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In a hole and shaft combination of 25 mm nominal size H7 hole limits are +0.021 mm and —0.000 mm E8 shaft limits are —0.040 mm and —0.073 mm The allowance (in microns) during the assembly is
Correct answer is '40'. Can you explain this answer?
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In a hole and shaft combination of 25 mm nominal size H7 hole limits ...
To determine the allowance during the assembly of a hole and shaft combination, we need to consider the hole limits and the shaft limits. The allowance is the difference between the maximum size of the hole and the minimum size of the shaft.
Given data:
Hole limits (H7): 0.021 mm (positive) and -0.000 mm (negative)
Shaft limits (E8): -0.040 mm (negative) and -0.073 mm (negative)
The maximum size of the hole is obtained by adding the positive hole limit to the nominal size:
Maximum hole size = Nominal size + Positive hole limit
Maximum hole size = 25 mm + 0.021 mm
Maximum hole size = 25.021 mm
The minimum size of the shaft is obtained by subtracting the negative shaft limit from the nominal size:
Minimum shaft size = Nominal size - Negative shaft limit
Minimum shaft size = 25 mm - (-0.073 mm)
Minimum shaft size = 25 mm + 0.073 mm
Minimum shaft size = 25.073 mm
The allowance is the difference between the maximum hole size and the minimum shaft size:
Allowance = Maximum hole size - Minimum shaft size
Allowance = 25.021 mm - 25.073 mm
Allowance = -0.052 mm
Since the allowance is negative, we need to consider its absolute value:
Allowance = |-0.052 mm|
Allowance = 0.052 mm
The allowance during the assembly is 0.052 mm or 52 microns.
Given data:
Hole limits (H7): 0.021 mm (positive) and -0.000 mm (negative)
Shaft limits (E8): -0.040 mm (negative) and -0.073 mm (negative)
The maximum size of the hole is obtained by adding the positive hole limit to the nominal size:
Maximum hole size = Nominal size + Positive hole limit
Maximum hole size = 25 mm + 0.021 mm
Maximum hole size = 25.021 mm
The minimum size of the shaft is obtained by subtracting the negative shaft limit from the nominal size:
Minimum shaft size = Nominal size - Negative shaft limit
Minimum shaft size = 25 mm - (-0.073 mm)
Minimum shaft size = 25 mm + 0.073 mm
Minimum shaft size = 25.073 mm
The allowance is the difference between the maximum hole size and the minimum shaft size:
Allowance = Maximum hole size - Minimum shaft size
Allowance = 25.021 mm - 25.073 mm
Allowance = -0.052 mm
Since the allowance is negative, we need to consider its absolute value:
Allowance = |-0.052 mm|
Allowance = 0.052 mm
The allowance during the assembly is 0.052 mm or 52 microns.
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Community Answer
In a hole and shaft combination of 25 mm nominal size H7 hole limits ...
Allowance = difference between maximum material limits = Lhole — Hshaft = 25.00 — 24.96 = 0.040 mm = 40 microns
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In a hole and shaft combination of 25 mm nominal size H7 hole limits are +0.021 mm and —0.000 mm E8 shaft limits are —0.040 mm and —0.073 mm The allowance (in microns) during the assembly isCorrect answer is '40'. Can you explain this answer?
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In a hole and shaft combination of 25 mm nominal size H7 hole limits are +0.021 mm and —0.000 mm E8 shaft limits are —0.040 mm and —0.073 mm The allowance (in microns) during the assembly isCorrect answer is '40'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In a hole and shaft combination of 25 mm nominal size H7 hole limits are +0.021 mm and —0.000 mm E8 shaft limits are —0.040 mm and —0.073 mm The allowance (in microns) during the assembly isCorrect answer is '40'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a hole and shaft combination of 25 mm nominal size H7 hole limits are +0.021 mm and —0.000 mm E8 shaft limits are —0.040 mm and —0.073 mm The allowance (in microns) during the assembly isCorrect answer is '40'. Can you explain this answer?.
In a hole and shaft combination of 25 mm nominal size H7 hole limits are +0.021 mm and —0.000 mm E8 shaft limits are —0.040 mm and —0.073 mm The allowance (in microns) during the assembly isCorrect answer is '40'. Can you explain this answer? for Mechanical Engineering 2024 is part of Mechanical Engineering preparation. The Question and answers have been prepared according to the Mechanical Engineering exam syllabus. Information about In a hole and shaft combination of 25 mm nominal size H7 hole limits are +0.021 mm and —0.000 mm E8 shaft limits are —0.040 mm and —0.073 mm The allowance (in microns) during the assembly isCorrect answer is '40'. Can you explain this answer? covers all topics & solutions for Mechanical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a hole and shaft combination of 25 mm nominal size H7 hole limits are +0.021 mm and —0.000 mm E8 shaft limits are —0.040 mm and —0.073 mm The allowance (in microns) during the assembly isCorrect answer is '40'. Can you explain this answer?.
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