© 2022 by R. C. Hibbeler. Published by Pearson Education, Inc., Hoboken, 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. B and a journal bearing at C. Determine the resultant internal loadings acting on the cross section at E.
B
A
4 ft
C
E
4 ft
4 ft
D
4 ft
400 lb 800 lb
SOLUTION Support Reactions: We will only need to compute Cy by writing the moment equation of equilibrium about B with reference to the free-body diagram of the entire shaft, Fig. a. a+ ΣMB = 0;
Cy(8) + 400(4) - 800(12) = 0
Cy = 1000 lb
Internal Loadings: Using the result for Cy, section DE of the shaft will be considered. Referring to the free-body diagram, Fig. b, + ΣFx = 0; S
NE = 0
+ c ΣFy = 0;
VE + 1000 - 800 = 0
Ans. Ans.
VE = -200 lb
a+ ΣME = 0; 1000(4) - 800(8) - ME = 0
ME = - 2400 lb # ft = - 2.40 kip # ft
Ans.
The negative signs indicates that VE and ME act in the opposite sense to that shown on the free-body diagram.
Ans:
Ans: Cy = 1000 =E 0,=VE-200 = 200lb, 2.40kip NElb,=N0, lb, MEM=E -= 2.40 kip #⋅ ftft E V
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