doubling-extension-doubles-bar-force - Elasticity and Nonlinear Dynamics Step by Step

Because bar force is linear in extension at fixed E, A, and L, doubling the stretch doubles the load through F = E * A * delta-L / L. The trend is exact only for the stated Young modulus, area, length, and extension in the ideal linear elastic model. Real materials add nonlinear response, geometry tolerance, temperature/loading history, fixture compliance, calibration, and measurement uncertainty.

For the same checked bar stiffness, force scales with extension. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Keep the same bar stiffness before changing extension

Young modulus, area, and length are unchanged. The larger extension is the only input that changes the force.

k=4\ \text{N/m}\qquad \Delta L=2\ \text{m}

Force follows extension after stiffness is fixed

Read the stiffness column first, then multiply by extension. The middle row is the doubled-extension case.

\begin{array}{c|c|c|c|c|c}E&A&L&\Delta L&k&F\\16\ \text{Pa}&2\ \text{m}^{2}&8\ \text{m}&1\ \text{m}&4\ \text{N/m}&4\ \text{N}\\16\ \text{Pa}&2\ \text{m}^{2}&8\ \text{m}&2\ \text{m}&4\ \text{N/m}&8\ \text{N}\\16\ \text{Pa}&4\ \text{m}^{2}&8\ \text{m}&1\ \text{m}&8\ \text{N/m}&8\ \text{N}\\\end{array}

Doubling extension doubles the bar force

The stiffness is unchanged, so the checked force follows the extension.

F=k\Delta L=4\ \text{N/m}\cdot2\ \text{m}=8\ \text{N}