Strength of Materials Test 2

Concept:

Volumetric strain \(= \frac{{{\sigma _x} + {\sigma _y} + {\sigma _z}}}{E}\left( {1 - 2\mu } \right)\)

For uniaxial, volumetric strain \(= \frac{\sigma }{E}\left( {1 - 2\mu } \right) = e\left( {1 - 2\mu } \right)\)

Calculation:

Given that:

σ₁ = 200 MPa ; σ₂ = 150 MPa, σ₃ = 200 MPa

E = 200 kN/mm² and μ  = 0.3

Volumetric strain, \({\varepsilon _v} = \left( {\frac{{{\sigma _1}\; + \;{\sigma _2}\; + \;{\sigma _3}}}{E}} \right)\left( {1 - 2\mu } \right)\)

\({\varepsilon _v} = \left( {\frac{{200\; + \;150\; + \;200}}{{210 \times {{10}^3}}}} \right)\left( {1 - 2 \times 0.3} \right)\)

\(= \left( {\frac{{2.62}}{{{{10}^3}}}} \right)\left( {0.4} \right) = 1.05 \times {10^{ - 3}}\)

∴ ε_v = 1.05 × 10^-3

Concept:

\(R = \;\frac{{EI}}{M}\;\; - - - \left( 1 \right)\)

Here, R = Radius of curvature of the beam

E = Elastic modulus, I = Area moment of inertia, M = Bending moment

We also know,

\({\rm{Shear\;force\;}}\left( {{\rm{SF}}} \right) = \;\frac{{dM\;}}{{dx\;}}\;\;\; - - - \;\left( 2 \right)\)

Option (a):

Let M = 0, then SF = 0, from equation (2)

From (1), R = Infinity, which implies shape is a straight line.

Option (b):

Let M = Constant, then SF = 0, from equation (2) i.e

From (1), R = Constant, which implies shape is a circular arc.

Option (c):

Let M = Variable w.r.t x, then SF exists according to (2)

Since R is varying, the shape is parabolic.

Option (d):

When bending Moment (M) is variable the shear force will exist and hence this option is incorrect.

Explanation:

From bending equation

\(\frac{σ }{y} = \frac{E}{R}\)

So, here

 \({σ _{max}} = \frac{{E{y_{max}}}}{{\left( {R + \frac{t}{2}} \right)}}\)

\({\rm{E}} = {\rm{\;}}70{\rm{\;GPa}},{\rm{\;\;}}{y_{max}} = \frac{t}{2} = \frac{1}{2}\;mm\)

\({σ _{max}} = \frac{{70\; \times \;{{10}^3}\; \times \;\left( {0.5} \right)}}{{500\; + \;0.5}}\)

∴ σ_max = 69.99 MPa

Concept:

\({\rm{Modulus\;of\;Elasticity}}\left( {\rm{E}} \right){\rm{\;}} = \;\frac{{Yield\;stress\left( {{\sigma _y}} \right)}}{{Yeild\;strain\left( {{ \in _y}} \right)}}\)

Calculation:

Given:

σ_e = 250 N/mm²

∈_e = 0.12%

For mild steel material, elastic stress and yield stress are close to each other and can be taken as equal. Similarly, elastic strain and yield strain are equal.

σ_e = σ_y = 250 N/mm²

∈_e = ∈_y = 0.12%

\(E = \frac{{{\sigma _y}}}{{{ \in _y}}} = \frac{{250}}{{0.0012}} = 2.0833 \times {10^5}N/m{m^2}\)

∴ E = 208.33 GPa

Concept:

Hoop strain in case of cylinder is given by:

\({\epsilon _{\rm{H}}} = \frac{{P \times {d}}}{{4 \times t \times E}} \times \left( {2 - \mu } \right)\)

Calculation:

Given:

t = 10 mm = 10 × 10^-3 m, Internal diameter (d) = 1.20 m, Internal pressure (P) = 1.5 MPa = 1.5 × 10⁶ Pa, Poisson’s ratio (μ) = 0.30

Modulus of elasticity (E) = 200 GPa = 200 × 10⁹ Pa

Hoop strain is:

\({\epsilon_H} = \frac{{{\rm{\Delta }}d}}{d} = \frac{{P \times {d}}}{{4 \times t \times E}} \times \left( {2 - \mu } \right)\)

\({\rm{\Delta }}d = \frac{{P \times {d^2}}}{{4 \times t \times E}} \times \left( {2 - \mu } \right)\)

\( {\rm{\Delta }}d= \frac{{1.5 \times {{10}^6} \times {{1.2}^2} \times \left( {2 - 0.3} \right)}}{{4 \times 10 \times {{10}^{ - 3}} \times 200 \times {{10}^9}}} = 4.59 \times {10^{ - 4}}\;m = 0.459\;mm\)

Concept:

As load is suddenly applied, the maximum stress produced is double the stress due to gradual loading.

\(Maximum\;stress = \;\sigma = \frac{{2P}}{A}\)

Calculation:

\(Area = \;\frac{\pi }{4} \times {50^2} = 1963.5\;m{m^2}\)

\(Maximum\;stress = \;\sigma = \frac{{2 \times 80000\left( N \right)}}{{1963.5\;(m{m^2})}} = 81.487\;\left( {\frac{N}{{m{m^2}}}} \right)\)

Maximum instantaneous elongation is given by,

\(\delta L = \;\frac{\sigma }{E} \times L\)

\(\delta L = \;\frac{{81.487\left( {\frac{N}{{m{m^2}}}} \right)}}{{200 \times 1000\left( {\frac{N}{{m{m^2}}}} \right)}} \times 4000\left( {mm} \right)\)

Concept:

\({\rm{Brinell\;hardness\;number\;}}\left( {{\rm{BHN}}} \right){\rm{\;}} = {\rm{\;}}\frac{{2P}}{{\pi D\left( {D - \;\sqrt {{D^2}\; - {d^2}\;} } \right)}}\)

P = load in kg

D = steel ball diameter

d = indentation diameter

Calculation:

Given:

P = 240 kg

D = 7 mm

d = 3 mm

Now, substituting the values in the formula

\({\rm{Brinell\;hardness\;number\;}}\left( {{\rm{BHN}}} \right){\rm{\;}} = {\rm{\;}}\frac{{2P}}{{\pi D\left( {D - \;\sqrt {{D^2}\; - {d^2}\;} } \right)}}\)

\(\therefore {\rm{Brinell\;hardness\;number\;}}\left( {{\rm{BHN}}} \right){\rm{}} = {\rm{}}\frac{{2\; \times \;240}}{{\pi\; \times \;7\left( {7\; - \;\sqrt {{7^2}\; - \;{3^2}\;} } \right)}}\)

\(\therefore {\rm{Brinell\;hardness\;number\;}}\left( {{\rm{BHN}}} \right){\rm{}} = {\rm{}}\frac{{480}}{{\pi\; \times \;7\left( {7\; - \;\sqrt {49 - 9\;} } \right)}}\)

\(\therefore {\rm{Brinell\;hardness\;number\;}}\left( {{\rm{BHN}}} \right){\rm{}} = \frac{{480}}{{\pi\; \times \;7\; \times \;\left( {7 - \;\sqrt {40} } \right)}}\)

∴ BHN = 32.31