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Beam Deflection (Simply Supported, Center Point Load)

Calculates the maximum midspan deflection of a simply supported beam under a single point load at its center.

CivilConstructionStructural Design

Beam Deflection CalculatorSimply Supported · Center Point Load

δ = P·L³ / (48·E·I)
P = load (N)  ·  L = span (m)  ·  E = elastic modulus (Pa)  ·  I = moment of inertia (m⁴)
⟹ δP, L, E, I
N
m
Pa
m⁴
m
Steel:
Solve for:
Deflection
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Deflection Level
Small (<1 mm) Moderate (1–5 mm) Large (5–20 mm) Very Large (>20 mm)
δ vs. Pfixed L, E, I
δ(P) = P·L³/(48·E·I) Computed point
δ = P·L³ / (48·E·I)  ·  Simply supported beam, center point load
δ = P*L³ / (48*E*I)
Beam Deflection (Simply Supported, Center Point Load)

Variables

SymbolQuantityUnit
δMaximum deflectionm
PPoint loadN
LBeam spanm
EModulus of elasticityPa
IMoment of inertiam4

What it means

This formula provides the maximum deflection at the mid‑span of a simply supported beam subjected to a concentrated load at its centre. It is derived from the elastic curve differential equation and assumes linear elastic material behaviour and small deflections. The deflection is inversely proportional to the beam's flexural rigidity (E·I), and directly proportional to the cube of the span length. This makes span length the most critical factor in controlling deflections. The formula is used to check serviceability limits (e.g., L/360 for floors) and to ensure that deflections do not cause damage to finishes or equipment. It is one of the standard beam deflection formulas used in structural engineering. For other loading conditions (uniformly distributed load, different support conditions), different constants apply. The formula is also used in machine design for shaft deflections. Accurate deflection prediction is essential for structural safety and performance.

Worked example

Beam Deflection – Two Examples

Real‑World
Scenario: A 4 m steel beam (E = 200 GPa) with point load 5 kN at mid‑span. I = 5.00×10⁻⁴ m⁴. Find deflection.
ParameterValue
P5000 N
L4 m
E2.0×10¹¹ Pa
I5.00×10⁻⁴ m⁴
1δ = 5000 × 4³ / (48 × 2e11 × 5e-4) = 320,000 / (48 × 1e8) = 6.667e-5 m = 0.067 mm
Result 0.067 mm ✓ Very small
Scenario: A 3 m timber beam (E = 10 GPa) with 3 kN load, I = 3.00×10⁻⁴ m⁴. Compute deflection.
ParameterValue
P3000 N
L3 m
E1.0×10¹⁰ Pa
I3.00×10⁻⁴ m⁴
1δ = 3000 × 27 / (48 × 1e10 × 3e-4) = 81,000 / 1.44e8 = 5.625e-4 m = 0.5625 mm
Result 0.56 mm ✓ Acceptable
Key insight: Deflection ∝ load × span³ / (E·I).

Common mistakes

  • Load case: This formula is for a central point load on a simply supported beam. For other loadings (uniform, off‑centre), use different factors (e.g., 5/384 for UDL).
  • Units: P in N, L in m, E in Pa, I in m⁴ → δ in m.
  • E and I: Use the modulus of elasticity (E) and the second moment of area (I) about the bending axis.
  • Superposition: For multiple loads, sum deflections from each.
  • Self‑weight: If the beam is heavy, include its own weight as a UDL – not covered by this formula.

Applications

Beam deflection at the centre of a simply supported beam under a central point load is calculated as δ = P·L³/(48·E·I). This formula is fundamental for ensuring that beams do not deflect excessively under service loads, which could cause damage to finishes, misalignment of equipment, or even structural distress. Civil and mechanical engineers use it to size beams for floors, bridges, crane girders, and machine supports. The deflection limit is often specified by building codes (e.g., span/360) to maintain serviceability. By controlling deflection through material selection and cross‑sectional geometry, engineers can achieve the required stiffness. The formula also provides insight into the relationship between load, span, and flexural rigidity.

  • Deflection checks for building floor beams
  • Design of bridge girders and crane runways
  • Sizing of beams for vibration-sensitive equipment
  • Serviceability limit state verification
  • Optimisation of beam shapes for stiffness