% transmission_tower.m % set up and solve the equations for a symmetric transmission tower % % Kaveh Ghayour, Matthias Heinkenschloss % Jan 26, 2001 % set number of bar (m) and number of node (n) m = 40; n = 19; % bar is an m times 2 array. The i-th row contains % information for bar i in the following form: % [ lower-left-node upper-right-node ] bar ... = [ 1 2; 1 4; 2 3; 3 4; 2 4; 1 3; 3 5; 3 6; 4 5; 4 6; 5 6; 5 8; 6 7; 7 8; 5 7; 6 8; 8 9; 7 10; 7 9; 8 10; 9 10; 10 12; 9 13; 9 12; 10 13; 10 14; 13 14; 11 12; 9 11; 12 13; 12 15; 13 15; 16 9; 16 11; 17 16; 17 11; 10 18; 18 14; 18 19; 19 14]; % node is an n times 2 array. The i-th row contains % information for node i in the following form: % [x-coordinate-of-node y-coordinate-of-node] % Coordinates are given in meters so that our units % are compatible. base = 10; % half the length of the base side_angle = 80.0*(pi/180.0); % angle between base and sides height = base * tan(side_angle); % height of the transmission tower temp = 1.0/ tan(side_angle); % cotan(side_angle) arm = base; % length of the top horizontal side of the hangover node ... = [ -base 0; base 0; -temp*5.*height/6. 1.*height/6.; temp*5.*height/6. 1.*height/6.; -temp*4.*height/6. 2.*height/6.; temp*4.*height/6. 2.*height/6.; -temp*3.*height/6. 3.*height/6.; temp*3.*height/6. 3.*height/6.; -temp*2.*height/6. 4.*height/6.; temp*2.*height/6. 4.*height/6.; -temp*1.*height/6.-arm 5.*height/6. -temp*1.*height/6. 5.*height/6.; temp*1.*height/6. 5.*height/6.; temp*1.*height/6.+arm 5.*height/6; 0.0 height; -temp*1.*height/6.-arm 4.*height/6; -temp*1.*height/6.-2.*arm 4.*height/6; temp*1.*height/6.+arm 4.*height/6; temp*1.*height/6.+2.*arm 4.*height/6]; % plot the undeformed truss truss_plot(bar, node, 1) % area is an m array. The i-th element contains the cross sectional % area of bar i. We assume that all bars have rectangular cross % section of size 0.01 [m] times 0.05 [m], i.e., area = 0.0005 [m^2]. area = 0.0005*ones(m,1); % young is an m array. The i-th element contains the Young's % modulus for bar i. We assume that all bars are made of the same % material with Young's modulus is 195 GPa young = 195.e9*ones(m,1); % set indices of fixed displacements fixed = [1 2 3 4 ]; % determine indices of free displacements free = []; for i = 1:2*n if (~any(fixed==i)) free = [free; i]; end end % determine the stiffness matrix [K] = stiff( bar, node, area, young, free); % factor the stiffness matrix [K, ipivt, iflag] = lu_pp( K ); disp('Hit return to apply a load.......'); pause; % right hand side b = zeros(2*n,1); b(34) = 80000*9.80665; b(38) = 80000*9.80665; u = b(free); [u, iflag] = lu_pp_sl( K, u, ipivt ); if( iflag ~= 0 ) error( [' lu_pp_sl returned with iflag = ', int2str(iflag)]) end % determine the locations of the node of the deformed truss dnode = node; for i = 1:size(free(:),1) j = free(i); if( mod(j,2) == 0 ) % j is even, i.e., u(i) is vertical displacement of node j/2 dnode(j/2,2) = node(j/2,2) - u(i); else % j is odd, i.e., u(i) is horizontal displacement of node (j+1)/2 dnode((j+1)/2,1) = node((j+1)/2,1) + u(i); end end % plot the undeformed and the deformed truss truss_plot(bar, node, 0, dnode) disp('Hit return to apply another load.......'); pause; % TRY OUT A NEW LOADING NOW! b = zeros(2*n,1); b(34) = 80000*9.80665; b(38) = 80000*9.80665; for i=3:n b(2*i-1) = 20000; end u = b(free); [u, iflag] = lu_pp_sl( K, u, ipivt ); if( iflag ~= 0 ) error( [' lu_pp_sl returned with iflag = ', int2str(iflag)]) end % determine the locations of the node of the deformed truss dnode = node; for i = 1:size(free(:),1) j = free(i); if( mod(j,2) == 0 ) % j is even, i.e., u(i) is vertical displacement of node j/2 dnode(j/2,2) = node(j/2,2) - u(i); else % j is odd, i.e., u(i) is horizontal displacement of node (j+1)/2 dnode((j+1)/2,1) = node((j+1)/2,1) + u(i); end end % plot the undeformed and the deformed truss truss_plot(bar, node, 0, dnode)