Question 1 |

A shaft AC rotating at a constant speed carries a
thin pulley of radius r = 0.4 m at the end C which
drives a belt. A motor is coupled at the end A of the
shaft such that it applies a torque M_z
about the shaft
axis without causing any bending moment. The
shaft is mounted on narrow frictionless bearings at
A and B where AB = BC = L = 0.5 m. The taut
and slack side tensions of the belt are T_1= 300 N
and T2
= 100 N, respectively. The allowable shear
stress for the shaft material is 80 MPa. The selfweights of the pulley and the shaft are negligible.
Use the value of \pi available in the on-screen virtual
calculator. Neglecting shock and fatigue loading
and assuming maximum shear stress theory, the
minimum required shaft diameter is _______ mm
(round off to 2 decimal places).

45.32 | |

32.78 | |

23.94 | |

18.25 |

Question 1 Explanation:

\begin{aligned} M_{max}&=400 \times L =400 \times 0.5 \times 10^3\\ &=200 \times 10^3 \; N.mm\\ T_{max}&=M_z\\ &=(T_1-T_1) \times r\\ &=(200 \times 0.4) \times 10^3\\ T_{max}&=80 \times 10^3 \; N.mm \end{aligned}

Here section-B is critical due to loading,

Critical particle :

According to maximum shear stress theory,

\begin{aligned} \frac{16}{\pi d^3}\sqrt{M_{max}^2+T_{max}^2}&=\frac{S_{{ys}}}{FOS}\\ \frac{16 \times 10^3}{\pi d^3}\sqrt{200^2+80^2}&=\frac{80}{1}\\ \Rightarrow d&=23.94mm \end{aligned}

Question 2 |

A shaft of length L is made of two materials, one
in the inner core and the other in the outer rim, and
the two are perfectly joined together (no slip at the
interface) along the entire length of the shaft. The
diameter of the inner core is d_i
and the external
diameter of the rim is d_o, as shown in the figure. The
modulus of rigidity of the core and rim materials are
G_i
and G_o, respectively. It is given that do
d_o=2d_i and
G_i=3G_o. When the shaft is twisted by application
of a torque along the shaft axis, the maximum shear
stress developed in the outer rim and the inner
core turn out to be \tau _o and \tau _i, respectively. All
the deformations are in the elastic range and stress
strain relations are linear. Then the ratio \tau _i /\tau_o is
______ (round off to 2 decimal places).

1.15 | |

2.65 | |

1.85 | |

1.5 |

Question 2 Explanation:

Given G_i=3G_o, d_o=2d_i, l_i=l_o,\theta _i=\theta _o \text{ (It is a rigid joint)}

Find \frac{T_i}{T_o}=?

\begin{aligned} \frac{T}{J}&=\frac{\tau }{r}=\frac{G\theta }{L}\\ \tau _{max}(\text{in core})&=\frac{G\theta \times r_{max}}{L}\\ &=\frac{G_i \times \theta _i \times \frac{d_i}{2}}{L_i}=\tau _i\\ \tau _{max}(\text{rim})&=\frac{G_o \times \theta _o\times \frac{d_o}{2} }{L_o}=\tau _o\\ \frac{T_i}{\tau _o}&=\frac{G_i \times \theta _i \times \frac{d_i}{2}}{L_i} \times \frac{L_o}{G_o \times \theta _o\times \frac{d_o}{2} }\\ \frac{T_i}{T_o}&=\frac{G_i}{G_o} \times \frac{d_i}{d_o}=3 \times \frac{1}{2}=1.5 \end{aligned}

Find \frac{T_i}{T_o}=?

\begin{aligned} \frac{T}{J}&=\frac{\tau }{r}=\frac{G\theta }{L}\\ \tau _{max}(\text{in core})&=\frac{G\theta \times r_{max}}{L}\\ &=\frac{G_i \times \theta _i \times \frac{d_i}{2}}{L_i}=\tau _i\\ \tau _{max}(\text{rim})&=\frac{G_o \times \theta _o\times \frac{d_o}{2} }{L_o}=\tau _o\\ \frac{T_i}{\tau _o}&=\frac{G_i \times \theta _i \times \frac{d_i}{2}}{L_i} \times \frac{L_o}{G_o \times \theta _o\times \frac{d_o}{2} }\\ \frac{T_i}{T_o}&=\frac{G_i}{G_o} \times \frac{d_i}{d_o}=3 \times \frac{1}{2}=1.5 \end{aligned}

Question 3 |

A short shoe external drum brake is shown in the figure. The diameter of the brake drum is 500 mm. The dimensions a= 1000 mm, b = 500 mm and c = 200 mm. The coefficient of friction between the drum and the shoe is 0.35. The force applied on the lever F= 100 N as shown in the figure. The drum is rotating anti-clockwise. The braking torque on the drum is______ N.m (round off to two decimal places).

12.75 | |

8.55 | |

20.35 | |

26.85 |

Question 3 Explanation:

W.r.t FBD of link

\mathrm{F}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{N}=0.35 \times \mathrm{N}

Taking summation of moment about O = zero

\begin{array}{l} \Sigma M_{0}=0 \\ N \times 500-F_{s} \times 200-F \times 1000=0 \\ N \times 500-0.35 \times N \times 200-100 \times 1000=D \\ N=\frac{100 \times 1000}{(500-70)}=232.558 \mathrm{N} \end{array}

With respect to FBD of drum

\text { Braking torque } \mathrm{M}=\mathrm{F}_{\mathrm{s}} \times \mathrm{r}=\frac{0.35 \times 232.558 \times 500}{2 \times 1000}

\mathrm{M}=20.35 \mathrm{N}-\mathrm{m}

\mathrm{F}_{\mathrm{s}}=\mu_{\mathrm{s}} \mathrm{N}=0.35 \times \mathrm{N}

Taking summation of moment about O = zero

\begin{array}{l} \Sigma M_{0}=0 \\ N \times 500-F_{s} \times 200-F \times 1000=0 \\ N \times 500-0.35 \times N \times 200-100 \times 1000=D \\ N=\frac{100 \times 1000}{(500-70)}=232.558 \mathrm{N} \end{array}

With respect to FBD of drum

\text { Braking torque } \mathrm{M}=\mathrm{F}_{\mathrm{s}} \times \mathrm{r}=\frac{0.35 \times 232.558 \times 500}{2 \times 1000}

\mathrm{M}=20.35 \mathrm{N}-\mathrm{m}

Question 4 |

A self-aligning ball bearing has a basic dynamic load rating ( C_{10} for 10^{6} revolutions) of 35 kN. If the equivalent radial load on the bearing is 45 kN, the excepted life (in revolutions) is

below 0.5 | |

0.5 to 0.8 | |

0.8 to 1.0 | |

above 1.0 |

Question 4 Explanation:

\begin{aligned} C &=35 \mathrm{kN} \\ \mathrm{P}_{C} &=45 \mathrm{kN} \\ L_{90} &=\left(\frac{C}{P_{C}}\right)^{3}=\left(\frac{35}{45}\right)^{3}=0.4705 \mathrm{MR} \end{aligned}

Question 5 |

Which of the bearings given below SHOULD NOT be subjected to a thrust load?

Deep groove ball bearing | |

Angular contact ball bearing | |

Cylindrical (straight) roller bearing | |

Single row tapered roller bearing |

Question 5 Explanation:

Correct option is (C)

There are 5 questions to complete.

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