 L p2 q0 L3 px x2 cos + C1 C2x + C3 L 2 p3 (1) (2) (3) (4) ...
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