# Problem 422 - Right-triangular Truss by Method of Sections

**Problem 422**

Refer to the truss described in Problem 412 and compute the force in members BD, CD, and CE by the method of sections.

**Solution 422**

$\Sigma M_F = 0$

$15R_A = 12(400 + 200) + 6(800 + 200)$

$R_A = 880 ~ \text{kN}$

**From section through MN**

$\Sigma M_C = 0$

$6(\frac{2}{\sqrt{5}}F_{BD}) = 3(880)$

$F_{BD} = 491.93 ~ \text{kN}$ compression *answer*

$\Sigma M_F = 0$

$12(\frac{1}{\sqrt{5}}F_{CD}) + 12(400 + 200) = 15(880)$

$F_{CD} = 1118.03 ~ \text{kN}$ compression *answer*

$\Sigma M_D = 0$

$3F_{CE} + 6(400 + 200) = 9(880)$

$F_{CE} = 1440 ~ \text{kN}$ tension *answer*