Matrix Stiffness Method – Axial bar element For a prismatic bar ab (area A, modulus E, length l) carrying only axial deformation, what is the horizontal stiffness coefficient K11 at node a?

Introduction / Context:
In the direct stiffness (matrix) method, an axial (truss) element has a simple 2×2 local stiffness matrix relating nodal axial displacements to nodal axial forces. Recognizing the form of this matrix is fundamental for assembling global stiffness matrices and solving indeterminate trusses and frames.

Given Data / Assumptions:

• Uniform bar with area A, modulus E, and length l.
• Only axial deformation; bending and shear are neglected.
• Local axis aligned with the member.

Concept / Approach:

The local axial stiffness matrix for a 2-node bar is k = (A E / l) * [[1, −1], [−1, 1]]. The diagonal term at node a is K11 = A E / l, representing the force that develops at node a due to a unit axial displacement at node a (with node b fixed) along the member axis.

Step-by-Step Solution:

Verification / Alternative check:

Energy method: axial strain ε = Δ/l; stress σ = E ε; force F = σ A = E A (Δ/l) → stiffness F/Δ = A E / l, confirming K11.

Why Other Options Are Wrong:

A E / l^2 and A E / 2 l mis-handle the length scaling; 2 A E / l doubles the correct stiffness without basis.

Common Pitfalls:

Confusing axial bar stiffness with bending stiffness (E I / l^3 terms) from beam elements, or forgetting sign symmetry in off-diagonal terms.

Final Answer:

A E / l