Tham khảo tài liệu 'engineering mechanics - statics episode 2 part 3', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Engineering Mechanics - Statics Chapter 6 Fbd -4 Ì Ffd -4 Ì Fcd 0 Fed 0 Fab -4 kN Fgf -4 kN Positvive means Tension Negative means Compression Fca 0 Feg 0 Fbc 10 J Ffe 10 J Problem 6-28 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used kN 103 N Given P1 2 kN P2 4 kN a 2 m Ỡ 15 deg Solution Take advantage of the symmetry. Initial Guesses Fbd 1 kN Fcd 1 kN Fca 1 kN Fbc 1 kN FAB 1 kN Given Joint D -p1 - FbD sin 2 ớ - Fcdsin 3 ớ 0 2 Joint B -P2 cos 2 ớ - Fbc 0 FBD - FAB - p2 sin 2ớ 0 Joint C Fcdcos ớ - Fca cos ớ 0 FCD FCA sin ớ FBC 0 482 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 6 FBD D C FAB FCA Find fbd FCD FAB FCA FBC Ffd FED FGF feg fbd d FCD Fz fca fbc ựFE ựBc fbd fcd fab kN fca fbc FFD y fed fgf kN feg ffe Positvive means Tension Negative means Compression Problem 6-29 Determine the force in each member of the truss and state if the members are in tension or compression. Units Used kip 103 lb Given F1 2 kip F2 kip F3 3 kip F4 3 kip a 4 ft b 10 ft Solution Ỡ atanl a I b. Initial Guesses 483 2007 R. C. Hibbeler. Published by 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 .
