Analysis of Thin Cylinder MCQ Quiz - Objective Question with Answer for Analysis of Thin Cylinder - Download Free PDF

Explanation:

Longitudinal Stress in a Thin Closed Cylinder:

• In the context of thin-walled pressure vessels, such as a thin closed cylinder containing a hydrostatic fluid, longitudinal stress refers to the stress experienced along the length of the cylinder due to the internal pressure exerted by the fluid. Understanding the nature of this stress is crucial for designing safe and efficient pressure vessels.
• When a thin-walled cylindrical vessel is subjected to internal hydrostatic fluid pressure, it experiences stress in both the longitudinal (axial) and circumferential (hoop) directions. The longitudinal stress is the stress along the length of the cylinder, and it is caused by the internal pressure pushing the ends of the cylinder apart.

In a thin closed cylinder with internal fluid pressure, longitudinal stress arises due to pressure acting on the end caps, trying to pull them apart.

Hence, the nature of longitudinal stress is tensile.

Formulas:

• Hoop stress: \( \sigma_h = \frac{p d}{2 t} \)
• Longitudinal stress: \( \sigma_l = \frac{p d}{4 t} \)

Explanation:

Given:

P = 0.2 MPa, D = 2 m, t = 10 mm

\(\sigma_{x}=\sigma_{L}=\frac{P D}{4 t}=\frac{0.2 \times 2000}{4 \times 10}=10 \mathrm{MPa}\)

\(\sigma_{y}=\sigma_{h}=\frac{P D}{2 t}=\frac{0.2 \times 2000}{2 \times 10}=20 \mathrm{MPa}\)

\(\text { Normal stress, }\left(\sigma_{n}\right)_{\theta}=\left(\frac{\sigma_{x}+\sigma_{y}}{2}\right)+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta+\tau_{x y} \sin 2 \theta\)

\(\left(\sigma_{n}\right)_{\theta=30^{\circ}}=\left(\frac{10+20}{2}\right)+\left(\frac{10-20}{2}\right) \cos \left(2 \times 30^{\circ}\right)=12.5 \mathrm{MPa}\)

Concept:

In the case of a thin cylinder, absolute shear stress is given by:

\(\tau_{max}=\frac{\sigma_1}{2}=\frac{pd}{4t}\)

In the case of a thin cylinder, in-plane of shear stress is given by:

\(\tau_{max}=\frac{\sigma_1 - \sigma_2}{2}=\frac{pd}{8t}\)

Calculation:

Given:

Pressure = 500 kPa, Inner radius = 2 m, Inner diameter = 4 m, Thickness = 10 mm.

In-plane shear stress:

\(\tau_{max}=\frac{\sigma_1 - \sigma_2}{2}=\frac{pd}{8t}\)

\(\tau_{max}=\frac{pd}{8t}\)

\(\tau_{max}=\frac{500\;\times\;10^3\;\times\;4\;\times\;10^3}{8\;\times\;10\;\times\;10^6}=25\;MPa\)

Explanation:

Hoop Stress in Thin Cylindrical Shells

When a thin cylindrical shell is subjected to internal pressure, the hoop stress is the stress that acts circumferentially in the wall of the cylinder. This stress is a result of the pressure trying to tear the cylinder apart in a circular manner. The hoop stress is given by the formula:

σ_h = (p * r) / t

Where:

• σ_h = Hoop stress

• p = Internal pressure

• r = Radius of the cylinder

• t = Thickness of the cylinder

Analyzing the Given Options

1. "It is uniform throughout." 

   • This is incorrect because the stress distribution in the cylindrical shell is not uniform. It varies from the inner surface to the outer surface.

2. "The mid-thickness." 

   • This is incorrect because the hoop stress is not at its maximum at the mid-thickness; it varies across the thickness of the shell.

3. "The outer surface." 

   • This is incorrect because the hoop stress is not maximum at the outer surface.

4. "The inner surface." 

   • This is correct because the hoop stress is maximum at the inner surface of the cylindrical shell. This is due to the fact that the internal pressure acts directly on the inner surface, causing the maximum stress at this point.

Concepts:

For thin spherical shell, the hoop stress and longitudinal stress is given as:

\(σ_h = σ_L = \frac{pd}{4t}\)

The hoop strain and longitudinal strains are given as:

\(ϵ_h = ϵ_L = \frac{pd}{4tE} (1-μ)\)

The volumetric strain is given as:

ϵV = 3 × ϵ­h

Concept:

Principal strain in the direction of σ₁ will be given as:

\({\epsilon_1} = \frac{{{\sigma _1}}}{E} - \mu \frac{{{\sigma _2}}}{E}\)

Principal strain in the direction of σ₂ will be given as:

\({\epsilon_2} = \frac{{{\sigma _2}}}{E} - \mu \frac{{{\sigma _1}}}{E}\)

Calculations:

Given:

As we know,

\({{\rm{\sigma }}_H} = \frac{{{\rm{pd}}}}{{2{\rm{t}}}} \Rightarrow {\rm{Hoop\;Stress}} = {\rm{Bring\;changes\;in\;diameter}}\)

\({{\rm{\sigma }}_L} = \frac{{{\rm{pd}}}}{{4{\rm{t}}}} \Rightarrow {\rm{Longitudinal\;Stress}} = {\rm{Bring\;changes\;in\;length}}\)

The longitudinal strain is:

\({\epsilon_L} = \frac{\delta{l}}{L}\Rightarrow\frac{{{\sigma _L}}}{E} - \mu \frac{{{\sigma _H}}}{E}\)

\(\therefore \frac{{\delta l}}{L} = \frac{{{\sigma _L}}}{E} - \mu \frac{{{\sigma _H}}}{E}\)

\(\frac{{\delta l}}{L} = \frac{{pd}}{{4tE}} - \mu \frac{{pd}}{{2tE}}\)

\(\frac{{\delta l}}{L} = \frac{{pd}}{{4tE}}\left( {1 - 2\mu } \right)\)
\(\therefore \delta l = \frac{{pdL}}{{2tE}}\left( {\frac{1}{2} - \mu } \right)\)

Concept:

For a thin cylinder:

Longitudinal stress: \({\sigma _L} = \frac{{pd}}{{4t}}\)

Hoop stress: \({\sigma _h} = \frac{{pd}}{{2t}}\)

Calculation:

Given:

Diameter, d = 200 cm = 2 m, Pressure head, h = 10,000 cm = 100 m, Thickness t = 0.75 cm = 0.0075 m

Both longitudinal and hoop stress are tensile in nature, But as per design consideration of thickness, we use hoop stress for calculation.

Pressure inside the cylinder, P = ρgh = 1000 × 10 × 100 = 10⁶ Pa

\({\sigma _h} = \frac{{pd}}{{2t}} \)

\({\sigma _h} = \frac{{10^6~\times~2}}{{2~\times~0.0075}} = 133.3~\times10^6\) = 133.3 MPa ∼ 130.5 MPa

Explanation:

For the thin spherical shell:

Hoop stress/Longitudinal stress:

\({\sigma _h} = {\sigma _L} = \frac{{pd}}{{4t}}\)

Hoop strain/longitudinal strain:

\({\epsilon_L} = {\epsilon_h} = \frac{{pd}}{{4tE}}\left( {1 - \mu } \right)\)

Volumetric strain:

\(\begin{array}{l} {\epsilon_V} = 3{\epsilon_h} = \frac{{3pd}}{{4tE}}\left( {1 - \mu } \right)\\ \frac{{{\epsilon_V}}}{{{\epsilon_D}}} = \frac{{{\epsilon_V}}}{{{\epsilon_h}}} = \frac{3}{1} \end{array}\)

Important Points
For thin-walled cylinder:

Circumferential stress/Hoop stress = \(\sigma _h=\frac{PD}{2t}\)

Longitudinal stress = \(\sigma _l=\frac{PD}{4t}\)

Concept:

• Circumferential stress is the stress acting along the circumferential direction, it is generally tensile in nature.
• Longitudinal stress is the stress which acts along the length and it is also tensile in nature whereas radial stress which acts in the direction of the radius is compressive in nature.

Circumferential stress/Hoop stress, \(\sigma _h=\frac{PD}{2t}\)

Longitudinal stress, \(\sigma _L=\frac{PD}{4t}\)

Where P = pressure due to flowing fluid, D = diameter, and t = thickness of shell

Calculation:

Ratio = \(\frac{\sigma _l}{\sigma _h}=\frac{PD/4t}{PD/2t}=\frac 12=0.5\)

Explanation

Maximum shear stress on the wall is maximum in-plane shear stress and is given by:

Max in Plane shear stress:

\(\begin{array}{l} \tau_{max}= \frac{{{\sigma _1} - {\sigma _2}}}{2}\\ {\sigma _1} = \frac{{Pd}}{{2t}};{\sigma _2} = \frac{{Pd}}{{4t}}\\ {\tau _{max}} = \frac{{\frac{{Pd}}{{2t}} - \frac{{Pd}}{{4t}}}}{2} = \frac{{Pd}}{{8t}} = \frac{{Pr}}{{4t}} \end{array}\)

Concept:

For a thin cylinder:

Longitudinal stress: \({\sigma _L} = \frac{{pd}}{{4t}}\)

Hoop stress: \({\sigma _h} = \frac{{pd}}{{2t}} = 2{\sigma _L}\)

Circumferential or hoop strain:

\({\epsilon_H} = \frac{1}{E}\left( {{\sigma _H} - ν {\sigma _L}} \right) = \frac{{{\sigma _L}}}{E}\left( {2 - ν } \right) = \frac{{pd}}{{4tE}}\left( {2 - ν } \right)\)

Longitudinal Strain:

\({\epsilon_L} = \frac{1}{E}\left( {{\sigma _L} - ν {\sigma _H}} \right) = \frac{{{\sigma _L}}}{E}\left( {1 - 2ν } \right) = \frac{{pd}}{{4tE}}\left( {1 - 2ν } \right)\)

Calculation:

Given:

Here ν = 1/m

Longitudinal strain \({{\epsilon }_{L}}=\frac{P d}{4tE}\left( 1-\frac{2}{m} \right)\)

Hoop strain, \({{\epsilon }_{h}}=\frac{P d}{4tE}\left( 2-\frac{1}{m} \right)\)

Concept:

When the thin cylinder is subjected to the internal pressure as we have already calculated that there is a change in the cylinder dimensions i.e., longitudinal strain and hoop strains come into the picture. Because of which there will be a change in the capacity of the cylinder or there is a change in the volume of the cylinder.

The volume of a cylinder:

\(V = \frac{\pi }{4}{d^2}L\)

\(\delta V = \frac{\pi }{4}{d^2} \times \delta L + \frac{\pi }{4}L \times 2d.\delta d\;\)

Volumetric strain = Change in volume/Original Volume

\({\epsilon_V} = \frac{{\delta V}}{V} = \frac{{\delta L}}{L} + 2\frac{{\delta d}}{d} = {\epsilon_L} + 2{\epsilon_h}\)

Concept:

• Consider a thin pressure vessel having closed ends and contains fluid under a gauge pressure P. Then the walls of the cylinder will have longitudinal stress, circumferential stress and radial stress.
• As shown in the figure, a point of the shell having stresses from all sides i.e. tri-axial stresses.
• σL =  longitudinal stress (tensile), σr = radial stress (compressive), σh = circumferential stress (tensile).

As, σr <<<< σL and σh, therefore we neglect σr and assumed the bi-axial stresses.

Circumferential or hoop stress: \({σ _h} = \frac{{Pd}}{{2t}}\)

Longitudinal or axial stress: \({σ _L} = \frac{{Pd}}{{4t}}\)

where d is the internal diameter and t is the wall thickness of the cylinder.

From the above-mentioned formulae, we can clearly see that Longitudinal stress is half of the circumferential stress

• For the spherical shell, longitudinal stress and circumferential stress both are equal,

σh = σL = \(\frac{Pd}{4t}\)

Concept:

Hoop stress is, \(\sigma _1=\frac{Pd}{2t}\)

Longitudinal stress is, \(\sigma _2=\frac{Pd}{4t}\)

Radial stress is, \(\sigma _3=0\)

And the Absolute maximum shear stress is, \(\tau _m=\frac{\sigma _1-\sigma _3}{2}=\frac{Pd}{4t}\)

Calculation:

Given:

d = 0.5 × 2 = 1 m = 1000 mm, P = 2 MPa, t = 0.52 - 0.5 = 0.02 m = 20 mm 

Therefore, \({\tau _{max}} = \frac{{2 ~\times~ 1000}}{{4~×~ 20}} = 25\;MPa\)

As, in option 25 MPa is not available so we need to choose more stress than 25 MPa.

Explanation:

When the thin cylinder is subjected to internal pressure, longitudinal strain and hoop strains come into the picture. Because of which there will be a change in the capacity of the cylinder or there is a change in the volume of the cylinder.

\(V = \frac{\pi }{4}{d^2}L\)

\(\delta V = \frac{\pi }{4}{d^2} \times \delta L + \frac{\pi }{4}L \times 2d.\delta d\;\)

Hoop strain is in the circumferential direction it is given by 

ϵh = \(\frac{{\delta d}}{d}\)

and linear strain is along the length of the cylinder.

ϵl = \(\frac{\delta l}{l}\)

Volumetric strain = Change in volume/Original Volume

\({\varepsilon _v} = \frac{{\delta V}}{V} = 2\frac{{\delta d}}{d} + \frac{{\delta l}}{l} = 2{\varepsilon _h} + {\varepsilon _l}\)

\(\epsilon_v = \frac{2}{E}\left( {{σ _h} - μ {σ _l}} \right) + \frac{1}{E}\left( {{σ _l} - μ {σ _h}} \right)\)

We khow that \(\sigma_l=\frac{\sigma_h}{2}\)

\(\epsilon_v = \frac{2}{E}\left( {{σ _h} - μ \frac{σ _h}{2}} \right) + \frac{1}{E}\left( {\frac{σ _h}{2} - μ {σ _h}} \right)\)

\(\epsilon_v=\frac{\sigma_h}{2E}(5-4\mu)\Rightarrow\frac{PD(5-4\mu)}{4tE}\)