# Two Member Frame - Fixed/Pin Side Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural Engineering

## Two Member Frame - Fixed/Pin Side Uniformly Distributed load formulas

 $$\large{ e = \frac{h}{L} }$$ $$\large{ \beta = \frac{I_h}{I_v} }$$ $$\large{ R_A = R_C = \frac{w\;h}{4} \; \left( \frac{\beta\;e^2 }{3\; \beta\;e \;+\; 4 } \right) }$$ $$\large{ H_A = \frac{w\;h}{2} \; \left( \frac{3\; \beta\;e \;+\; 5 }{3\; \beta\;e \;+\; 4 } \right) }$$ $$\large{ H_C = \frac{3\;w\;h}{2} \; \left( \frac{ \beta\;e \;+\; 1 }{3\; \beta\;e \;+\; 4 } \right) }$$ $$\large{ M_A = \frac{w\;h^2}{4} \; \left( \frac{ \beta\;e \;+\; 2 }{3\; \beta\;e \;+\; 4 } \right) }$$ $$\large{ M_B = \frac{w\;h^2}{4} \; \left( \frac{\beta\;e }{3\; \beta\;e \;+\; 4 } \right) }$$

### Where:

$$\large{ h }$$ = height of frame

$$\large{ H }$$ =  horizontal reaction load at bearing point

$$\large{ w }$$ = load per unit length

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

$$\large{ A, B, C }$$ = points of intersection on frame

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

$$\large{ I }$$ = second moment of area (moment of inertia)

$$\large{ I_h }$$ = horizontal second moment of area (moment of inertia)

$$\large{ I_v }$$ = vertical second moment of area (moment of inertia)

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