MecMovies 4.0 : M5.5: Rod and Post in Series

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The figure shows the free body diagram around flange B. The free body diagram cuts through members 1 and 2. The load on the flange is P which equals 110 kilo newtons directed vertically downwards. The reaction force on member 1 is F subscript 1 directed vertically upwards. The reaction force on member 2 is F subscript 2 directed vertically downwards. The parameters corresponding to member 1 are A subscript 1, L subscript 1 and E subscript 1 and the parameters for member 2 are A subscript 2, L subscript 2 and E subscript 2.

Two equations have been derived for this structure. An equilibrium equation

$\sum_{} F_{y} = F_{1} - F_{2} - P = 0$

and a compatibility equation

$\frac{F_{1} L_{1}}{A_{1} E_{1}} + \frac{F_{2} L_{2}}{A_{2} E_{2}} = 0$

These two equations can be solved simultaneously to determine the internal forces in members $(1)$ and $(2)$ .