Simply Supported Beam Deflection — Worked Example

1. Problem

A simply supported steel beam, length L = 2000 mm, carries a central point load P = 15 kN. The cross-section is a rectangular bar 60 mm wide × 30 mm deep. Calculate the maximum deflection, maximum bending stress, and compare with an FEA result.

2. Given

3. Required

• Maximum deflection at midspan
• Maximum bending stress
• Margin of safety against yielding
• Comparison with FEA result

4. Assumptions

• Euler-Bernoulli beam theory (plane sections remain plane)
• Linear elastic material
• Small deflections (slope < 5°)
• Prismatic constant cross-section
• Self-weight neglected
• Supports are ideal pins (no rotational restraint)

5. Governing Equations

Max deflection (central point load):
  δ_max  =  P·L³ / (48·E·I)

Max bending moment:
  M_max  =  P·L / 4

Max bending stress:
  σ_max  =  M·c / I  =  M / Z

Section properties (rectangle):
  I  =  b·h³/12
  Z  =  b·h²/6
  c  =  h/2

6. Calculation

I  =  60 × 30³ / 12  =  60 × 27,000 / 12  =  135,000 mm⁴
Z  =  60 × 30² / 6  =  60 × 900 / 6  =  9000 mm³
c  =  30 / 2  =  15 mm

M_max  =  15,000 × 2000 / 4  =  7,500,000 N·mm  =  7.5 kN·m

σ_max  =  7,500,000 / 9000  =  833 MPa

δ_max  =  15,000 × 2000³ / (48 × 200,000 × 135,000)
       =  15,000 × 8 × 10⁹ / (48 × 2.7 × 10¹⁰)
       =  1.2 × 10¹⁴ / 1.296 × 10¹²
       =  92.6 mm

7. Result

σ_max = 833 MPa — exceeds Fy = 355 MPa. The beam yields. δ_max = 92.6 mm — a very large deflection (L/21.6). The beam is grossly under-sized.

8. Check

• Deflection ratio: δ/L = 92.6/2000 = 1/21.6 — far exceeds typical serviceability limit of L/250
• Stress: σ = 833 MPa >> Fy = 355 MPa — confirms yielding
• Sanity: a 60×30 mm bar spanning 2 m under 15 kN is clearly too small — the numbers are physically plausible
• Dimensional check: mm⁴ × GPa = N·mm² → deflection in mm — consistent

9. Interpretation

The beam fails both strength and serviceability. The section needs to be increased. For a target deflection of L/250 = 8 mm, the required I = PL³/(48Eδ) = 1.56 × 10⁶ mm⁴. A 100×50 mm rectangular section (I = 1.04 × 10⁶ mm⁴) is still insufficient; a standard I-section (e.g. 152×152 UC 30, I = 22.3 × 10⁶ mm⁴) would give δ = 0.56 mm and σ = 16 MPa — well within limits.

FEA comparison: a beam model with 20 B31 elements gives δ_max = 92.4 mm (0.2% below the analytical value). A solid model with a coarse mesh (5 mm elements) gives 89 mm — the slight underestimate is due to shear deformation in the solid model (Timoshenko effect), which the Euler-Bernoulli beam formula neglects.

10. Limitations

• Euler-Bernoulli theory neglects shear deformation — significant for deep beams (h/L > 0.1)
• No plasticity — the beam would actually form a plastic hinge at a load lower than that causing first yield
• No self-weight included
• No lateral-torsional buckling check — relevant for slender beams in bending
• The FEA comparison assumes the beam is modelled as a 2D beam, not a 3D solid

11. Related Resources

• Common Beam Deflection Formulae (Reference Library)
• Section Properties for Common Shapes (Reference Library)
• How to Compare FEA with Hand Calculations
• How to Verify an FE Model