Strength of Materials Test 6

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:

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:

Slenderness Ratio, \(S=\frac{effective~length~of~column}{Radius~of~Gyration}=\frac{L_e}{k}\)

Radius of Gyration \(k=\sqrt{\frac{{{I}_{min}}}{A}}\)

Calculation:

Length of column = 5 m

Side of the square (b) = 50 mm = 0.05 m 

\(\therefore k=\sqrt{\frac{{{b}^{4}}}{12\times {{b}^{2}}}}=\frac{b}{\sqrt{12}}\)

Now, given condition is both ends fixed

\(\therefore L_e=\frac{L}{2}\)

\(\Rightarrow S=\frac{L_e}{k}\)

Concept:

\({{P }_{e}}=\frac{{{\pi }^{2}}E{{I}_{min}}}{L{{e}^{2}}}\Rightarrow {{P}_{e}}\propto {{I}_{min}}\)

\({{I}_{min}}=\frac{\pi }{64}~{{d}^{4}}\Rightarrow {{I}_{min}}\propto {{d}^{4}}\)

\(\therefore \frac{{{P}_{2}}}{{{P}_{1}}}=\frac{{{I}_{2}}}{{{I}_{1}}}={{\left( \frac{{{d}_{2}}}{{{d}_{1}}} \right)}^{4}}\)

Calculation:

And d₂: 0.75 d₁

\(\therefore \frac{{{P}_{2}}}{{{P}_{1}}}=0.3164\)

Concept:

\({\rm{Hoop\;stress}},{\rm{\;}}({{\rm{\sigma }}_{\rm{h}}}) = \frac{{{\rm{Pd}}}}{{2{\rm{t}}}}\)

\({\rm{Longitudinal\;stress}}\left( {{{\rm{\sigma }}_{\rm{c}}}} \right) = \frac{{{\rm{Pd}}}}{{4{\rm{t}}}}\)

Calculation:

Considering plane stress conditions, we have

\({\rm{Hoop\;stress}},{\rm{\;}}({{\rm{\sigma }}_{\rm{h}}}) = \frac{{{\rm{Pd}}}}{{2{\rm{t}}}}\)

\({\rm{Hoop\;stress}},{\rm{\;}}({{\rm{\sigma }}_{\rm{h}}}) = \frac{{1 \times 10 \times 1000}}{{2 \times 30}}\)

∴ Hoop stress, σ_h = 166.67 MPa

Now,

\({\rm{Longitudinal\;stress}}\left( {{{\rm{\sigma }}_{\rm{c}}}} \right) = \frac{{{\rm{Pd}}}}{{4{\rm{t}}}}\)

\({\rm{Longitudinal\;stress}}\left( {{{\rm{\sigma }}_{\rm{c}}}} \right) = \frac{{1 \times 10 \times 1000}}{{4 \times 30}}\)

∴ Longitudinal stress (σ_c) = 83.33 MPa

Now,

Maximum shear stress in-plane is given as,

\({\tau _{max}} = \frac{{{\sigma _h} - {\sigma _l}}}{2}\)

τ_max = 41.67

Concept:

Euler buckling load for a column is given by,

\({{\rm{P}}_{\rm{E}}}{\rm{\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}_{\rm{e}}^2}}\)

Where, Le is the effective length of the column that depends on the end support conditions, and EI is the flexural rigidity of the column.

Now, effective length (L_e) for different end conditions in terms of actual length (L) are listed in the following table:

Calculation:

Given, Euler’s buckling load is 50 kN for a strut with one end fixed and the other end is free.

∴ L_e = 2L

\(50{\rm{\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{{\left( {2{\rm{L}}} \right)}^2}}}{\rm{\;\;}}\therefore \frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{{\rm{L}}^2}}}{\rm{\;}} = {\rm{\;}}50 \times 4{\rm{\;}} = {\rm{\;}}200\)

For the second case, the end conditions are given as both ends are fixed.

∴ L_e = L/2

\({{\rm{P}}_{\rm{E}}}{\rm{\;}} = {\rm{\;}}\frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{{\left( {\frac{{\rm{L}}}{2}} \right)}^2}}}{\rm{\;}} = {\rm{\;}}4 \times \frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{{\rm{L}}^2}}}{\rm{\;}} = {\rm{\;}}4 \times 200{\rm{\;}} = {\rm{\;}}800{\rm{\;kN}}\)

∴ The Euler buckling load for the strut with both ends fixed is 800 kN.

Note: Actual length = 5 m is not required for solving this question.

Concept:

For a thin cylinder:

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

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

Longitudinal Strain:

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

Circumferential or hoop strain:

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

Calculation:

Given, l = 2.5 m = 2500 mm, d = 1.25 m = 1250 mm, t = 20 mm

P = 1.5 MPa, E = 200 GPa = 200 × 10³ MPa, μ = 0.3

In cylindrical shell Hoop strain \({{\epsilon }_{H}}=\frac{\delta d}{d}=\frac{Pd}{4tE}~\left( 2-\mu \right)\)

\(\Rightarrow \delta D=\frac{P{{d}^{2}}}{4tE}\left( 2-\mu \right)\)

\(\delta D=\frac{1.5\times {{1250}^{2}}}{4\times 20\times 200\times {{10}^{3}}}\left( 2-0.3 \right)\)