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ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 5 © 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. *1–8. Determine the resultant internal loadings on the cross section through point C. Assume the reactions at the supports A and B are vertical. 6 kN 3 kN/m Referring to the FBD of the entire beam, Fig. a, a+©M B = 0; -A y (4) + 6(3.5) + 1 2 (3)(3)(2) = 0 A y = 7.50 kN A C D B Referring to the FBD of this segment, Fig. b, 0.5 m 0.5 m 1.5 m 1.5 m : + ©F x = 0; N C = 0 Ans. + c ©F y = 0; 7.50 - 6 - V C = 0 V C = 1.50 kN Ans. a+©M C = 0; M C + 6(0.5) - 7.5(1) = 0 M C = 4.50 kN # m Ans. •1–9. Determine the resultant internal loadings on the cross section through point D. Assume the reactions at the supports A and B are vertical. 6 kN 3 kN/m Referring to the FBD of the entire beam, Fig. a, a+©M A = 0; B y (4) - 6(0.5) - 1 2 (3)(3)(2) = 0 B y = 3.00 kN A 0.5 m 0.5 m C 1.5 m D 1.5 m B Referring to the FBD of this segment, Fig. b, : + ©F x = 0; N D = 0 Ans. + c ©F y = 0; V D - 1 2 (1.5)(1.5) + 3.00 = 0 V D = -1.875 kN Ans. a +©M D = 0; 3.00(1.5) - 1 2 (1.5)(1.5)(0.5) - M D = 0 M D = 3.9375 kN # m = 3.94 kN # m Ans. 5

01 Solutions 46060 5/6/10 2:43 PM Page 6 © 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. 1–10. The boom DF of the jib crane and the column DE have a uniform weight of 50 lb/ft. If the hoist and load weigh 300 lb, determine the resultant internal loadings in the crane on cross sections through points A, B, and C. 5 ft D B A 2 ft 8 ft 3 ft F C 300 lb 7 ft E Equations of Equilibrium: For point A ; + © F x = 0; N A = 0 Ans. + c © F y = 0; V A - 150 - 300 = 0 V A = 450 lb Ans. a +©M A = 0; -M A - 150(1.5) - 300(3) = 0 M A = -1125 lb # ft = -1.125 kip # ft Ans. Negative sign indicates that M A acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point B ; + © F x = 0; N B = 0 Ans. + c © F y = 0; V B - 550 - 300 = 0 V B = 850 lb Ans. a +© M B = 0; -M B - 550(5.5) - 300(11) = 0 M B = -6325 lb # ft = -6.325 kip # ft Ans. Negative sign indicates that M B acts in the opposite direction to that shown on FBD. Equations of Equilibrium: For point C ; + © F x = 0; V C = 0 Ans. + c © F y = 0; -N C - 250 - 650 - 300 = 0 N C = -1200 lb = -1.20 kip Ans. a +©M C = 0; -M C - 650(6.5) - 300(13) = 0 M C = -8125 lb # ft = -8.125 kip # ft Ans. Negative signs indicate that N C and M C act in the opposite direction to that shown on FBD. 6

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