Linear 1D bar-element stiffness matrix

Element strain depends on u₂−u₁. Stored energy is ½(EA/L)(u₂−u₁)²; its second derivatives with respect to nodal displacements give EA/L[[1,−1],[−1,1]]. Positive diagonal and negative coupling terms create equal and opposite end forces. A common rigid translation creates no strain and is not penalized. Linear two-node bar with constant EA and small axial strain; supports are applied globally. Application: Understand linear finite-element modeling of axial bars and trusses. The numerical example With E=200 GPa, A=1.0×10⁻⁴ m² and L=2 m, EA/L=1.0×10⁷ N/m. For u=[0,0.001 m]ᵀ, f=[−10,000,10,000] N: equal and opposite end forces. A common translation creates no extension. Units check: (N/m)·m=N anchors the idea in a concrete measurement case.

If both nodes receive the same displacement, the bar does not deform and there is no restoring force. That rigid-body mode makes the element matrix noninvertible by itself. Appropriate global supports remove free translations and, where relevant, rotations. A remaining unconstrained mode makes the assembled system singular; it does not mean the local stiffness is wrong. Linear two-node bar with constant EA and small axial strain; supports are applied globally. The corresponding units are E in Pa; A in m²; L in m; matrix entries in N/m; displacement in m and force in N. With E=200 GPa, A=1.0×10⁻⁴ m² and L=2 m, EA/L=1.0×10⁷ N/m. For u=[0,0.001 m]ᵀ, f=[−10,000,10,000] N: equal and opposite end forces. A common translation creates no extension. Units check: (N/m)·m=N. This checks that the reference quantities are compatible.

Axial stiffness is EA/L=200·10⁹·10⁻⁴/2=10⁷ N/m. A relative extension of 0.001 m produces a force magnitude of 10⁴ N. Nodal forces have opposite signs because they balance the member’s internal pull. Check both N/m times m equals N and that the two nodal forces sum to zero. Linear two-node bar with constant EA and small axial strain; supports are applied globally. Scope: Linear interpolation of axial displacement and constant strain gives f_e=k_eu_e for a straight two-node bar with constant EA. Assumptions are small deformation, linear elasticity and axial loading only. The element matrix has a rigid-translation null mode; supports and assembly produce a solvable global system. Varying section, bending and geometric nonlinearity are excluded. A practical application is Understand linear finite-element modeling of axial bars and trusses.

The displayed 2×2 matrix uses local axial degrees of freedom. A globally inclined bar couples horizontal and vertical nodal displacements through direction cosines. Transform local stiffness to global coordinates before assembly. Otherwise loads and deformation are incorrectly treated as aligned only with global axes. Linear two-node bar with constant EA and small axial strain; supports are applied globally. Check the calculation with this case: With E=200 GPa, A=1.0×10⁻⁴ m² and L=2 m, EA/L=1.0×10⁷ N/m. For u=[0,0.001 m]ᵀ, f=[−10,000,10,000] N: equal and opposite end forces. A common translation creates no extension. Units check: (N/m)·m=N. Keep the model limits and units in view.

A free bar can translate as a whole without changing strain. The global model therefore needs suitable boundary conditions to remove rigid-body motion and determine displacements uniquely. Supports must represent the real restraint rather than being arbitrary. Excess constraints can create artificial reactions or overconstrain the model. Linear two-node bar with constant EA and small axial strain; supports are applied globally. The relation describes The two-node linear bar element relates axial nodal displacements and forces for a uniform elastic member in the finite element method. This helps distinguish its target quantity from neighbouring quantities.