More Stiffness Method, Trusses During the last class, the stiffness matrix was determined, in local coordinates, for a truss member. A E 1 1 (21.1) k L 1 1 Except for different values of A, E, and L, this will be the local stiffness matrix for every member of any truss. The displacements at the ends of a truss member, in local coordinates, are denoted by d N and d F . A column matrix, d, was defined. d d N d F (21.2) The forces exerted at the ends of a truss member, in local coordinates, are denoted by q N and q F . A column matrix, q, was defined. q q N q F (21.3) A matrix equation was developed that determines local forces in terms of local displacements. q k d (21.4) MAE3501 - 21 - 1 The angles between the local and global coordinate systems were defined. X Y Y X X The cosines of those angles and a transformation matrix, T, were defined. X cos X (21.5) Y cos Y (21.6) X T 0 Y 0 0 X 0 Y (21.7) Except for the numerical values of X and Y , this will be the transformation matrix for every member of any truss. The displacements, in global coordinates, of the near and far ends of a truss member were defined. A column matrix, D, was defined. D NX D D NY D FX D FY (21.8) With these definitions, a matrix equation was developed that determines local displacements of the ends of truss member in terms of global displacements of the ends of a truss member. d TD (21.9) MAE3501 - 21 - 2 Develop a matrix equation that determines global forces in terms of local forces. The order is now reversed. Q NX , Q NY , Q FX , and Q FY denote the forces exerted, in global coordinates, on the near and far ends of a truss member. Define a column matrix, Q. Q NX Q Q NY Q FX Q FY (21.10) Consider the forces applied on the near end, in both local and global coordinates. qN Q NY Y X Q NX Q NX q N cos X X q N (21.11) Q NY q N cos Y Y q N (21.12) Consider the forces applied on the far end. Q FX X q F (21.13) Q FY Y q F (21.14) The following matrix equation determines global forces in terms of local forces. Q NX X Q NY Y Q FX 0 Q FY 0 0 0 X Y q N q F Q TT q (21.15) (21.16) MAE3501 - 21 - 3 Substitute equation (21.9) into equation (21.4). q k T D (21.17) Substitute equation (21.17) into equation (21.16). Q TT k T D (21.18) Define k, a global stiffness matrix for each member k TT k T X k Y 0 0 (21.19) 0 0 A E 1 1 X X L 1 1 0 Y X2 AE X Y k L X2 X Y X Y Y2 X Y Y2 Y 0 X2 X Y X2 X Y 0 Y (21.20) X Y Y2 X Y Y 2 (21.21) 0 X Equation (21.20) shows the individual terms in equation (21.19). Equation (21.21) is the result when all the terms are multiplied out. Except for the numerical values of A, E, L, X and Y , this 4 by 4 matrix is the global stiffness matrix for every member of any truss. For members parallel to a global-coordinate axis, there will be twelve zeroes in this matrix. The indices of the rows in the 4 by 4 matrix in equation (21.21) are, top-to-bottom, the subscripts of the elements of D or Q, in order. The indices of the columns in the 4 by 4 matrix in equation (21.21) are, left-to-right, the subscripts of the elements of D or Q, in order. Substitute equation (21.19) into equation (21.18) and develop a matrix equation that calculates the forces exerted on the ends of a member, in terms of the displacements of the ends of the member, both expressed in global coordinates. Q kD (21.22) Return to the truss as a whole. MAE3501 - 21 - 4 The next step is to assemble the five individual, 4 by 4, k matrices into K, an 8 by 8 global stiffness matrix for the entire truss. The assembly is accomplished by superposition. Begin with k 1 , the global stiffness matrix for member number 1. Determine the sixteen individual terms for this member, and arrange them in a 4 by 4 matrix, but use indices for the rows and columns of that matrix that match, in the proper order, the global displacements of the ends of member number 1. 1 2 3 4 1 2 k1 3 4 MAE3501 - 21 - 5 (21.23) Insert the sixteen terms into K in the following locations. 1 2 3 4 5 6 7 8 1 2 3 4 K 5 6 7 8 x x x x x x x x x x x x x x x x (21.24) Repeat the process for the remaining four members of the truss. When k 2 is created, the indices will be, in order, 3, 4, 5, and 6. 3 4 5 6 3 4 k2 5 6 (21.25) When these sixteen terms are inserted into the K matrix, four of the locations will already be occupied, those with both indices equal to either 3 or 4. Simply add the four terms from k 2 to the values that are already in the K matrix. When the process has been completed for all 5 members, regardless of whether the individual global stiffness matrices contain zeroes, the structure global stiffness matrix will contain zeroes. No member connects displacements 1 and 2 with displacements 5 and 6, so there will be no terms that have one index equal to 1 or 2 and the other index equal to 5 or 6. MAE3501 - 21 - 6 Homework There is no written assignment. However, it will be necessary on the next assignment for you to determine K, the stiffness matrix in global coordinates, of the truss. I strongly suggest you perform that calculation prior to the next class. Determine the k matrices for the individual truss members, then determine K. A 6.25 (10) 4 m 2 E 200 (10) 6 kPa Leave the A E multiplier outside the matrix. Express lengths of members in m. You may check your work by comparing with the following result. 0.354 0.354 0.354 0.854 0.354 0.354 0.354 6 0.354 K 0.125 (10) 0 0 0 0 0 0 0.5 0 0.354 0.354 1.061 0.354 0.354 0.354 0.354 0.354 0.354 0 0 0 0 0.354 0 0 0 0.5 0.354 0.354 0.354 0.354 0.354 1.061 0.354 0.354 0.354 0.354 0.354 0.854 0.354 0.5 0 0.354 0.354 0.354 0 0 0.354 0.5 0 0.854 0.354 0.354 0 0 0.354 0.854 MAE3501 - 21 - 7