Stress and deformation of a cylinder

[zDrajCa]

A hollow cylinder with thin wall, of internal radius r=30 mm, of thickness h=3 mm and subjected to a pressure p of 1 bar inside.

The radial stress road is equal to 0 on the surface. The surface of this solid is thus submitted to a state plan of stress.

1) Calculate the circonferencial and longitudinal stress and write them in the form of tensor of the constraints in the base(basis) (0, r, c, l) and to say if the latter is main.
2) A bow in 45 ° is stuck on the surface of this cylinder without particular precaution, such as the following drawing: (see attachment)

The measures are the following ones: Ja=33114.10 6, Jb=6701^10-6, Jc=4384.10^-6

Jc is on y and Ja is on x.

Determine the deformations(distortions) in the main base(basis) and the main angle.

3) Determine the coeff of fish and the module of elasticity.

-> I have right on the first question
[ 0 0 0 ]
[ 0 1 0 ] (MPa)
[ 0 0 0.5]

On the second I have found εx=Ja, εy=Jc, εxy=εb-(εa+εc)/2=-12048 x 10^-6.

The principal base : the angle is -20° (not sure : tg(2α)=2εxy/(εx-εy)

εI=37497 x 10^-6. εII= 0.466 x10^-6 Is it that ? (εI (or II)=(εa+εc)/2+( or -) sqrt((εa-εc)²+(εa+εc-2εb)²)

εIII we can't know.
εI=εc ? and εII=εl ?? It is that ?

3) haven't got idea

 

[KataKoniK]

Hello,

Thank you for your question. I would like to provide some clarification and calculations to address the questions presented in the forum post.

1) The circumferential stress (σc) and longitudinal stress (σl) can be calculated using the following equations:

σc = pr/t = (1 bar)(30 mm)/(3 mm) = 10 bar = 10 MPa
σl = pr/2t = (1 bar)(30 mm)/(2*3 mm) = 5 bar = 5 MPa

In tensor form, the stresses can be represented as:

[0 0 0]
[0 10 0] (MPa)
[0 0 5]

Since the stresses are non-zero in the circumferential and longitudinal directions, this is not a principal stress state.

2) To determine the deformations in the principal base (eigenbasis), we can use the equations:

εI = (εa + εc)/2 + √((εa - εc)^2 + εb^2)
εII = (εa + εc)/2 - √((εa - εc)^2 + εb^2)

Substituting the values from the given data, we get:

εI = (33114.10^-6 + 4384.10^-6)/2 + √((33114.10^-6 - 4384.10^-6)^2 + (-12048.10^-6)^2) = 37496.96.10^-6
εII = (33114.10^-6 + 4384.10^-6)/2 - √((33114.10^-6 - 4384.10^-6)^2 + (-12048.10^-6)^2) = 0.466.10^-6

Therefore, the principal deformation angles are given by:

tan(2α) = 2εxy/(εx - εy) = 2*(-12048.10^-6)/(33114.10^-6 - 4384.10^-6) = -0.571
α = -16.74°

3) The coefficient of Poisson (ν) can be calculated using the equation:

ν = -εy/εx = -4384.10^-6/33114.10^-6 = -0.132