The Problem

sin  1 = ( ) / (20) = -0.6 cos  1 = (16 - 0) / (20) = 0.8 sin  2 = (12 - 0) / (15) = 0.8 cos  2 = (9 - 0) / (15) = 0.6 Sines and Cosines

Element Matrices [S]

System Stiffness Matrix

Element Matrices [S]

Summing Element Stiffnesses

Summing Element Stiffnesses

Two Matrix Contributions

Final [K]

Final Equation P=KX

System Stiffness Matrices

Solving the System of Equations

Modify for Known Loads

Modify for Boundary Conditions

Modify to Ease Solution

Return Symmetry

Modified Equations

Recap

Initial Matrix

Loads

Boundary Conditions

Symmetry

Solution 10 = AE/L X1 100 = AE/L X2 0 = AE/L X3 0 = AE/L X4 0 = AE/L X5 0 = AE/L X6

Force Calculation (f=sbX) {(X3i-X1i) cos  i + (X4v-X2v) sin  i} is simply the change in length t1 = AE/L {(10L/AE - 0)(0.8) + (100L/AE - 0)(-0.6)} + (0) t2 = AE/L {(0 - 10L/AE - 0)(0.6) + ( L/AE - 0)(0.8)} + (0) t1 = f2 = -f1 = -52 kips t2 = f4 = -f3 = -86 kips