Strength of Materials Test 1

Concept:

\({{P}_{e}}=\frac{{{\pi }^{2}}E~{{I}_{min}}}{L_{e}^{2}}\)

\({{P}_{e}}\propto \frac{1}{L_{e}^{2}}\)

Calculation:

L₁ = L

L₂ = 3L

If other parameters are treated constant

\({{P}_{e}}\propto \frac{1}{{{\left( 3L \right)}^{2}}}\)

\(\frac{{{P}_{e}}_{2}}{{{P}_{e}}_{1}}=\frac{1{{L}^{2}}~}{9{{L}^{2}}}\)

\({{P}_{e2}}=\frac{1}{9}~{{P}_{e1}}\)

Concept:

Hoop Stress or Circumferential Stress:

It is the stress-induced along the circumference of a thin-walled tube or cylindrical shell when it is subjected to internal fluid pressure. It is always tensile in nature.

For thin cylindrical shell, the hoop stress or circumferential stress is given by, \({{\rm{\sigma }}_{\rm{h}}}{\rm{\;}} = {\rm{\;}}\frac{{{\rm{pd}}}}{{2{\rm{t}}}}\)

Where,

p = internal fluid pressure,

D = inner diameter of the cylindrical shell, and

t = Thickness of the cylindrical shell

Calculation:

Given: p = 2.5 MPa, (D/t) = 50

\(\therefore {\rm{\;Hoo}}{\rm{p\;Stress\;}}\left( {{{\rm{\sigma }}_{\rm{h}}}} \right) = \frac{{{\rm{pD}}}}{{2{\rm{t}}}} = \frac{{2.5}}{2} \times 50{\rm{\;MPa}} = 62.5{\rm{\;MPa}}\)

Mistake Point:

For the thin cylindrical shell, the axial stress or longitudinal stress is given by, \({{\rm{\sigma }}_{\rm{L}}}{\rm{\;}} = {\rm{\;}}\frac{{{\rm{pd}}}}{{4{\rm{t}}}}\)

Concept:

Change in length of a prismatic bar when subjected to load P is given by,

\(\Delta L=\frac{PL}{AE}\)

Where, A = cross-sectional area, E = Young’s modulus, L = initial length

Calculation:

Given, the solid and hollow bars of same length, made of same materials are subjected to same load, therefore, the elongation is the function of area only.

For solid bar

\(\Delta L=\frac{PL}{AE}\)

\(A=\frac{\pi }{4}{{d}^{2}}\)       ----  (1)

For hollow bar

\(A=\frac{\pi }{4}\left( {{D}^{2}}-{{\left( 0.5D \right)}^{2}} \right)=\frac{\pi }{4}~\left( 0.75{{D}^{2}} \right)\)       ----  (2)

Since they have same extensions

Equating (1) and (2)

\({{d}^{2}}=0.75{{D}^{2}}\) 

\(\frac{d}{D}=~\sqrt{3/4}=\frac{\sqrt{3}}{2}\)

Concept:

• Tensile failure in ductile materials:

• In ductile materials, shear strength is less than that of tensile strength. Generally, shear strength is 57% of tensile strength.
• Ductile material shows neck formation and breaks in cup and cone structure.

• Tensile failure in brittle materials:

• In brittle materials, tensile strength is less than that of shear strength.
• Brittle material fails in tension fracture without much changing cross-section area and with the rough surface perpendicular to the loading direction.

• Compression failure in ductile materials:

• Long compression ductile members fail in buckling and short members fail in compression yielding.
• Plane of failure is near to perpendicular to direction loading.

• Compression failure in brittle materials:

• In a compression test, a short member fails in shear and shear cracks are formed at 45° to the direction of loading.

Concept:

Power (P) = \(\frac{{2\pi NT}}{{60}}\)

T → Torgue (kN – m)

P → Power (kw)

Also,

\(\frac{{\rm{\tau }}}{{\left( {\left( {{{\rm{D}}_{\rm{o}}}} \right)/2} \right)}} = \frac{{\rm{T}}}{{{{\rm{I}}_{\rm{P}}}}}\)

τ → allowable shear stress in the shaft material, D_O → External diamer of shaft

I_P → Polar moment of Inertia of Shaft, Angular Velocity ω = \(\frac{{2\pi N}}{{60}}\), N → Rotation of Shaft (rpm)

ω → angular velocity (red/sec)

Calculations:

Given:

DO = 100mm

τ = 70 N/mm2

\({{\rm{I}}_{\rm{P}}} = \frac{{\rm{\pi }}}{{32}}\left( {{\rm{D}}_O^4 - {\rm{D}}_{in}^4} \right)\)

\({{\rm{I}}_{\rm{P}}} = \frac{{\rm{\pi }}}{{32}}\left( {{{100}^4} - {{50}^4}} \right)\)

I_P = 9.20 × 10⁶ mm⁴

Now,

\(\frac{{\rm{\tau }}}{{\left( {\left( {{{\rm{D}}_{\rm{o}}}} \right)/2} \right)}} = \frac{{\rm{T}}}{{{{\rm{I}}_{\rm{P}}}}}\)

\(\frac{{70}}{{\left( {\frac{{100}}{2}} \right)}} = \frac{T}{{9.20 \times {{10}^6}}}\)

T = 12880 N – m

∴ T = 12.88 kN – m

Now,

P = \(\frac{{2{\rm{\pi NT}}}}{{60}}\)

4000 = P = \(\frac{{2{\rm{\pi }} \times {\rm{N}} \times 12.88}}{{60}}\)

N = 2965.62 rpm

ω = \(\frac{{2{\rm{\pi N}}}}{{60}}\)

ω = \(\frac{{2{\rm{\;}} \times {\rm{\;\pi \;}} \times {\rm{\;}}2965.62}}{{60}}\)

∴ ω = 310.56 rad/sec

Concept:

For a Spherical pressure vessel:

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

Hoop stress: \({\sigma _h} = \frac{{pd}}{{4t}} \)

Both are the same for spherical pressure vessels.

Calculation:

Given: d = 500 mm, t = 10 mm, p = 4000 kPa, σy = 200 MPa

\({\sigma _1} = {\sigma _h} = \frac{{Pd}}{{4t}} = \frac{{4000 \times 500}}{{4 \times 10}} = 50000\;kPa\)

σ_y = 200 MPa = 200 × 10³ kPa

\(FOS = \frac{{{\sigma _y}}}{{{\sigma _h}}} = \frac{{200 \times {{10}^3}}}{{50 \times {{10}^3}}} = 4\)

Concept:

Change in volume = e_v × Original volume

\({{e}_{v}}=\frac{change~in~volume}{original~volume}=\frac{P}{K}\)

Where, P = pressure and K = Bulk Modulus

Calculation:

d = 50 mm, h = 750 m

Specific weight of seawater (γ) = 10.3 KN/m³

Pressure on sphere at depth of 750 m is given as,

P = γh

P = 10.3 × 750

∴ P = 7.725 MPa

Volumetric strain (e_v):

\({{e}_{v}}=\frac{change~in~volume}{original~volume}=\frac{P}{K}\)

\({{e}_{v}}=\frac{7.725}{200\times {{10}^{3}}}\)

\({{e}_{v}}=3.86\times {{10}^{-5}}a\)

\({e_v} = \frac{{change\;in\;volume}}{{original\;volume}}\)

\({{e}_{v}}=3.86\times {{10}^{-5}}\)

Now,

\(V = volume\;of\;sphere = \frac{4}{3}\pi {r^3} = 65449.85\;m{m^3}\)

Change in volume = e_v × Original volume

∴ Change in volume = 65449.85 × 3.86 ×10^-5

∴ Change in volume = 2.526 mm³

Concept:

Bending equation: \(\frac{M}{I} = \frac{{\rm{\sigma }}}{Y} = \frac{E}{R}\) 

\(\Rightarrow M = \frac{{EI}}{R}\)

Calculation:

Given: M_Al = M_Cu

\(\therefore {\left( {\frac{{EI}}{R}} \right)_{Al}} = {\left( {\frac{{EI}}{R}} \right)_{Cu}}\)

E_Al = 70 GPa, E_Cu = 120 GPa, R_Al = 2m, d_Al = 5 mm, d_Cu = 4 mm

For circular road, \(I = \frac{\pi }{{64}}{d^4}\) 

\(\therefore {\left( {\frac{{EI}}{R}} \right)_{Al}} = {\left( {\frac{{EI}}{R}} \right)_{Cu}}\)

\(\Rightarrow \frac{{70 \times \pi \times {5^4}}}{{64 \times 2}} = \frac{{120 \times \pi \times {4^4}}}{{64 \times {R_{cu}}}}\)

∴ R_cu = 1.404 m

Concept:

Hoop stress, in this case, is given as,

\({\sigma _h} = \;\frac{{E\left( {D - d} \right)}}{d}\)

Calculation:

Given:

Rigid wheel diameter (d) = 800 mm, Hoop stress (σ_h) = 120 MPa

Coefficient of thermal expansion (α) = 12 × 10^-6/℃, Modulus of elasticity (E) = 210 GPa

Now,

\({\sigma _h} = \frac{{E\left( {D - d} \right)}}{d} = 120\)

\(\therefore \frac{{D - d}}{d} = \frac{{120}}{{210 \times {{10}^3}}}\)

\(\therefore \frac{{D - d}}{d} = 5.71 \times {10^{ - 4}}\)

The perimeter of the wheel after an increase in temperature by ∆T is given as,

πD = πd (1 + α∆T)

\(\alpha \Delta T = \;\frac{{D - d}}{d} = 5.71 \times {10^{ - 4}}\)

\(\Delta T = \;\frac{{5.71 \times {{10}^{ - 4}}}}{{12 \times {{10}^{ - 6}}}}\)

∆T = 47.58 ℃

Concept:

Strain recorded by a gauge at an angle

\({\epsilon_n} = \;{\epsilon_x}{\cos ^2}\theta + \;{\epsilon_y}{\sin ^2}\theta + \;{\epsilon_{xy}}sin2\theta \)

Where,

\({\epsilon_x}\;and\;{\epsilon_y}\;are\;strain\;in\;x\;and\;y\;directions\)

\({\epsilon_{xy}} = \frac{{{\gamma _{xy}}}}{2},\;corresponds\;to\;shear\;strain\;\)

Calculation:

Angle made by diagonal with horizontal

\(Tan\theta = \frac{4}{6}\)

∴ θ = 33.69°

\({\epsilon_n} = 500\;\times \;{10^{ - 6}} \times \;0.6923 + 76.923 \times {10^{ - 6}}\)

∴ ϵ_n = 423.0768 × 10^-6

Now,

Change in diagonal = ϵ_n × length

\({\rm{Change\;in\;diagonal}} = \left[ {423.0768 \times {{10}^{ - 6}} \times \sqrt {{6^2} + {4^2}\;} \times 1000\;} \right]mm\)

∴ Change in diagonal = 3.05 mm