| 재료역학4 : 재료역학 7판 솔루션 (James M. Gere & Goodno 7th ed | ||
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재료역학4 : 재료역학 7판 솔루션 (James M. Gere & Goodno 7th edi Down 재료역학4 : 재료역학 7판 솔루션 (James M. Gere & Goodno 7th edi [솔루션] 재료역학7판 (James M. Gere & Goodno 7th edition) 솔루션 ch1- 11로 구성 저자 : James M. Gere & Goodno (영문판) [솔루션] 재료역학7판 (James M. Gere & Goodno 7th edition) 솔루션 ch1- 11로 구성 저자 : James M. Gere & Goodno (영문판)Draft solutions manual for 7th edition of Mechanics of Materials, J. Gere & B. Goodno, 2009 -- see also updated Answers to Problems -- see also errata table NOTE: This file contains only solutions for those problems which are either new or revised versions of those in the 6e of this text; please consult the 6e ISM for any problem solutions not shown here ========================================================================================= SOLUTION Part (a) P1 = 1700 dAB = 1.25 tAB = 0.5 dBC = 2.25 tBC = 0.375 2 2 π ? ? dAB ? dAB ? 2? tAB ? ? ? AAB = 4 ( ) AAB = 1.178 σ AB = P1 AAB σ AB = 1.443 × 10 1443 psi 3 psi Part (b) 2 2 π ? ? dBC ? dBC ? 2? tBC ? ? ? ABC = 4 ( ) ABC = 2.209 P2 = σ AB? ABC ? P1 P1 + P2 P2 = 1.488 × 10 1488 lbs 3 3 lbs CHECK: Part (c) P2 = 2260 P1 + P2 σ AB 2 ABC = 1.443 × 10 = ABC P1 + P2 σ AB = 2.744 green box contains answer in accordance with significant digits rule (Se 자료출처 : http://www.allreport.co.kr/search/detail.asp?pk=11031100&sid=knp868group1&key=%C0%E7%B7%E1%BF%AA%C7%D04 [문서정보] 문서분량 : 597 Page 파일종류 : PDF 파일 자료제목 : 파일이름 : [솔루션] 재료역학7판 (James M. Gere & Goodno 7th edi.pdf 키워드 : 솔루션,재료역학7판,James,M,Gere,&,Goodno,7th,edition,재료역학4 - [재료역학] 재료 역학 6판 솔루션 ChapGere 챕터 2부터 12까지 - gere 재료역학 6판 솔루션 (materials of mechanics) Jame.Gere 재료역학solu |
