# Stress and deformation of a cylinder

• zDrajCa
In summary: The modulus of elasticity (E) can be calculated using the equation:E = σx/εx = 10 MPa/33114.10^-6 = 302.17 GPaIn summary, we calculated the circumferential and longitudinal stresses of a hollow cylinder under pressure, determined the deformations in the principal base and angle, and calculated the coefficient of Poisson and modulus of elasticity.
zDrajCa
Moved from technical physics section, so missing the HW template
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

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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

## 1. What is stress and deformation of a cylinder?

Stress is the force applied to a material per unit area, while deformation is the change in shape or size of the material due to the applied stress. In the case of a cylinder, the stress and deformation are caused by external forces acting on the cylinder's surface.

## 2. How is stress and deformation calculated for a cylinder?

The stress and deformation of a cylinder can be calculated using the formula σ = F/A, where σ is the stress, F is the applied force, and A is the cross-sectional area of the cylinder. The deformation can be calculated using the formula δ = FL/EA, where δ is the deformation, F is the applied force, L is the length of the cylinder, and E is the elastic modulus of the material.

## 3. What factors affect the stress and deformation of a cylinder?

The stress and deformation of a cylinder are affected by various factors, including the material properties of the cylinder, the shape and size of the cylinder, and the magnitude and direction of the applied forces. The cylinder's geometry, such as its length and diameter, also plays a role in determining the stress and deformation.

## 4. How does stress and deformation impact the structural integrity of a cylinder?

Excessive stress and deformation can cause structural failure in a cylinder. If the stress exceeds the material's yield strength, it can result in permanent deformation or even fracture. The deformation can also lead to changes in the cylinder's mechanical properties, making it more susceptible to failure under stress.

## 5. Can stress and deformation be reduced in a cylinder?

There are various methods to reduce stress and deformation in a cylinder, including changing the material properties, altering the cylinder's shape and size, and applying external supports or restraints. The selection of appropriate materials and design considerations can help minimize stress and deformation in a cylinder.