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
| [1] | [2] | [3] | [4] | [5] | [6] | [7] | [8] | [9] | [10] |
자료출처 : http://www.ALLReport.co.kr/search/detail.asp?pk=11031100&sid=qjfRnf777&key=
댓글 없음:
댓글 쓰기