Problem 431 - Members in the Third Panel of a Parker Truss

$10.5(\frac{25}{\sqrt{634}}F_{DF}) + 22.5(472.5) = 15(135) + 7.5(135)$

$F_{DF} = -728.40 ~ \text{kN}$

$F_{DF} = 728.40 ~ \text{kN compression}$       answer
 

$\Sigma M_D = 0$

$9.6F_{EG} + 7.5(135) = 15(472.5)$

$F_{EG} = 632.81 ~ \text{kN tension}$       answer
 

$\Sigma F_V = 0$

$\frac{32}{\sqrt{1649}}F_{DG} + 2(135) = 472.5 + \frac{3}{\sqrt{634}}F_{DF}$

$\frac{32}{\sqrt{1649}}F_{DG} + 270 = 472.5 + \frac{3}{\sqrt{634}}(-728.40)$

$F_{DG} = 146.84 ~ \text{kN tension}$       answer
 

Checking:
$\Sigma F_H = 0$

$F_{EG} + \frac{25}{\sqrt{1649}}F_{DG} + \frac{25}{\sqrt{634}}F_{DF} = 0$

$632.81 + \frac{25}{\sqrt{1649}}(146.84) + \frac{25}{\sqrt{634}}(-728.40) = 0$

$0 = 0$       (check!)