# Problem 435 - Transmission Tower by Method of Sections

**Problem 435**

For the transmission tower shown in Fig. P-435, determine the force in member CJ.

**Solution 435**

## Click here to expand or collapse this section

$a = \frac{1}{2}[ \, 2(8.5) - 5 \, ] = 6 ~ \text{m}$

$b = \frac{1}{3}a = \frac{1}{3}(6) = 2 ~ \text{m}$

$c = \frac{1}{2}(5) + b = 2.5 + 2 = 4.5 ~ \text{m}$

Section Through M-M

$\Sigma M_G = 0$

$2(4.5 + 2)\left( \frac{3}{\sqrt{10}}F_{BC} \right) = 12(7.8)$

$F_{BC} = \frac{12}{5}\sqrt{10} ~ \text{kN tension}$

Section Through N-N

$\Sigma M_F = 0$

$9\left( \frac{4}{5} \right)F_{CJ} + 9\left( \frac{3}{\sqrt{10}} \right)\left( \frac{12}{5}\sqrt{10} \right) = 6(7.8)$

$F_{CJ} = -2.5 ~ \text{kN} = 2.5 ~ \text{kN compression}$ *answer*

- Log in to post comments