Engineer's Notebook §2.3

Beam deflection

A simply supported beam with one load at mid-span sags most at the centre. The maximum deflection is:

δmax = P·L³48·E·I
δ
deflection
m
P
point load
N
L
span
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 on a pin support at the left and a roller at the right, with a point load at mid-span. The deflected shape is drawn below the unloaded position. P = 2 kN L = 1.50 m deflection drawn ×1
Fig. 2.3-1Deflection exaggerated; scale noted under the beam.
I = bh³/12—
Mmax = PL/4—
θend = PL²/16EI—
δmax = PL³/48EI—

Worked example

Uses the values in Fig. 2.3-1. Change them and the working updates.

Given
  • Ppoint load at mid-span2 kN
  • Lspan between supports1.5 m
  • Esteel200 GPa
  • b×hrectangular section40 × 80 mm
Find

The maximum deflection δ at mid-span.

Solution
I = bh³/12 = 1.707 × 10⁻⁶ m⁴ δ = PL³/48EI = 2,000 N × (1.5 m)³ / (48 × 2 × 10¹¹ Pa × 1.707 × 10⁻⁶ m⁴) = 4.12 × 10⁻⁴ m
Answer
δ = 0.412 mm

Span over deflection: L/3,641.

Other load cases

Same symbols. M is the maximum moment; θ is the slope at a support for simply supported beams, and at the free end for cantilevers.

  • Simply supported, uniform load

    δ = 5wL⁴384EIM = wL²8θ = wL³24EI

  • Cantilever, end load

    δ = PL³3EIM = PLθ = PL²2EI

  • Cantilever, uniform load

    δ = wL⁴8EIM = wL²2θ = wL³6EI

Assumptions
  • The beam is straight, prismatic and made of one linear-elastic material.
  • Deflections are small compared with the span, so the slope squared is negligible.
  • Shear deformation is ignored. This is reasonable when the span is more than about ten times the depth.
  • The beam's own weight is ignored. Add it as a uniform load if it matters.

Derivation

Measure x from the left support, with y positive upwards. Each support carries P/2, so for the left half of the beam the bending moment is M = Px/2.

  1. Start from the bending equation: EI·y″ = Px/2.
  2. Integrate once: EI·y′ = Px²/4 + C₁. By symmetry the slope is zero at mid-span, so C₁ = −PL²/16.
  3. Integrate again: EI·y = Px³/12 − PL²x/16 + C₂. The deflection is zero at the support, so C₂ = 0.
  4. At mid-span, x = L/2: EI·y = −PL³/48.

The minus sign means the beam deflects downwards, so its magnitude is δ = PL³/48EI. The slope at the support is C₁/EI, which gives θ = PL²/16EI.