Hey Kun,
Below is a basic example of how you might use MATLAB to calculate the joint displacement from the joint rotation value for a simple truss structure. This example assumes a basic configuration and uses some simplifying assumptions.
Let's assume a simple truss with two members meeting at a joint, and we know the rotation at that joint. We'll calculate the resulting displacements.
% Given data
L = 5; % Length of truss members in meters
theta = 0.1; % Rotation at the joint in radians
E = 210e9; % Young's modulus in Pascals
A = 0.01; % Cross-sectional area in square meters
% Stiffness of the truss members
k = E * A / L; % Axial stiffness of each member
% Geometry of the truss (angles in radians)
alpha = pi / 4; % Angle of member AB with the horizontal
beta = -pi / 4; % Angle of member AC with the horizontal
% Rotation matrix to find displacements
R = [cos(alpha), sin(alpha); cos(beta), sin(beta)];
% Compatibility equations for small rotations
% Assuming small rotation, the displacements u and v can be approximated
u = L * theta * [sin(alpha); sin(beta)];
% Displacements at the joint
displacements = R \ u;
% Display the results
disp('Joint Displacements (in meters):');
disp(['Horizontal displacement: ', num2str(displacements(1))]);
disp(['Vertical displacement: ', num2str(displacements(2))]);
Explanation
- Inputs: The length of the truss members (L), rotation at the joint (theta), Young's modulus (E), and cross-sectional area (A).
- Stiffness Calculation: The axial stiffness of each member is calculated using the formula ( k = \frac{EA}{L} ).
- Geometry: The angles of the truss members with respect to the horizontal axis are defined as alpha and beta.
- Rotation Matrix: A rotation matrix R is created to relate the displacements to the rotation.
- Compatibility Equations: The displacements u are calculated using the known rotation and the lengths of the members.
- Solve for Displacements: The joint displacements are found by solving the linear system R \ u.
- Display Results: The horizontal and vertical displacements are displayed.
This example assumes small rotations and linear behavior. For more complex structures or larger rotations, the analysis may require more advanced methods, including finite element analysis.
Hope it helps
