Simple Beam - Load Increasing Uniformly to Center

Written by Jerry Ratzlaff on 19 April 2018. Posted in Structural

\(\large{ R = V = \frac {W } {2} }\)

\(\large{ V_x \; }\) when \(\large{ \left( x < \frac {L}{2} \right) = \frac {W} {2L^2} \left( L^2 - 4x^2 \right) }\)

\(\large{ M_{max} }\) (at center) \(\large{ = \frac {W L} {6} }\)

\(\large{ M_x \; }\) when \(\large{ \left( x < \frac {L}{2} \right) = Wx \left( \frac {1}{2} - \frac {2x^2}{3L^2} \right) }\)

\(\large{ \Delta_{max} }\) (at center) \(\large{ = \frac {W L^3} {60 \lambda I } }\)

\(\large{ \Delta_x = \frac {W x} {480 \lambda I L^2} \left( 5L^2 - 4x^2 \right)^2 }\)

Where:

\(\large{ I }\) = moment of inertia

\(\large{ L }\) = span length of the bending member

\(\large{ M }\) = maximum bending moment

\(\large{ R }\) = reaction load at bearing point

\(\large{ V }\) = shear force

\(\large{ w }\) = load per unit length

\(\large{ W }\) = total load from a uniform distribution

\(\large{ x }\) = horizontal distance from reaction to point on beam

\(\large{ \lambda }\) (Greek symbol lambda) = modulus of elasticity

\(\large{ \Delta }\) = deflection or deformation