Design of Machine Elements Test 2

Concept:

Maximum Tensile stress at point P is \(= {\sigma _0}\left( {1 + \frac{{2a}}{b}} \right)\)

where \(\left( {1 + \frac{{2a}}{b}} \right)\) is Theoretical stress concentration factor.

From figure it is clear that a > b

\(\therefore 1 + \frac{{2a}}{b} > 3\)

Hence maximum tensile stress in more than 3σ.

If a = b then the stress at point P will be equal to 3 times the applied stress.

300 × (100+ 2l) = 15000 or l = 200

Concept:

\(\mathop L\nolimits_{90} = \mathop {\left( {\frac{C}{P}} \right)}\nolimits^k \)

L90 = Life of the bearing for 90% reliability

C= Basic dynamic capacity

P = Equivalent load

k = 3 for ball bearing

For 90% reliability,

Given, P1 = 10 kN, P2 = 22kN, L1 = 400 M.Rev

\(\begin{array}{l} \frac{{\mathop L\nolimits_2 }}{{\mathop L\nolimits_1 }} = \mathop {\left( {\frac{{\mathop P\nolimits_1 }}{{\mathop P\nolimits_2 }}} \right)}\nolimits^3 \\ \mathop L\nolimits_2 = 400\mathop {\left( {\frac{{10}}{{22}}} \right)}\nolimits^3 \end{array}\)

L2 = 37.565 M.Rev

For 60% reliability under an equivalent load of 22 kN

\(\frac{L}{{\mathop L\nolimits_{90} }} = \mathop {\left( {\frac{{\ln \left( {\frac{1}{R}} \right)}}{{ln\left( {\frac{1}{{\mathop R\nolimits_{90} }}} \right)}}} \right)}\nolimits^{\frac{1}{{1.17}}} \)

\(\frac{L}{{37.565}} = \mathop {\left( {\frac{{\ln \left( {\frac{1}{{0.6}}} \right)}}{{ln\left( {\frac{1}{{0.9}}} \right)}}} \right)}\nolimits^{\frac{1}{{1.17}}} \)

L=144.79 M.Rev

Given that :

Radial clearance, λ = 50 μm

Load, F = 2 kN, N = 20 rps 

diameter, d = 50 mm

viscosity, μ = 20 mPa-s

length, L = 50 mm                 

\(P = \frac{F}{A} = \frac{F}{{L \times d}} \\= \frac{{2000}}{{50 \times 50}} = 0.8\;\frac{N}{{m{m^2}}}\)

sommerfield number

\(Z = \frac{{\mu N}}{P}{\left( {\frac{r}{\lambda}} \right)^2}\)

\(= \frac{{\left( {20 \times {{10}^{ - 3}}} \right) \times 20}}{{\left( {0.8 \times {{10}^6}} \right)}} \times {\left( {\frac{{25\times10^{-3}}}{{50\times10^{-6}}}} \right)^2}\)

= 0.125

Concept:

Number of pair in contact, N = n1 + n2 – 1

where n1 is the number of discs on driving shaft and n2 is the number of discs on the driven shaft.

Calculation:

Given:

n₁ = 6, n₂ = 5

Given that,

\(\frac{{{T_1}}}{{{T_2}}} = 2\),

\({T_{max}} = 150\ N/c{m^2} = 150 \times 31.25 = 4687.5\ N\)

For Maximum power, velocity of belt

\(V = \sqrt {\frac{T_{max}}{{3 \times m}}} =\sqrt {\frac{{4687.5 \times 9.81}}{{3 \times 1400}}} = 3.308\ m/sec\)

For maximum power

\({T_c} = \frac{{{T_{max}}}}{3} = \frac{{4687.5}}{3} = 1562.5\ N\)

Power transmitted \(= \left( {{T_1} - {T_2}} \right)V\)

\(\left( {{T_1} - \frac{{{T_1}}}{2}} \right)V = {T_1}\left( {1 - \frac{1}{2}} \right)V = \left( {{T_{max}} - {T_c}} \right)\left( {1 - \frac{1}{2}} \right)V \)

\(= \left( {{4687.5} - {1562.5}} \right)\left( {1 - \frac{1}{2}} \right)3.308= 5168.75 W = 5.168 kW \)

Given that, w = 14 mm, t = 8 mm, d = 30 mm

\({\tau _{shaft}} = \frac{{16T}}{{\pi {d^3}}} = {\tau _{key}}\)

Torque in the key for shear failure, is given by

\(\begin{array}{l}T = {\tau _{key}} \times l \times w \times \frac{d}{2} \Rightarrow T = \frac{{16T}}{{\pi {d^3}}} \times l \times w \times \frac{d}{2}\\ \Rightarrow l = \frac{{\pi {d^3}}}{{8 \times wd}} \Rightarrow l = \frac{{\pi \times {{\left( {30} \right)}^3}}}{{8 \times 14 \times 30}} = 25.25\ mm\end{array}\)

Load allowed = 100 x 0.707 x 10 x15 = 10.6 kN

Concept:

\(\begin{array}{l} \mathop \sigma \nolimits_{\max } = 100MPa\\ \mathop \sigma \nolimits_{\min } = - 130MPa \end{array}\)

Stress ratio = \(\frac{{\mathop \sigma \nolimits_{\min } }}{{\mathop \sigma \nolimits_{\max } }} = \frac{{ - 130}}{{100}} = - 1.3\)

Variable stress(\(\mathop \sigma \nolimits_v \) = \(\frac{{\mathop \sigma \nolimits_{\max } - \mathop \sigma \nolimits_{\min } }}{2} = \frac{{100 - ( - 130)}}{2} = 115MPa\)

Mean stress \(\mathop \sigma \nolimits_m \)= \(\frac{{\mathop \sigma \nolimits_{\max } + \mathop \sigma \nolimits_{\min } }}{2} = \frac{{100 + ( - 130)}}{2} = - 15MPa\)

Amplitude ratio = \(\frac{{\mathop \sigma \nolimits_v }}{{\mathop \sigma \nolimits_m }} = \frac{{115}}{{ - 15}} = \frac{{ - 23}}{3}\)

Shear is the dominant stress on the key

Concept:

The power transmitted by the shaft is, \(P=\frac{2\pi NT}{60}\)

where, N = rpm of the shaft, T = torque transmitted by shaft.

The shear stress in shaft is, \(τ =\frac{16T}{\pi d^3}\)...................(1)

Apply Soderburg theory for ductile material, \(\frac{τ _m}{τ _{yt}}+\frac{τ _v}{τ _{e}}=\frac{1}{FOS}\)....................(2)

where, τ_m = mean shear stress, τ_v = variable shear stress, τ_yt = yield strength, τ_e = endurance limit of the material, N = factor of safety

Calculation:

Given:

Mean stress:

\(τ _m=\frac{τ~+~(-τ)}{2}=0\)

Variable stress:

 \(τ _v=\frac{τ~+~(τ)}{2}=τ\)

τ_e = 0.5σ_e = 0.5 × 210 = 105 N/mm², FOS = 3, P = 1600 kW, N = 500 rpm 

Therefore, from equation (2), \(\frac{τ_v}{105}=\frac{1}{3}\)

Therefore, τ = 35 N/mm²

Torque:

\(T=\frac{60~×~1600~×10^3}{2\pi~×~500}=\frac{96000}{\pi}~N-m\)

Shear stress:

From equation (1), \(τ =\frac{16T }{\pi d^3}\)

⇒ \(d=(\frac{16T}{\pi \tau})^\frac{1}{3}\) ⇒ \(d=(\frac{16~×~96000}{\pi^2 ~×~35~×~10^6})^\frac{1}{3}\)

\(d=3\sqrt {\frac{{96}}{{4375{\pi ^2}}}} \)

Concept:

The equivalent load for a cyclic loading condition is given by:

\({P_e} = \sqrt[3]{{\frac{{{N_1}{P_1}^3 + {N_2}{P_2}^3}}{{{N_1} + {N_2}}}}}\)

The load capacity can be determined by:

\(L\left( {in\;million\;rev} \right) = {\left( {\frac{C}{F}} \right)^3}\)

Calculation:

Given, N₁ = 5 rev, N₂ = 10 rev, P₁ = 2500 N, P₂ = 1500 N

\({P_e} = \sqrt[3]{{\frac{{{N_1}{P_1}^3 + {N_2}{P_2}^3}}{{{N_1} + {N_2}}}}} = \sqrt[3]{{\frac{{5 \times {{2500}^3} + 10 \times {{1500}^3}}}{{5 + 10}}}} = 1953.8\;N\)

\(20 = {\left( {\frac{C}{{1953.8}}} \right)^3} \Rightarrow C = {\left( {20} \right)^{\frac{1}{3}}} \times 1953.8 = 5303.43\;N\;\)

Key is made the weakest so that it is cheap and easy to replace in case of failure.

Concept:

Let P_P be the primary shear force on the rivet.

And P_s be the secondary shear force on the rivet.

Resultant shear force is \({R^2} = P_s^2 + P_P^2 + 2{P_s}{P_P}\cos 45 \)

Calculation:

Given:

Primary shear force,

P_P1 = P_P2 = P_P3 = P_P4

Primary shear on one rivet

\({P_p} = \frac{{840}}{4} = 210\;N\)

For secondary shear force,

Secondary shear on top right corner rivet

The distance between the center of 4 rivets and the rivet on the top right corner is given as

\(r = \sqrt {{{30}^2} + {{30}^2}} = 30\sqrt 2 \) mm

Taking moment about the center of 4 rivets, we get

\(840 \times 300 = \left( {{P_s} \times 30\sqrt 2 } \right) \times n\)

\(840 \times 300 = \left( {{P_s} \times 30\sqrt 2 } \right) \times 4\)

P_s = 1484.92 N

The angle between primary and secondary shear is given as θ = 45°

Resultant shear force on C is given by

\({R^2} = P_8^2 + P_P^2 + 2{P_s}{P_P}\cos 45 = 17\;kN\)

\({R^2} = {1484.92^2} + {210^2} + 2 \times 1484.92 \times 210\cos 45\)

\({R^2} = 2690086.15\;N\)

R = 1640.148 N

So, the shear load on the worst loaded rivet (1 or 2) is 1640 N

Concept:

\({\eta _{tearing}} = \frac{{\left( {P - d} \right)t \times \frac{{{S_y}t}}{{FOS}}}}{{pt \times \frac{{{S_y}t}}{{FOS}}}}\)

\({\eta _{tearing}} = \frac{{\left( {P - d} \right)t}}{{Pt}}\)

Calculation:

Given:

 \(\frac{d}{P} = 0.25\)

\({\eta _{tearing}} = \frac{{P - d}}{P}\)

\({\eta _{tearing}} = 1 - \frac{d}{P}\)

∴ η_tearing = 1 – 0.25

η_tearing = 0.75 = 75%

Parallel sunk key. The parallel sunk keys may be of rectangular or square section uniform in width and thickness throughout. It may be noted that a parallel key is a taperless and is used where the pulley, gear or other mating piece is required to slide along the shaft. In designing a key, forces due to fit of the key are neglected and it is assumed that the distribution of forces along the length of key is uniform.

Concept:

In uniform wear theory, Pr = constant

Calculation:

Given:

R₀ = 50 mm, R_i = 20 mm

\(\mu \) = 0.4

Now,

Torque transmitted

\(T = \int\limits_{\mathop R\nolimits_i }^{\mathop R\nolimits_0 } {\mu P(2\pi r)rdr} \)

\(\begin{array}{l} T = \mathop P\nolimits_i \mathop R\nolimits_0 \int\limits_{\mathop R\nolimits_i }^{\mathop R\nolimits_0 } {\mu (2\pi )rdr} \\ = \mathop P\nolimits_i \mathop R\nolimits_0 \mu (2\pi )\left( {\frac{{\mathop R\nolimits_0^2 - \mathop R\nolimits_i^2 }}{2}} \right)\\ = 2 \times 50 \times 0.4 \times 2\pi \times \left( {\frac{{\mathop {0.05}\nolimits^2 - \mathop {0.02}\nolimits^2 }}{2}} \right)\\ = 263.89Nm \end{array}\)

\(\begin{array}{l} W = \int\limits_{\mathop R\nolimits_i }^{\mathop R\nolimits_0 } {P(2\pi )rdr} \\ = \mathop P\nolimits_i \mathop R\nolimits_0 (2\pi )(\mathop R\nolimits_0 - \mathop R\nolimits_i )\\ = 2 \times 50 \times 2\pi \times 30\\ = 18.849kN \end{array}\)

Concept:

Given:

T_min = 200 Nm, T_max = 600 Nm, Factor of safety (N) = 2.5

S_yt = 200 MPa, S_ut = 400 MPa, K_t = 2, q = 0.8

Calculation:

T_m = \(\frac{{200 + 600}}{2} = 400\) N.m

T_v = 400 - 200 = 200 N.m

\(\begin{array}{l} \mathop K\nolimits_f = 1 + q\left( {\mathop K\nolimits_t - 1} \right)\\ \mathop K\nolimits_f = 1 + 0.8\left( {2 - 1} \right) = 1.8 \end{array}\)

\(\begin{array}{l} \mathop \tau \nolimits_m = \frac{{16\mathop T\nolimits_m }}{{\pi \mathop D\nolimits^3 }} = \frac{{16 \times 400}}{{\pi \mathop D\nolimits^3 }}\\ \mathop \tau \nolimits_v = \frac{{16\mathop T\nolimits_v }}{{\pi \mathop D\nolimits^3 }} = \frac{{16 \times 200}}{{\pi \mathop D\nolimits^3 }} \end{array}\)

\(\begin{array}{l} \mathop \tau \nolimits_e = \mathop \sigma \nolimits_e^ * \mathop K\nolimits_a \mathop K\nolimits_b \mathop K\nolimits_c \\ \mathop \tau \nolimits_e = 0.5\mathop S\nolimits_{ut} \mathop K\nolimits_a \mathop K\nolimits_b \mathop K\nolimits_c \\ \mathop \tau \nolimits_e = 200MPa \end{array}\)(Assume K_a = K_b = K_c = 1, Since they are not given)

\(\mathop \tau \nolimits_{ys} = \mathop S\nolimits_{ys} = 0.5 \times \mathop S\nolimits_{yt} = 100MPa\)

\(\begin{array}{l} \frac{1}{N} = \frac{{\mathop \tau \nolimits_m }}{{\mathop \tau \nolimits_{ys} }} + \left( {\frac{{\mathop \tau \nolimits_v }}{{\mathop \tau \nolimits_e }}} \right)\mathop k\nolimits_f \\ \frac{1}{{2.5}} = \frac{{\left( {16 \times 400} \right)}}{{\pi \mathop D\nolimits^3 (100)}} + \left( {\frac{{16 \times 200 \times 1.8}}{{\pi \mathop D\nolimits^3 (200)}}} \right)\\ D = 41.95mm \end{array}\)

Equivalent shear stress = \(\frac{{\mathop \tau \nolimits_{ys} }}{N} = \frac{{100}}{{2.5}} = 40MPa\)

The bolts are subjected to shear and bearing stresses while transmitting torque.

Given : m = 3 mm, Z = 16, b = 36 mm, ϕ = 20°, 3 kW, n = 20 rev/s, C_V = 1.5, y = 0.3

Lewis equation for beam strength :

S_b = mbσ_b γ

Where S_b is beam strength of gear tooth (N), σ_b is permissible bending stress (N/mm²) and Y is Lewis form factor

As S_b = m.b.σ_b.γ

Where S_b is beam strength of gear tooth (N), σ_b is permissible bending stress (N/mm²) and Y is Lewis form factor

As \(F=P/v; \\v=\frac{πDN}{60\times1000} =\frac{πDn}{1000};\\ m=\frac DZ⇒D=mZ\)

\(S_b=\frac{P}{\frac{πDn}{1000}} = \frac{3000}{\frac{π.mZ.n}{1000}}=\frac{3000}{\frac{π\times3\times16\times20}{1000}}=994.72 N\)

\(σ_b=\frac{S_b}{m.b.y}=\frac{994.72}{3×36×0.3}=30.70 N/mm^2 \)

Permissible working stress (by considering dynamic load)

σ_W = σ_b × C_V = 30.70 × 1.5 = 46.05 N/mm²