Question

1 m 1000 kN - Determine displacements and reactions E = 210 GPa 1 for 1 and 2 A=6x10-4 m| E = 210 GPa 1 m →X A=672x10-4 m2 for 3

Determine the nodal displacements and find the reaction forces using the finite element method.

Correct Answer:

Answer 1

Solution - DOO KN S 20 Given : - E=210 Gpa for Ifa A = 6x10-4 m² E = 210 G Pa A = 6/2 x 104 m2 Now, 'Stiffness matrix is, K = AE is s² - c² -cs -cs c² -s² cs. L-cs -52 cs s2 * The global elemental stiffness matrix for element 1 is: : . Cosco, 0 sino = 1. v. ₂ 1 So oo olu, 567. - (210 x 106 kN/m2) (6x154 m2) lo io lu Im * The is :- global elemental stiffness matrix COS Q₂ = 1 , sino₂ = 0 for

10 -1 0 (210x10 kN/m²) (6x15 4m2). [k], -20 -1 0 1 0 im 5000 0 * The global elemental stiffness matrix for element 3 is to50- .sin(0) = uu, u V3 1 -1 -170 (210 x 10 kN/m?) (682x104m²) 11.1 - 1 v 252 m -1 -1 SKI Now, the global stiffness matrix is, [k] - [k], +[k], +[k], Tois os oo Tois I.S o -1 [k ] = 1260 X405N/ M 0 0 0 0 1 - 0.5-0.5 - Ós 1-o's -os 0 0 0.5 0.5 we must transform the global displacement into Co-ordinates. . The The to transformation CT, ] is in local [1] [t,] =/ Co], [o] [o] [1] [o] [o] 1,010 oloo olo o - -- ooooo 01 Oo oo Falz 0 0 0 0 2/2 512 [o] [t ] ! o

. Now, The Product of [T] [k ] is, [1][] = 1260x105 N/m os os 0 0 Tos 1. 5 0 - 0 0 1 0 0 -1 0 1 - 707 -0707 - 0707 0 0 0 0 707 0 -0.5 -0.5 -ois nos -1 0 0 0 1'414 0707 0707 Now, we have to find product of [1][*] [137 Tos os 00-05 -0.57700 000 os is on -os os o loo oo 0. ml 0 1 0 1 = 1260 x 10 0 0 - 0 1 OO 1000 0 0 1 0 0 0 1 0 0 -0.707 07070707 0 1.414 0707 OOOO 52/2 -5212 o o 0.707 0070T Olloooo 52/2 542 Tosos = 12 60x10 N/m lois iso o o - 1 0 0 1 0 0 -1 0 1 -0-707-01707 -0.707 o 0 0 707 0 0707 20707 -0.707 0707 0 0 1.5 -0.5 -0.5 0.5

Now, The boundary conditions, 4, = U, = V₂ = Uzl=0 1-os os Tos is 0 0 o o o t 0 0.707 07 -0. 107 to -0 707 0707 RUTAS = 1260 1-0107 - 0707 - 0707 o ins -os o o 700 o -ois os "By Applying boundary ment equations the global force - displace- conditions are, 12600 10 SNI 0.7078 2 ) Shu=1000kn) -0.707 1.51 The By solving above equations, we get. U₂ = 119 linn 43 = 5.61 mm uz = 1)-91 X16?m -31x10 m (43= 5.61x10 me. . global Nodal (Ans) forces are calculated as os os 0 0 – 0:707 7 os is o - , -0.707 o = 1260x10² N/mm o o l o - 0707 0 TOT 0-1 0 1 0 0 Fo70707070707 0 1.5 –o's 10.00 707 0-OS OS FO Fax 11.91 10ookN . Fix = - sooKN y = - SOORN 707 kn Fay= 0 F 3y = 707 KN. = 707 KN - 500KN rookn