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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 1 © 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. 1–1. Determine the resultant internal normal force acting on the cross section through point A in each column. In (a), segment BC weighs 180 lb>ft and segment CD weighs 250 lb>ft. In (b), the column has a mass of 200 kg>m. 5 kip B 200 mm 8 kN 200 mm (a) (b) + c ©F y = 0; F A - 1.0 - 3 - 3 - 1.8 - 5 = 0 F A = 13.8 kip + c ©F y = 0; F A - 4.5 - 4.5 - 5.89 - 6 - 6 - 8 = 0 F A = 34.9 kN Ans. Ans. 10 ft 8 in. 3 kip 4 ft 8 in. 3 kip C 6 kN 200 mm 4.5 kN 6 kN 3 m 200 mm 4.5 kN 4 ft A D A 1 m (a) (b) 1–2. Determine the resultant internal torque acting on the cross sections through points C and D.The support bearings at A and B allow free turning of the shaft. A 250 Nm ©M x = 0; T C - 250 = 0 ©M x = 0; T D = 0 T C = 250 N # m Ans. Ans. C 300 mm 200 mm 150 mm 200 mm 150 Nm 250 mm 400 Nm D B 150 mm 1–3. Determine the resultant internal torque acting on the cross sections through points B and C. ©M x = 0; T B + 350 - 500 = 0 T B = 150 lb # ft ©M x = 0; T C - 500 = 0 T C = 500 lb # ft Ans. Ans. A 3 ft 1 ft 600 lbft B 350 lbft 2 ft C 500 lbft 2 ft 1

01 Solutions 46060 5/6/10 2:43 PM Page 2 © 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. *1–4. A force of 80 N is supported by the bracket as shown. Determine the resultant internal loadings acting on the section through point A. 0.1 m 0.3 m 30 A 80 N 45 Equations of Equilibrium: + Q©F x¿ = 0; N A - 80 cos 15° = 0 N A = 77.3 N Ans. a + ©F y¿ = 0; V A - 80 sin 15° = 0 V A = 20.7 N Ans. a + ©M A = 0; M A + 80 cos 45°(0.3 cos 30°) - 80 sin 45°(0.1 + 0.3 sin 30°) = 0 M A = -0.555 N # m Ans. or a + ©M A = 0; M A + 80 sin 15°(0.3 + 0.1 sin 30°) -80 cos 15°(0.1 cos 30°) = 0 M A = -0.555 N # m Ans. Negative sign indicates that M A acts in the opposite direction to that shown on FBD. 2

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