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JNTU B.Tech III Year ENGINEERING METROLOGY Previous Question Paper (August - 2022 - R13)

B.Tech | III Year | III-I | R13 | August - 2022

Answer PaperPaper Code: 115AJMaximum Marks: 75Time: 3 hours

Answer any five questions

All questions carry equal marks

Main

115 marks
a
A hole and mating shaft are to have a nominal assembly size of 50 mm. the assembly is to have a maximum clearance of 0.15 mm and a minimum clearance of 0.05 mm. The hole tolerance is 1.5 times the shaft tolerance. Determine the limits for both hole and shaft using hole basis and shaft basis system.
0 marks

Answer

Given:

Nominal size = 50 mm

Maximum clearance = 0.15 mm

Minimum clearance = 0.05 mm

Hole tolerance (TH) = 1.5 * Shaft tolerance (TS)

1. Hole Basis System

In the hole basis system, the hole is kept at the basic size, and the shaft varies within its tolerance.

Maximum clearance = Hole maximum size (Hmax) - Shaft minimum size (Smin)

Minimum clearance = Hole minimum size (Hmin) - Shaft maximum size (Smax)

Since the hole is at the basic size, Hmax = Hmin = 50 mm.

So, 0.15 = 50 - Smin => Smin = 50 - 0.15 = 49.85 mm

And, 0.05 = 50 - Smax => Smax = 50 - 0.05 = 49.95 mm

Shaft tolerance, TS = Smax - Smin = 49.95 - 49.85 = 0.10 mm

Hole tolerance, TH = 1.5 * TS = 1.5 * 0.10 = 0.15 mm

Limits for Hole:

Hmax = 50 mm

Hmin = Hmax - TH = 50 - 0.15 = 49.85 mm

Limits for Shaft:

Smax = 49.95 mm

Smin = 49.85 mm

2. Shaft Basis System

In the shaft basis system, the shaft is kept at the basic size, and the hole varies within its tolerance.

Maximum clearance = Hole maximum size (Hmax) - Shaft minimum size (Smin)

Minimum clearance = Hole minimum size (Hmin) - Shaft maximum size (Smax)

Since the shaft is at the basic size, Smax = Smin = 50 mm.

So, 0.15 = Hmax - 50 => Hmax = 50 + 0.15 = 50.15 mm

And, 0.05 = Hmin - 50 => Hmin = 50 + 0.05 = 50.05 mm

Hole tolerance, TH = Hmax - Hmin = 50.15 - 50.05 = 0.10 mm

Shaft tolerance, TS = 1.5 * TH (given TH = 1.5 * TS, so TS = TH / 1.5)

TS = 0.10 / 1.5 = 0.0667 mm (approx)

Limits for Hole:

Hmax = 50.15 mm

Hmin = 50.05 mm

Limits for Shaft:

Smax = 50 mm

Smin = Smax - TS = 50 - 0.0667 = 49.9333 mm (approx)

b
Distinguish between allowance and Tolerance.
0 marks

Answer

Tolerance is the total permissible variation allowed in a dimension. It is the difference between the maximum and minimum limits of size.

Allowance is the intentional difference between the maximum material limits of two mating parts. It is the minimum clearance or maximum interference between mating parts.

In simpler terms:

  • Tolerance defines the acceptable size variation for a single part.
  • Allowance defines the intended space or interference between two mating parts when they are assembled at their extreme limits.

Formulaic Representation:

For clearance fits:

Maximum Clearance = Hole Maximum Size - Shaft Minimum Size

Minimum Clearance = Hole Minimum Size - Shaft Maximum Size

Allowance (for clearance fit) = Minimum Clearance

Tolerance (Hole) = Hole Maximum Size - Hole Minimum Size

Tolerance (Shaft) = Shaft Maximum Size - Shaft Minimum Size

215 marks
a
Explain the unilateral and bilateral system of writing tolerances using examples.
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b
Define fit. Explain various types of fits with neat sketches.
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315 marks
a
With the help of a neat sketch, explain the principle and use of a spirit level.
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b
What is wringing? Explain the procedure for wringing of slip gauges.
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415 marks
a
What is a dial indicator? Enumerate its desirable qualities along with advantages and limitations.
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b
Distinguish between line standard and end standard with limitations.
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5
Explain with the help of a neat sketch, the construction, working, advantages and limitations of a Toolmaker’s microscope.
15 marks

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615 marks
a
Describe a surface plate with reference to its instruction, use and material.
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b
What are straight edges? Explain its types along with the uses.
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715 marks
a
In the measurement of surface roughness, heights of 10 successive peaks and troughs were measured from a datum and were 33, 25, 30, 19, 22, 18, 24, 27, 29 and 20 microns. If these measurements were obtained on 10 mm length, determine CLA and RMS values of surface roughness.
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b
With the help of neat sketch, describe the construction and working of profilometer.
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815 marks
a
Describe any one method of measuring the effective diameter of internal threads.
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b
Describe with neat sketches, ‘spindle centre runout’ and ‘chuck runout’ tests on lathe.
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About this paper

This page contains the question paper of ENGINEERING METROLOGY from JNTU for B.Tech ME III Year III-I, August - 2022, R13.

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