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Questions and Answers
What is the purpose of tolerances in engineering design?
What is the purpose of tolerances in engineering design?
If a hole has a nominal size of 30 mm with a tolerance of ±0.5 mm, what are the limits of this hole dimension?
If a hole has a nominal size of 30 mm with a tolerance of ±0.5 mm, what are the limits of this hole dimension?
What is the allowable stress in the analysis of machine elements primarily intended for?
What is the allowable stress in the analysis of machine elements primarily intended for?
In a positive allowance system, how should the shaft be related to the hole?
In a positive allowance system, how should the shaft be related to the hole?
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Given a shaft with an upper limit of 25.00 mm and a lower limit of 24.90 mm, what is the shaft tolerance?
Given a shaft with an upper limit of 25.00 mm and a lower limit of 24.90 mm, what is the shaft tolerance?
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What is the calculated stress for a rod under a 2000 lbf axial force given an area of 1 in²?
What is the calculated stress for a rod under a 2000 lbf axial force given an area of 1 in²?
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If a material has a strength of 24 kpsi and a factor of safety of 3.0, what is the allowable stress?
If a material has a strength of 24 kpsi and a factor of safety of 3.0, what is the allowable stress?
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In what situation would a higher factor of safety be required?
In what situation would a higher factor of safety be required?
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Which type of force typically does not require a higher factor of safety?
Which type of force typically does not require a higher factor of safety?
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Given a rod subjected to a maximum stress of $\sigma = 8 kpsi$, what would be the minimum diameter to ensure a factor of safety of 3.0 with a material strength of 24 kpsi?
Given a rod subjected to a maximum stress of $\sigma = 8 kpsi$, what would be the minimum diameter to ensure a factor of safety of 3.0 with a material strength of 24 kpsi?
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Study Notes
Intended Learning Outcomes
- Determine tolerances and allowances essential for mating parts in engineering.
- Calculate design factors and allowable stresses for machine elements.
- Analyze machine components subjected to axial, shear, bending, and torsional loads.
Engineering Tolerances and Allowances
- Nominal size: The designated dimension of a part.
- Tolerance: The permissible variation in dimensional limits.
- Limits: The extreme permissible dimensions, e.g., Ø28 ± 0.2 indicates a range of Ø27.8 to Ø28.2.
- Allowance: The difference between mating parts’ dimensions leading to either a clearance fit (shaft < hole) or interference fit (shaft > hole).
- Tolerances ensure functionality, minimize costs, and prevent delays.
Sample Problem on Tolerances
- Given hole dimensions 25.00 mm to 25.02 mm and shaft dimensions 24.95 mm to 24.97 mm:
- Hole tolerance: 0.02 mm (25.02 mm - 25.00 mm)
- Shaft tolerance: 0.02 mm (24.97 mm - 24.95 mm)
- Allowance: 0.03 mm (25.00 mm - 24.97 mm)
Stress Analysis
- Strength (S): Derived from structure and treatment of materials.
- Stress (σ): Caused by thermal or mechanical loading and is typically calculated.
Factor of Safety (FS)
- Defined as FS = Failure strength / Allowable stress.
- A higher FS is necessary for scenarios with significant risks, dynamic loads, or corrosive environments.
- Important for ensuring reliability, particularly in cost-sensitive components.
Sample Problem on Stress
- For a rod with 2000 lbf axial force, material strength of 24 kpsi and FS of 3.0:
- Calculated stress (σ) is 8000 psi.
- Minimum diameter calculated as 0.564 inches.
Stress Forms
- Normal Stress (σ): σ = F / A⊥.
- Shear Stress (τ): τ = F / A∥.
-
Bending Stress:
- Solid section: σ = (32M)/(πd³).
- Hollow section: σ = (32Mdₒ)/(π(dₒ⁴ - dᵢ⁴)).
-
Torsional Shear Stress:
- Solid section: τ = (16T)/(πd³).
- Hollow section: τ = (16Tdₒ)/(π(dₒ⁴ - dᵢ⁴)).
Sample Problems
- When a force of 80 kN is applied through a bar and pin, calculate diameters considering factors of safety.
- For the bar with yield strength of 200 MPa, diameter (dₗ) is calculated to be 32 mm.
- For the pin with shear yield strength of 160 MPa, diameter (dₚ) is calculated to be 25.2 mm.
- For a lever with forces of 25 kN and 35 kN:
- Calculate the necessary shaft diameter to not exceed stress limit of 100 MPa, yielding a diameter of 86.3 mm.
Knowledge Check Example
- Allowable stress for a failure strength of 300 MPa with FS of 1.5 is 200 MPa.
- Diameter of a pin transmitting a pull of 70 kN with permissible shear stress of 75 MPa calculates to 24.38 mm.
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Description
This quiz focuses on the fundamental concepts covered in Module 1 of Tolerances and Stress Analysis. It involves determining tolerances and allowances, calculating design factors, and analyzing machine components under various loads. Prepare to assess your understanding of engineering tolerances and stress analysis concepts.
