Design of Machine Elements Test 2

Concept:

For uniform strength of bolt,

Threaded area = Hole area

\(\frac{\pi }{4}\left( {{d^2} - d_c^2} \right) = \frac{\pi }{4}d_h^2\)

Where,

d_c = core diameter, d_h = hole diameter

\(\therefore {d_h} = \sqrt {{d^2} - d_c^2} \) 

Calculation:

\(\therefore {d_h} = \sqrt {{d^2} - d_c^2} \) 

d = 24 mm

d_c = 0.84 × 24 = 20.16 mm

∴ d_h = 13.022 mm

Explanation:

F_T = 7.5 kN, ϕ = 22°

Normal lead,

\({F_n} = \frac{{{F_T}}}{{\cos \phi }}\)

\({F_n} = \frac{{7.5}}{{\cos 22}}\)

Concept:

The equivalent torque on the shaft with combined shock and fatigue factors is:

\({T_e} = \sqrt {{{({k_b}M)}^2} + {{({k_t}T)}^2}} \)

Calculation:

A shaft subjected to a loading system has a torque of 1000 N-m and a bending moment of 2000 N-m acting on it simultaneously. The combined shock and fatigue factors for bending and torsion are 2 and 1.4 respectively.

Given: T = 1000 Nm, M = 2000 Nm, k_b = 2, k_t = 1.4

Putting the values, we get:

\({T_e} = \sqrt {{{(2 \times 2000)}^2} + {{(1.4 \times 1000)}^2}} = 4237.92\;Nm\)

Concept:

The maximum stress on a hollow shaft is on the outer surface, given by:

\(\tau = \frac{{16T}}{{{\rm{\pi }}{{\rm{D}}^3}\left[ {1 - {{\rm{k}}^4}} \right]}},\;where\;k = \;\frac{d}{D}\)

The minimum stress on the shaft will be at the inner radius

and as τ ∝ R

\(\frac{{{\tau _{max}}}}{{{\tau _{min}}}} = \frac{D}{d}\)

Calculation:

A hollow shaft of outer diameter 180 mm and 100 mm inner diameter is subjected to a torque of 93.24 kN-m. The minimum stress the shaft is subjected to in MPa is

d_o = 180 mm, d_i = 100 mm, T = 93.24 kN.m = 93.24 × 10⁶ N.mm

k = d/D = 100/180 = 5/9

\({\tau _{max}} = \frac{{16T}}{{{\rm{\pi }}{{\rm{D}}^3}\left[ {1 - {{\rm{k}}^4}} \right]}} = \frac{{16 \times 93.24 \times {{10}^6}}}{{{\rm{\pi }} \times {{180}^3} \times \left[ {1 - {{\left( {\frac{5}{9}} \right)}^4}} \right]}} = 90\;MPa\)

Thus, 

\(\frac{{{\tau _{max}}}}{{{\tau _{min}}}} = \frac{D}{d} \Rightarrow \frac{{90}}{{{\tau _{min}}}} = \frac{{180}}{{100}} \Rightarrow \;{\tau _{min}} = \frac{{90 \times 100}}{{180}} = 50\;MPa\)

Concept:

Normal load exerted by pinion on shaft F_N = ?

Let F_T = Tangential load

\(\therefore {F_N} = \frac{{{F_T}}}{{\cos \phi }}\)

\({F_T} = \frac{{Power}}{{Velocity}}\)

Calculation:

ω = 20 rad/sec, T_p = 20, P = 20 kW, ϕ = 20°, m = 20 mm

Now,

\(m = \frac{{{D_p}}}{{{T_p}}}\)

\(20 = \frac{{{D_p}}}{{20}} \Rightarrow {D_p} = 400\;mm\) 

∴ r_p = 20 mm

Velocity = V = r_p × ω = 0.2 × 20

V = 4 m/s

Now,

\({F_T} = \frac{{20 \times {{10}^3}}}{4} = 5000\;N = 5\;kN\) 

\({F_N} = \frac{5}{{\cos 20^\circ }}\)

∴ F_N = 5.32 kN 

Concept:

Let tangential force be given by F_T, Normal force be given by F_N, Radial Force be given by F_R,

Pressure angle be given by ø

Power transmitted is given by (2πNT/60)

T is the torque in Nm

T = Tangential force × Pitch radius

Tangential force: F_T = F_N cos ø

Radial or normal force: F_R = F_N sin ø

F_R / F_T = tan ø

Calculation:

Given: d = 150 mm, N = 600 rpm, F_N = 17 kN,  ø = 20°

 F_T = 17000 × cos 20

F_T = 15.974 kN

T = F_T × pitch radius

T = 15.974 × 0.075 kN-m = 1.198 kN.m

Power = 2πNT/60 = 2π × 600 × 1.198/60

P = 75.27 kW

Important Point:

Tangential force: F_T = F_N cos ϕ

Radial force: F_R = F_N sin ϕ

Torque exerted on the gear shaft: T = F_T × r