MEC-35

Cantilever bending energy

U = P² L³ /(6 EI) for a tip-loaded cantilever. Castigliano δ = ∂U/∂P.

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EnergyCastigliano

Governing equation

U=\dfrac{P^2 L^3}{6EI},\quad\delta=\dfrac{P L^3}{3EI}

where

P: Tip load (kN)
L: Length (m)
E: Modulus (GPa)
I: Inertia (cm^4)
U: Stored energy (J)
\delta: Tip deflection (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-35 — Cantilever bending energy) is the form associated with Castigliano. Working symbols: $P$ , $L$ , $E$ , $I$ $\rightarrow$ $U$ , $\delta$ . Bending energy is ∫ M² dx / 2EI. For M = P(L−x) the integral is P²L³/6EI, so δ = PL³/3EI.

Purpose

Purpose: compute

U

\delta

from

P

L

E

I

in Mechanical via

U=\dfrac{P^2 L^3}{6EI},\quad\delta=\dfrac{P L^3}{3EI}

U = P² L³ /(6 EI) for a tip-loaded cantilever. Castigliano δ = ∂U/∂P. 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 = 5.000\,\mathrm{kN}

L = 1.200\,\mathrm{m}

E = 200.000\,\mathrm{GPa}

I = 800.000\,\mathrm{cm^4}

, the governing relation

U=\dfrac{P^2 L^3}{6EI},\quad\delta=\dfrac{P L^3}{3EI}

yields

U = 4.50\,\mathrm{J}

\delta = 1.800\,\mathrm{mm}

. A cantilever, a tip load, a stored-energy bar. Move a slider: the numbers are this situation, not a canned story.

Calculator

Inputs

Tip load P5 kNLength L1.2 mModulus E200 GPaInertia I800 cm^4

Outputs

Stored energy U4.50 J
Tip deflection \delta1.800 mm

Reading speed

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Narration of this film

A cantilever, a tip load, a stored-energy bar.

Bending energy is ∫ M² dx / 2EI. For M = P(L−x) the integral is P²L³/6EI, so δ = PL³/3EI.

Reading speed

Watch on YouTube