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 direction within the FINITE point approach offers an easy, easy method of the direction fabric that may be understood by means of either undergraduate and graduate scholars with out the standard must haves (i.e. structural analysis). The ebook is written basically as a simple studying instrument for the undergraduate scholar in civil and mechanical engineering whose major curiosity is in rigidity research and warmth move. The textual content is aimed toward those that are looking to follow the finite point technique as a device to resolve sensible 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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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. 0144 0  0 ¹ 0667 (3) 0. 05 u2 + 0 v2 + 0 u3 + 0 v3] AE 51 © 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.

14 Given u = a + bx2 for 2 noded bar ∈= du = 2bx dx u(0) = u1 = a u(L) = u2 = u1 + bL2 ∴ b= u2  u1 L2 u = u1 + Ë u2  u1 Û 2 x ÌÍ L2 Ü Ý This displacement function allows for a rigid body displacement as the a = u1 term does this. Also should allow for constant strain, but have ∈ = 2bx or a linear strain. Therefore, not complete. Need to complete 2nd degree polynomial and 3rd node for compatible function. Try u = a1 + a2 x + a3 x2 du = a2 + 2a3x dx ‘a2’ allows for constant strain term. 15 (a) .

All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 6857 × 10–3 in. 2857 × 10–3 in. 6857 × 10–3) 12 One element È A = A0 ÉÊ1  L 2Ø Ù LÚ = A0 (1 + 1 3 )= A0 2 2 34 © 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. 667 × 10–3 in. 13 l 2 1 l 2 3 2 x l/2 u = a1 + a2x + a3x2 (A) u(0) = u2 = a1 u(– u( (1) l l l ) = u1 = u2 + a2 (– ) + a3 (– )2 2 2 2 (2) l l l ) = u3 = u2 + a2 ( ) + a3 ( )2 2 2 2 (3) Solving for a2 and a3 from (2) and (3) a2 = u3  u1 2(u1  u3  2u2 ) , a3 = l l2 (4) By (1) and (4) into (A) u = u2 +  u1 Ø È u3 Ê l 2(u1  u3  2u2 ) x+ Ú l2 x2 u = [N] {d} (6) Ë x 2x2 u= Ì  2 Í l l {°} = (5) 1 4x2 l2 Îu1 Þ x 2x2 Û Ñ Ñ  2 Ü Ïu2 ß l l ÝÑ Ñ Ðu3 à ˜u ˜N = [B] {d} = {d} ˜x ˜x (7) (8) Using (7) in (8) 1 4x {°} = ËÌ   2 Í l l ∴ [B] = Ë 1   Ì Í l [K] = A l /2 8 x l2 4x 8x 2 l2 l Îu1 Þ 1 4x Û Ñ Ñ  2 Ü Ïu2 ß l l ÝÑ Ñ Ðu3 à 1 4x Û  l l 2 ÜÝ Ô–l /2 [B ] E [B] dx T (9) (10) (11) A = cross sectional area of the bar E = Young’s Modulus of the bar 35 © 2012 Cengage Learning.

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