# Problem 427 - Interior Members of Nacelle Truss by Method of Sections

**Problem 427**

Determine the force in bars BD, CD, and DE of the nacelle truss shown in Fig. P-427.

**Solution 427**

$9R_F + 3(60) = 9(120) + 12(120)$

$9R_F = 2340$

$R_F = 260 ~ \text{kN}$

**Section to the right of M-M**

$\Sigma F_V = 0$

$F_{DE}\left( \frac{1}{\sqrt{10}} \right) + 120 + 120 = 260$

$\frac{1}{\sqrt{10}}F_{DE} = 20$

$F_{DE} = 20\sqrt{10} ~ \text{kN} = 63.24 ~ \text{kN tension}$ *answer*

$\Sigma M_E = 0$

$6F_{CF} + 3(120) = 0$

$6F_{CF} = -360$

$F_{CF} = -60 ~ \text{kN} = 60 ~ \text{kN compression}$

$\Sigma F_H = 0$

$F_{BE} + F_{CF} + \frac{3}{\sqrt{10}}F_{DE} = 0$

$F_{BE} - 60 + \frac{3}{\sqrt{10}}(20\sqrt{10}) = 0$

$F_{BE} - 60 + 60 = 0$

$F_{BE} = 0$

**Section to the right of N-N**

$\Sigma M_C = 0$

$\left( \frac{3}{\sqrt{13}} F_{BD} \right)(6) + 260(9) = 120(9) + 120(12)$

$\frac{18}{\sqrt{13}} F_{BD} = 180$

$F_{BD} = 10\sqrt{13} ~ \text{kN} = 36.06 ~ \text{kN tension}$ *answer*

$\Sigma M_B = 0$

$\left(\frac{3}{5}F_{CD} \right)(6) + 120(9) + 120(12) = 60(6) + 260(9)$

$\frac{18}{5}F_{CD} = 180$

$F_{CD} = 50 ~ \text{kN tension}$ *answer*