Beam Fixed at Both Ends - Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural Uniformly Distributed Load Formula

$$\large{ R = V = \frac {w L} {2} }$$

$$\large{ V_x = w \left( \frac {L} {2} - x \right) }$$

$$\large{ M_{max} }$$ (at ends)  =  $$\large{ \frac {w L^2} {12} }$$

$$\large{ M_1 }$$ (at center)  =  $$\large{ \frac {w L^2} {24} }$$

$$\large{ M_x = \frac {w}{12} \left( 6Lx - L^2 - 6x^2 \right) }$$

$$\large{ M_{max} }$$ (at center)  $$\large{ = \frac {w L^4} {384 \lambda I} }$$

$$\large{ \Delta_x = \frac {w x^2} {24 \lambda I} \left( L - x \right) ^2 }$$

$$\large{ x }$$ (points of contraflexure)  $$\large{ = \left( 0.2113 \right) L }$$

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