Machine Design
Exam Duration: 45 Mins Total Questions : 30
The stress which vary from a minimum value to a maximum value of the same nature (i.e., tensile or compressive) is called
- (a)
repeated stress
- (b)
yield stress
- (c)
alternating stress
- (d)
fluctuating stress
Failure of a material is called fatigue when it fails
- (a)
at the elastic limit
- (b)
below the elastic limit
- (c)
above the yield point
- (d)
below the yield point
The notch sensitivity q is expressed in terms of fatigue stress concentration factor kf and theoretical stress concentration factor kt as
- (a)
\(\frac { { k }_{ f }+1 }{ { k }_{ t }+1 } \)
- (b)
\(\frac { { k }_{ f }-1 }{ { k }_{ t }-1 } \)
- (c)
\(\frac { { k }_{ t }-1 }{ { k }_{ f }-1 } \)
- (d)
\(\frac { { k }_{ f }-1 }{ { k }_{ t }+1 } \)
A loaded semi-infinite flat plate is having an elliptical hole (A/B = 2) in the middle as show in the figure. The stress concentration factor at points either X or Y is
- (a)
1
- (b)
5
- (c)
3
- (d)
7
What is the value of stress at point A?
- (a)
8 N/m2 (compressive)
- (b)
4 N/m2 (tensile)
- (c)
4 N/m2 (compressive)
- (d)
8 N/m2 (tensile)
In below figure a bi-axial stress acting on the block. The maximum yield stress of the material is 40 N/m2. The value of factor of safety is (according to maximum distortion energy theorem)
- (a)
1.88
- (b)
1.72
- (c)
1.80
- (d)
1.81
What is the value of hoop stress in cylinder at point A, if thickness of cylinder is 20 cm?
- (a)
2 MPa
- (b)
3.15 MPa
- (c)
4 MPa
- (d)
2.5 MPa
Stress concentration is 2.5 and notch sensitivity is 2 then, the value of fatigue stress concentration is
- (a)
62.5 MPa
- (b)
60 MPa
- (c)
50 MPa
- (d)
25 MPa
In above question value of factor of safety according to Goodman theory is
- (a)
8.69
- (b)
7.14
- (c)
6.59
- (d)
7.35
A cylindrical shaft is subjected to an alternating stress of 150 MPa. Fatigue strength to sustain 1000 cycle is 600 MPa. If corrected endurance limit is 100 MPa, what is the life of shaft?
- (a)
213796 cycle
- (b)
198576 cycle
- (c)
200000 cycle
- (d)
210813 cycle
A flat foot step bearing, 200 mm in diameter supports a load of 5000 N. If coefficient of friction is 0.08 and shaft rotates at 60 rpm then, power lost in friction is
- (a)
0.072 kW
- (b)
0.290 kW
- (c)
0.167 kW
- (d)
1.23 kW
The contact surfaces in a cone clutch have an effective diameter of 100 mm. The angle of cone is 30°. If an axial force of 200 N is applied then, torque required to produce slipping of clutch is (μ=0.3)
- (a)
11.60 N-m
- (b)
8.93 N-m
- (c)
12.70 N-m
- (d)
4.27 N-m
A single block brake has a drum brake diameter of 1 m and angle of contact is 30°. It takes 200 N-m torque at 100 rpm (μ = 0.35).
This arrangement is shown in figure. Calculate the required force P to be applied when rotation of drum is clockwise.
- (a)
240 N
- (b)
210 N
- (c)
287.02 N
- (d)
236.5 N
A band brake acts on 2/3 rd circumference of drum of 100 mm diameter. The band brake provides a braking torque of 100 N-m. One end of band is attached to a fulcrum pin of the lever and other end to a pin, 100 mm from the fulcrum. (μ = 0.25)
(I) Angle of lap is
- (a)
4.18 rad
- (b)
2.72 rad
- (c)
3.27 rad
- (d)
None of these
A band brake acts on 2/3 rd circumference of drum of 100 mm diameter. The band brake provides a braking torque of 100 N-m. One end of band is attached to a fulcrum pin of the lever and other end to a pin, 100 mm from the fulcrum. (μ = 0.25)
If operating force is applied at 1000 mm from the fulcrum which rotates drum clockwise then, value of operating force is
- (a)
308.6 N
- (b)
501.2 N
- (c)
491.3 N
- (d)
621.01 N
Consider a hand operated brake (differential band brake) as shown in figure. If radius of gyration of wheel is 45 cm then, evaluate the value of braking torque. Given, a = 5 cm, b = 10 cm, 1= 25 cm, and P = 100 N, μ = 0.20.
- (a)
99.72 N-m
- (b)
100.01 N-m
- (c)
89.13 N-m
- (d)
92.3 N-m
In a band brake system, braking torque is 103.26 N-m. Mass of drum is 250 kg and radius of gyration is 350 mm. If drum rotates with 300 rpm then, the number of revolutions of drum before it comes to
- (a)
26.09
- (b)
25.71
- (c)
23.29
- (d)
28.72
A bolt of M 24 x 2 means that
- (a)
the pitch of the thread is 24 mm and depth is 2 mm
- (b)
the cross-sectional area of the threads is 24 mm2
- (c)
the nominal diameter of bolt is 24 mm and the pitch is 2 mm
- (d)
None of the above
A bolt of uniform strength can be developed by
- (a)
keeping the core diameter of threads equal to the diameter of unthreaded portion of the bolt.
- (b)
keeping the core diameter of threads smaller than the diameter of unthreaded portion of the bolt
- (c)
keeping the core diameter of threads higher than the diameter of unthreaded portion of bolt
- (d)
None of the above
The shock absorbing capacity of a bolt may be increased by
- (a)
increasing its shank diameter
- (b)
decreasing its shank diameter
- (c)
making the shank diameter equal to core diameter of thread
- (d)
tightening the bolt properly
When a nut is tightened by placing a washer below it, the bolt will be subjected to
- (a)
tensile stress
- (b)
compressive stress
- (c)
shear stress
- (d)
All of these
A double strap butt joint (with equal straps) is
- (a)
always in single shear
- (b)
always in double shear
- (c)
either in single shear or double shear
- (d)
None of the above
The longitudinal joint in boilers is used to get the required
- (a)
length and diameter of boiler
- (b)
efficiency of boiler
- (c)
diameter of boiler
- (d)
length of boiler
The transverse fillet welded joints are designed for
- (a)
tensile strength
- (b)
compressive strength
- (c)
shear strength
- (d)
bending strength
The parallel fillet welded joint is designed for
- (a)
bending strength
- (b)
shear strength
- (c)
tensile strength
- (d)
compressive strength
For a parallel load on a fillet weld of equal legs, the plane of maximum shear occurs at
- (a)
45°
- (b)
22.5°
- (c)
30°
- (d)
60°
The size of the weld in butt welded joint is equal to
- (a)
0.5 x throat of weld
- (b)
2 x throat of weld
- (c)
\(\sqrt{2}\) x throat of weld
- (d)
throat of weld
A double fillet welded joint with parallel fillet weld of length L and leg s is subjected to a tensile force P. Assuming uniform stress distribution, the shear stress in the weld is given by
- (a)
\(\frac { \sqrt { 2 } P }{ s.L } \)
- (b)
\(\frac { P }{ \sqrt { 2 } sL } \)
- (c)
zero
- (d)
\(\frac { 2P }{ sL } \)
Diameter of the rivet will be
- (a)
12 mm
- (b)
20.78 mm
- (c)
15.78 mm
- (d)
19.78 mm
If fillet leg is 3 mm and shear stress of weld material is 60 MPa
The strength of weld material is
- (a)
7144.64 N
- (b)
8144.64 N
- (c)
7544.64 N
- (d)
6844.64 N