Question

1. (60%) For the truss system shown below (a) (12%) Determine the element stiffness matrix w.r.t. the global coordinate system for all elements. (b) (10%) Determine the global stiffness matrix, [K]. (c) (5%) List all the boundary conditions. (d) (33%) Determine the internal force, elongation, stress, and strain for each element. Indicate whether it is under tension or compression. My LLLLLLLL 1 0 1-2=45° \ 30° 30° / 14116 141 16 12 2-3 = 30 3 1-4=300 Join But 45=225 Let E = 10 psi, A= 3 inand L=150 in.

Answer 1

OW Given, E = 100 psí A= 3 in² L = 150 in Let le cose X2-31 le | 30 / 141 lb 3830 m = sine 4,-4, 600 La... -60° 1545 (lee length of element) 970 Element I leo l Cin) m 0.866 173.2 150 600 90° -600 0.5 0 -0.5 ė 0.25 0 0.25 m² 0.7499 1 0.7499 lm 0.433 0 -0.433 173.2 0.866 Elemental stiffness matrix: ko. Arte ez lim -e² - em 7 le me m' -lm m² -12 -em e em 1.lm -me em me 4x4 1 2 3 4 0.25 0.433 -0.25 -0.433 0.433 0.7499 -0.433 -0.7499 -0,25 -0.433 0.25 0.433 -0.433 -0.1499 0.433 0.7499 k. 3x106 173,2 3X10° 0 1 0 ī - 2 K₂= 150 LCM ( 50, R3,4) = 2s980 • Multiply all ke matrices with 25980 or 150x173,2 and assemble to get Global Stiffness matrix - 3x106/ K₃ = 173,2 0,25 -0.433 -0.25 0.4331 -0.433 0,7499 0.433 -0,7499 -0.25 0.433 0.25 -0,433 0.433 -0,7499 -0.433 0,7499 - Sito

Global Shiffness Matrix : 2 75 o 3 4 -37,5 -64,95 5 0 6 7 8 0 -37.5 64.95 348.2 - 64.95 - 112.5 0 -173,2 64,95 -119.5 k=30000 - 37.5 - 64.95 37,5 64.95 w – 112.5 64.95 112. 5 o oo -64,95 0 o - 37,5 64.95 0 173,2 64.95 -112,5 0 0 0 0 0 0 0 0 0 0 0 0 0 o o 0 0 0 173.2 0 0 37.5 -64,95 0 -64.95 119,5 + 0 Boundary conditions: (At fixed ends) Q=Q4=0 Qs = 96=0 Oz = Qg=0 Fie141 cosu5 = 99.7 lb Fa= 141 sinus = 99,7 eb Considex KQ = F After applying Boundary conditions, we get 1. 99.7 3x106 398.2 99.7 on Solving, 9 = 8,345x10 in Q, . 4.431 x10+ in Nodal displacements Elongation. on )

Strain: & I [ -e mem] 15x108 • E, 3,2 (-0.5 -0.866 0.5 0.866 78.345 14,431x10 771 8.345x1082 E, : -1.6964 x109 54.431x10771 • €* [O- O ] 8345x1682 -5.563x16" 150 Ez = [0,5 -0.866 -0.5 0.860] [ 4.431x1077 -10 8.345x168 = 8.619 x 10 Stresses: = Eee 02 100x (-1,6964X169) - -1.696x16"psi 02= 106x (-5,563x16") = -5.563x16 psi 03 - 100x (8.619 x 16") = 8.619 x 10"psi compression) (compression) (Tensim)

1 x lo Internal force Reaction : (R, eR220) R = KQ-F. a & Ko 1-37,5 -64,95 37,5 64,95 o o o of [ 4.4 -64,95 -112,5 64,95 112.5 o o o ol 3x18 0 0 0 0 0 0 0 0 o -73,2 0 0 0 143,2 0 0 -37,5 64,95 o o o o 37,5 -64,95 64.95 -112,5 0 0 0 0 6495 117,5 0 0 0 8,345x108 0 0 Oo oo 0 (+ R, and R220, only those rows of k needed are considered) corresponding to Reactions are On solving R3 = –66, 108 lb Ru= -114,502 lb RS² o R6 = -43,36 lb Rz = - 33,588 eb R8 = 58. 173 lb