Limit, Fit & Tolerance – Numerical Problems

A medium force fit on a 75 mm shaft requires a hole tolerance and shaft tolerance each equal to .225 mm and an maximum interference of 0.0375 mm. Determine the proper hole and shaft dimension with the basis hole standard.

A medium force fit on a 75 mm shaft requires a hole tolerance and shaft tolerance each equal to .225 mm and an maximum interference of 0.0375 mm. Determine the proper hole and shaft dimension with the basis hole standard. Minimum diameter (lower limit) of hole: Dmin = 75 mm Maximum diameter (higher limit) of hole: Dmax = 75 +0.225 = 75.225 mm Minimum diameter (lower limit) of shaft: dmin = Dmax + maximum interference dmin = 75.225 + 0.0375 = 75.2625 mm Maximum diameter (higher limit) of shaft: dmax = dmin + tolerance = 75.2625 + .225 = 75.4875 mm

2. A 75 mm shaft rotates in a bearing 2. A 75 mm shaft rotates in a bearing. The tolerance for both shaft and bearing is 0.075 mm and the required allowance is 0.10 mm. Determine the dimension of the shaft, and the bearing bore with the basis hole standard.

2. A 75 mm shaft rotates in a bearing 2. A 75 mm shaft rotates in a bearing. The tolerance for both shaft and bearing is 0.075 mm and the required allowance is 0.10 mm. Determine the dimension of the shaft, and the bearing bore with the basis hole standard.

3. In a limit system, the following limits are specified to give a clearance fit between a shaft and a hole: Determine:(a)basic size (b) shaft and hole tolerance (c) the maximum and minimum clearance

3. In a limit system, the following limits are specified to give a clearance fit between a shaft and a hole: Determine:(a)basic size (b) shaft and hole tolerance (c) Shaft and hole limit (d) the maximum and minimum clearance Basic Size: 30 mm Shaft tolerance: Tolerance = upper limit – lower limit = -0.005 – (-0.018) = 0.013 mm Hole Tolerance = 0.02 mm (c) Shaft Limit: High limit (dmax) = 30-0.005 = 29.995 mm Low Limit(dmin) = 30-0.018 = 29.982 mm Hole Limit: High Limit (Dmax) = 30+0.02 = 30.02 mm Low Limit (dmin) = 30 mm (d) Maximum Clearance = Dmax- dmin = 30.02 – 29.982 = 0.038 mm Minimum Clearance = Dmin – dmax = 30 – 29.995 = 0.005 mm

4. . A hole and shaft have a basic size of 25 mm, and are to have a clearance fit with a maximum clearance of 0.02 mm and a minimum clearance of 0.01 mm. The hole tolerance is to be 1.5 times the shaft tolerance. Determine: limit for both hole and shaft (a) using a hole basis system (b) using a shaft basis system.

4. . A hole and shaft have a basic size of 25 mm, and are to have a clearance fit with a maximum clearance of 0.02 mm and a minimum clearance of 0.01 mm. The hole tolerance is to be 1.5 times the shaft tolerance. Determine: limit for both hole and shaft (a) using a hole basis system (b) using a shaft basis system.