Analysis of Thin Sphere MCQ Quiz in मराठी - Objective Question with Answer for Analysis of Thin Sphere - मोफत PDF डाउनलोड करा

Concept:

Let the internal dia be D.

\(\therefore {\sigma _h} \times \pi Dt = {\rm{\Delta }}P \times \frac{\pi }{4}{D^2}\)

\( \Rightarrow {\sigma _h} = \frac{{PD}}{{4t}}\)

If we cut the sphere vertically we will get the same results, therefore hoop stress and longitudinal stress both are same.

\( \Rightarrow {\sigma _h}={\sigma _l} = \frac{{PD}}{{4t}}\)

Hoop strain:

\(\epsilon_h=\frac{\sigma_h}{E}-\mu\frac{\sigma_l}{E}\)

\({\varepsilon _h} = \frac{{dD}}{D} = \frac{{PD}}{{4tE}}\;\left( {1 - \mu } \right)\)

Volumetric strain:

\(V = \frac{4}{3}\pi \left( {\frac{{{D^3}}}{8}} \right) = \frac{{\pi {D^3}}}{6}\)

\(\Rightarrow {\varepsilon _v} = \frac{{dV}}{V} = \frac{{\left( {\frac{\pi }{6} \times3{D^2} dD} \right)}}{{\left( {\frac{{\pi {D^3}}}{6}} \right)}} = 3\frac{{dD}}{D} = 3{\varepsilon _h}\)

\(\therefore {\varepsilon _v} = 3{\varepsilon _h}\)

Calculation:

Given:

ε_h = 0.35, D = 70 mm

We know that;

\(\therefore {\varepsilon _v} = 3 \times {\varepsilon _h}\)

\( = 3 \times 0.35 = 1.05\)

D₁ = 1.1 D

T₁ = 0.9 t

\(\begin{array}{l} {\sigma _h} = \frac{{PD}}{{2t}}\\ {\sigma _{h1}} = \frac{{P{D_1}}}{{2{t_1}}} = \frac{{P \times 1.1D}}{{2 \times 0.9t}} = 1.22\left( {\frac{{PD}}{{2t}}} \right) \end{array}\)

% change in hoop stress = 22%

Concept:

Fig: Bursting of pressure vessel circumferentially when (σ_L > S_ut)

Fig: Bursting of pressure vessel longitudinally when (σ_h > S_ut)

Concept:

For a thin spherical pressure vessel, both longitudinal and hoop stress is the same and given as

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

Whereas for thin Cylindrical pressure vessel

Longitudinal Stress = \({\sigma _L} = \frac{{Pd}}{{4t}} \)

and Hoop Stress = \({\sigma _h} = \frac{{Pd}}{{2t}}\)

Calculation:

There, σ_t = σ_h = σ_l

Given, σ_t = 45 MPa, d = 2 m, t = 2 cm = 2 × 10^-2 m

\(\Rightarrow \frac{{P d}}{{4t}} = 45\)

Concept:

For thin spherical vessel:

For a thin spherical pressure vessel, both longitudinal and hoop stress is the same and given as

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

Calculation:

Given:

d = 2 m, σL = 500 kg/cm2, P = 50 kg/cm2, t = ?

\( {σ _L} = \frac{{pd}}{{4t}} \)

\(500 = \frac{{50 \times 2 \times 100}}{{4 \times t}} \)

t = 5 cm

The thickness of a shell to withstand a pressure of 50 kg/cm2 is 5 cm. 

Concept:

Thin spherical walls.

The hoop stress/longitudinal stress of a spherical shell is given by

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

The hoop strain / longitudinal strain of a spherical shell is given by

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

The volumetric strain is given by 3 times the hoop strain as the strain is same in all directions because sphere is same about all the axes. (Option 4)

\({\epsilon_V} = 3 \times {\epsilon_H} = \frac{{3pd}}{{4tE}}\left( {1 - \mu } \right)\)

Calculation:

Given:

P = 5 MPa; t = 10 mm; d = 400 mm; E = 20 GPa; μ = 0.3;

Now

\({\sigma _H} = \frac{{5\; \times \;400}}{{4\; \times \;10}} = 50\;MPa\)      (Option 1)

\({\varepsilon _H} = \frac{{pd}}{{4tE}}\left( {1 - \mu } \right) = \frac{{5\; \times \;400\; \times \;0.7}}{{4\; \times \;10\; \times \;200\; \times \;{{10}^3}}} = 1.75 \times {10^{ - 4}}\)

ϵ_V = 3 × ϵ_H = 3 × 1.75 × 10^-4 = 5.25 × 10^-4 (Option 3 wrong)

Due to symmetricity the strain of the sphere in all directions is same.(option 4)

For thin spherical vessel, the maximum hoop stress and maximum longitudinal stress is same.

\(\begin{array}{l} {\sigma _t} = \frac{{Pd}}{{4t}} \le 125\\ P \le 125 \times \frac{{4t}}{d} \end{array}\)

Or \(P \le 125 \times \frac{{4 \times 10}}{{20 \times 1000}}\)

Or P ≤ 0.25 MPa

Explanation:

Effect of Poisson’s Ratio on Longitudinal Strain in a Thin Sphere:

In mechanics of materials, the Poisson’s ratio (denoted as ν) is a fundamental material property that relates the longitudinal strain to the lateral strain in a material subjected to stress. It is defined as the negative ratio of the lateral strain to the longitudinal strain:

Poisson’s Ratio (ν) = - (Lateral Strain / Longitudinal Strain)

For a thin-walled spherical vessel subjected to internal pressure, the stresses induced are the hoop stress (circumferential stress) and the longitudinal stress. The longitudinal strain (ε_l) in such a structure is influenced by both the applied hoop stress and the material’s Poisson’s ratio. The relationship for longitudinal strain can be expressed as:

ε_l = (σ_h / E) × (1 - ν)

Where:

• ε_l = Longitudinal strain
• σ_h = Hoop stress
• E = Modulus of elasticity of the material
• ν = Poisson’s ratio

From the above equation, it is clear that the longitudinal strain is inversely proportional to Poisson’s ratio. This means that as the Poisson’s ratio increases, the longitudinal strain decreases. This happens because the material becomes less prone to elongation in the longitudinal direction as it exhibits more lateral contraction when subjected to the same stress.

Why is Option 3 Correct?

Option 3 states: “The longitudinal strain decreases”. This is the correct answer because, as derived above, an increase in Poisson’s ratio (ν) directly reduces the magnitude of the longitudinal strain (ε_l) in a thin spherical vessel. The material’s response to stress becomes more dominated by lateral contraction, leaving less deformation in the longitudinal direction.

Important Information:

Let us analyze why the other options are incorrect:

Option 1: Increases

This option is incorrect because the longitudinal strain is inversely related to Poisson’s ratio. An increase in Poisson’s ratio causes the material to exhibit greater lateral contraction and thus less longitudinal elongation, resulting in a decrease, not an increase, in longitudinal strain.

Option 2: Does not change

This option is incorrect because longitudinal strain is explicitly dependent on Poisson’s ratio as shown in the relationship: ε_l = (σ_h / E) × (1 - ν). Any change in Poisson’s ratio will directly affect the magnitude of longitudinal strain.

Option 4: Increases curvilinearly

This option is incorrect because the relationship between longitudinal strain and Poisson’s ratio is linear, not curvilinear. The term (1 - ν) ensures a direct linear dependency, where the strain decreases linearly as the Poisson’s ratio increases.

Concept:

The longitudinal and hoop stress in the thin spherical vessel is given by,

\({\varepsilon _H} = {\varepsilon _L}=\varepsilon = \frac{{pd}}{{4tE}}\left( {1 - \mu } \right)\)

Calculation

Given

d = 400 mm, t = 20 mm, E = 200 GPa, μ = 0.3

Maximum strain, ε_max = 100 × 10^-7

When strain in the sphere exceeds the maximum allowable strain, There will be a failure.

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

\(100 \times {10^{ - 7}} = \frac{{p \times 400}}{{4 \times 20 \times 200 \times {{10}^3}}}\left( {1 - 0.3} \right)\)

P = 0.57 MPa