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.

Concept:

The maximum shear load acting on the cantilevered bar is at the joint of the bar and the maximum shear load is equal to the applied load.

The direct shear stress acting in the bar due to shear load is given by \(\tau = \frac{P}{A}\)

Where A is the total weld area which is given by A = 4 × a × t

Where ‘a’ is the size of the bar and ‘t’ is the thickness of the throat of the weld

Calculation:

Given, a = 150 mm, τ_per = 20 MPa, P = 25 kN

Let ‘t’ is the throat size of the weld

∴ Total weld area = 4 × 150 × t = 600 t mm²

Direct shear stress, \(\tau = \frac{P}{A} = \frac{{25000}}{{600t}} = \;\frac{{41.67}}{t}\;N/m{m^2}\)

Permissible direct shear stress, \(\frac{{41.67}}{t} = 20\)

∴ t = 2.1 mm = 3 mm

Concept:

For a rectangular key Allowable shear stress \(\tau = \frac{F}{A} = \frac{F}{{b\; \times \;l}}\)

The torque transmitted by the shaft is given by, \(T = F \times \frac{d}{2}\)

Calculation:

Given, T = 800 N-m, d = 50 mm, b = 10 mm, τ_per = 40 MPa

\(T = F \times \frac{d}{2}\)

\( \Rightarrow 800 = F \times \frac{{50}}{2} \times {10^{ - 3}}\)

∴ Force on key F = 800/25 = 32 kN

Allowable shear stress \(\tau = \frac{F}{A} = \frac{F}{{b \times l}}\)

\( \Rightarrow 40 = \frac{{32000}}{{10\; \times \;l}}\)

⇒ l = 3200/40 = 80 mm

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:

Cotter Joint: A cotter joint is used to connect the two shafts which are either subjected to tensile or compressive axial force. It is not used to connect the shafts that are rotating or transmitting torque. e.g. Joint between the piston rod and the crosshead of a steam engine.

The cotter in the cotter joints fails due to double shear.

Shear failure of cotter joint

2 × t × b × τs = F

Where t = thickness of the cotter, b = width of the cotter, τs = shear stress developed in the cotter due to load F. 

Calculation: 

Given, t = 1.2 cm = 12 mm, b = 5 cm = 50 mm, F = 60 kN = 60 × 10³ N

Then, 

2 × 12 × 50 × τs = 60 × 10​ 3

\({\tau _s}\; = \;\frac{{60 × 1000}}{{2 × 12 × 50}}\; =50\;N/mm^2\)

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

It is a case of double shear.

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

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

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

Figure II is better because the weld is in tension and safe stress of weld in tension is greater than shear.

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

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

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

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

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

Concept:

T₁ = Tension in the tight side of the band

T₂ = Tension in the slack side of the band

r = Radius of the drum

t = Thickness of the band

w = width of the band

The maximum tension in the band is:

T₁ = σ (w × t)

Limiting ratio of the tensions is given by

\(\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }}\)

The torque capacity of the brake is:

\({T_B} = \left( {{T_1} - {T_2}} \right) \times R\)

Calculation:

Given: w = 150 mm = 0.15 m, t = 5 mm = 0.005 m, σ₁ = 75 MPa = 75 N/mm² = 75 × 10⁶ N/m², μ = 0.5, r = 300 mm = 0.3 m

θ = 240° = (π/180) × 240 = 4π/3 rad

\({T_1} = {\sigma _1}wt = 75 \times 150 \times 5 = 56250\;N\)

\(\frac{{{T_1}}}{{{T_2}}} = {e^{\mu \theta }} \Rightarrow \frac{{56250}}{{{T_2}}} = {e^{\left( {0.5 \times \frac{{4{\rm{\pi }}}}{{3\;}}} \right)}} = 8.12 \Rightarrow {T_1} = 8.12\;{T_2}\)

\({T_2} = \frac{{56250}}{{8.12}} = 6927.34\;N\)

Thus, torque capacity is:

\({T_B} = \left( {{T_1} - {T_2}} \right) \times R = \left( {8.12\;{T_2} - {T_2}} \right) \times R = 7.12\;{T_2}R = 7.12 \times 6927.34 \times 0.3\;N.m\)

\({T_B} = 14796.8\;N = 14.7\;kN.m\)