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

Written by Jerry Ratzlaff on . Posted in Structural

### Three Member Frame - Fixed/Fixed Top Uniformly Distributed Load Formula

$$\large{ e = \frac{h}{L} }$$

$$\large{ \beta = \frac{I_h}{I_v} }$$

$$\large{ R_A = R_E = \frac{ w\;L }{ 2 } }$$

$$\large{ H_A = H_E = \frac{w\;L^2}{4\;h\; \left( \beta\;e\;+\;2 \right) } }$$

$$\large{ M_A = M_E = \frac{w\;L^2}{12\;h\; \left( \beta\;e\;+\;2 \right) } }$$

$$\large{ M_B = M_D = \frac{w\;L^2}{6\;h\; \left( \beta\;e\;+\;2 \right) } }$$

$$\large{ M_C = \frac{w\;L^2}{24} \; \left( \frac{3\; \beta\;e\;+\;2 }{ \beta\;e\;+\;2 } \right) }$$

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