Beam Deflection

Goal: Determine the deflection and slope at specified points of beams and shafts Solve statically indeterminate beams: where the number of reactions at the supports exceeds the number of equilibrium equations available. Maximum deflection of the beam: Design specifications of a beam will generally include a maximum allowable value for its deflection.

Sign Conventions

From ref pages

Boundary Conditions

From ref pages

Moment-Curvature Equation

Elastic Curve of a Beam:

Taken from TAM251 Lecture Notes - L12S3

Moment-curvature equation:

$$ M(x) = \frac{E(x)I(x)}{\rho(x)}\ $$

$$ \kappa = \frac{1}{\rho} = \frac{M(x)}{EI}\ $$

Taken from TAM251 Lecture Notes - L12S3

Governing equation of the elastic curve:

$$ \frac{d^2y}{dx^2} = \frac{M(x)}{EI}\ $$

Assumptions

$ y(x) $ is the vertical direction
Bending only: we will neglect effects of transverse shear
Small deflection angles

Integration Methods

Elastic curve equation for constant $ E $ and $ I $: $ EIy'' = M(x) $ Differentiating both sides gives: $ EIy''' = \frac{dM(x)}{dx} = V(x) $ Differentiating again: $ EIy'''' = \frac{dV(x)}{dx} = w(x) $ In summary, we have:

$$ V(x) = \int w(x) dx\ $$

$$ M(x) = \int V(x)dx\ $$

$$ \frac{dy}{dx} = \int \frac{1}{EI}M(x)dx\ $$

$$ y(x) = \int y'(x) dx\ $$

Where: $ y(x): $ deflection $ y'(x): $ slope $ EIy''(x): $ bending moment $ EIy '''(x): $ shear force $ EIy''''(x): $ distributed load Example: Overhanging Beam

Taken from TAM251 Lecture Notes - L12S8

Example: Cantilever Beam

Taken from TAM251 Lecture Notes - L12S8

Beam Solutions

Common beam deflection solutions have been worked out.

Taken from the formula sheet

To solve loadings that are not in the table, use superposition to get the resulting deflection curve.

Example: Moment and distributed load

Taken from TAM251 Lecture Notes - L12S16

Solid Mechanics Reference