________________________________________________________________________ PROBLEM (10.37) A cantilever of variable cross section supports a uniformly distributed load w (Fig. P10.37). Verify, using the last of Eqs. (10.4), that the expression for the deflection curve is v= wL (− x 3 + 3L2 x − 2 L3 ) Eb0 h3 (P10.37) Here h represents the depth of the beam and b0 is the width at the fixed end. SOLUTION 1 x ( b0 )h3 12 L The last of Eq. (10.4): (b0 x L)h3 d 2 v d2 d 2v d2 ( ) [ ] = −W EI = E dx 2 dx 2 dx 2 12 dx 2 Integrating twice (b x L)h3 d 2 v d [E 0 ] = − wx + C1 = −V dx 12 dx 2 I= (b0 x L)h3 d 2 v 1 E = − wx 2 + C1 x + C2 = M 2 12 2 dx Boundary conditions: M (0) = 0 : C2 = 0 V (0) = 0 : C1 = 0 Equation (1) becomes then d 2v 6wL dv 3wL 2 =− x =− x + C3 2 3 dx Eb0 h dx Eb0 h3 wL 3 v=− x + C3 x + C4 Eb0 h3 We have 3wL3 2 wL4 v '( L) = 0 : C3 = , v ( L ) = 0 : C4 = − Eb0 h3 Eb0 h3 Substitution of these into Eq (2) results in Eq. (P10.37). (1) (2) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.