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.
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.
2.12 | |
6.25 | |
4.11 | |
8.25 |
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
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?
S_{ut} | |
S_{2} | |
S_{3} | |
S_{4} |
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).
45.32 | |
32.78 | |
23.94 | |
18.25 |
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}
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 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).
1.15 | |
2.65 | |
1.85 | |
1.5 |
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}
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
70 Mpa | |
90 Mpa | |
120 Mpa | |
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}
\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.
NYC approach
question 101 ans is worng