**Problem 423**

Use the method of sections to determine the force acting in members DF, EF, and EG of the Howe truss described in Problem 409.

**Solution 423**

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$\Sigma M_A = 0$

$12R_H = 3(2.7) + 6(4.5) + 9(1.8)$

$R_H = 4.275 ~ \text{kN}$

From the section to the right of M-N

$\Sigma M_E = 0$

$6(F_{DF} \sin 30^\circ) + 3(1.8) = 6(4.275)$

$F_{DF} = 6.75 ~ \text{kN compression}$ *answer*

$\Sigma M_H = 0$

$6(F_{EF} \sin 30^\circ) = 3(1.8)$

$F_{EF} = 1.8 ~ \text{kN compression}$ *answer*

$y = 3 \tan 30^\circ = \sqrt{3}$

$\Sigma M_F = 0$

$y F_{EG} = 3(4.275)$

$\sqrt{3} F_{EG} = 3(4.275)$

$F_{EG} = 7.404 ~ \text{kN tension}$ * answer*