________________________________________________________________________ PROBLEMS (10.60 and *10.61) A beam is loaded and supported as shown in Fig. P10.60 and P10.61. Determine the deflection at the free end . SOLUTION (10.60) Q P a a A a C B D Actual Loading = Q P a a A C + B D A B Equivalent Loadings Use Table B.14 vB = (vB ) P + ( vB ) Q where, (vB ) P = (vC ) P + (θC ) P (2a) =− Pa 3 Pa 2 4 Pa 3 − (2a) = − 3EI 2 EI 3 EI (vB )Q = (vD )Q + (θ D )Q (a) =− Q(2a)3 Q(2a) 2 14 Qa 3 − (a) = − 3EI 2 EI 3 EI Thus, 2a 3 vB = (2 P + 7Q) ↓ 3EI SOLUTION (*10.61) P M0 B A L/2 C L/2 L/2 D Actual Loading = P A C (θ B ) M P M0 a + B A C B M0 Continued on next slide 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. Use Table B.14 (cases 7, 9, and 2): vD = ( vB ) M , P + (vD ) M where, L (vD ) M , P = (θ B ) M , P ( ) 2 2 PL L M 0 L L =− ( )− ( ) 16 EI 2 3EI 2 (vD )Q = (vD )Q + (θ D )Q (a) ( vD ) M = − M 0 ( L 2) 2 2 EI Hence, vD = L2 (3PL − 28M 0 ) 96 EI 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.