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Elastic Beam Deflection

δ_{m a x} = \frac{P L ^{3}}{48 E I} (simple, mid P), δ_{m a x} = \frac{5 w L ^{4}}{384 E I} (simple, uniform)

Small-deflection Euler–Bernoulli theory: EI y'' = M(x). Use Handbook tables rather than integrating in the exam. Stiffer (larger EI) and shorter beams deflect less.

E

(GPa)

Modulus of elasticity

I

(m⁴)

Moment of inertia

δ

(mm)

Transverse deflection

Step-by-step solved example

Simple steel beam, L = 8 m, I = 200×10⁶ mm⁴, E = 200 GPa, mid-point 40 kN. Midspan δ?

δmax = 5wL⁴/(384 EI) downward at midspan for a simple uniform beam.

1. Units
EI = 200e9 × 200e−6 = 4.0×10⁷ N·m².
2. Table
δ = P L³ / 48 EI = 40000 × 512 / (48 × 4e7) = 0.0107 m = 10.7 mm.
$δ = 10.7 mm$

Answer: 10.7 mm downward

10 practice questions

0/10 correct

1.Cantilever end load P: δ_tip =

2.If L doubles, midspan δ of a simple uniform beam becomes

3.Doubling I will

4.The moment-area first theorem states that the change in slope is

5.Live-load deflection limit for floors is often

6.Conjugate beam method is useful because

7.A 10% increase in E reduces δ by about

8.Superposition of deflections is valid when

9.Simple span, uniform w, δ_max occurs at

10.Comparing wood (E≈10 GPa) to steel (E=200 GPa) same I, L, P: wood δ is