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Prob.
NOTE: This problem was taken from Text Example 3.2. Stiffness Matrix:
^
K
Sym
Displacement Vector:
The displacement response of the structure is related to the applied loads through the stiffness matrix as follows: K^ ^ P Thus the force vector is given by: (^) (^) 6.348 1.912 3.536 0.871 383. 100 1.912 4.473 3.536 1.244 321. 3.536 3.536 6.830 0.193 0. ^ ^ ^ ^ (^) ^ ^ ^ ^ P K K
P
MATLAB script file used for verification of hand calculations: % HW#1, Problem # K=100[6.348 -1.912 -3.536;... -1.912 4.473 3.536;... -3.536 3.536 6.83] d=[0.871; 1.244; -0.193] P=Kd Output from running script file: K = 634.8000 -191.2000 -353. -191.2000 447.3000 353. -353.6000 353.6000 683. d =
-0. P =
Output from running script file: K = 1.0e+003 * 0.5044 -0.0225 -0. -0.0225 0.2294 0. -0.3522 0.0421 1. P = 200 600
Kinv = 0.0023 0.0001 0. 0.0001 0.0044 -0. 0.0005 -0.0001 0. d =
-0. Displacement vector:
Prob.
NOTE: This problem was taken from Text Example 3.1. Stiffness Matrix:
K
Sym
MATLAB script file: % HW#1 - Problem # K=100*[6.348 -1.912 -3.536 3.536 -2.812 -1.624;... -1.912 4.473 3.536 -3.536 -1.624 -0.938;... -3.536 3.536 6.83 -3.536 -3.294 0;... 3.536 -3.536 -3.536 3.536 0 0;... -2.812 -1.624 -3.294 0 6.107 1.624;... -1.624 -0.938 0 0 1.624 0.938] detK = det(K) Kinv=inv(K) Output from running script file: K = 634.8000 -191.2000 -353.6000 353.6000 -281.2000 -162. -191.2000 447.3000 353.6000 -353.6000 -162.4000 -93. -353.6000 353.6000 683.0000 -353.6000 -329.4000 0 353.6000 -353.6000 -353.6000 353.6000 0 0 -281.2000 -162.4000 -329.4000 0 610.7000 162. -162.4000 -93.8000 0 0 162.4000 93.
detk = 1.8538e- Warning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 5.929069e-029.
In Prob3 at 14 Kinv = 1.0e+024 * -2.7118 -0.0000 -2.7118 0.0000 -2.7118 -0. -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0. -2.7118 -0.0000 -2.7118 0.0000 -2.7118 -0. 0.0000 -0.0000 0.0000 -0.0000 0.0000 -0. -2.7118 -0.0000 -2.7118 0.0000 -2.7118 -0. -0.0000 -0.0000 -0.0000 -0.0000 -0.0000 -0. The determinant is now essentially equal to zero and Matlab indicates that the matrix may be singular. The inverse is computed but the values do not make sense in that 3 of the rows and 3 of the columns have all zeros. Further, the non‐zero entries are all equal to the same value which is a very large number relative to the values in the original stiffness matrix. These results provide a clear indication that the structure is unstable. Furthermore, the fact that the determinant and inverse are so sensitive to a very small change in a single entry in the stiffness matrix indicates that there is a "problem" with the stiffness matrix. In fact, this stiffness matrix was taken from Text Example 3.1 wherein the stiffness matrix is developed for a truss that has no supports. Thus, it is an unstable system.
