Pressure Vessels MCQ Quiz - Objective Question with Answer for Pressure Vessels - Download Free PDF

Concept:

We use strain relations in a thin cylindrical shell to determine the ratio of longitudinal strain to volumetric strain under internal pressure.

Given:

• Diameter of the cylinder, \( d \)
• Thickness of the cylinder, \( t \)
• Internal pressure, \( P \)
• Poisson's ratio, \( \frac{1}{m} \)

Step 1: Calculate Hoop Stress and Longitudinal Stress

Hoop stress, \( \sigma_h = \frac{P d}{2 t} \)

Longitudinal stress, \( \sigma_l = \frac{P d}{4 t} \)

Step 2: Calculate Longitudinal Strain

Longitudinal strain, \( \epsilon_l = \frac{\sigma_l}{E} - \frac{1}{m} \cdot \frac{\sigma_h}{E} \)

Substituting the stresses:

\( \epsilon_l = \frac{P d}{4 t E} \left(1 - \frac{2}{m}\right) \)

Step 3: Calculate Hoop Strain

Hoop strain, \( \epsilon_h = \frac{\sigma_h}{E} - \frac{1}{m} \cdot \frac{\sigma_l}{E} \)

Substituting the stresses:

\( \epsilon_h = \frac{P d}{4 t E} \left(2 - \frac{1}{m}\right) \)

Step 4: Calculate Volumetric Strain

Volumetric strain, \( \epsilon_v = \epsilon_l + 2 \epsilon_h \)

Substituting the strains:

\( \epsilon_v = \frac{P d}{4 t E} \left(1 - \frac{2}{m} + 4 - \frac{2}{m}\right) = \frac{P d}{4 t E} \left(5 - \frac{4}{m}\right) \)

Step 5: Determine the Ratio of Longitudinal Strain to Volumetric Strain

Ratio, \( \frac{\epsilon_l}{\epsilon_v} = \frac{\frac{P d}{4 t E} \left(1 - \frac{2}{m}\right)}{\frac{P d}{4 t E} \left(5 - \frac{4}{m}\right)} = \frac{m - 2}{5m - 4} \)

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: \( \sigma_l = \frac{p d}{4 t} \)

Concept:

In a thin-walled cylinder under internal pressure, two principal stresses act on the wall: hoop stress and longitudinal stress.

Given:

• Hoop stress, σ_h = 40 MPa
• Longitudinal stress, σ_l = 20 MPa
• Radial stress = 0 MPa (negligible in thin cylinders)

Formula for Maximum Absolute Shear Stress:

\(\tau_{\text{max}} = \frac{1}{2}(σ_{\text{max}} - σ_{\text{min}})\)

Calculation:

\(\tau_{\text{max}} = \frac{1}{2}(40 - 0) = 20~\text{MPa}\)

Concept:

We use strain relations in a thin cylindrical shell to determine the increase in capacity under internal pressure.

Given:

• Internal diameter, \( D = 800 \, \text{mm} \)
• Internal radius, \( r = \frac{D}{2} = 400 \, \text{mm} \)
• Wall thickness, \( t = 10 \, \text{mm} \)
• Internal pressure, \( p = 1.5 \, \text{MPa} \)
• Modulus of elasticity, \( E = 200 \times 10^3 \, \text{MPa} \)
• Poisson's ratio, \( \nu = 0.3 \)
• Initial volume, \( V = 1 \, \text{m}^3 = 1 \times 10^9 \, \text{mm}^3 \)

Step 1: Calculate circumferential (hoop) stress and strain

\( \sigma_c = \frac{pr}{t} = \frac{1.5 \times 400}{10} = 60 \, \text{MPa} \)

Longitudinal stress, \( \sigma_l = \frac{pr}{2t} = \frac{1.5 \times 400}{2 \times 10} = 30 \, \text{MPa} \)

Hoop strain, \( \varepsilon_c = \frac{\sigma_c}{E} - \nu \times \frac{\sigma_l}{E} = \frac{60}{200 \times 10^3} - 0.3 \times \frac{30}{200 \times 10^3} \)

\( \varepsilon_c = 3 \times 10^{-4} - 0.045 \times 10^{-3} = 2.55 \times 10^{-4} \)

Step 2: Calculate longitudinal strain

\( \varepsilon_l = \frac{\sigma_l}{E} - \nu \times \frac{\sigma_c}{E} = \frac{30}{200 \times 10^3} - 0.3 \times \frac{60}{200 \times 10^3} \)

\( \varepsilon_l = 1.5 \times 10^{-4} - 0.9 \times 10^{-4} = 0.6 \times 10^{-4} \)

Step 3: Calculate volumetric strain

\( \varepsilon_v \approx 2 \varepsilon_c + \varepsilon_l = 2 \times 2.55 \times 10^{-4} + 0.6 \times 10^{-4} = 5.7 \times 10^{-4} \)

Step 4: Calculate increase in volume

\( \Delta V = \varepsilon_v \times V = 5.7 \times 10^{-4} \times 1 \times 10^9 = 570000 \, \text{mm}^3 \)

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:

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:

Hoop stress:

\({\sigma _1} ={\sigma _h} = \frac{{pd}}{{2t}} =\sigma \)

Longitudinal stress:

\({\sigma _2} ={\sigma _L} = \frac{{pd}}{{4t}} =\frac{\sigma}{2} \)

As this is the case of a thin cylinder:

\({\sigma _r} =0\)

All these above stress can be considered principal stresses as no shear stress is acting.

Maximum shear stress,

= \(\tau _{max} = max\ [\frac{{\sigma _{max}}\;-\;{\sigma _{min}}}{{2}}=\frac{{\sigma _1}\;-\;0 }{2} =\frac{{pd}}{{4t}} \)

\(\tau _{max} = max\ [| \frac{{\sigma _{1}}\;-\;{\sigma _{2}}}{{2}}|, | \frac{{\sigma _{2}}\;-\;{\sigma _{3}}}{{2}}|,| \frac{{\sigma _{3}}\;-\;{\sigma _{1}}}{{2}}|]\)

\(\Rightarrow\tau_{max} = max\ [\frac{\sigma }{4}, \ \frac{\sigma }{4}, \ \frac{\sigma }{2}] \)

.\(\Rightarrow \tau_{max} = \frac{\sigma}{2}\)

\(\Rightarrow {\tau _{max}}= \frac{{pd}}{{4t}} \)

Mistake Points 

• Whenever we are told maximum shear stress at any point or in-plane shear stress there we have to ignore the third principal stress and the value of maximum shear stress will be pd/8t.
• When we asked absolute maximum shear stress then the third principal stress will also be in there, whose value will be zero. In that case the maximum shear stress is pd/4t and in this case, the following analysis will be done.

Concept:

Hoop stress (σ_h) for a thin spherical shell is given by:

\({{σ _h}} = \frac{{Pd}}{{4t}}=\frac{{Pr}}{{2t}}\)

Hoop stress (σh) for thin cylindrical shell is given by:

\({{σ _h}} = \frac{{Pd}}{{2t}}=\frac{{Pr}}{{t}}\)

Calculation:

Given:

Now r^' = 1.01r, t^' = 0.99t

\({{σ _h'}} = \frac{{Pd'}}{{4t'}}=\frac{{Pr'}}{{2t'}}= \frac{{P\times 1.01r}}{{2\times 0.99t}}=\frac{{1.01}}{{0.99}}\times \frac{{P r}}{{2 t}}=1.0202 \;σ_h\)

Percentage change in hoop stress is:

\(\% \;change = \frac{{\sigma _h' - {\sigma _h}}}{{{\sigma _h}}} \times 100 = \frac{{1.0202\sigma _h - {\sigma _h}}}{{{\sigma _h}}} \times 100 = 2.02\% \)

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 = \(\sigma _l=\frac{PD}{4t}\)

Explanation

Let εL and εh be the longitudinal strain and hoop strain and μ is the Poisson ratio and E is the modulus of elasticity.

We know,

Longitudinal strain:

\(\begin{array}{l} {\varepsilon _L} = \frac{{pd}}{{4tE}} - \mu\frac{{pd}}{{2tE}} \\ {\varepsilon _L} = \frac{{pd}}{{4tE}}\left( {{1} - 2\mu } \right) \end{array}\)

Circumferential/Hoop strain:

\(\begin{array}{l} {\varepsilon _h} = \frac{{pd}}{{2tE}} - \mu\frac{{pd }}{{4tE}}\\ {\varepsilon _h} = \frac{{pd}}{{2tE}}\left( {1 - \frac{1}{2}\mu } \right)\\ {\varepsilon _h} = \frac{{pd}}{{4tE}}\left( {2 - \mu } \right) \end{array}\)

Now,

Hoop strain = change in circumference or circumferential original circumference strain.

\({\varepsilon _h} = \frac{{\pi δ d}}{{\pi d}} = \frac{{δ d}}{d}\) (δd ischange in diameter)

And also, the longitudinal strain is \(= \frac{{δ \ell }}{\ell } = {\varepsilon _L}\)

We know for a cylinder Volumetric strain = ϵ_l + 2ϵ_h

\(\frac{\delta V}{V} = \frac{{pd}}{{4tE}}\left( {{1} - 2\mu } \right)\;+\;2\times\frac{{pd}}{{4tE}}\left( {2 - \mu } \right)\)

\(\frac{\delta V}{V} = \frac{{pd}}{{4tE}}\left( {{1} - 2\mu \;+\;4\;-\;2\mu} \right)\)

\(\frac{\delta V}{V} = \frac{{pd}}{{4tE}}\left( {5\;-\;4\mu} \right)\)

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.

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

σh = σL = \(\frac{Pd}{4t}\)

Concept:

Thin spherical Shell:

The hoop stress of a spherical shell is given by

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

The hoop strain of a spherical shell is given by

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

Calculation:

Given:

P = 2 MPa; t = 0.25 cm = 2.5 mm, d = 15 cm = 150 mm

Now

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

\({\sigma _L} = {\sigma _H} = \frac{{2 \times 150}}{{4\times 2.5}}=30~MPa\)      

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:

Thin pressure vessel:

If the Diameter of the vessel is greater than or equal to 20 times the thickness of the vessel then the vessel is called a thin cylinder.

\(\frac{D}{t}\geqslant 20\) 

where,

D = diameter of the cylinder, t = thickness of the cylinder.

A thin pressure vessel having closed ends and contain fluid under gauge pressure ‘P’. Then the walls of the cylinder will have longitudinal stress as well as circumferential or hoop stress.

Circumferential or hoop stress: 

​\({\sigma _h} = \frac{{PD}}{{2t}}=X\) ..... (1)

Longitudinal or axial stress:

\({\sigma _L} = \frac{{PD}}{{4t}}=Y\) ......(2)

X =  2Y or Y = X/2