Support Reactions
$R = \frac{1}{2}(135 \times 7) = 472.5 ~ \text{kN}$
From the Section to the Left of M-M
$\Sigma M_E = 0$
$9.6(\frac{5}{\sqrt{26}}F_{BD}) + 15(472.5) = 7.5(135)$
$F_{BD} = -645.34 ~ \text{kN}$
$F_{BD} = 645.34 ~ \text{kN compression}$ answer
$\Sigma M_B = 0$
$8.1F_{CE} = 7.5(472.5)$
$F_{CE} = 437.5 ~ \text{kN tension}$ answer
$\Sigma F_V = 0$
$\frac{27}{\sqrt{1354}}F_{BE} + 135 = 472.5 + \frac{1}{\sqrt{26}}F_{BD}$
$\frac{27}{\sqrt{1354}}F_{BE} + 135 = 472.5 + \frac{1}{\sqrt{26}}(-645.34)$
$F_{BE} = 287.48 ~ \text{kN tension}$ answer
Checking:
$\Sigma F_H = 0$
$F_{CE} + \frac{25}{\sqrt{1354}}F_{BE} + \frac{5}{\sqrt{26}}F_{BD} = 0$
$437.5 + \frac{25}{\sqrt{1354}}(287.48) + \frac{5}{\sqrt{26}}(-645.34) = 0$
$0 = 0$ (check!)
From Joint E
$\Sigma F_V = 0$
$F_{DE} + \frac{27}{\sqrt{1354}}F_{BE} = 135$
$F_{DE} + \frac{27}{\sqrt{1354}}(287.48) = 135$
$F_{DE} = -75.94 ~ \text{kN}$
$F_{DE} = 75.94 ~ \text{kN compression}$ answer
