Question

finite element methods question. I need the answer immediately pls help

Question 2) 50 Points: Consider a truss structure that is built 300 N from two identical truss elements. For the truss elements, the modulus of elasticity is E = 69 GPa, and the cross-section area is A = 968 mm². Assume that the origin (0,0) is at Node 1 and C (2000) elements 1 and 2 are connected as 1-3 and 3-2, respectively. 500N Considering the forces 300 N and 500 N which are aplied to the 3rd node; 10, 40) BH B E KS D (a) Find the nodal displacements at Node 3. (b) Find the stresses in each element. (c) Calculate the reaction forces. MY A х k, 2000000 -2000000 - 2000000 2000000 N/m n W ?

Answer 1

Page No.: I Date : BOON 4₂=0, x=0 (2) 43²9 SooN Y5-9 2 face) 21,50 =0 here two truss element & spring Ks stiffness atteched. 1-3 3-2 and a - for finding the displacement at node 3 first we have to find stiff global stiffress in which add three Stiffress matrica global stiffness matriz[k]; A] = [ki] + [ki +/ks here given that. Ks = 2000000 2000000 -2000000 M 2000000 -2.000000 N/m.

Page No.: 2 Date : here there is two displacement direction, and y We can curite the stiffness matrix for the Ispring element [ks] V h 2000000 g -2000000 O Q M2 2000000 O - 2.000000 Tool 2.000000 (: Spring is O 20000oo V₂ CD stretch in y direction only). stiffness matria node Element length for AE etement 1 & 2 element o ایم کا C.S mm) I (N/m 1 34 = 56-56 11180905.23 45 otot otot/0.5 0.5 0.5 (2 3 2 L₂=40 1669800 180 1 Tolo

(3) for element 1 [ke 2 C cs -cs ki - AECS عی -CS 2 -CS L-2² -CS 2 -S OS 2 s 2ی۔ [kiel 0.5 - 11 80.905.230.5 +6 X lot -05 -0.5 0.5 0.5 -0.5 -0.5 -0.5 -0.5 0.5 -0.5 -0.5 6.5 N/m

Page No.: 4 Date : [kl] = 1590.45 v N/m 590.45 1-590.45 - 596.45 Tool Vz -590.45 1 -590.45 590.454 59045 590.45 590.45 -590.45 -Syaus -590.45 -590.45 590.45 590.45 Stiffness matrix of element 12 [K] O O kit 0 O 1669800 xlo o o V O Us ณ 2 11669.8 o #1669.8 - 16698 O 1669.8 O Nm 12 bor Vz L By adding heq. (1, lid & we get global stiffness matria.

Page No. 5 No. 7 Date : V OP 59045to 590.457 -590.45 590.45to 1590.45 to -590.45 to ICKI = -590a5 to 1o oto oto 43 Vz -590.45 ~5gaus -590.45 -59a45 o -2000000 546, 45+(1.69.9) 590.45+0 -1669.8 59o.usto 590.45+ 0 -1669.8 1669.870 oto o to -59045 0 O - 2000000

here ks=2.0

Page No.: 6 Date : SE} =[k] {d} ज force fyo Soo -2.0 23 -590.45 -590.us 2260.25 590.45 -1669.8 1590.45 59o.us 159ous 592.05 x = -59005 - Sons If zy 2300 -590.45 -590.05 fazo -2.0 -590.45 -590.45 o 5gaus -1669.8 59o.us O 1669.8 23 6 o kisso] bo 2.010 500= 106 [22 60.25/213) + 590.45 (V) 300=100 [590.46 (213) + syGas (vs) - 1.1977 x 107 m m V3 = 3.88xlot mm U= 1.1977x164 za 3.88 x 104 mm.

Page No.: 7 Date : stress For plement 1. -5 С S प V 4 -6.7071 - 0.7071 0.7071 o = 69x100 [-6. 56.66 0.7071 O 1.1g 777164 3.888154 8 = 69x106 [(0.7071) [1.197164 7 3.88x164) 5656 o 6 = 437.34 N/mm2 (or mpa or= 69x106 40 to 1.19x104 3. 88x104 lor= 205.27 N/mm2

Page No.: 8 Date : reaction forces at node z 590.45 590.us 590.45 590.45 -590.45 -64645 -590.45 -syo.us fir sgous 0 -2 11.19x107 13.884107 fix = 10" [f-sgo, us2619217+) + (540.45X3 88 x 177]] =-182993.58 Sie = -299-35 Fy= 10°C-596.013)(1-19,167) + (sya ug (388x167 167) - 299.35 N If y=-299.35 - At node! - 2 6 O fort Kfzy O - 1669.8 =100 1669.8 o 2 O -2 0 11.19x10 1388x107

(9) tot 108 [-1669.8 x 1.19x107 167] fix = -198.7 W] y = ON

if we neglect the spring then also we get same answer. because both end of the spring is fixed.