**Member ***JK*
Equation to use: Σ*M*_{C} = 0

Location of Unit Load = Point *C*

$U_{JK} = -\dfrac{ab}{Ld} = -\dfrac{8(8)}{16(8\sin 60^\circ)}$

$U_{JK} = -\sqrt{3}/3$

**Influence Diagram for Member ***JK*

$F_{JK} = 10\left[ \frac{1}{2}(16)\left( -\sqrt{3}/3 \right) \right] + 30\left( -\sqrt{3}/3 \right)$

$F_{JK} = -63.51 ~ \text{kN}$ ← [ B ] *answer for part 1*

**Member ***BC*

Equation to use: Σ*M*_{J} = 0

Location of Unit Load = Point *B*

$U_{BC} = +\dfrac{ab}{Ld} = \dfrac{4(12)}{16(8\sin 60^\circ)}$

$U_{BC} = \sqrt{3}/4$

**Influence Diagram for Member ***BC*

$F_{BC} = 10\left[ \frac{1}{2}(16)\left( \sqrt{3}/4 \right) \right] + 30\left( \sqrt{3}/4 \right)$

$F_{BC} = 47.63 ~ \text{kN}$ ← [ A ] *answer for part 2*

**Member ***CG*

Equation to use: Σ*F*_{V} = 0

Location of Unit Load = Point *B* (Within the Section)

$U_{CGv} = -\dfrac{a}{L}$

$U_{CG}\sin 60^\circ = -\dfrac{4}{16}$

$U_{CG} = -\sqrt{3}/6$

Location of Unit Load = Point *C* (Outside the Section)

$U_{CGv} = +\dfrac{b}{L}$

$U_{CG}\sin 60^\circ = \dfrac{8}{16}$

$U_{CG} = \sqrt{3}/3$

**Influence Diagram for Member ***CG*

$y = \sqrt{3}/3 - \left( -\sqrt{3}/6 \right) = \sqrt{3}/2$

$\dfrac{x - 8}{\sqrt{3}/3} = \dfrac{4}{\sqrt{3}/2}$ → $x = 32/3$

Maximum Compression Force:

$F_{CG} = 10\left[ \frac{1}{2}\left(16 - \frac{32}{3} \right)\left( -\sqrt{3}/6 \right) \right] + 30\left( -\sqrt{3}/6 \right)$

$F_{CG} = -16.36 ~ \text{kN}$

Maximum Tension Force:

$F_{CG} = 10\left[ \frac{1}{2}\left(\frac{32}{3} \right)\left( \sqrt{3}/3 \right) \right] + 30\left( \sqrt{3}/3 \right)$

$F_{BC} = 47.63 ~ \text{kN}$

Answer for part 3 = [ C ]