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:

Strain energy is defined as the capacity of energy absorption of a material when it is subjected to strain.

Mathematically,

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.

Since stress is in x direction, longitudinal strain will develop in x direction. and lateral strain will be in y direction. Poisson's ratio= lateral strain/ longitudinal strain. i.e. 

€v = 0. (1-2¥) × (£x + £y + £z) / E = 0.

Since body is constrained in all directions there will be equal stress in all directions.

£x = £y = £z

1-2¥ = 0

¥ = 1/2 = 0.5

Concept:

For a bi-axial state of stress normal stress at any plane is given by,

\({{\rm{\sigma }}_{\rm{n}}} = \frac{{{{\rm{\sigma }}_{\rm{x}}} + {{\rm{\sigma }}_{\rm{y}}}}}{2} + \frac{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}{2}\cos 2{\rm{\theta }} + {{\rm{\tau }}_{{\rm{xy}}}}\sin 2{\rm{\theta }}\)

Where is the normal stress in a plane which is inclined at an angle θ with the horizontal axis;

\({{\rm{\sigma }}_{\rm{x}}}{\rm{\;and\;}}{{\rm{\sigma }}_{\rm{y}}}\) are the normal stress in the direction of X-axis and Y-axis respectively;

\({{\rm{\tau }}_{{\rm{xy}}}}\) is the shear stress acting in the XY plane.

Tensile normal stress is taken as positive, compressive normal stress is taken as negative, and anticlockwise shear stress is taken as positive.

Calculations:

Given, \({\rm{\;}}{{\rm{\sigma }}_{\rm{x}}} = 100{\rm{\;N}}/{\rm{m}}{{\rm{m}}^2}\;and\;{{\rm{\sigma }}_{\rm{y}}} = 0\;and\;{{\rm{\tau }}_{{\rm{xy}}}} = 0\)

The angle 45° is given with the vertical. But from geometry it is found that the angle with the horizontal direction is also 45°. So, θ = 45°.

\(\therefore {{\rm{\sigma }}_{\rm{n}}} = \frac{{100 + 0}}{2} + \frac{{100 - 0}}{2}\cos 2 \times 45 + 0 \times \sin 2 \times 45 = 50{\rm{\;N}}/{\rm{m}}{{\rm{m}}^2}\)

Important Point:

For a bi-axial state of stress shear stress along any plane is given by,

\({{\rm{\tau }}_{\rm{n}}} = - \frac{{{{\rm{\sigma }}_{\rm{x}}} - {{\rm{\sigma }}_{\rm{y}}}}}{2}\sin 2{\rm{\theta }} + {{\rm{\tau }}_{{\rm{xy}}}}\cos 2{\rm{\theta }}\)

Where, \({{\rm{\tau }}_{\rm{n}}}\) is the shear stress along a plane which is inclined at an angle θ with the horizontal axis;

\({{\rm{\sigma }}_{\rm{x}}}{\rm{\;and\;}}{{\rm{\sigma }}_{\rm{y}}}\) are the normal stress in the direction of X axis and Y axis respectively;

\({{\rm{\tau }}_{{\rm{xy}}}}\) is the shear stress acting in XY plane.

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

Proof resilience:

The strain energy stored in a body due to external loading, within the elastic limit, is known as resilience and the maximum energy which can be stored in a body up to the elastic limit is called proof resilience.

Modulus of resilience:

Modulus of resilience is defined as proof resilience per unit volume. It is the area under the stress-strain curve up to the elastic limit.

Toughness:

It is defined as the ability of the material to absorb energy before fracture takes place.

• This property is essential for machine components which are required to withstand impact loads.
• Tough materials have the ability to bend, twist or stretch before failure takes place.
• Toughness is measured by a quantity called modulus of toughness. Modulus of toughness is the total area under the stress-strain curve in a tension test.

Flexure rigidity:

• It is a measure of the resistance of a beam to bending, that is, the larger the flexural rigidity, the smaller the curvature for a given bending moment.
• EI = Flexural rigidity

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:

From the bending equation, \({\bf{\sigma }} = \frac{{{\bf{My}}}}{{{{\bf{I}}_{{\bf{NA}}}}}}\)

where M = Moment applied at the cross-section, y = Distance of fiber from the neutral axis, I_NA = M.I. about neutral axis

So, it is clear from the bending equation that the stress is directly proportional to the distance of fiber from the neutral axis and does not depend on other parameters.

Hence the bending stress will vary linearly from top to bottom fiber and will be zero at the neutral axis.

Also for other cross-sections, bending stress varies linearly:

Tensile stress in the pipe wall = Circumferential stress in pipe wall = Pd/2t

Where,

Concept:

The both wires having equal deformation due to load of 20 kN.

Therefore, δ_Al = δ_St

And the deformation is given as, \(\delta=\frac{PL}{AE}\)

Calculation:

Given:

L_Al = L_St = 2 m, A_Al = 120 × 10^-6 m², E_Al = 70 GPa, A_St = 60 × 10^-6 m², E = 200 GPa 

\( \Rightarrow \frac{{{P_{Al}}{L_{Al}}}}{{{A_{Al}}{E_{Al}}}} = \frac{{{P_{st}}{L_{st}}}}{{{A_{st}}{E_{st}}}}\)

\(\frac{{{P_{Al}}}}{{{P_{st}}}} = \frac{{{L_{st}}}}{{{L_{Al}}}} × \frac{{{A_{Al}}}}{{{A_{st}}}} × \frac{{{E_{Al}}}}{{{E_{st}}}}\)

\(\frac{P_{Al}}{P_{St}} = \frac{2}{2} × \frac{{120 × {{10}^{ - 6}}}}{{60 × {{10}^{ - 6}}}} × \frac{{70}}{{200}}=\frac{7}{10}\)....................(1)

But we know that, P_Al + P_St = 20..........................(2)

From equation (1), \(P_{St}=\frac{10}{7}P_{Al}\)

Therefore, from equation (1) and (2), we will get,

\(P_{Al}+\frac{10}{7}P_{Al}=20\)

\({P_{Al}} = \frac{{140}}{{17}}~kN\)

\({\sigma _{Al}} = \frac{{{P_{Al}}}}{{{A_{Al}}}} = \frac{{140 × {{10}^3}}}{{17 × 120 × {{10}^{ - 6}}}}\)

\(\sigma _{Al} = \frac{{14}}{{12 × 17}} × {10^9}\frac{N}{{{m^2}}}\)

\({\sigma _{Al}} = \frac{7}{{6 × 17}}~GPa = \frac{7}{{102}}~GPa\)

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

• The volumetric strain in a thin cylindrical shell due to internal pressure is the change in volume per unit volume.


• Consider the hoop stress, σₕ = (p * d) / (2 * t), and longitudinal stress, σₗ = (p * d) / (4 * t).


• Hoop strain, εₕ = σₕ / E - μ * σₗ / E.


• Longitudinal strain, εₗ = σₗ / E - μ * σₕ / E.


• Volumetric strain, εᵥ = εₕ + 2 * εₗ.


• Substitute values, εᵥ = (p * d) / (2 * t * E) * (3 - 2μ).


• Correct answer: Option C.

Concept:

Since the bars are in parallel:

\({\delta _{ad1}} + {\delta _{th1}} = {\delta _{ad2}} + {\delta _{th2}}\)

∴ \({\delta _{th1}} - {\delta _{th2}} = {\delta _{ad1}} - {\delta _{ad2}}\)

where, δ_th1 and δ_th2 are the change in the length of copper bar and steel bar respectively due to thermal stress.

And δ_ad1 and δ_ad2 are the change in length due to axial stress in Cu bar and steel bar respectively.

The deformation of the bar is given as, 

α_cL_cT - α_sL_sT = \(\frac{{{{\rm{\sigma }}_c}{{\rm{L}}_c}}}{{{{\rm{E}}_c}}} + \frac{{{{\rm{\sigma }}_s}{{\rm{L}}_s}}}{{{{\rm{E}}_s}}}\)     …(1)

Calculation:

Given:

L_c = L_s = 1 m, T = 80, E_c = 1 × 10⁵ N/mm², E_s = 2 × 10⁵ N/mm²

Assuming coefficient of thermal expansion ,

α_c = 16 × 10^-6/˚C, α_s = 12 × 10^-6/˚C  

Put in equation (1),

80 × (16 – 12)× 10^-6 = \(\frac{{{{\rm{\sigma }}_c}}}{{1 × {{10}^{ - 5}}}} + \frac{{{{\rm{\sigma }}_s}}}{{2 × {{10}^{ - 5}}}}\)

2σ_c + σ_s = 64     … (2)

Also since The stress in the bar is \(\sigma=\frac{P}{A}\)

\(\frac{{{{\rm{\sigma }}_{\rm{c}}}}}{{{{\rm{\sigma }}_{\rm{s}}}}} = \frac{{{{\rm{A}}_{\rm{s}}}}}{{{{\rm{A}}_{\rm{c}}}}}\)......(Here, P is the same for both bars)

\(\frac{{{{\rm{\sigma }}_c}}}{{{{\rm{\sigma }}_s}}} = \frac{{600}}{{200}} = 3\)

σ_c = 3σ_s     …(3)

Solving (2) and (3), we get

σ_s = 9.14 MPa (Tensile)

σ_c = 27.42 MPa (compressive)

Note:

Since the coefficient of thermal expansion not mentioned answer differs from the option. But as we know Copper tube has a greater coefficient of thermal expansion so compressive stress will be developed and steel having a lesser coefficient of thermal expansion will have tensile stresses developed.