Cantilever Deflection — End Load

δ=PL33EI\delta = \frac{P L^{3}}{3 E I}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

The cantilever is the shape Galileo drew in 1638 and the one every diving board, balcony and pipe-support bracket still copies. With the load at the tip, δ = PL³/3EI — sixteen times the sag a simply supported beam of the same span, load and section would show, because the fixed end must resist the full moment PL rather than sharing it with a second support. A 2 kN load on the end of a 2 m steel cantilever with I = 4.167 × 10⁻⁶ m⁴ droops δ = 2000 × 8 ÷ (3 × 200 × 10⁹ × 4.167 × 10⁻⁶) = 0.0064 m, 6.4 mm.

Everything hinges on the word fixed. The formula assumes the built-in end has zero rotation, and in real life that is the hardest thing to achieve — a bracket bolted to a flexible column, a beam pocketed into masonry, or a weld that is not full-strength will all rotate, and the tip deflection then comes out well above prediction. Measure the true rotation, or design conservatively. If instead the load is spread uniformly along the cantilever, the coefficient becomes wL⁴/8EI; and enter I as a plain number in m⁴, since this engine has no m⁴ unit.

Cantilever Deflection — End Load
δ=PL33EI\delta = \frac{P L^{3}}{3 E I}
Where
  • δ\delta= Tip deflection
  • PP= End load
  • LL= Cantilever length
  • EE= Young's modulus
  • II= Area moment of inertia