Solid Mechanics Reference

    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