# Machine Design

 Question 1
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).
 A 163840 B 124589 C 365247 D 457812
GATE ME 2021 SET-2      Fatigue Strength and S-N Diagram
Question 1 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 2
The von Mises stress at a point in a body subjected to forces is proportional to the square root of the
 A total strain energy per unit volume B plastic strain energy per unit volume C dilatational strain energy per unit volume D distortional strain energy per unit volume
Question 2 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.
 Question 3
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).
 A 6.6 B 2.8 C 4.6 D 8.2
GATE ME 2021 SET-1      Bolted, Riveted and Welded Joint
Question 3 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 4
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).
 A 1.2 B 2 C 2.5 D 2.9
Question 4 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 5
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).
 A 150 B 175 C 200 D 250
GATE ME 2021 SET-1      Brakes and Clutches
Question 5 Explanation:

\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}
 Question 6
Shear stress distribution on the cross-section of the coil wire in a helical compression spring is shown in the figure. This shear stress distribution represents

 A direct shear stress in the coil wire cross-section B torsional shear stress in the coil wire cross-section C combined direct shear and torsional shear stress in the coil wire cross-section D combined direct shear and torsional shear stress along with the effect of stress concentration at inside edge of the coil wire cross-section
GATE ME 2021 SET-1      Springs
Question 6 Explanation:

Fig. : Direct shear stress distribution across the cross-section of wire

Fig. : Torsional shear stress distribution across the cross-section of wire

Fig. : Resultant shear stress variation across the cross-section of wire (without considering stress concentration effect)

Fig. : Resultant shear stress variation considering stress concentration effect
 Question 7
A steel spur pinion has a module (m) of 1.25 mm, 20 teeth and $20^{\circ}$ pressure angle. The pinion rotates at 1200 rpm and transmits power to a 60 teeth gear. The face width (F) is 50 mm, Lewis form factor Y = 0.322 and a dynamic factor $K_v = 1.26$. The bending stress ($\sigma$) induced in a tooth can be calculated by using the Lewis formula given below.

If the maximum bending stress experienced by the pinion is 400 MPa. the power transmitted is __________ kW (round off to one decimal place).

Lewis formula: $\sigma =\frac{K_vW^t}{FmY}$, where $W^t$ is the tangential load acting on the pinion.
 A 10 B 20 C 30 D 40
GATE ME 2020 SET-2      Gears
Question 7 Explanation:
\begin{aligned} F_{t} \times c_{v} s &=b m y\left[\sigma_{b}\right]_{\max } \\ c_{v} &=1.26 \\ F_{t} \times 1.26 \times 1 &=50 \times 1.25 \times 0.322 \times 400 \\ F_{t} &=6388.88 \mathrm{N} \\ P_{\text {angle }} &=F_{t} \times v=\frac{F_{t} \times \pi D_{p} N}{60} \\ &=\frac{6388.88 \times \pi \times 1.25 \times 20 \times 1200}{60}=10 \mathrm{kW} \end{aligned}
 Question 8
A helical spring has spring constant k. If the wire diameter, spring diameter and the number of coils are all doubled then the spring constant of the new spring becomes
 A k/2 B k C 8k D 16k
GATE ME 2020 SET-2      Brakes and Clutches
Question 8 Explanation:
\begin{aligned} k_{(\text {sping })} &=\frac{G d^{4}}{8 D^{3} n} \\ k_{\text {new }} &=\frac{G(2 d)^{4}}{8(2 D)^{3}(2 n)}=\frac{G d^{4}}{8 D^{3} n} \\ \text{Hence}\qquad k_{\text {new }} &=k \end{aligned}
 Question 9
A bolt head has to be made at the end of a rod of diameter d = 12 mm by localized forging (upsetting) operation. The length of the unsupported portion of the rod is 40 mm. To avoid buckling of the rod, a closed forging operation has to be performed with a maximum die diameter of ________ mm.
 A 12 B 18 C 40 D 24
GATE ME 2020 SET-2      Bolted, Riveted and Welded Joint
Question 9 Explanation:
\begin{array}{l} \text { If } l \gt 3 d \text { then } \\ \qquad \begin{aligned} \text { Die dia } &=1.5 d \\ &=1.5(12) \\ &=18 \mathrm{mm} \end{aligned} \end{array}
 Question 10
A machine member is subjected to fluctuating stress $\sigma =\sigma _0 \cos (8 \pi t)$. The endurance limit of the material is 350 MPa. If the factor of safety used in the design is 3.5 then the maximum allowable value of $\sigma _0$ is __________ MPa (round off to 2 decimal places).
 A 68.5 B 78.2 C 122.5 D 100
GATE ME 2020 SET-2      Fatigue Strength and S-N Diagram
Question 10 Explanation:
\begin{aligned} \text{Fluctuating stress, } \quad \sigma&=\sigma_{o} \cos (8 \pi t) \\ \sigma_{\max } &=\sigma_{o} \\ \sigma_{\min } &=-\sigma_{o} \\ \sigma_{\text {man }} &=\frac{\sigma_{\max }+\sigma_{\min }}{2}=0 \\ \sigma_{v} &=\frac{\sigma_{\max }-\sigma_{\min }}{2}=\frac{2 \sigma_{o}}{2}=\sigma_{o} \\ \sigma_{e} &=350 \mathrm{MPa} \\ \mathrm{FOS} &=3.5 \end{aligned}
From strength criteria, $\frac{\sigma_{v}}{\sigma_{e}} \leq \frac{1}{\mathrm{FOS}}$
\begin{aligned} \frac{\sigma_{o}}{350} & \leq \frac{1}{3.5} \\ \sigma_{o} & \leq 100 \mathrm{MPa} \end{aligned}

There are 10 questions to complete.

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