Question

Problem 6.101: A structure is composed of two CST elements, and is subjected two concentrated forces at the center node of the structure as shown in the figure. Please take advantages of the symmetry of structure to simplify the analysis of the problem with dimension D-10. Answer the following questions (hand-calculation only): 1) Show your FE models with proper boundary conditions and applied loads; 2) Compute reduced stiffness matrix and loads vector of your models; and 3) Compute the nodal displacements and element stresses in your model. (Note: The problem needs to be solved with structure symmetry twice with proper boundary conditions, one for symmetric loading, i.e., P1, and another one for anti- symmetric loading, i.e., P2. Then, superimposing displacements and stresses from two results for the final solution.) 4 5 Elm 2 E-10000 3 v- 0.25 t=0.1 P1-100 P2-80 1x Elm 1 2

Answer 1

A Given D=10 - (2-D) metal Pi=100 Pa=80 E=10000 V=0.25 t = 0.1 Hore Hoe stru cture is composed of two CST elements, it is subjected two forces (concentrated) at the center node of the structure

L Stifness matrix: ke Ae cose cossine -COSO -cos. sine 7 reize] cosa.sino sinza -cos, sino - Sinyoluy - cos.sinal -coso -coso, sino costo L-cos.sine - sinre cosesino sinza y lazy Here Ux = displacement in a direction Jy = displacement in 'y' direction A= Area ad uToz . Ez young's module 1. L = Length => A= .004 - X100 = 25 m A = 10000 Here P2 if we coprider any syngnetic y corresponding any symme toic in the given if we consider the lower point the load will get half

* Here model is fixed so ug = 012=0 + model is filed so vy=62=0 + mis madea is a roller so vag-o from this symmetric Uut = Quy - UZR=U2y = U5+ - 054 -0 @ every node has an all member of nodel = 6 stifnese mat sit = 12x12 * computing (oad Stitness + An elemente alle similaen and length too. they have same angles C i 6-68 & 5.6 Stifness matrix ... , K 14 = Ae | 0 25 09 33 -6 25 -0.4 331 0.433 0.15 -0.433 -0.15 70.25 |-0.433 -0.423 0.25 0.4331 -0.75 0.433 0.75 louy] => Kiz = -A viy 0.25 0.433 -0.25 -0.433) 0.433 0.75 -0.433 -0.751 -0.25 -0.433 0.25 0.433 (-0.43 -0.75 0.433 0.75/ for all nodes will be same The local matrix

F=k u uix viy Uzt. Uzy Ust Ug uur. Uyy Use aby The Global matrix (k) uru/ k: He x ury Unx by ugt uz, uz buy in Displacement = F= ko uz u ni va Uzz Vey lux vuy unt оу их жу ((( fiy а) I r fix uuy | Oі น U31 this is the process) (arrangment of matrix) E (3 (зу] [ 34 ]. uz = uy - о гO, Ugt Out Us Uь1 U9) (ох у then 03 ч . auto ust to ory P 04 I uby JV

= G) - 68T0F displacemont 2999 = X99 ] 290 pobom wo) on bo us for apnalogns © [usy = 19.28) 0.4999 064 = 9.64 tego se han stoot van een o što hoone : hombontot * yn SF50 Soxtos : fon sot zyn 993.0 980x t on hyn 92 50+ .30 so o 0 - SHOX non Lo 0.50 0.5 J Luty 778.0 5.00 21 solving this we get Uzt, 067 ay ) (9