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01 Solutions 46060 5/6/10 2:43 PM Page 15 © 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. *1–24. The machine is moving with a constant velocity. It has a total mass of 20 Mg, and its center of mass is located at G, excluding the front roller. If the front roller has a mass of 5 Mg, determine the resultant internal loadings acting on point C of each of the two side members that support the roller. Neglect the mass of the side members. The front roller is free to roll. G C 2 m B A 1.5 m 4 m Support Reactions: We will only need to compute N A by writing the moment equation of equilibrium about B with reference to the free-body diagram of the steamroller, Fig. a. a+©M B = 0; N A (5.5) - 20(10 3 )(9.81)(1.5) = 0 N A = 53.51(10 3 ) N Internal Loadings: Using the result for N A , the free-body diagram of the front roller shown in Fig. b will be considered. ; + ©F x = 0; 2N C = 0 N C = 0 Ans. + c ©F y = 0; 2V C + 53.51(10 3 ) - 5(10 3 )(9.81) = 0 V C = -2229.55 N = -2.23 kN Ans. a+©M C = 0; 53.51(10 3 )(2) - 5(10 3 )(9.81)(2) - 2M C = 0 M C = 4459.10 N # m = 4.46 kN # m Ans. 15
01 Solutions 46060 5/6/10 2:43 PM Page 16 © 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. •1–25. Determine the resultant internal loadings acting on the cross section through point B of the signpost.The post is fixed to the ground and a uniform pressure of 7 lb> ft 2 acts perpendicular to the face of the sign. 3 ft z 2 ft ©F x = 0; (V B ) x - 105 = 0; (V B ) x = 105 lb Ans. ©F y = 0; (V B ) y = 0 Ans. 3 ft ©F z = 0; (N B ) z = 0 Ans. 7 lb/ft 2 ©M x = 0; (M B ) x = 0 ©M y = 0; (M B ) y - 105(7.5) = 0; (M B ) y = 788 lb # ft Ans. Ans. B 6 ft ©M z = 0; (T B ) z - 105(0.5) = 0; (T B ) z = 52.5 lb # ft Ans. A 4 ft x y 1–26. The shaft is supported at its ends by two bearings A and B and is subjected to the forces applied to the pulleys fixed to the shaft. Determine the resultant internal loadings acting on the cross section located at point C.The 300-N forces act in the z direction and the 500-N forces act in the x direction. The journal bearings at A and B exert only x and z components of force on the shaft. A z 400 mm 150 mm x 300 N 300 N 500 N 200 mm C 250 mm B 500 N y ©F x = 0; (V C ) x + 1000 - 750 = 0; (V C ) x = -250 N ©F y = 0; (N C ) y = 0 ©F z = 0; (V C ) z + 240 = 0; (V C ) z = -240 N ©M x = 0; (M C ) x + 240(0.45) = 0; (M C ) x = -108 N # m ©M y = 0; (T C ) y = 0 ©M z = 0; (M C ) z - 1000(0.2) + 750(0.45) = 0; (M C ) z = -138 N # m Ans. Ans. Ans. Ans. Ans. Ans. 16
01 Solutions 46060 5/6/10 2:43 PM P
02 Solutions 46060 5/6/10 1:45 PM P
03 Solutions 46060 5/7/10 8:45 AM P
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