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

Fig: signConvention1

From ref pages

Boundary Conditions

Fig: BoundryConditions

From ref pages

Moment-Curvature Equation

Elastic Curve of a Beam:

Fig: ElasticCurveBeam

Taken from TAM251 Lecture Notes - L12S3

Moment-curvature equation:

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

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

Fig: ElasticCurveGraph

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

Fig: Overhang Ex

Taken from TAM251 Lecture Notes - L12S8

Example: Cantilever Beam

Fig: Cantilever Ex

Taken from TAM251 Lecture Notes - L12S8

Beam Solutions

Common beam deflection solutions have been worked out.

Fig: Common Solutions

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

Fig: Superposition Ex

Taken from TAM251 Lecture Notes - L12S16

TAM 2xx References