Problem 428 - Howe Truss by Method of Sections

$\Sigma M_L = 0$

$12R_A + 1(20 \sin \theta) + 2(20 \sin \theta) = 10(20 \cos \theta) + 8(20 \cos \theta)$
 

 

$12R_A + 1(20)\left( \frac{1}{\sqrt{5}} \right) + 2(20)\left( \frac{1}{\sqrt{5}} \right) = 10(20)\left( \frac{2}{\sqrt{5}} \right) + 8(20)\left( \frac{2}{\sqrt{5}} \right)$

$12R_A = \frac{660}{\sqrt{5}}$

$R_A = 11\sqrt{5} ~ \text{kN}$
 

By inspection:

At joint K, $JK = 0$

At joint J, $IJ = 0$

At joint I, $HI = 0$

At joint H, $GH = 0$

 

$\Sigma M_G = 0$

$6F_{DF}\left( \frac{1}{\sqrt{5}} \right) + 6\left( 11\sqrt{5} \right) + 1(20)\left( \frac{1}{\sqrt{5}} \right) + 2(20)\left( \frac{1}{\sqrt{5}} \right) = 4(20)\left( \frac{2}{\sqrt{5}} \right) + 2(20)\left( \frac{2}{\sqrt{5}} \right)$

$\frac{6}{\sqrt{5}}F_{DF} = -30\sqrt{5}$

$F_{DF} = -25 ~ \text{kN} = 25 ~ \text{kN compression}$       answer
 

 

$\Sigma M_A = 0$

$6F_{FG} = \sqrt{5}(20) + 2\sqrt{5}(20)$

$6F_{FG} = 60\sqrt{5}$

$F_{FG} = 10\sqrt{5} ~ \text{kN} = 22.36 ~ \text{kN tension}$       answer
 

$\Sigma M_F = 0$

$3F_{GI} + 2\sqrt{5}(20) + \sqrt{5}(20) = 6\left( 11\sqrt{5} \right)$

$3F_{GI} = 6\sqrt{5}$

$F_{GI} = 2\sqrt{5} ~ \text{kN} = 4.47 ~ \text{kN tension}$       answer