# Three Member Frame - Fixed/Fixed Side Uniformly Distributed Load

Written by Jerry Ratzlaff on . Posted in Structural

## Three Member Frame - Fixed/Fixed Side Uniformly Distributed Load formulas

 $$\large{ e = \frac{h}{L} }$$ $$\large{ \beta = \frac{I_h}{I_v} }$$ $$\large{ R_A = R_D = \frac{ w\;h\; \beta\; e^2 }{ 6\; \beta \;e\;+\;1 } }$$ $$\large{ H_A = \frac{w\;h}{4} \; \left( \frac{8\; \beta\;e\;+\;17 }{2\; \left( \beta\;e\;+\;2 \right) } - \frac{4\; \beta\;e\;+\;3 }{6\; \beta\;e\;+\;1 } \right) }$$ $$\large{ H_D = \frac{w\;h}{4} \; \left( \frac{4\; \beta\;e\;+\;3 }{6\; \beta\;e\;+\;1 } - \frac{1 }{2\; \left( \beta\;e\;+\;2 \right) } \right) }$$ $$\large{ M_A = \frac{w\;h^2}{4} \; \left( \frac{4\; \beta\;e\;+\;1 }{6\; \beta\;e\;+\;1 } - \frac{\beta \;e \;+\; 3 }{6\; \left( \beta\;e\;+\;2 \right) } \right) }$$ $$\large{ M_B = \frac{w\;h^2 \; \beta \;e}{4} \; \left( \frac{ 6 }{6\; \beta\;e\;+\;1 } - \frac{ 1 }{6\; \left( \beta\;e\;+\;2 \right) } \right) }$$ $$\large{ M_C = \frac{w\;h^2 \; \beta \;e}{4} \; \left( \frac{ 2 }{6\; \beta\;e\;+\;1 } - \frac{ 1 }{6\; \left( \beta\;e\;+\;2 \right) } \right) }$$ $$\large{ M_D = \frac{w\;h^2}{4} \; \left( \frac{4\; \beta\;e\;+\;1 }{6\; \beta\;e\;+\;1 } - \frac{\beta \;e \;+\; 3 }{6\; \left( \beta\;e\;+\;2 \right) } \right) }$$

### 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

$$\large{ A, B, C, D, E }$$ = 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