Engineer's Notebook §2.4

Cantilevers

A cantilever is fixed at one end and free at the other. With a point load at the free end, the tip deflects most:

δmax =P·L³3·E·I
δ
deflection
m
P
point load
N
L
length
m
E
Young's modulus
Pa
I
second moment of area
m⁴
M
bending moment
N·m
b
section width
m
h
section depth
m
θ
slope
rad
w
load per length
N/m
A beam built into a wall at the left with a point load at its free right end. The deflected shape is drawn below the unloaded position. P = 1 kN L = 1.00 m deflection drawn ×50
Fig. 2.4-1Deflection exaggerated; scale noted under the beam.
I = bh³/121.71 × 10⁶ mm⁴
Mwall = PL1 kN·m
θtip = PL²/2EI1.46 mrad
δmax = PL³/3EI0.977 mm

Worked example

Uses the values in Fig. 2.4-1. Change them and the working updates.

Given
  • Ppoint load at the free end1 kN
  • Llength from the wall1 m
  • Esteel200 GPa
  • b×hrectangular section40 × 80 mm
Find

The deflection δ at the free end.

Solution
I = bh³/12 = 1.707 × 10⁻⁶ m⁴ δ = PL³/3EI = 1,000 N × (1 m)³ / (3 × 2 × 10¹¹ Pa × 1.707 × 10⁻⁶ m⁴) = 9.77 × 10⁻⁴ m
Answer
δ = 0.977 mm

Length over tip deflection: L/1,024.

Other load cases

Same symbols. δ and θ are at the free end; M is at the wall.

  • Uniform load

    δ = wL⁴8EIM = wL²2θ = wL³6EI

  • Point load at a from the wall

    δtip = Pa²(3L − a)6EIM = Pa

  • End moment M0

    δ = M0L²2EIθ = M0LEI

Assumptions
  • The root is fully fixed: no movement or rotation at the wall. Real fixings rotate a little, which adds deflection.
  • The beam is straight, prismatic and made of one linear-elastic material.
  • Deflections are small compared with the length, so the slope squared is negligible.
  • Shear deformation and the beam's own weight are ignored.

Derivation

Measure x from the wall, with y positive upwards. An end load gives a hogging moment M = −P(L − x) at x.

  1. Start from the bending equation: EI·y″ = −P(L − x).
  2. Integrate once: EI·y′ = −P(Lx − x²/2) + C₁. The slope is zero at the wall, so C₁ = 0.
  3. Integrate again: EI·y = −P(Lx²/2 − x³/6) + C₂. The deflection is zero at the wall, so C₂ = 0.
  4. At the tip, x = L: EI·y = −PL³/3.

So δ = PL³/3EI, sixteen times the mid-span deflection of the same beam simply supported over the same length with a central load. The tip slope is θ = PL²/2EI.