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.

Explanation:

The engineering stress-strain curve for mild steel is

Poisson's ratio is the ratio of transverse strain to longitudinal strain.

For perfectly isotropic elastic material, Poisson’s ratio is 0.25 but for most of the materials, the value of Poisson's ratio lies in the range of 0 to 0.5.

Poisson's ratio for various materials are:

• Cork: 0.0
• Aluminium: 0.31
• Cast iron: 0.21 – 0.26
• Steel: 0.27 – 0.30
• Stainless steel: 0.30 – 0.31
• Copper: 0.33
• Rubber: 0.5

A stress-strain curve represents the relationship between the stress applied to a material and the resulting strain (deformation) experienced by the material. The curve can be divided into several regions, each corresponding to a different behavior of the material under stress.

1. Proportionality limit: This is the region where the stress-strain curve is linear, and the material follows Hooke's Law, which states that the stress is proportional to the strain. In this region, the material will return to its original shape and size when the stress is removed.

2. Elastic limit: This is the point on the stress-strain curve immediately after the proportionality limit. Up to the elastic limit, the material will still return to its original shape and size when the stress is removed, but the relationship between stress and strain is no longer linear. Beyond the elastic limit, the material will enter the plastic region, where it will experience permanent deformation even when the stress is removed.

3. Lower yield point: This is the point on the stress-strain curve where the material starts to yield or undergo plastic deformation. The material will not return to its original shape and size when the stress is removed at this point.

4. Upper yield point: This is the point on the stress-strain curve where the material has reached its maximum resistance to plastic deformation. Beyond this point, the material will continue to deform with little or no increase in stress.

5. Ultimate point: This is the point on the stress-strain curve where the material experiences its maximum stress before failure. Beyond this point, the material will begin to fracture and eventually break under the applied stress.

Since the elastic limit occurs immediately after the proportionality limit, it is the correct answer.

• For longitudinal strain we need Young's modulus and for calculating transverse strain we need Poisson's ratio.
• We may calculate Poisson's ratio from E = 2G(1+μ) for that we need Shear modulus.

When shaft rotates at constant ω, each fibre of the shaft will undergo tensile and compressive loads. Therefore, the shaft is under the action of cyclic load.Hence the design should be for fatigue loading.

Composite Bars

• A composite bar is made of two bars of different materials rigidly fixed together.
• The strain in both bars /materials is the same under external load. 
• Since strains in the two bars are the same, the stresses in the two bars/materials depends on their young’s modulus of elasticity.
• Examples are Reinforced cement concrete, Flitched beam.

RCC is a composite material made of steel and concrete in which concrete's relatively low tensile strength is compensated for by the inclusion of steel bars having higher tensile strength as reinforcement.

Concept:

When the normal stresses on the two mutually perpendicular planes are equal and alike then the radius of the Mohr circle will be zero.

\({\rm{Radius\;of\;Mohr's\;Circle}},R = \sqrt {{{\left( {\frac{{{\sigma _x} - {\sigma _y}}}{2}} \right)}^2} + {\tau _{xy}}^2} \)

Only in figure 3, we can see that 

 \({\sigma _x} = {\sigma _y}\;and\;{\tau _{xy}} = 0\)

∴ option 3 is the correct answer

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

A cantilever-loaded rotating beam, showing the normal distribution of surface stresses. (i.e., tension at the top and compression at the bottom)

Net stress pattern obtained when loading a surface treated beam. The reduced magnitude of the tensile stresses contributes to increased fatigue life.

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

Elongation of bar due to its own weight=Pl/2AE

Elongation of bar due to axial load=Pl/AE

Now comparing both..we will take the ratio.. (Pl/2AE)÷(Pl/AE)=1/2...i.e half

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

Consider a typical stress-strain curve of the material:

1. Proportionality Limit: It is the limit where stress is directly proportional to strain i.e. Hook’s law is valid. (Point A in the diagram)
2. Elastic limit: It is the limit up to which material can regain its original position after unloading. (Point B in the diagram).
3. Yielding: The slow growth of strain or increase in strain under steady stress or load is called yielding. (region BC in the diagram)
4. Strain Hardening: After yielding, there is an increase in the deformation of material that starts again on an increase in load called strain hardening (region CD in the diagram).
5. Necking: It is deformation in material that continues even after slow removal of load. (Region DE in the diagram).
6. Fracture/Breaking: It is the point up to which material can sustain the load. If the load is applied beyond this point material will fail ultimately. (Point E in the diagram)
7. Creep is the tendency of a solid material to deform permanently under the influence of sustained load for a very large period of time.