# Beam Fixed at One End - Concentrated Load at Any Point

Written by Jerry Ratzlaff on . Posted in Structural

### Beam Fixed at One End - Concentrated Load at Center Formula

$$\large{ R_1 = V_1 = \frac {Pb^2} {2L^3} \left( a + 2L \right) }$$

$$\large{ R_2 = V_2 = \frac {Pa} {2L^3} \left( 3L^2 - a^2 \right) }$$

$$\large{ M_1 }$$  (at point of load)  $$\large{ = R_1 a }$$

$$\large{ M_2 }$$  (at fixed end)  $$\large{ = \frac {Pab} {2L^2} \left( a +L \right) }$$

$$\large{ M_x \; }$$  when $$\large{ \left( x < a \right) = R_1 x }$$

$$\large{ M_x \; }$$  when $$\large{ \left( x > a \right) = R_1 x - P \left( x - a \right) }$$

$$\large{ \Delta_{max} \; }$$  when  $$\large{ \left( a < .414L \right) \; }$$  at  $$\large{ L \frac { L^2 + a^2 } { 3L^2 - a^2 } = \frac {Pa} {3 \lambda I } \frac { \left( L^2 - a^2 \right) ^3 } { \left( 3L^2 - a^2 \right) ^2 } }$$

$$\large{ \Delta_{max} \; }$$  when  $$\large{ \left( a > .414L \right) \; }$$  at  $$\large{ L \sqrt{ \frac { a } { 2L + a } } = \frac {Pab^2} {6 \lambda I } \sqrt{ \frac { a } { 2L + a } } }$$

$$\large{ \Delta_a \; }$$  (at point of load)  $$\large{ = \frac { Pa^2 b^3} {12 \lambda I L^3} \left( 3L + a \right) }$$

$$\large{ \Delta_x \; }$$  when $$\large{ \left( x < a \right) = \frac { Pb^2 x} {12 \lambda I L^3} \left( 3aL^2 - 2Lx^2 - ax^2 \right) }$$

$$\large{ \Delta_x \; }$$  when $$\large{ \left( x > a \right) = \frac { Pa} {12 \lambda I L^3} \left( L - x \right)^2 \left( 3L^2 x - a^2 x - 2a^2 L \right) }$$

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

$$\large{ V }$$ = 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