User:Egm4507.s13.team4.colocar/FEA.s13.R6.3.cc

Problem Statement
Redo R3.1. Compute the reactions, and compare the reactions in the case of non-zero prescribed disp dofs with the reactions in the case of zero prescribed disp dofs.



Solution
On my honor, I have neither given nor received unauthorized aid in doing this assignment.

Given zero displacement DOFs, the following MATLAB program solves for the unknown global displacement DOF's, unknown for components (reactions), and member forces in the system.

function R6p3 F = 20000; %N E = 206000000000; %Pa L = 1; %m A = .0001; %m^2 % displacement in meters d3 = 0; % =u2 d4 = 0; % =v2 d5 = 0; % =u3 d6 = 0; % =v3 % Table 2.1 KS 2008 p.82 % Element  EA/L      i-->j    phi    l=cos(phi)  m=sin(phi) % 1      206x10^5    1-->3   -pi/6    0.866       -0.5 % 2      206x10^5    1-->2   -pi/2      0           1 % 3      206x10^5    1-->4   -5pi/6  -0.866       -0.5 l1 = 0.866; m1 = -0.5; l2 = 0; m2 = 1; l3 = -0.866; m3 = -0.5; % element stiffness matrices % E2.46 % element 1 stiffness matrix 1-->3 k1 = [l1^2 l1*m1 0 0 -l1^2 -l1*m1 0 0; l1*m1 m1^2 0 0 -l1*m1 -m1^2 0 0; 0 0 0 0 0 0 0 0;      0 0 0 0 0 0 0 0;       -l1^2 -l1*m1 0 0 l1^2 l1*m1 0 0; -l1*m1 -m1^2 0 0 l1*m1 m1^2 0 0; 0 0 0 0 0 0 0 0;      0 0 0 0 0 0 0 0]; % element 1 stiffness matrix 1-->2 k2 = [l2^2 l2*m2 -l2^2 -l2*m2 0 0 0 0; l2*m2 m2^2 -l2*m2 -m2^2 0 0 0 0; -l2^2 -l2*m2 l2^2 l2*m2 0 0 0 0; -l2*m2 -m2^2 l2*m2 m2^2 0 0 0 0; 0 0 0 0 0 0 0 0;      0 0 0 0 0 0 0 0;       0 0 0 0 0 0 0 0;       0 0 0 0 0 0 0 0]; % element 1 stiffness matrix 1-->4 k3 = [l3^2 l3*m3 0 0 0 0 -l3^2 -l3*m3; l3*m3 m3^2 0 0 0 0 -l3*m3 -m3^2; 0 0 0 0 0 0 0 0;      0 0 0 0 0 0 0 0;       0 0 0 0 0 0 0 0;       0 0 0 0 0 0 0 0;       -l3^2 -l3*m3 0 0 0 0 l3^2 l3*m3; -l3*m3 -m3^2 0 0 0 0 l3*m3 m3^2]; %global stiffness matrix K = E*A/L*(k1 + k2 + k3) % Force matrix Fpost = [F*cos(pi/4); F*sin(pi/4); 0; 0; 0; 0; 0; 0] % global disp dofs dispg = K\Fpost % u2=v2=u3=v3=u4=v4=0, so delete rows/columns 3 through 8 Knew = 1*10^7*[3.0898 0; 0 3.0900]; % displacement values u1, v1 Fcut = [F*cos(pi/4); F*sin(pi/4)]; disp = Knew\Fcut % forces in each element P1 = E*A/L*(0.866*(d5-disp(1))-0.5*(d6-disp(2))) P2 = E*A/L*(0*(d3-disp(1))+1*(d4-disp(2))) P3 = E*A/L*(-0.866*(0-disp(1))-0.5*(dispg(4)-disp(2))) stress1 = P1/A stress2 = P2/A stress3 = P3/A Global Displacements dispg = 1.0e+12 * NaN NaN NaN 3.5927    3.7905     6.5693     2.3727     3.0716

Unknown disp at node 1(m) disp = 1.0e-03 * 0.4577    0.4577

Element forces (N) P1 = -3.4512e+03 P2 = -9.4281e+03 P3 = 1.2879e+04

Element stresses (Pa) stress1 = -3.4512e+07 stress2 = -9.4281e+07 stress3 = 1.2879e+08

Verifying with Calfem and plotting

function R6p3calfem F = 20000; %N E = 206000000000; %Pa L = 1; %m A = .0001; %m^2 % Checking with CALFEM edof = [1 1 2 5 6; 2 1 2 3 4;        3 1 2 7 8]; coord = [cos(pi/6) sin(pi/6);cos(pi/6) 1+sin(pi/6);2*cos(pi/6) 0;0 0]; dof = [1 2;3 4;5 6;7 8]; [ex,ey] = coordxtr(edof,coord,dof,2); ep = [E A]; K = zeros(8); Fo = zeros(8,1); Fo(1) = F*cos(pi/4); Fo(2) = F*sin(pi/4); for i = 1:3 Ke = bar2e(ex(i,:),ey(i,:),ep); K = assem(edof(i,:),K,Ke); end bc = [3 0; 4 0; 5 0; 6 0; 7 0; 8 0]; Q = solveq(K,Fo,bc); ed = extract(edof,Q); for i = 1:3 N(i)=bar2s(ex(i,:),ey(i,:),ep,ed(i,:)); end plotpar=[1 4 0];scale=100; eldraw2(ex,ey); eldisp2(ex,ey,ed,plotpar,scale); grid on N1 = N(1) N2 = N(2) N3 = N(3) stress1c = N(1)/A stress2c = N(2)/A stress3c = N(3)/A

OUTPUT N1 = -3.4509e+03 N2 = -9.4281e+03 N3 = 1.2879e+04

stress1c = -3.4509e+07 stress2c = -9.4281e+07 stress3c = 1.2879e+08



Original values of non-zero displacement degrees of freedom: (R3.1) Global displacement DOF's (m) $$ u_1 = -0.005 $$ $$ v_1 = -0.0103 $$ $$ u_2 = 0.02 $$ $$ v_2 = -0.01 $$ $$ u_3 = -0.03 $$ $$ v_3 = 0.05 $$ $$ u_4 = -0.0257 $$ $$ v_4 = 0.0764 $$

Element Reaction Forces (N) $$ P^1 = -854900 $$ $$ P^2 = -418180 $$ $$ P^3 = 740079 $$

Element Stresses (GPa) $$ \sigma^1 = -8.5490 $$ $$ \sigma^2 = -4.1818 $$ $$ \sigma^3 = 7.4008 $$