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Àç·á¿ªÇÐ4 : Àç·á¿ªÇÐ 7ÆÇ ¼Ö·ç¼Ç (James M. Gere & Goodno 7th edi Down Àç·á¿ªÇÐ4ÀÚ·á (¾ÐÃàÆÄÀÏ).zip Àç·á¿ªÇÐ4 : Àç·á¿ªÇÐ 7ÆÇ ¼Ö·ç¼Ç (James M. Gere & Goodno 7th edi [¼Ö·ç¼Ç] Àç·á¿ªÇÐ7ÆÇ (James M. Gere & Goodno 7th edition) ¼Ö·ç¼Ç ch1- 11·Î ±¸¼º ÀúÀÚ : James M. Gere & Goodno (¿µ¹®ÆÇ) [¼Ö·ç¼Ç] Àç·á¿ªÇÐ7ÆÇ (James M. Gere & Goodno 7th edition) ¼Ö·ç¼Ç ch1- 11·Î ±¸¼º ÀúÀÚ : James M. Gere & Goodno (¿µ¹®ÆÇ)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 ¥ð ? ? dAB ? dAB ? 2? tAB ? ? ? AAB = 4 ( ) AAB = 1.178 ¥ò AB = P1 AAB ¥ò AB = 1.443 ¡¿ 10 1443 psi 3 psi Part (b) 2 2 ¥ð ? ? dBC ? dBC ? 2? tBC ? ? ? ABC = 4 ( ) ABC = 2.209 P2 = ¥ò AB? ABC ? P1 P1 + P2 P2 = 1.488 ¡¿ 10 1488 lbs 3 3 lbs CHECK: Part (c) P2 = 2260 P1 + P2 ¥ò AB 2 ABC = 1.443 ¡¿ 10 = ABC P1 + P2 ¥ò AB = 2.744 green box contains answer in accordance with significant digits rule (Se ÀÚ·áÃâó : http://www.allreport.co.kr/search/detail.asp?pk=11031100&sid=knp868group1&key=%C0%E7%B7%E1%BF%AA%C7%D04 [¹®¼Á¤º¸] ¹®¼ºÐ·® : 597 Page ÆÄÀÏÁ¾·ù : PDF ÆÄÀÏ ÀÚ·áÁ¦¸ñ : ÆÄÀÏÀ̸§ : [¼Ö·ç¼Ç] Àç·á¿ªÇÐ7ÆÇ (James M. Gere & Goodno 7th edi.pdf Ű¿öµå : ¼Ö·ç¼Ç,Àç·á¿ªÇÐ7ÆÇ,James,M,Gere,&,Goodno,7th,edition,Àç·á¿ªÇÐ4 - [Àç·á¿ªÇÐ] Àç·á ¿ªÇÐ 6ÆÇ ¼Ö·ç¼Ç ChapGere éÅÍ 2ºÎÅÍ 12±îÁö - gere Àç·á¿ªÇÐ 6ÆÇ ¼Ö·ç¼Ç (materials of mechanics) Jame.Gere Àç·á¿ªÇÐsolu
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