Answer:
41.015
Step-by-step explanation:
Solution
Given that:
Th first step to take is to find the longitudinal stress in the cylinder
σl = PD/4t
P = the pressure
D = the diameter
t = the thickness
Thus,
σl = 1.25 * ^ 6 * 0.45 / 4 * 6 * 10 ^ ⁻3
=23.475Mpa
Now. we find the hoop stress in the cylinder∠
σh = PD/2t
σh = 1.25 * ^ 6 * 0.45/ 2 * 6 * 10 ^ ⁻3
σh =46.875 Mpa
Then
we find the normal stress in the line of the 30° angle with the longitudinal axis stated below:
σab = σh + σl/2 + ( σh - σl/2) cos 2θ + t sin 2θ
So,
σab =46.875 + 23.4375/2 + ( 46.875 - 23.4375/2) cos 2(30°) + 0
σab= 41.015
Therefore the normal stress to line A-B is 41.015