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% STK_TESTFUN_TRUSS3_BB computes displacements and stresses for 'truss3'
%
% CALL: Z = stk_testfun_truss3_bb (X, CONST)
%
% See also: stk_testcase_truss3, stk_testfun_truss3_vol
% Copyright Notice
%
% This file: stk_testfun_truss3_bb.m was written in 2017
% by Julien Bect <julien.bect@centralesupelec.fr>.
%
% To the extent possible under law, the author(s) have dedicated all
% copyright and related and neighboring rights to this file to the pub-
% lic domain worldwide. This file is distributed without any warranty.
%
% This work is published from France.
%
% You should have received a copy of the CC0 Public Domain Dedication
% along with this file. If not, see
% <http://creativecommons.org/publicdomain/zero/1.0/>.
% Copying Permission Statement (STK toolbox as a whole)
%
% This file is part of
%
% STK: a Small (Matlab/Octave) Toolbox for Kriging
% (https://github.com/stk-kriging/stk/)
%
% STK is free software: you can redistribute it and/or modify it under
% the terms of the GNU General Public License as published by the Free
% Software Foundation, either version 3 of the License, or (at your
% option) any later version.
%
% STK is distributed in the hope that it will be useful, but WITHOUT
% ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
% or FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public
% License for more details.
%
% You should have received a copy of the GNU General Public License
% along with STK. If not, see <http://www.gnu.org/licenses/>.
function z = stk_testfun_truss3_bb (x, const)
% Convert input to double-precision input data
% (and get rid of extra structure such as table or stk_dataframe objects)
x_ = double (x);
% Check input size
[n, dim] = size (x_);
switch dim
case 4
% Use nominal loads
F = zeros (n, 2);
F(:, 1) = const.F1_mean;
F(:, 2) = const.F2_mean;
x_ = [x_ F];
case 6
% Loads have been provided as well
otherwise
error ('Incorrect number of variables.');
end
% Extract variables
a = x_(:, 1:3); % Cross-sections of the bars [m^2]
w = x_(:, 4); % Horizontal position of bar #2 [m]
F = x_(:, 5:6); % Horizontal and vertical loads [N]
% Extract constants
D = const.D; % Total width of the structure [m]
L = const.L; % Length of the vertical bar [m]
E = const.E; % Young's modulus [Pa]
% Check w values
D_w = D - w;
if any (w < 0) || any (D_w < 0)
error ('w should be between 0 and D.')
end
% Lengths
LL = repmat (L, [n 3]);
LL(:, 1) = sqrt (L ^ 2 + w .^ 2);
LL(:, 3) = sqrt (L ^ 2 + D_w .^ 2);
% Sines and cosines
sin_theta = L ./ LL(:, 1);
cos_theta = w ./ LL(:, 1);
sin_alpha = L ./ LL(:, 3);
cos_alpha = D_w ./ LL(:, 3);
% Linear relation between tensile forces and elongations,
% assuming linear elasticity (Hooke's law)
C = E * a ./ LL;
% Compute displacement of node P and stresses
y = zeros (n, 2); % Displacement of node P
s = zeros (n, 3); % Tensile stress in the bars
for i = 1:n
% Rectangular matrix A for Equ. 9.1 in Das (1997) p.65, gives the
% equilibrium relation between tensile forces and loads (small displacements)
A = [cos_theta(i) sin_theta(i); 0 1; -cos_alpha(i) sin_alpha(i)];
% Stiffness matrix
K = A' * (diag (C(i, :))) * A;
% Compute the displacement of node P
y(i, :) = F(i, :) / K;
% Bar elongations
delta = y(i, :) * A';
% Tensile stresses (Hooke's law)
s(i, :) = E * delta ./ LL(i, :);
end
% Output: return displacement of node P and tensile stresses (five outputs)
z = [y s];
% df-in/df-out
if isa (x, 'stk_dataframe')
z = stk_dataframe (z, {'y1' 'y2' 'sigma1' 'sigma2' 'sigma3'}, x.rownames);
end
end % function
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