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

scene 9 of 14

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.

Solve for $F_{1}$ :

$F_{1} - F_{2} - P = 0$

$F_{1} = F_{2} + P$

and substitute the expression for $F_{1}$ into the compatibility equation:

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