811_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

811_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 10Ch10.qxd 9/27/08 7:29 AM Page 805 805 SECTION 10.3 Differential Equations of the Deflection Curve DEFLECTION CURVE (EQ. 5) REACTIONS RA 13 qL 30 0 V(0) RB ; 7 qL 30 0 V(L) x q0 a 2 MA 4 13 b+ q Lx 30 0 12L2 x 1 q0 L2 (counter-clockwise) 15 MA 1 q L2 (clockwise) 20 0 MB x4 24 x6 13 x3 q0 L b+ 30 6 360L2 1 x2 q0 L2 , or 15 2 ; From equilibrium 2 q0 a EIv v 1 q L2 15 0 q0 360L2EI [x 6 + 26L3x 3 15L2x 4 12L4x 2] ; ; ; Problem 10.3-9 A fixed-end beam of length L is loaded by triangularly q0 y distributed load of maximum intensity q0 at B. Use the fourth-order differential equation of the deflection curve to solve for reactions at A and B and also the equation of the deflection curve. MB A MA L x B RB RA Solution 10.3-9 Triangular load q q0 x L B.C. B.C. DIFFERENTIAL EQUATION EIv –– EIv –¿ q0 x L (1) x2 + C1 2L (2) q EIv – EIv ¿ q0 x3 q0 + C1x + C2 6L (3) x4 x2 + C1 + C2x + C3 24L 2 (4) M q0 5 3 4 (L) 0 0 ‹ C1L + 2C2 ‹ C1L + 3C2 Solve Eqs. (6) and (7): C1 C2 3 q0 L 20 1 q L2 30 0 SHEAR FORCE (EQ. 2) V 2 x x x + C1 + C2 + C3x + C4 120L 6 2 3 ¿ (L) (5) q0 x2 3 + qL 2L 20 0 EIv q0 B.C. 1 ¿ (0) 0 ‹ C3 0 REACTIONS B.C. 2 v(0) 0 ‹ C4 0 RA V(0) 3 qL 20 0 ; q0 L2 12 q0 L2 20 (6) (7) ...
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