# Two Member Frame - Fixed/Fixed Top Point Load

Written by Jerry Ratzlaff on . Posted in Structural

## two Member Frame - Fixed/Fixed Top Point Load formulas

 $$\large{ e = \frac{h}{L} }$$ $$\large{ \beta = \frac{I_h}{I_v} }$$ $$\large{ R_A = \frac{P\;x^2}{2\;L^3 \; \left( \beta \; e \;+\; 1 \right) } \; \left[ \beta \; e \; \left( 3 \; L - x \right) + 2 \;\left( 3 \; L - 2 \; x \right) \right] }$$ $$\large{ R_D = P - R_A }$$ $$\large{ H_A = H_D = \frac{3\;P\;x^2}{2\;h\;L^2} \; \left( \frac{L\;-\;x}{ \beta\;e \;+\; 1} \right) }$$ $$\large{ M_A = \frac{P\;x^2}{2\;L^2} \; \left( \frac{L\;-\;x}{ \beta\;e \;+\; 1} \right) }$$ $$\large{ M_B = \frac{P\;x^2}{L^2} \; \left( \frac{L\;-\;x}{ \beta\;e \;+\; 1} \right) }$$ $$\large{ M_C = R_B\;x - M_B }$$ $$\large{ M_D = \frac{P\;x \;\left( L \;-\; 1 \right) }{ 2\;L^2 } \; \left( \frac{ \beta\;e\;\left( 2\;L \;-\; x \right) \;+\;2\;\left( L \;-\; x \right) }{\beta \;e\;+\;1} \right) }$$

### Where:

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

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

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

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

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

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