Download A First Course in the Finite Element Method (5th Edition): by Daryl L. Logan PDF

By Daryl L. Logan

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A FIRST path within the FINITE aspect technique offers an easy, simple method of the direction fabric that may be understood by way of either undergraduate and graduate scholars with no the standard necessities (i.e. structural analysis). The e-book is written basically as a easy studying device for the undergraduate pupil in civil and mechanical engineering whose major curiosity is in tension research and warmth move. The textual content is aimed toward those that are looking to observe the finite aspect process as a device to resolve useful actual difficulties.

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Extra resources for A First Course in the Finite Element Method (5th Edition): Instructor Solution Manual

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23 12,000 lb L = 100 in. E = 10 ¥ 10 6 psi A = 1 in2. 24 in. 24Ñ 0 Ñà 47 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 48 48 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 0192 Ñ Ì ? 0 Ñ Ì Ñ Ì Í ? 0 à ? Ñ ? 204 v3] AE ⇒ v3 = 105021 AE (6) 49 © 2012 Cengage Learning. All Rights Reserved.

36 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. (b) . C= [K] = 1 2 Ë 1 4 Ì 3 15 – 106 – 1 Ì  4 Ì 15 Ì 1 Ì 4 Ì 3 Í 4 Ë 6 Ì [K] = 3 2 ,S= 10 4  1 Ì 3 Ì 1 Ì Í 3 3 Û 4  43  14 3 4 3 4 3 4 1 4  43  43 Ü  34 Ü Ü  43 Ü Ü 3 Ü 4 Ý 3Û Ü 3 3 Ü lb  3 Ü in. 1 Ü  3 3 Ý 1 3 3 3 3 (c) C= [K] = 3 , 2 S=– Ë 3 4 Ì 4 Ì 3 6 (210 – 10 )(4 – 10 ) 4 Ì 3 Ì 3 Ì 4 Ì 3 Í 4 1 2  43  34 1 4 3 4 3 4 3 4  14  43 3 Û 4 Ü  14 Ü Ü  43 Ü Ü 1 Ü 4 Ý 37 © 2012 Cengage Learning.

283 in. 2 43 © 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 21 141 lb Element 1–2 2 L1–2 = 3 L; θ = 60° Ë 1 4 Ì Ì 3 Ì 4 Ì 1 Ì 4 Ì 3 Í 4 3 AE 2L [k1–2] = 3 4  41 3 4  43  43 1 4  14 3 4  43 Û Ü  43 Ü Ü 3 Ü 4 Ü 3 Ü 4 Ý Element 1–3 L1–3 = L; θ = 90° [k1–2] = AE L Ë0 Ì Ì0 Ì0 Ì Í0 0Û 0 0 1 0 0 0 1 0 Ü 1Ü 0Ü Ü 1Ý Element 1–4 2 L1–4 = 3 L; θ = 120° 3 AE 2L [k1–4] = Ë 1 4 Ì Ì 3 4 Ì Ì 1 4 Ì Ì 3 Í 4  43  14 3 4 3 4 3 4 1 4 3 4  43 3 Û 4 Ü  34 Ü Ü  43 Ü Ü 3 Ü 4 Ý Applying the boundary conditions u2 = v2 = u3 = v3 = u4 = v4 = 0 AE Ë Ì [K] = L Ì 3 1 ( ) 2 4 3 Í 2 = 0 3 1 ( ) 2 4 43  0  23  43 AE Ë Ì L Ì Í 3 4 0 43  0  23  43 ÜÛ 3 3 ( )  1  23 ( 34 ) 2 4 Ü Ý Û Ü 1  3 4 3 ÜÝ 0 Î F1x Þ AE Ë 4 Ì Ï ß= L Ì ÐF1 y à Í0 3 Û Îu Þ 1 ÜÏ ß 3 3 Ü Ð v1 à 1 4 Ý AE Ë 4 Î100Þ Ì ⇒ Ï ß = L Ì Ð100à Í0 Û Îu Þ 1 ÜÏ ß 3 3 Ü Ð v1 à 1 4 Ý 3 3 2 0 0 44 © 2012 Cengage Learning.

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