Moment of guy wire forces about tower base

Assumptions from the figure: the tower is along z, height 600 m, top T(0, 0, 600). The three wires are anchored on the ground at 300 m from O, 120° apart in plan. The y axis is collinear with AO, so A is on the −y axis: A(0, −300, 0). Wire C is 30° above the x axis in plan, and B is symmetric with it: C(300 cos 30°, 300 sin 30°, 0) = (259.8, 150, 0), B(−259.8, 150, 0). Tensions: T_A = T_B = 60 kN, T_C = 30 kN.

Length of each wire

L = √(300² + 600²) = 670.8 m

Wire forces at the top (direction from top to anchor)

F_A = 60 (0, −300, −600)/670.8      = (  0.00, −26.83, −53.67) kN
F_B = 60 (−259.8, 150, −600)/670.8  = (−23.24,  13.42, −53.67) kN
F_C = 30 (259.8, 150, −600)/670.8   = ( 11.62,   6.71, −26.83) kN

Resultant: ΣF = (−11.62, −6.71, −134.16) kN

Moment about the base O

All forces act at the top, r = (0, 0, 600) m, so M_O = r × ΣF = (0, 0, 600) × (−11.62, −6.71, −134.16)

Mx = 0×(−134.16) − 600×(−6.71)  = +4025 kN·m
My = 600×(−11.62) − 0×(−134.16) = −6971 kN·m
Mz = 0

|M_O| = √(4025² + 6971²) = 8050 kN·m

Answer: M_O = 4025 i − 6971 j kN·m (magnitude 8050 kN·m). The moment vector lies in the horizontal xy plane, so there is no twisting moment about the tower axis.