By Daryl L. Logan

Observe: top of the range local PDF. info refers back to the textbook that accompanies this answer guide.

A FIRST direction within the FINITE point procedure offers an easy, simple method of the direction fabric that may be understood through either undergraduate and graduate scholars with out the standard must haves (i.e. structural analysis). The booklet is written basically as a easy studying device for the undergraduate pupil in civil and mechanical engineering whose major curiosity is in rigidity research and warmth move. The textual content is aimed at those that are looking to follow the finite aspect approach as a device to resolve useful actual difficulties.

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

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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. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 36Ý Ñu4 Ð f2 y à ÑÐv4 Þ Ñ Ñ ß 0Ñ 0 Ñà ⇒ f2x = 1333 lb (C) 50 © 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. 25 The global stiffness matrix is changed since matrix [k2–4] is not incorporated in (1) (2) F 0 Î 1x ÑF Ñ 1y Ñ F2 x Ñ Ñ F2 y Ï Ñ F3 x Ñ F3 y Ñ Ñ F4 x ÑF Ð 4y Þ Ñ Ñ 0 Ñ Ñ 1000Ñ ß = AE 0 Ñ 1000 Ñ Ñ 0 Ñ 0 Ñ à 0 Ë 0.

May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2 × 10 ÑF Ñ Ð 3x ? à ⇒ Ë 1 1 0 Û Îu1 0 Þ Ñ Ì 1 2 1Ü Ñu ? 10 kN m Ë 1 1Û [k(1)] = [k(2)] = 7000 Ì Í1 1ÜÝ Ë 1 1Û [k(2)] = 2000 Ì Í1 1ÜÝ {F} = [K] {d} ÎF1x ? Þ Ë 7 7 0 0Û Îu1 ÑÑF Ñ Ì 8kN Ñ 0Ü ÑÑu2 2x 3 7 14 7 ÜÏ Ï ß = 10 Ì F 0 0 7 9 2 Ì   Ü Ñu3 3 x Ñ Ñ Ì 0 0 2 2 Ü Ñu ÑÐF4 x ? Ñà Í ÝÐ 4 ⇒ 8 = 10 3 [14 u2 – 7 u3] 0 = 10 3 [– 7 u2 + 9 u3] (1) 7 u2 9 Substituting (2) into (1) ⇒ ⇒ u3 = 8 3 10 = 14 u2 – 7 × 0Þ ?

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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