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Computational Engineering / Second-order, buckling and dynamic methods

Geometric stiffness: why load changes stiffness

The second matrix. Where it comes from, why it scales with force rather than with EA, and what that costs you.

Lesson review: expert verification remains open

This lesson has received a full-context editorial review and a check of its technical explanations. Selected arithmetic has been checked separately. This is not independent engineering sign-off.

Checks needed for project use

  • A computational engineering reviewer must check algorithms, convergence claims, statistical assumptions and numerical examples beyond the cases explicitly recorded.
  • Confirm current adopted code edition and national provisions for stability screening; SCI reference documents are explanatory and not a substitute for the governing standard.
  • Audit buckling/eigenvalue helper implementations and lab mode-classification weighting; translations and rotations must not be combined unscaled.
  • Numerical mesh table and all dynamic modelling choices require specialist verification.

These are educational models and exercises. Check their assumptions before using the results. The full technical content and every diagram have not yet been independently verified.

Concrete and steel calculations retain their stated first-generation Eurocode teaching basis. Use the standards and amendments specified for your project; do not mix generations without an appropriate design basis.

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Why this matters

Tension increases the transverse stiffness of a taut string; compression reduces the lateral stiffness of a slender column. These effects arise from equilibrium in the displaced geometry. Their importance depends on the force state, geometry and response being assessed. A first-order model omits them, so assess their significance before choosing the analysis.

By the end of this lesson you should be able to

  • Explain tension stiffening and compression softening from a free body
  • Write the effective stiffness of a two-bar column with axial load
  • Say why Kg is proportional to P/L and not to EA/L
  • Explain why P-delta results cannot be superposed

What you should already know

  • The elastic stiffness matrix and its assembly (Module 9)
  • Euler buckling of a strut — the classical result, recapped here

A two-bar model

Consider two rigid bars of length L, pinned together, with the outer ends restrained laterally and axial shortening permitted at the loaded end. A spring of stiffness k restrains the middle joint. For small lateral displacement δ and compressive dead load P, the transverse destabilising force is approximately 2Pδ/L. The effective lateral stiffness is therefore keff = k − 2P/L. Tension reverses the geometric contribution and increases stiffness.

The geometric term depends on the axial force and geometry. At fixed P and L it is independent of EA, although changing section properties in an indeterminate structure can also change the force distribution.

Forming the analysis

A common approximate procedure finds member forces from a first-order solution, forms a geometric stiffness and solves the modified equations. More general methods update forces and stiffness iteratively or incrementally; there is no universal two-pass rule.

Load combinations

When stiffness changes with the combined load, separately analysed nonlinear case results generally cannot be scaled and added to recover the combination. Analyse the required combined loading with a suitable procedure. Linear perturbation analyses about the same fixed prestressed state can still obey superposition within their assumptions. Envelopes may be formed from separately valid results.

Effective stiffness of the two-bar column
keff=k−2PL,Pcrit=kL2k_{\text{eff}} = k - \frac{2P}{L}, \qquad P_{\text{crit}} = \frac{kL}{2}

What it calculates: The lateral stiffness that remains once the axial load is accounted for

k
Lateral spring stiffness at the joint (kN/m)
L
Length of each bar (m)
P
Axial load, positive in compression (kN)
Pcrit
Load at which the effective stiffness reaches zero (kN)

This assumes

  • The bars are rigid, so all the flexibility is in the spring
  • Small angles, so the transverse component is 2Pδ/L rather than a trigonometric function of δ

In plain terms: For this idealised small-angle model, keff reaches zero at P = kL/2. The bifurcation load is independent of the lateral forcing within these assumptions. Finite displacements, imperfections and material nonlinearity can change the real load–displacement path.

Try it

Geometric stiffness explorer

Two rigid bars, a lateral spring at mid-height and an axial load. Positive P is compression; drag it negative for tension.

kN/m
m
kN

Positive is compression. Negative is tension, and it stiffens.

Elastic stiffness k
250 kN/m
Geometric contribution −2P/L
-40.0 kN/m

compression softens

Effective stiffness
210.0 kN/m
Critical load Pcrit = kL/2
625 kN
Load factor αcr
6.25
Deflection amplifier α/(α−1)
1.190
Imperfection amplifier 1/(α−1)
0.190

Ignoring second-order effects here costs 19.0 %, which is above the 10 % line. Second-order analysis is required, not optional.

What to notice

  • Change k and Pcrit moves with it. Change L and Pcrit moves with it. Change nothing else and the material never enters.
  • The geometric contribution is proportional to P/L. Doubling the section area would double k and leave it untouched.
  • αcr = 11 is where ignoring second-order effects costs exactly 10 %. The convention quoted in practice is 10, where the cost is 11.1 %.

What this shows: Load changes stiffness. Tension adds 2P/L, compression takes it away, and no material property appears anywhere in it.

Predict first

A cable net is analysed linearly and comes out unrealistically flexible. An engineer suggests running a P-delta analysis. Will that make it stiffer or softer?

Practice

A two-bar column has a lateral spring of 250 kN/m and bars of 5 m. Compute the critical compression load in kN.

Practice

The same column carries 100 kN of compression. What is its remaining effective lateral stiffness, in kN/m?

Worked example

Why P-delta results cannot be added together

Given

  • A frame is analysed for gravity alone and for wind alone, both with P-delta
  • The design combination is 1.35 × gravity + 1.5 × wind

Find

Whether factoring and adding the two results is valid

    Worked example

    Reading an amplification factor

    Given

    • An idealised single-mode elastic sway model has a critical load factor of 5

    Find

    How much the first-order sway is amplified, and what to conclude

      Check yourself

      What does a negative geometric stiffness contribution represent?

      Summary

      • Tension stiffens, compression softens: keff = k − 2P/L
      • Geometric stiffness scales with load over length, not with EA over length
      • Geometric stiffness requires a force state; the solution procedure may be approximate, iterative or incremental
      • The critical load in the idealised two-bar model is independent of lateral forcing
      • Do not superpose load-dependent results unless the chosen linearisation explicitly permits it