Question

A shaft has diameter $20^{+0.05}_{-0.15}$ mm and a hole diameter $20^{+0.20}_{-0.10}$ mm. When these are assembled, then what is the nature of fit yield?

• A. clearance fit
• B. interference fit
• C. transition fit
• D. best fit

Hint:

To classify fits: - Clearance Fit: Always positive clearance. This means the smallest hole is larger than the largest shaft. - Interference Fit: Always negative clearance (i.e., positive interference). This means the largest shaft is smaller than the smallest hole. - \textbf{Transition Fit:} Can be either a clearance or an interference. This occurs when the tolerance zones of the hole and shaft overlap, allowing for both possibilities.

Answer 1

The Correct Option is C

Step 1: Determine the maximum and minimum dimensions for the shaft and the hole.
Given data from the problem:
Shaft diameter: $20^{+0.05}_{-0.15}$ mm
Hole diameter: $20^{+0.20}_{-0.10}$ mm
For the shaft:
Maximum shaft diameter ($D_{s,max}$) = Basic size + Upper deviation = $20 + 0.05 = 20.05$ mm
Minimum shaft diameter ($D_{s,min}$) = Basic size + Lower deviation = $20 - 0.15 = 19.85$ mm
For the hole:
Maximum hole diameter ($D_{h,max}$) = Basic size + Upper deviation = $20 + 0.20 = 20.20$ mm
Minimum hole diameter ($D_{h,min}$) = Basic size + Lower deviation = $20 - 0.10 = 19.90$ mm
Step 2: Calculate the maximum clearance and maximum interference.
Maximum Clearance: Occurs when the hole is at its maximum size and the shaft is at its minimum size.
Maximum Clearance = $D_{h,max} - D_{s,min}$
Maximum Clearance = $20.20 - 19.85 = 0.35$ mm
Maximum Interference: Occurs when the shaft is at its maximum size and the hole is at its minimum size.
Maximum Interference = $D_{s,max} - D_{h,min}$
Maximum Interference = $20.05 - 19.90 = 0.15$ mm
Step 3: Determine the nature of the fit.
Clearance Fit: All possible assemblies result in a clearance. This means the minimum hole size is always larger than the maximum shaft size. If $D_{h,min} - D_{s,max}$ is positive, it's a clearance fit.
Minimum Clearance = $D_{h,min} - D_{s,max}$
Minimum Clearance = $19.90 - 20.05 = -0.15$ mm (Since this is negative, it's not a pure clearance fit).
Interference Fit: All possible assemblies result in an interference.
This means the maximum shaft size is always larger than the maximum hole size. If $D_{s,max} - D_{h,max}$ is positive, and $D_{s,min} - D_{h,min}$ is positive, it's an interference fit.
Maximum Interference = $0.15$ mm (Positive, indicating interference is possible).
Minimum Interference = $D_{s,min} - D_{h,max} = 19.85 - 20.20 = -0.35$ mm (This is negative, indicating clearance is possible, so it's not a pure interference fit). Transition Fit: This type of fit can result in either a clearance or an interference, depending on the actual sizes of the shaft and hole within their respective tolerance zones. This occurs when the maximum clearance is positive and the maximum interference is also positive (or equivalently, minimum clearance is negative and maximum clearance is positive).
In our case: Maximum Clearance = $0.35$ mm (Positive, indicating clearance is possible).
Maximum Interference = $0.15$ mm (Positive, indicating interference is possible).
Since both clearance and interference are possible depending on the actual dimensions, the fit is a transition fit. The final answer is \( \boxed{\text{3}} \).