The permissible angle of twist=1.30 in length of 6 m and the shear stress is not exceed 61 MPa. Take Modulus of Rigidity G=90 MPa.

Data: Diameter (d) = 260 mm.

$r = \frac{260}{2} = 130 \ mm$

Angle of twist $(\theta) = 1.3° = (1.3 \times \frac{\pi}{180})$ (radians)

Length (L) = 6m = 6000 mm

Shear stress $\zeta = 61 \ Mpa$ and G = 90 Mpa

$T_{max} = ?$

1] Torque from strength:

$\frac{T}{J_p} = \frac{\zeta}{r}$

$\frac{T}{\frac{\pi}{32} \times 260^4} = \frac{61}{130}$

$\therefore $ $T = 210.51 \times 10^6 \ Nmm$

2] Torque from stiffness:

$\frac{T}{J_p} = \frac{G \theta}{l}$

$\frac{T}{\frac{\pi}{32} \times 260^4} = \frac{90}{6000} \times (1.3 \times \frac{\pi}{180})$

$\therefore$ $T = 152.68 \times 10^3 \ Nmm$

$\therefore$ $T_{max}$ is minimum of above two values = $152.68 \times 10^3 \ Nmm$