Problem 430 - Parker Truss by Method of Sections | Engineering Mechanics Review at MATHalino

Problem 430
The loads on the Parker truss shown in Fig. P-430, determine the forces in members BD, BE, CE, and DE.

Solution 430

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$R = \frac{1}{2}(135 \times 7) = 472.5 ~ \text{kN}$

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

Problem 430 - Parker Truss by Method of Sections