# Simple Beam - Central Point Load and Variable End Moments

Written by Jerry Ratzlaff on . Posted in Structural

### Simple Beam - Central Point Load and Variable End Moments Formula

(Eq. 1)  $$\large{ R_1 = V_1 = \frac { P } { 2 } + \frac { M_1 - M_2 } { L } }$$

(Eq. 2)  $$\large{ R_2 = V_2 = \frac { P } { 2 } - \frac { M_1 - M_2 } { L } }$$

(Eq. 3)  $$\large{ M_3 }$$  (at center)  $$\large{ = \frac { PL } { 4 } - \frac { M_1 + M_2 } { L } }$$

(Eq. 4)  $$\large{ M_x \left( x < \frac{L} {2} \right) = \left( \frac { P } { 2 } + \frac { M_1 - M_2 } { L } \right) x - M_1 }$$

(Eq. 5)  $$\large{ M_x \left( > \frac{L} {2} \right) = \frac {P}{2} \left( L - x \right) + \frac { \left( M_1 - M_2 \right) x } { L } - M_1 }$$

(Eq. 6)  $$\large{ \Delta_x \left( x < \frac{L} {2} \right) = \frac { Px } { 48 \lambda I } \left[ 3L^2 - 4x^2 - \frac { 8 \left( L - x \right) } { PL } \left[ M_1 \left( 2L - x \right) + M_2 \left( L + x \right) \right] \right] }$$

(Eq. 7)  $$\large{ x }$$  (first point of contraflexure)  $$\large{ = \frac { 2L M_1 } { LP + 2M_1 - 2M_2 } }$$

(Eq. 8)  $$\large{ x }$$  (second point of contraflexure)  $$\large{ = \frac { L \left( LP - 2M_1 \right) } { LP + 2M_1 + 2M_2 } }$$

Where:

$$\large{ I }$$ = moment of inertia

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

$$\large{ M }$$ = maximum bending moment

$$\large{ P }$$ = total concentrated load

$$\large{ R }$$ = reaction load at bearing point

$$\large{ V }$$ = maximum shear force

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

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

$$\large{ \Delta }$$ = deflection or deformation