How this instrument works
The stiffness matrix collapses, for the simplest possible structural element, to a single number: k = AE ⁄ L, the force it takes to stretch or compress an axially loaded bar by one metre. E, Young's modulus, converts strain into stress — how much a material resists being stretched at the molecular level. Multiplying by the cross-sectional area A turns that stress into an actual force, since stress is force per unit area. Dividing by the length L captures a real, everyday fact: a longer bar of the same material and cross-section stretches more under the same load, so it takes less force to move its end by a given amount. Put the three together and k tells you, in newtons per metre, exactly how hard the element pushes back.
That scalar is also the entire local stiffness matrix for a two-node axial element, written [[k, −k], [−k, k]]. Displace node 1 by δ while node 2 stays fixed and the internal force needed is kδ at node 1, balanced by −kδ at node 2 — Newton's third law written into a 2×2 array. A finite-element solver never treats a truss or frame as one continuous object; it treats it as many such elements, each with its own tiny stiffness matrix, rotated into the structure's global coordinates and summed wherever elements share a node. The outcome is one large stiffness matrix K that the software inverts to solve Kd = F for every joint's displacement under load.
The formula only holds inside the linear-elastic range, where stress stays proportional to strain and Hooke's law applies — push a bar past its yield point and E stops being constant, so k stops meaning anything useful. It also covers axial behaviour alone: a slender column can fail by buckling at a fraction of the load AE ⁄ L implies is safe, because buckling is governed by bending stiffness EI and effective length, a separate check entirely. An engineer sizing a bracing rod or a hanger cable uses this figure for elastic deflection and for feeding into a larger frame analysis, then checks yield and buckling as independent limits.
- Enter the Cross-sectional area of the element — 500 mm² for the reference bar; the unit menu also offers cm² if that matches your drawing.
- Enter the material's Young's modulus in MPa — 200,000 MPa (200 GPa) is typical structural steel.
- Set the Element length to the unsupported distance between the two nodes, in metres.
- Read Axial stiffness, N/m off the k field — the value that fills every entry of the element's local 2×2 stiffness matrix.
Worked example — a 500 mm² steel bar, 2 m long
Take a steel truss element with a 500 mm² cross-section (0.0005 m²), Young's modulus 200 GPa (200,000,000,000 Pa), and a length of 2 m between its two nodes. The formula gives k = 0.0005 × 200,000,000,000 ⁄ 2 = 50,000,000 N/m — fifty meganewtons per metre, the axial stiffness a finite-element solver assigns to this member.
That single scalar populates the element's local 2×2 stiffness matrix as [[50,000,000, −50,000,000], [−50,000,000, 50,000,000]], since holding one node fixed while the other moves takes a restoring force of equal size and opposite sign at each end. A frame or truss solver rotates this matrix into global coordinates for every angled member and sums the overlapping entries at shared nodes, building the structure's full stiffness matrix — the object it actually inverts to find displacements under load.
Questions
Why does doubling the length halve the stiffness?
Because elongation for a given force scales with length: a longer bar stretches more under the same axial load, so less force is needed to produce a given displacement. Halving L directly halves the denominator in k = AE ⁄ L, doubling the result — which is why a short bracing rod feels rigid while a long guy-wire of the same section visibly stretches under load.
Is this the whole stiffness matrix, or just one number?
Just the scalar that fills it. For a 1D axial element with two end nodes and one degree of freedom each, the full local stiffness matrix is [[k, −k], [−k, k]] — every entry is plus or minus k. A beam element carrying bending adds terms built from EI and length raised to higher powers; this calculator covers only the pure axial case, the building block for truss analysis.
What happens if I enter a non-steel material?
The formula does not care what the material is; swap Young's modulus and the same arithmetic applies to aluminium (about 69 GPa) or timber along the grain (around 11 GPa). What changes is the physical result: an aluminium bar with the same 500 mm² area and 2 m length as the steel example returns roughly 17,250,000 N/m, about a third of the steel figure.
Does k = AE ⁄ L account for buckling?
No. This is the linear axial stiffness from Hooke's law alone; it says nothing about whether a slender member under compression will buckle before reaching that load. Buckling is governed separately by Euler's formula, which depends on bending stiffness EI and effective length, not on A and L the way axial stiffness does. Check both for any compression member.
Why is the off-diagonal term negative?
It encodes equilibrium. If node 1 moves by δ while node 2 stays fixed, holding node 1 there takes a force kδ, and by Newton's third law node 2 feels −kδ from that same stretch. Writing the matrix as [[k, −k], [−k, k]] captures both nodes' forces from both displacements at once, and is why the matrix is symmetric and singular — rigid-body translation costs zero force.
How does this scalar fit into a larger structure?
Each element's local 2×2 matrix gets rotated into the structure's global coordinate system using the member's direction cosines, then added into a global stiffness matrix at the rows and columns for its two nodes, with overlapping entries from shared nodes summed. Solving Kd = F for that assembled matrix gives every node's displacement under the applied loads.