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 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 3 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}

Question 4 |

A bracket is attached to a vertical column by
means of two identical rivets U and V separated
by a distance of 2a = 100 mm, as shown in the
figure. The permissible shear stress of the rivet
material is 50 MPa. If a load P = 10 kN is applied
at an eccentricity e=3\sqrt{7}a, the minimum crosssectional area of each of the rivets to avoid failure
is ___________ mm^2

.

.

800 | |

25 | |

100 \sqrt{7} | |

200 |

Question 4 Explanation:

The given load is eccentric lateral load which results
in

(i) primary shear due to direct loading

(ii) secondary shear due to eccentricity

(i) Primary shear:

F_p=\frac{F}{n}=\frac{10kN}{2}=5kN

(ii) Secondary shear:

F_s=\frac{m}{r_1^2+r_2^2} \times r =\frac{10 \times 3\sqrt{7}a}{a^2+a^2} \times a=15\sqrt{7}kN \; \; \; \; (\because a=50mm)

Finding resultant: R=\sqrt{F_p^2+F_s^2+2F_pF_s \cos \theta }

Here, \theta \text{ is }90^{\circ}

As secondary load is same on both rivets. Both are critical due to loading.

\therefore \;\;R_{max}=\sqrt{F_p^2+F_s^2}=\sqrt{5^2+(15\sqrt{7})^2}=40kN

Design of Rivet: \begin{aligned} \tau _{max} &=\frac{S_{ys}}{FOS} \\ \frac{R_{max}}{A}&= \frac{50}{1}\\ \frac{40 \times 10^3}{A} &=50 \\ A&= 800 mm^2 \end{aligned}

As FOS is considered as 1, A represents the minimum cross section area required.

(i) primary shear due to direct loading

(ii) secondary shear due to eccentricity

(i) Primary shear:

F_p=\frac{F}{n}=\frac{10kN}{2}=5kN

(ii) Secondary shear:

F_s=\frac{m}{r_1^2+r_2^2} \times r =\frac{10 \times 3\sqrt{7}a}{a^2+a^2} \times a=15\sqrt{7}kN \; \; \; \; (\because a=50mm)

Finding resultant: R=\sqrt{F_p^2+F_s^2+2F_pF_s \cos \theta }

Here, \theta \text{ is }90^{\circ}

As secondary load is same on both rivets. Both are critical due to loading.

\therefore \;\;R_{max}=\sqrt{F_p^2+F_s^2}=\sqrt{5^2+(15\sqrt{7})^2}=40kN

Design of Rivet: \begin{aligned} \tau _{max} &=\frac{S_{ys}}{FOS} \\ \frac{R_{max}}{A}&= \frac{50}{1}\\ \frac{40 \times 10^3}{A} &=50 \\ A&= 800 mm^2 \end{aligned}

As FOS is considered as 1, A represents the minimum cross section area required.

Question 5 |

A square threaded screw is used to lift a load W
by applying a force F. Efficiency of square threaded
screw is expressed as

The ratio of work done by W per revolution to work done by F per revolution | |

W/F | |

F/W | |

The ratio of work done by F per revolution to work done by W per revolution |

Question 5 Explanation:

\text{Screw efficiency}=\frac{\text{Work done by the applied force/rev}}{\text{Work done in lifting the load/rev}}

Efficiency of screw jack \eta =\frac{\tan \alpha }{\tan(\alpha +\phi )}

Efficiency depends on helix angle and friction angle.

Efficiency of screw jack \eta =\frac{\tan \alpha }{\tan(\alpha +\phi )}

Efficiency depends on helix angle and friction angle.

Question 6 |

The figure shows the relationship between fatigue strength (S) and fatigue life
(N) of a material. The fatigue strength of the material for a life of 1000 cycles is 450 MPa, while its fatigue strength for a life of 10^6 cycles is 150 MPa.

The life of a cylindrical shaft made of this material subjected to an alternating stress of 200 MPa will then be cycles (round off to the nearest integer).

The life of a cylindrical shaft made of this material subjected to an alternating stress of 200 MPa will then be cycles (round off to the nearest integer).

163840 | |

124589 | |

365247 | |

457812 |

Question 6 Explanation:

Equation of line A \bar{B}:

\begin{aligned} y-y_{1} &=\frac{\left(y_{2}-y_{1}\right)}{\left(x_{2}-x_{1}\right)}\left[x-x_{1}\right] \\ \log _{10} 200-\log _{10} 450 &=\frac{\log _{10} 150-\log _{10} 450}{(6-3)}\left[\log _{10} N-3\right] \\ N &=163840.580 \text { cycles } \end{aligned}

Question 7 |

The von Mises stress at a point in a body subjected to forces is proportional to the square root of the

total strain energy per unit volume | |

plastic strain energy per unit volume | |

dilatational strain energy per unit volume | |

distortional strain energy per unit volume |

Question 7 Explanation:

Condition for failure as per M.D.E.T.

Distortion energy per unit volume under tri-axial state of stress > Distortion energy per unit volume under uni-axial state of stress.

\begin{aligned} &\text { Hence, }\left(\frac{1+\mu}{6 E}\right)\left[\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}\right]>\left(\frac{1+\mu}{3 E}\right)\left(S_{y t}\right)^{2}\\ &\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}>2\left(S_{y t}\right)^{2}\\ &\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2}+\sigma_{1} \sigma_{2}-\sigma_{2} \sigma_{3}-\sigma_{1} \sigma_{3}}>\left(S_{y t}\right) \text { (Von Mises effective stress) } \end{aligned}

S_{y t} = Von Mises effective stress is defined as the uni-axial yield stress that would create same distortion energy created by the tri-axial state of stress.

Distortion energy per unit volume under tri-axial state of stress > Distortion energy per unit volume under uni-axial state of stress.

\begin{aligned} &\text { Hence, }\left(\frac{1+\mu}{6 E}\right)\left[\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}\right]>\left(\frac{1+\mu}{3 E}\right)\left(S_{y t}\right)^{2}\\ &\left(\sigma_{1}-\sigma_{2}\right)^{2}+\left(\sigma_{2}-\sigma_{3}\right)^{2}+\left(\sigma_{1}-\sigma_{3}\right)^{2}>2\left(S_{y t}\right)^{2}\\ &\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}+\sigma_{3}^{2}+\sigma_{1} \sigma_{2}-\sigma_{2} \sigma_{3}-\sigma_{1} \sigma_{3}}>\left(S_{y t}\right) \text { (Von Mises effective stress) } \end{aligned}

S_{y t} = Von Mises effective stress is defined as the uni-axial yield stress that would create same distortion energy created by the tri-axial state of stress.

Question 8 |

A cantilever beam of rectangular cross-section is welded to a support by means of two fillet welds as shown in figure. A vertical load of 2 kN acts at free end of the beam.

Considering that the allowable shear stress in weld is 60 N/mm^2, the minimum size (leg) of the weld required is _______-mm (round off to one decimal place).

Considering that the allowable shear stress in weld is 60 N/mm^2, the minimum size (leg) of the weld required is _______-mm (round off to one decimal place).

6.6 | |

2.8 | |

4.6 | |

8.2 |

Question 8 Explanation:

\begin{aligned} \tau_{\max }=\frac{2 \times 10^{3}}{0.707 t(40) \times 2}&=\frac{35.36}{t} \mathrm{MPa} \\ \sigma_{\max }=\frac{M_{\max }}{I_{N A}} \cdot \tau_{\max } &=\frac{2000 \times 150 \times 20}{\frac{0.707 t(40)^{3} \times 2}{12}} \\ \sigma_{\max }&=\frac{795.615}{t} \mathrm{MPa}\\ \text { MSST, } \quad \sqrt{\sigma_{\max }^{2}+4 \tau^{2}} &\leq 2\left(\frac{S_{y s}}{N}\right)\\ \sqrt{\left(\frac{795.615}{t}\right)^{2}+4\left(\frac{35.36}{t}\right)^{2}} & \leq 2 \times 60 \\ \frac{798.752}{t} & \leq 2(60) \\ t &=6.65 \mathrm{~mm} \end{aligned}

Question 9 |

A machine part in the form of cantilever beam is subjected to fluctuating load as shown in the figure. The load varies from 800 N to 1600 N. The modified endurance, yield and ultimate strengths of the material are 200 MPa, 500 MPa and 600 MPa, respectively.

The factor of safety of the beam using modified Goodman criterion is _______ (round off to one decimal place).

The factor of safety of the beam using modified Goodman criterion is _______ (round off to one decimal place).

1.2 | |

2 | |

2.5 | |

2.9 |

Question 9 Explanation:

A : Critical Point

\begin{aligned} \sigma_{\max , A} &=\sigma_{b, \max } \text { at } A \text { due to } 1600 \mathrm{~N}=\frac{6 \mathrm{M}}{b d^{2}}=\frac{6 \times 1600 \times 200}{12 \times(2 \sigma)^{2}} \\ \sigma_{\max } &=200 \mathrm{MPa} \\ \sigma_{\min , A} &=\sigma_{b, \max } \text { at } A \text { due to } 800 \mathrm{~N} \\ &=\frac{6 \times 800 \times 200}{12 \times(20)^{2}}=100 \mathrm{MPa} \\ \text { Modified Goodman } &=\frac{\sigma_{m}}{S_{y t}}+\frac{\sigma_{a}}{\sigma_{e}} \leq \frac{1}{N} \\ \sigma_{m} &=\left|\frac{\sigma_{\max }+\sigma_{\min }}{2}\right|=150 \mathrm{MPa} \\ \sigma_{a} &=\left|\frac{\sigma_{\max }-\sigma_{\min }}{2}\right|=50 \mathrm{MPa} \\ \frac{150}{600}+\frac{50}{200} & \leq \frac{1}{N} \\ N & \leq 2 \\ N & \approx 2\\ \text { Langer, } \qquad \frac{\sigma_{m}}{S_{y t}}+\frac{\sigma_{a}}{S_{y t}} & \leq \frac{1}{N} \\ \frac{150}{500}+\frac{50}{500} & \leq \frac{1}{N} \qquad \qquad \qquad \qquad (N\leq2.5)\\ N &=2.5 \end{aligned}

Modified Goodman = Safe result of [Goodman or Langer]

Question 10 |

A short shoe drum (radius 260 mm) brake is shown in the figure. A force of 1 kN is applied to the lever. The coefficient of friction is 0.4.

The magnitude of the torque applied by the brake is ________N.m (round off to one decimal place).

The magnitude of the torque applied by the brake is ________N.m (round off to one decimal place).

150 | |

175 | |

200 | |

250 |

Question 10 Explanation:

Taking moment about 'O'

\begin{aligned} R_{N}(500)+F_{\gamma}[310-260] &-1000 \times 1000=0 \\ R_{N}(500)+0.4\left(R_{N}\right)(50) &-1000 \times 1000=0 \\ R_{N} &=1923.076 \mathrm{~N} \\ F_{r} &=\mu R_{N}=769.23 \mathrm{~N} \\ T_{f} &=F_{r} \times R=200 \mathrm{~N}-\mathrm{m} \end{aligned}

There are 10 questions to complete.

NYC approach

question 101 ans is worng