$$\sigma_h = \fracP \cdot Rt, \quad \sigma_a = \fracP \cdot R2t$$
Unlike a segmented pipe, a continuous tube has no axial stress discontinuity at joints, so no stress concentration factor ($K_t \approx 1$) in the longitudinal direction. This significantly improves burst strength and fatigue life. When a continuous tube is bent to a radius $R_b$, the bending strain is:
Bending strain: $\epsilon = \fracD2R_b = \frac25.42 \times 1200 = 0.01058$ (1.06%)
$$\epsilon = \fracD2 R_b$$