Design of Machine Elements Test 1

Stress concentration: Stress concentration is defined as the localization of high stresses due to irregularities present in the component and abrupt changes in the cross-section. 

Causes of stress concentration:

Variation in properties of the material: In general the material is no homogenous throughout, there are some variations in the material properties due to the following factors:

• Internal cracks and flaws like blowholes
• Cavities in welds
• Air holes in steel components
• Nonmetllic inclusions

These variations act as discontinuities in the component and cause stress concentration. 

Load application: The machine components are subjected to forces. These forces act either at a small or point area on the component. Since the area is small, the pressure at these points is excessive. This results in stress concentration. 

Abrupt changes in section: The abrupt changes are due to steps cut on the shafts to accommodate the bearings, pulleys sprockets. These create change in cross-section of the shaft ad results in the stress concentration. 

Discontinuities in the component: There are some features of machine components such as oil holes, keyways, and splines, and screw threads result in a discontinuity in the cross-section of the component. There is stress concentration in the vicinity of these discontinuities. 

Machining Scratches: Machining scratches stamp marks or inspection marks are surface irregularities which cause stress concentration. 

Explanation:

We have to use Miner’s equation

\(\frac{{{n_1}}}{{{N_1}}} + \frac{{{n_2}}}{{{N_2}}} = \frac{1}{N}\)

Where, n → % of time machine operates

N → Fatigue life for that time

\(\frac{{0.6}}{{{{10}^6}}} + \frac{{0.4}}{{4\; \times\; {{10}^4}}} = \frac{1}{N}\)

∴ N = 9.4339 × 10⁴ cycles.

Explanation:

Since,

\(L = {\left( {\frac{C}{P}} \right)^3}\)

Thus,

\(\frac{{{L_2}}}{{{L_1}}} = {\left( {\frac{{{P_1}}}{{{P_2}}}} \right)^3}\)

\(\frac{{{L_2}}}{{700}} = {\left( {\frac{5}{{10}}} \right)^3}\)

\({L_2} = \frac{{700}}{{{2^3}}} = 87.5\)

Concept:

The tearing efficiency of a rivet joint is given by:

\(η = 1 - \frac{d}{P}\)

where, P = pitch, d = diameter of rivet

Calculation:

Given:

P = 15 mm, η = 70%

\(0.7 = 1 - \frac{d}{{15}}\)

\(\frac{d}{{15}} = 0.3 \Rightarrow d = 4.5\;mm\)

Concept:

The Wahl’s factor for the spring is given by:

\(K = \frac{{4C - 1}}{{4C - 4}} + \frac{{0.615}}{C}\)

Calculation:

\(K = \frac{{4 \times 6 - 1}}{{4 \times 6 - 4}} + \frac{{0.615}}{6} = 1.2525\)

Concept:

The Sommerfield Number is given by,

\(S = \frac{{\mu {N_s}}}{P}{\left( {\frac{r}{c}} \right)^2}\)

where, μ = Oil Viscosity, N_s is the revolution per second, P is the unit bearing pressure, it is load per unit projected area, r is the radius of the journal and c is the radial clearance

Projected area = L × D

\(P = \frac{W}{{L \times D}}\)

Calculation:

Given, load on bearing W = 3672 N, Bearing Diameter D = 50 mm, Length of bearing L = 0.1 m, Diametral clearance C = 0.1 mm, Oil Viscosity, μ  = 0.06 kg/m.s, N = 500 r.p.m

P = \(\frac{W}{{L \times D}} = \frac{{3672}}{{100 \times 50}} = 0.7344\;N/{m^2}\)

\({N_s} = \frac{N}{{60}} = \frac{{500}}{{60}}\)

\(S = \frac{{0.06\; \times \;500}}{{0.7344\; \times 60\;}}{\left( {\frac{{50}}{{0.1}}} \right)^2} = 10.212 \times {10^6}\;\)

Concept:

Tension: Tearing resistance or pull required to tear off the plate per pitch length

P_t = σ_t (p - d).t

Shear: Shearing resistance or pull required to shear off the rivet per pitch length

\({P_s} = n \times \frac{\pi }{4} \times {d^2} \times \tau \) {single shear}

\({P_s} = n \times 2 \times \frac{\pi }{4} \times {d^2} \times \tau \) {double shear}

Crushing: crushing resistance or pull required to crush the rivet per pitch length:

\({P_c} = n{\sigma _c}dt\)

The rivet diameter can be found by equating the strength of the rivet in shear and crushing

\(2 \times \frac{\pi }{4} \times {d^2} \times \tau = 2 \times {\sigma _c}dt \Rightarrow \;d = \frac{{4{\sigma _c}t}}{{\pi \tau }}\)

Calculation:

Given: t = 5 mm, n = 2, σ_c = 100 MPa, τ = 70 MPa

Putting the given values, we get:

\(d = \frac{{4{\sigma _c}t}}{{\pi \tau }} = \frac{{4 \times 100 \times 5}}{{70{\rm{\pi }}}} = 9.094\;mm\)

Concept:

For uniform wear,

\(T = \mu W\left( {\frac{{{r_1} + {r_2}}}{2}} \right)\)

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

W = 2πC (r₁ – r₂)

Calculation:

Given:

P = 75 × 10³ W, μ = 0.25, P_max = 0.825 MPa, N = 1000 rpm

\(\frac{D}{d} = \frac{4}{3}\)

\(T = \frac{{60\;P}}{{2\pi N}}\)

\(T = \frac{{60 \times 75 \times {{10}^3}}}{{2\pi \times 1000}}\)

∴ T = 716.2 N.m

Now,

For uniform wear,

\(T = \mu W\left( {\frac{{{r_1} + {r_2}}}{2}} \right)\)

\(T = \mu \times 2\pi C\left( {{r_1} - {r_2}} \right)\;\frac{{({r_1} + {r_2})}}{2}\)

\(T = \mu \times 2\pi \times {P_{max}} \times {r_{min}}\;\frac{{\left( {{r_1} - {r_2}} \right)\left( {{r_1} + {r_2}} \right)}}{2}\)

\(716.2 = 0.25 \times 2\pi \times 0.825 \times \frac{{{{10}^6}}}{2}\left( {\frac{{16}}{9}r_2^2 - r_2^2} \right) \times {r_2}\)

∴ r₂ = 112 m

∴ d₂ = 224 mm

Concept:

\(\frac{T}{J} = \frac{{G\theta }}{L} = \frac{\tau }{R}\)

Calculation:

\(\tau = \frac{{T \times R}}{J}\)

J = polar moment of inertia = 2πr³t

R = radius = r

\(\tau = \frac{{T \times r}}{{2\pi {r^3}t}} = \frac{T}{{2\pi {r^2}t}}\)

For weld size of s.

We have, t = 0.707 s

\(\therefore \tau = \frac{{2T}}{{\pi {d^2}\left( {0.7075} \right)}}\)

\(\therefore \tau = \frac{{2.83\;T}}{{\pi {d^2}s}}\)

Where, s is size of weld and t = thickness of throat. 

Explanation:

Given:

P_min = -30 kN, P_max = +130 kN, σ_e = 180 Mpa, σ_yt = 400 MPa, σ_ut = 650 MPa, F.O.S = 1.5

\({\sigma _{max}} = \frac{{{P_{max}}}}{{\frac{\pi }{4}{d^2}}}\)

\({\sigma _{max}} = \frac{{4 \times 130 \times {{10}^3}}}{{\pi {d^2}}} = \frac{{52 \times {{10}^4}}}{{\pi {d^2}}}\)

\({\sigma _{max}} = \frac{{4 \times 130 \times {{10}^3}}}{{\pi {d^2}}} = \frac{{52 \times {{10}^4}}}{{\pi {d^2}}}\)

\({\sigma _{min}} = \frac{{{P_{min}}}}{{\frac{\pi }{4}{d^2}}} = \frac{{4 \times \left( { - 30} \right) \times {{10}^3}}}{{\pi {d^2}}}\)

\(\therefore {\sigma _{min}} = \frac{{ - 12 \times {{10}^4}}}{{\pi {d^2}}}\)

Now,

\({\rm{Mean\;stress\;}}\left( {{\sigma _m}} \right) = \frac{{{\sigma _{max}} + {\sigma _{min}}}}{2} = \frac{{\left( {52 - 12} \right) \times {{10}^4}}}{{2 \times \pi {d^2}}}\)

\({\sigma _m} = \frac{{20 \times {{10}^4}}}{{\pi {d^2}}}\)

\({\rm{Amplitude\;stress\;}}\left( {{\sigma _a}} \right) = \frac{{{\sigma _{max}} - {\sigma _{min}}}}{2} = \frac{{\left( {52 + 12} \right) \times {{10}^4}}}{{2 \times \pi {d^2}}}\)

\({\sigma _a} = \frac{{32 \times {{10}^4}}}{{\pi {d^2}}}\)

According to Goodman’s Criterion

\(\frac{{{\sigma _m}}}{{{\sigma _{ut}}}} + \frac{{{\sigma _a}}}{{{\sigma _e}}} = \frac{1}{{\left( {F.O.S} \right)}}\)

\(\frac{{20 \times {{10}^4}}}{{\pi {d^2} \times 650}} + \frac{{32 \times {{10}^4}}}{{\pi {d^2} \times 180}} = \frac{1}{{1.5}}\)

∴ d = 31.55 mm

Concept:

Soderberg’s criteria:

\(\left( {\frac{{{\sigma _a}}}{{{S_e}}}} \right) + \left( {\frac{{{\sigma _m}}}{{{S_{yt}}}}} \right) = \frac{1}{N}\)

For static failure, \(N = \frac{{{S_{yt}}}}{{{\sigma _{max}}}}\) 

Calculation:

Given:

σ_min = 40 N/mm², σ_max = 100 N/mm², S_e = 270 N/mm², S_ut = 600 N/mm², S_yt = 450 N/mm²

Mean stress,

\({\sigma _m} = \frac{{{\sigma _{max}} + {\sigma _{min}}\;}}{2}\)

\({\sigma _m} = \frac{{100 + 40}}{2} = 70\;N/m{m^2}\)

Amplitude stress:

\({\sigma _a} = \frac{{{\sigma _{max}} - {\sigma _{min}}}}{2}\)

\({\sigma _a} = \frac{{100 - 40}}{2} = 30\;N/m{m^2}\)

Now, according to soderberg line,

\(\frac{{30}}{{270}} + \frac{{70}}{{450}} = \frac{1}{{{N_1}}}\)

N₁ = 3.75

Factor of safety against static failure,

\({N_2} = \frac{{{S_{yt}}}}{{{\sigma _{max}}}} = \frac{{450}}{{100}} = 4.5\)

Now,

\({\rm{Required\;ratio\;}} = \frac{{{N_1}}}{{{N_2}}} = \frac{{3.75}}{{4.5}}\)

∴ Required ratio = 0.833 

Concept:

The beam strength of the pinion as per Lewis equation is:

S_b = bmσ_bY × C_v {as C_v < 1}

Where b = face width, m = module, Y = form factor, C_v = velocity factor

S_b = Beam strength of gear tooth, σ_b = Permissible bending stress

Calculation:

Given: m = 3 mm, ϕ = 20°, T_P = 20, N = 900 rpm, P = 15 kW = 15000 W, T_G = 60, b = 40 mm, C_V = 0.55, Y = 0.32

Torque transmitted is:

\(T = \frac{{60P}}{{2{\rm{\pi N}}}} = \frac{{60 \times 15000}}{{2{\rm{\pi }} \times 900}} = 159.15\;Nm\)

Tangential load on the tooth is:

\({F_t} = \frac{T}{R} = \frac{T}{{D/2}} = \frac{{2T}}{D} = \frac{{2 \times 159.15}}{{0.003 \times 20}} = 5305\;N\)

(Used m = d/T ⇒ d = mZ)

Using Lewis equation:

S_b = bmσ_bY × C_v

5305 = 40 × 3 × σ_b × 0.55 × 0.32

σ_b = 251.18 MPa

Concept:

F_W = k D_pwQ

Where,

k = 1.5 MPa (given),

D_p = diameter of pinion, w = Width of pinion

\({\rm{Q\;}} = {\rm{\;Ratio\;factor\;}} = \frac{{2\;{T_g}}}{{{T_p} + {T_g}}}\)

Calculation:

\(Q = \frac{{2\;{T_g}}}{{{T_p} + {T_s}}} = \frac{2}{{\frac{{{T_p}}}{{{T_s}}} + 1}}\)

\(\frac{{{T_p}}}{{{T_s}}} = \frac{{{V_s}}}{{{V_p}}} = \frac{1}{2}\)

\(\therefore Q = \frac{2}{{\frac{1}{2} + 1}} = \frac{4}{3}\)

\(\therefore {F_W} = 1.5 \times {10^6} \times 0.2 \times 0.06 \times \frac{4}{3}\) 

∴ F_W = 24000 N = 24 kN

Concept:

The equivalent load on a bearing simultaneously subjected to axial and radial forces is:

\({P_e} = \;\left( {X{F_r} + Y{F_a}} \right)\)

F_r is radial load (N), F_a is axial or thrust load (N), X and Y are radial and thrust factor respectively.

The relationship between the dynamic load carrying capacity, the equivalent dynamic load and the bearing life is:

\({L_{10}} = {\left( {\frac{C}{P}} \right)^k}\)

where L₁₀ is rated bearing life (in million revolutions), C is basic dynamic load capacity (in N), P is equivalent dynamic load

k = 3 for ball bearing, k = 10/3 for roller bearing

The relationship between life in million revolutions (L₁₀) and life in working hours (L_10h) is given by:

\({L_{10}} = \frac{{60N{L_{10h}}}}{{{{10}^6}}}\)

N is speed of rotation in rpm.

\({L_{10}} = \frac{{60N{L_{10h}}}}{{{{10}^6}}} = {\left( {\frac{C}{P}} \right)^k}\)

Calculation:

\({L_{10}} = \frac{{60N{L_{10h}}}}{{{{10}^6}}} = {\left( {\frac{C}{{{P_e}}}} \right)^3}\)

F_r = 2500 N, F_a = 1000 N, C = 7350 N, X = 0.56, Y = 1.6, N = 720 rpm

\({P_e} = \left( {X{F_r} + Y{F_a}} \right) = 0.56 \times 2500 + 1.6 \times 1000 = 3000\;N\)

\(\frac{{60N{L_{10h}}}}{{{{10}^6}}} = {\left( {\frac{C}{{{P_e}}}} \right)^3} \Rightarrow \frac{{60 \times 720 \times {L_{10h}}}}{{{{10}^6}}} = {\left( {\frac{{7350}}{{3000}}} \right)^3}\)

Concept:

Shear stress for solid shaft,

\({\tau _s} = \frac{{16\;T}}{{\pi {D^3}}}\)

For hollow shaft,

\({\tau _h} = \frac{{16\;T}}{{\pi D_0^3\left( {1 - {k^4}} \right)}}\)

Where, \(k = \frac{{{D_i}}}{{{D_0}}}\) 

Where,

D = diameter of solid shaft

D₀ = outside diameter of Hollow shaft

D_i = inner diameter of Hollow shaft

\(\frac{{{\tau _s}}}{{{\tau _h}}} = \frac{{D_0^3\left( {1 - {k^4}} \right)}}{{{D^3}}}\)

\(k = \frac{{{D_i}}}{{{D_0}}} = \frac{{D/2}}{D} = 0.5\)

\(\therefore \frac{{{\tau _s}}}{{{\tau _h}}} = \frac{{{D^3}}}{{{D^3}}}\left[ {1 - {{\left( {\frac{1}{2}} \right)}^4}} \right]\)

\(\therefore \frac{{{\tau _s}}}{{{\tau _h}}} = \frac{{15}}{{16}}\)