# Machine Design

 Question 1
The braking system shown in the figure uses a belt to slow down a pulley rotating in the clockwise direction by the application of a force $P$. The belt wraps around the pulley over an angle $\alpha = 270$ degrees. The coefficient of friction between the belt and the pulley is 0.3. The influence of centrifugal forces on the belt is negligible.
During braking, the ratio of the tensions $T_1$ to $T_2$ in the belt is equal to __________. (Rounded off to two decimal places)
Take $\pi =3.14$. A 2.12 B 6.25 C 4.11 D 8.25
GATE ME 2023      Brakes and Clutches
Question 1 Explanation:
\begin{aligned} \mu & =0.3 \\ \alpha & =270^{\circ}=\frac{3 \pi}{2} \\ \frac{T_{1}}{T_{2}} & =\mathrm{e}^{\mu \alpha} \end{aligned}
So, $\quad \frac{T_{1}}{T_{2}}=e^{0.3 \times \frac{3 \pi}{2}}=4.11$
 Question 2
The S-N curve from a fatigue test for steel is shown. Which one of the options gives the endurance limit? A $S_{ut}$ B $S_{2}$ C $S_{3}$ D $S_{4}$
GATE ME 2023      Fatigue Strength and S-N Diagram
Question 2 Explanation:
For some ferrous (iron base) and titanium alloys, the S-N curve becomes horizontal at higher N values; or there is a limiting stress level, called the fatigue limit (also sometimes the endurance limit), below which fatigue will not occur. This fatigue limit represents the largest value of fluctuating stress that will not cause failure for essentially an infinite number of cycles. For many steels, fatigue limits range between 35% and 60% of the tensile strength. Question 3
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). A 45.32 B 32.78 C 23.94 D 18.25
GATE ME 2022 SET-2      Bearings, Shafts and keys
Question 3 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}

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 4
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). A 1.15 B 2.65 C 1.85 D 1.5
GATE ME 2022 SET-2      Bearings, Shafts and keys
Question 4 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}
 Question 5
A structural member under loading has a uniform state of plane stress which in usual notations is given by $\sigma _x=3P,\sigma _y=-2P,\tau _{xy}=\sqrt{2}P$, where $P \gt 0$. The yield strength of the material is 350 MPa. If the member is designed using the maximum distortion energy theory, then the value of $P$ at which yielding starts (according to the maximum distortion energy theory) is
 A 70 Mpa B 90 Mpa C 120 Mpa D 75 Mpa
Question 5 Explanation:
Given,
$\sigma _x=3P$
$\sigma _y=-2P$
$\tau =\sqrt{2}P$
According to maximum distortion energy theory,
\begin{aligned} \sqrt{\sigma _x^2-\sigma _x\sigma _y+\sigma _y^2+3\tau _{xy}^2} &=\frac{S_{yt}}{FOS} \\ P\sqrt{3^2-3(-2)+(-2)^2+3(\sqrt{2})^2}&= \frac{350}{1}\\ P \times 5&=350 \\ P&=70\; MPa \end{aligned}

There are 5 questions to complete.

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