MEC-38

Castigliano axial deflection

δ = ∂U/∂P = P L /(A E) for an axial bar. U = P² L /(2 A E).

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EnergyCastigliano 1873

Governing equation

\delta=\dfrac{\partial U}{\partial P}=\dfrac{PL}{AE}

where

P: Axial load (kN)
L: Length (m)
A: Area (cm^2)
E: Modulus (GPa)
\delta: Extension (mm)

Lecture brief

Historical brief

Machine design grew from Coulomb torsion and Hertz contact (1881) through Soderberg fatigue and the heat-engine cycle. The lab writes shaft, bearing, contact and thermodynamic limits in SI. This sheet (MEC-38 — Castigliano axial deflection) is the form associated with Castigliano 1873. Working symbols: $P$ , $L$ , $A$ , $E$ $\rightarrow$ $\delta$ . Castigliano's second theorem: displacement conjugate to a force is the derivative of complementary energy.

Purpose

Purpose: compute

\delta

from

P

L

A

E

in Mechanical via

\delta=\dfrac{\partial U}{\partial P}=\dfrac{PL}{AE}

δ = ∂U/∂P = P L /(A E) for an axial bar. U = P² L /(2 A E). Use it when a real mechanical question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given

P = 80.000\,\mathrm{kN}

L = 2.000\,\mathrm{m}

A = 12.000\,\mathrm{cm^2}

E = 200.000\,\mathrm{GPa}

, the governing relation

\delta=\dfrac{\partial U}{\partial P}=\dfrac{PL}{AE}

yields

\delta = 0.6667\,\mathrm{mm}

. A bar, an axial P, a stretch. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Axial load P80 kNLength L2 mArea A12 cm^2Modulus E200 GPa

Outputs

Extension \delta0.6667 mm

Reading speed

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Narration of this film

A bar, an axial P, a stretch.

Castigliano's second theorem: displacement conjugate to a force is the derivative of complementary energy.

Reading speed

Watch on YouTube