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Cantilever Tip Deflection

A steel cantilever beam of length $L = 2.0\ \text{m}$ is rigidly built in at one end and carries a concentrated load $P = 5\ \text{kN}$ at its free tip.

The beam has Young's modulus $E = 200\ \text{GPa}$ and second moment of area $I = 8.3 \times 10^{6}\ \text{mm}^4$ .

Find the vertical deflection at the free tip.

Given

P	5 kN	Tip load
L	2 m	Cantilever length
E	200 GPa	Young's modulus
I	8300000 mm^4	Second moment of area

Hint 1

This is a standard beam case — identify which one before doing any arithmetic. Is the load distributed or concentrated? Where is it applied?

Hint 2

The formula is δ = PL³/(3EI). The hard part is not the formula.

Hint 3

Watch the conversion on I: mm⁴ to m⁴ is a factor of 10⁻¹², because length is raised to the fourth power.

Worked solution — try the problem first

For a cantilever with a point load at the free end, the standard result is

$\delta = \frac{PL^3}{3EI}$

Convert everything to base SI before substituting — this is where most of the marks are lost.

$P = 5\ \text{kN} = 5000\ \text{N}$ $E = 200\ \text{GPa} = 200 \times 10^{9}\ \text{Pa}$ $I = 8.3\times10^{6}\ \text{mm}^4 = 8.3\times10^{6} \times (10^{-3})^4 = 8.3\times10^{-6}\ \text{m}^4$

Note the conversion factor on $I$ is $10^{-12}$ , not $10^{-3}$ — it is a fourth power of length.

Numerator: $PL^3 = 5000 \times (2.0)^3 = 5000 \times 8 = 4.0\times10^{4}\ \text{N·m}^3$

Denominator: $3EI = 3 \times 200\times10^{9} \times 8.3\times10^{-6} = 3 \times 1.66\times10^{6} = 4.98\times10^{6}\ \text{N·m}^2$

$\delta = \frac{4.0\times10^{4}}{4.98\times10^{6}} = 8.03\times10^{-3}\ \text{m}$

$\boxed{\delta = 8.03\ \text{mm downward}}$

Sanity check: span/250 is a common deflection limit; $2000/250 = 8\ \text{mm}$ . This beam sits right at that limit, which is a plausible design point rather than an absurd answer.

Concepts:Beam deflectionCantilever bendingUnit consistencySecond moment of area

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