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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 19 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. •1–29. The curved rod has a radius r and is fixed to the wall at B. Determine the resultant internal loadings acting on the cross section through A which is located at an angle u from the horizontal. B r A U Equations of Equilibrium: For point A R+©F x = 0; P cos u - N A = 0 P N A = P cos u Ans. Q+©F y = 0; V A - P sin u = 0 V A = P sin u Ans. d+©M A = 0; M A - P[r(1 - cos u)] = 0 M A = Pr(1 - cos u) Ans. 19

01 Solutions 46060 5/6/10 2:43 PM Page 20 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1–30. A differential element taken from a curved bar is shown in the figure. Show that dN>du = V, dV>du = -N, dM>du = -T, and dT>du = M. M dM V dV T dT N dN ©F x = 0; N cos du 2 ©F y = 0; N sin du 2 ©M x = 0; T cos du 2 ©M y = 0; + V sin du 2 - V cos du 2 + M sin du 2 - (N + dN) cos du 2 + (N + dN) sin du 2 - (T + dT) cos du 2 T sin du du du - M cos + (T + dT) sin 2 2 2 du du Since is can add, then sin , 2 = du 2 2 + (V + dV) sin du 2 = 0 + (V + dV) cos du 2 = 0 + (M + dM) sin du 2 = 0 + (M + dM) cos du 2 = 0 cos du 2 = 1 (1) (2) (3) (4) T M V N du Eq. (1) becomes Vdu - dN + dVdu 2 = 0 Neglecting the second order term, Vdu - dN = 0 dN QED du = V Eq. (2) becomes Neglecting the second order term, Ndu + dV = 0 dV QED du = -N Eq. (3) becomes Neglecting the second order term, Mdu - dT = 0 dT QED du = M Eq. (4) becomes Ndu + dV + dNdu 2 Mdu - dT + dMdu 2 Tdu + dM + dTdu 2 = 0 = 0 = 0 Neglecting the second order term, Tdu + dM = 0 dM QED du = -T 20

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