# Applied Mechanics and Design (Test 5)

## Gate Exam : Me Mechanical Engineering

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Applied Mechanics and Design
| Applied Mechanics and Design |
Q.1
Two identical ball bearings P and Q are operating at loads 30kN and 45kN respectively. The ratio of the life of bearing P to the life of bearing Q is
A. 81/16
B. 27/8
C. 9/4
D. 3/2
Explaination / Solution:

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Q.2
A triangular–shaped cantilever beam of uniform–thickness is shown in the figure. The Young’s modulus of the material of the beam is E. A concentrated load P is applied at the free end of the beam.

The area moment of inertia about the neutral axis of a cross-section at a distance x measured from the free end is
A. bxt3/6I
B. bxt3/12I
C. bxt3/24I
D. bxt3/12
Explaination / Solution:
No Explaination.

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Q.3
A mass m is attached to two identical springs having spring constant k as shown in the figure. The natural frequency 𝜔 of this single degree of freedom system is

A. √2k/m
B. √k/m
C. √k/2m
D. √4k/m
Explaination / Solution:

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Q.4
What are the upper and lower limits of the shaft represented by 60 f8
Use the following data:
Diameter 60 lies in the diameter step of 50 -80mm.
Fundamental tolerance unit,
i = in μm = 0.45D1/3 + 0.001D, where D is the representative size in mm;
Tolerance value for IT8 = 25i. Fundaental deviation for ‘f’ shaft = -5.5D0.41
A. Lower limit = 59.924mm, Upper Limit = 59.970mm
B. Lower limit = 59.954mm, Upper Limit = 60.000mm
C. Lower limit = 59.970mm, Upper Limit = 60.016mm
D. Lower limit = 60.00mm, Upper Limit = 60.046mm
Explaination / Solution:
No Explaination.

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Q.5
A cantilever beam having square cross-section of side a is subjected to an end load. If a is increased by 19%, the tip deflection decreases approximately by
A. 19%
B. 29%
C. 41%
D. 50%
Explaination / Solution:

From the deflection formula, for cantilever beam having point load at its free end,

Hence, decrement in the deflection in percentage will be

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Q.6
The tool life equation for HSS tool is 𝑉𝑇0.14𝑓0.7𝑑0.4 = Constant. The tool life (T) of 30 min is obtained using the following cutting conditions: V = 45 m/min, f = 0.35 mm, d = 2.0 mm If speed (V), feed (f) and depth of cut (d) are increased individually by 25%, the tool life (in min) is
A. 0.15
B. 1.06
C. 22.50
D. 30.0
Explaination / Solution:

Given tool lite equation
𝑉𝑇0.14𝑓0.7𝑑0.4 = C
T = 30 min
And V = 45 m/min, f = 0.35 mm, d = 2.0 mm
After increasing 25%, we get
v’ = 56.25 m/min,
f’ = 0.4375 mm
d’= 2.5 mm

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Q.7
A massless beam has a loading pattern as shown in the figure. The beam is of rectangular cross-section with a width of 30mm and height of 100mm.

The maximum bending moment occurs at
A. Location B
B. 2675mm to the right of A
C. 2500mm to the right of A
D. 3225mm to the right of A
Explaination / Solution:
No Explaination.

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Q.8
In the assembly shown below, the part dimensions are:
L1 = 22.0±0.01mm,
L2 = L3 = 10.0±0.005mm.
Assuming the normal distribution of part dimensions, the dimension L4 in mm for assembly condition would be:

A. 2.0±0.008
B. 2.0±0.012
C. 2.0±0.016
D. 2.0±0.020
Explaination / Solution:

Since all dimensions have bilateral tolerances

Tolerance is calculated assuming L4 to be sink and tolerance of sink will be cumulative sum of all tolerances

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Q.9
In an interchangeable assembly, shafts of size  mm mate with holes of size  mm. The maximum interference (in macrons) in the assembly is
A. 40
B. 30
C. 20
D. 10
Explaination / Solution:

+ 0.040
– 0.010
Shaft size 25.000 mm
24 mm 90 µm < d< 25 mm 40 µm
+ 0.030
+ 0.020
Hole size
25 mm 20 µm < d< 25 mm 30 µm
Max Interference
= Max shaft size – Min Hole size
= 25 mm 40 µm – 25 mm 20 µm
= 20 µm

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Q.10
A natural feed journal bearing of diameter 50 mm and length 50 mm operating at 20 revolution/ second carries a load of 2 kN. The lubricant used has a viscosity of 20 mPas. The radial clearance is 50 μm . The Sommerfeld number for the bearing is
A. 0.062
B. 0.125
C. 0.250
D. 0.785