Jawaharlal Nehru Technological University Hyderabad B.Tech III ENGINEERING METROLOGY Previous Question Paper (September - 2021 - R13)

Interference fits are a type of mechanical fit where the shaft's size is larger than the hole's size, resulting in a force-fit connection. This type of fit is used when a permanent or semi-permanent assembly is required, and it relies on the elastic deformation of the components or the application of force to assemble them. The classification of interference fits is based on the degree of interference, which dictates the assembly method and the strength of the joint.

The complete classification of interference fits can be categorized as follows:

• Light Drive Fit (or Easy Force Fit): This is the mildest form of interference fit. The shaft is slightly larger than the hole, requiring only light pressure or a gentle tap to assemble. The interference is small, typically in the range of 0.005 mm to 0.025 mm for shafts up to 25 mm in diameter.
• Example: Pressing a bearing into a housing where a light interference is sufficient to hold the bearing securely without excessive stress on the housing.
• Medium Drive Fit (or Force Fit): This fit requires a moderate amount of force to assemble, often using a hydraulic press. The interference is larger than in a light drive fit, typically ranging from 0.013 mm to 0.050 mm for shafts up to 25 mm in diameter.
• Example: Pressing a gear onto a shaft or a pulley onto a shaft where a secure connection is needed to transmit torque.
• Heavy Drive Fit (or Tight Fit): This fit requires significant force for assembly, often involving heating the outer component or cooling the inner component. The interference is substantial, typically ranging from 0.025 mm to 0.125 mm for shafts up to 25 mm in diameter.
• Example: Pressing a crankshaft into an engine block or fitting a wheel hub onto an axle where a very strong, permanent connection is essential.
• Shrink Fit: In this method, the outer component (e.g., a sleeve or gear) is heated, causing it to expand. The inner component (e.g., a shaft) is then inserted into the expanded outer component. As the outer component cools, it contracts, creating a tight interference fit. This is a very common and effective method for achieving strong interference fits, especially for large components.
• Example: Mounting large gears onto shafts, fitting liners into engine cylinders, or creating press-fit assemblies in heavy machinery.
• Press Fit: This is a general term that can encompass drive fits. It refers to assembling components by applying external force, usually with a press. The degree of interference determines the type of press fit (light, medium, or heavy).
• Example: Assembling pins into holes, fitting bushings, or securing components in precision instruments.

For a precision running fit designated by 40H7/g6, we need to determine the tolerances and actual sizes of the hole and shaft.

Given:

• Nominal size (D) = 40 mm
• The size range 40 mm lies between 30 mm and 50 mm.
• Fundamental deviation for 'g' shaft (es) = +0.34 mm.
• Formula for 'g' shaft tolerance (ITg) = 2.5 * D^0.34
• Formula for IT7 tolerance (IT7) = 10 * D^0.34

1. Determine the IT grades and tolerance values:

For H7 hole, the tolerance grade is IT7.

For g6 shaft, the tolerance grade is IT6.

We need to find the tolerance values for IT7 and IT6 for a nominal size of 40 mm.

Using the given formulas:

Calculating the D^0.34 term:

Now, calculating the tolerance values:

2. Determine the limits for the hole (40H7):

The 'H' in H7 indicates that the fundamental deviation for the hole is zero, meaning the basic hole is at its maximum material limit.

Upper limit of hole (Hole max) = Basic size + Upper deviation (es) = 40 + 0 = 40.000 mm

Lower limit of hole (Hole min) = Basic size + Lower deviation (ei) = Basic size + (es - IT7) = 40 + (0 - 0.03035) = 40 - 0.03035 = 39.96965 mm

3. Determine the limits for the shaft (40g6):

The fundamental deviation for 'g' shaft is given as es = +0.34 mm.

Upper limit of shaft (Shaft max) = Basic size + Upper deviation (es) = 40 + 0.34 = 40.34 mm

Lower limit of shaft (Shaft min) = Basic size + Lower deviation (ei) = Basic size + (es - IT6) = 40 + (0.34 - 0.0075875) = 40 + 0.3324125 = 40.3324125 mm

Summary of Limits:

• Hole Max Size: 40.000 mm
• Hole Min Size: 39.96965 mm
• Shaft Max Size: 40.34 mm
• Shaft Min Size: 40.3324125 mm

4. Determine the clearances:

Clearance is the difference between the hole size and the shaft size.

Maximum Clearance = Hole Max Size - Shaft Min Size

Minimum Clearance = Hole Min Size - Shaft Max Size

Note: The calculated clearances are negative. This indicates that the shaft is always larger than the hole, resulting in an interference fit, which is expected for a running fit designation like H7/g6 where 'g' is a clearance shaft designation. However, the problem statement specifies a 'precision running fit' and provides a fundamental deviation for 'g' shaft. Let's re-evaluate based on standard ISO tolerance tables for H7/g6 fit.

Re-evaluation using standard ISO tolerance values for 40H7/g6:

For a nominal size of 40 mm:

H7 tolerance (from ISO tables) = +0.025 mm / -0.000 mm (IT7 = 25 μm)

g6 tolerance (from ISO tables) = +0.018 mm / +0.006 mm (IT6 = 12 μm)

Revised Limits:

• Hole Max Size: 40 + 0.025 = 40.025 mm
• Hole Min Size: 40 + 0.000 = 40.000 mm
• Shaft Max Size: 40 + 0.018 = 40.018 mm
• Shaft Min Size: 40 + 0.006 = 40.006 mm

Revised Clearances:

Maximum Clearance = Hole Max Size - Shaft Min Size

Minimum Clearance = Hole Min Size - Shaft Max Size

Conclusion based on standard ISO values:

The fit 40H7/g6 is a clearance fit, as the minimum clearance is positive (0.019 mm) and the maximum clearance is also positive (0.019 mm). The minimum clearance is 0.019 mm and the maximum clearance is 0.019 mm. The shaft is always smaller than the hole, allowing for running movement.

Addressing the provided formula and fundamental deviation:

If we strictly follow the provided formulas and fundamental deviation for 'g' shaft (es = +0.34), the results indicate an interference fit, which contradicts the typical designation of H7/g6 as a running fit. It's possible the question intended a different fit designation or provided non-standard formulas for illustrative purposes.

Assuming the question's formulas are to be used:

Final Answer based on provided formulas:

Tolerance on the hole (40H7): 0.03035 mm

Tolerance on the shaft (40g6): 0.0075875 mm

Actual Maximum and Minimum Sizes:

• Hole Max Size: 40.000 mm
• Hole Min Size: 39.96965 mm
• Shaft Max Size: 40.34 mm
• Shaft Min Size: 40.3324125 mm

Maximum and Minimum Clearances (Interference):

• Maximum Clearance (Interference) = Hole Max Size - Shaft Min Size = 40.000 - 40.3324125 = -0.3324125 mm
• Minimum Clearance (Interference) = Hole Min Size - Shaft Max Size = 39.96965 - 40.34 = -0.37035 mm

This implies a significant interference, which is unusual for a 'running fit'. The term 'precision running fit' usually implies a clearance. The fundamental deviation of 'g' is typically for clearance shafts.