function TrussElementStiffnessMatrixMethod
elasticity = 10200000; % aluminium modulus
elasticity
area = 200; % area cross-section
L = 1; % bar length
elasticity_area = elasticity*area;
nodes = [1 4; 2 4; 3 4; 4 6; 3 6; 3 5; 5 6; 5 7; 6
7];
coordinates = [0 1 ;0 0 ;1 1 ;1 0 ;2 1 ;2 0 ;3 1
];
elements = size(nodes,1);
totalNodes = size(coordinates,1);
x_coord = coordinates(:,1);
y_coord = coordinates(:,2);
degree_freedom = 2 * totalNodes;
u_displacement = zeros(degree_freedom,1); %
displacement
reactionForces = zeros(degree_freedom,1); % solution
for (b)
% gravitational force loads
reactionForces(4)= -2500;
reactionForces(5)= -2500;
reactionForces(6)= -2500;
reactionForces(7)= -5000;
% stiffnessMatrix matrix
[stiffnessMatrix]=formStiffness2Dtruss(degree_freedom,elements,nodes,totalNodes,coordinates,x_coord,y_coord,elasticity_area);
% boundary-conditions
prescribed_freedom = [0 -100 0 0 50 0 10]';
% a) final displacements answer
displacements =
stiffnessMatrix\prescribed_freedom;
disp('displacements: ')
disp(displacements)
fprintf('reactionForces: \n')
disp(reactionForces)
end
function stiffnessMatrix =
formStiffness2Dtruss(degree_freedom,elements,nodes,totalNodes,coordinates,x_coord,y_coord,elasticity_area)
stiffnessMatrix = zeros(totalNodes);
for e=1:elements;
% elementDof: element degrees of
freedom (Dof)
degree_freedom=nodes(e,:) ;
stiffnessMatrix(degree_freedom,degree_freedom)=stiffnessMatrix(degree_freedom,degree_freedom)+[1
-1;-1 1];
end
end
%caling method
TrussElementStiffnessMatrixMethod();
OUTPUT
The plane truss is made of three members and is loaded with force P as shown....
Finite Element Method 5.17 Displacements of the three-member truss shown are confined to the plane of the figure, and points 1, 2 and 3 are fixed to the stationary rim. All members have the same A, E, and L a) Obtain the 2x2 stiffness matrix that operates on the horizontal and vertical degrees of freedom of the central node. b) Obtain the corresponding global force vector c) Solve for the displacements and for axial stress in member (2-4), when the...
The plane truss is subjected to a load as shown in Figure 4. Take E = 200 GPa and cross sectional areas of members 1, 2 and 3 as 150, 250 and 200 mm2 respectively a) Assemble the upper triangular part of the global stiffness matrix for the truss b) Determine the horizontal and vertical displacements at node 4 c) Calculate the forces in each member of the truss. (25 marks) 20 kN 3 60° 4 1.5m 2 2 20m...
Question 4 The plane truss is subjected to a load as shown in Figure 4. Take E = 200 GPa and cross sectional areas of members 1, 2 and 3 as 150, 250 and 200 mm2 respectively a) Assemble the upper triangular part of the global stiffness matrix for the truss. b) Determine the horizontal and vertical displacements at node 4. c) Calculate the forces in each member of the truss. (25 marks) 20 kN 3 600 4 3 1.5m...
Figure Q5(a) shows a plane truss supported by a horizontal spring at the top node. The truss members are of a solid circular cross section having a diameter of 20 mm and an elastic modulus (E) of 80 GPa (10° N/m2). The spring has a stiffness constant of k-2000 kN/m. A point load of 15 kN is applied at the top node. The direction of the load is indicated in the figure. The code numbers for elements, nodes, DOFS, and...
For the truss shown in the figure below, develop element stiffness matrices in the global co-ordinate system. AE 200 [MN] is the same for all members. Use the direct stiffness matrix method to: i. Establish all element stiffness matrices in global coordinates ii.Find the displacements in node 3 ii. Calculate the member stresses 4m 3m 20kN 2 2 Use HELM resources on Moodle to find required determinant and inverse matrix. Answer 9.6x103 [MPa] 0.24mmm u3-0.20mm 0.45mm 16x10-3 MPa σ2-3- 1...
The plane truss shown in Figure is composed of members having a square 15 mm × 15 mm cross section and modulus of elasticity E = 69 GPa. a. Assemble the global stiffness matrix. b. Compute the nodal displacements in the global coordinate system for theloads shown. c. Compute the axial stress in each element 3 kN 3 5 kN 2 1.5 m 4. 1.5 m
For the 3-D indeterminate (4-member) TRUSS structure shown in Figure 2A. Given that Px 10K (in X-direction); Py none (in Y-direction); E 30,000 ksi; A 0.2 square inches. The nodal coordinates, the earth-quake displacement/settlement, and members' connectivity information are given aS Applied Load! Earth-Quake MEMBER #1 NODE # X node-i node-j 120.00" 160.00"| 80.00"| Px=-10 Kips none Py- none 120.00" 160.00"0.00"none 120.00"0.00" 0.00" none 0.00" 0.00"0.00" none 0.00" 0.00" 80.00" none none 2 none 4 4 none 4 +2.00" (in...
A plane structure consists of three truss elements connected to four nodes, as shown below. All trusses have cross sectional area A -7.104 m2 and elastic modulus E = 210 GPa. The length of each truss element is L = 1 m. A point force, P -5 kN, is acting on node 4 L/2 3.1 Calculate the displacements at the nodes 3.2 Calculate the reaction forces 3.3 Calculate the stress in each bar A plane structure consists of three truss...
matrix structural Problem #1: Solve for nodal displacements, reactions, and member forces of the truss shown. The support at node 1 displaces down 0.6 in and node 4 displaces to the left 0.3 in. All areas are 2 in2 and E- 29 x 10° psi. Use the stiffness matrix method. 30000 All areas 2 in2 E-29x106 psi 21 ② 3 10 ft Problem #1: Solve for nodal displacements, reactions, and member forces of the truss shown. The support at node...
i need help with c and d but explain why Question 1 (10 marks). Assembly A model consists of two 1D trusses with dimensions as given in Figure 1. Element 1 runs angle, connecting parallel to the x-axis, connecting node 1 and 2. Element 2 is running at an node 1 and 3. Node 1 has an applied force in the negative y-direction. Node 1 can only in y-direction, while nodes 2 and 3 are fixed in both x and...