The stiffness matrix method, step by step
The direct stiffness method behind every structural analysis program, on a chain of three springs: assemble the global stiffness matrix, solve K·u = F for the displacements, and recover the force in each spring.
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The calculation
k_1 := 2000 kN/m
k_2 := 1500 kN/m
k_3 := 1000 kN/m
K_s := [k_1 + k_2, -k_2; -k_2, k_2 + k_3]/(1 kN/m) =
F_s := [12; 4]
u := lusolve(K_s, F_s) =
u_1 := u[1,1]*1 m = mm
u_2 := u[2,1]*1 m = mm
N_1 := k_1*u_1 = kN
N_2 := k_2*(u_2 - u_1) = kN
N_3 := -k_3*u_2 = kN
K_s*u =Results for these inputs
- Displacements 5.54 mm and 4.92 mm.
- Spring forces 11.08 kN (tension), −0.92 kN and −4.92 kN (compression).
- The check K·u gives back the loads 12 and 4 kN.
How it works
Each spring adds its stiffness to the rows and columns of the two nodes it joins. The supports remove rows and columns, leaving a square matrix that lusolve solves for the free displacements. Member forces follow from the stiffness times the difference of the end displacements.
These calculations use classical textbook mechanics, not a design code. For design, use the method, factors and checks that apply to your work and have the result checked by a competent person.
From springs to frames
Beam and frame elements work the same way with 4×4 or 6×6 element matrices. Block matrices, [A, B; C, D], make assembly easy; see the matrix calculator and truss analysis.
Questions
What is the direct stiffness method?
A matrix method that assembles the stiffness of every element into one global matrix, then solves K·u = F for the unknown displacements; forces follow from the displacements.
Why divide by 1 kN/m?
lusolve works on plain numbers, so the matrix is made dimensionless first; the units are put back on the results with 1 m.
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