**Problem 431**

Determine the force in the members DF, DG, and EG for the Parker truss shown in Fig. P-430.

**Solution 431**

## Click here to show or hide the solution

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

From the Section to the Left of N-N

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

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

$F_{DF} = 728.40 ~ \text{kN compression}$ *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!*)