**Problem 418**

The Warren truss loaded as shown in Fig. P-418 is supported by a roller at C and a hinge at G. By the method of sections, compute the force in the members BC, DF, and CE.

**Solution 418**

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$\Sigma M_G = 0$

$12R_C + 6(60) = 6(100) + 9(80) + 18(40)$

$R_C = 140 ~ \text{kN}$

**At section through M-M**

$\frac{2}{\sqrt{5}}F_{BC} = 40$

$F_{BC} = 44.721 ~ \text{kN}$ compression *answer*

**At section through N-N**

$6F_{CE} + 9(40) = 3(140)$

$F_{CE} = 10 ~ \text{kN}$ tension *answer*

$6F_{DF} + 3(80) + 12(40) = 6(140) + 6(60)$

$F_{DF} = 80 ~ \text{kN}$ compression *answer*