Tolerances and Stress Analysis Module 1 PDF

Tolerances and Stress Analysis Module 1 INTENDED LEARNING OUTCOMES By the end of this module, the learner is intended to: 1. Determine the tolerances and allowances of mating parts. 2. Calculate design factors and allowable stresses in the analysis of machine elements. 3. Analyze machine compon...

Tolerances and Stress Analysis Module 1 INTENDED LEARNING OUTCOMES By the end of this module, the learner is intended to: 1. Determine the tolerances and allowances of mating parts. 2. Calculate design factors and allowable stresses in the analysis of machine elements. 3. Analyze machine components under axial, shear, bending, and torsional loads.  Subtopic 1 Engineering Tolerances and Allowances TERMINOLOGY Nominal size: Designated dimension Tolerance: Permissible variation in dimension Limits: Extreme permissible size Example: Ø28 ± 0.2 Part can be anywhere between limits Ø27.8 and Ø28.2 TERMINOLOGY Allowance: Difference between dimensions of mating parts Positive: Shaft < Hole (Clearance Fit) Negative: Shaft > Hole (Interference Fit) WHY TOLERANCE? Ensures functionality Minimizes cost Prevents delays SAMPLE PROBLEM The dimensions of the mating parts are given as follows: Hole: 25.00 mm Shaft: 24.97 mm 25.02 mm 24.95 mm Find the hole tolerance, shaft tolerance and allowance. SAMPLE PROBLEM The dimensions of the mating parts are given as follows: Hole: 25.00 mm Shaft: 24.97 mm 25.02 mm 24.95 mm Find the hole tolerance, shaft tolerance and allowance. Solution: Hole tolerance = Upper limit of hole – Lower limit of hole = 25.02 𝑚𝑚 − 25.00 𝑚𝑚 = 0.02 𝑚𝑚 Shaft tolerance = Upper limit of shaft – Lower limit of shaft = 24.97 𝑚𝑚 − 24.95 𝑚𝑚 = 0.02 𝑚𝑚 Allowance = Lower limit of hole – Upper limit of hole = 25.00 𝑚𝑚 − 24.97 𝑚𝑚 = 0.03 𝑚𝑚 KNOWLEDGE CHECK 1. Find the total tolerance on the following dimension: 1.134 ± 0.008. Answer: 0.016 2. Determine the limits of the following dimension: 3.375+0.015 −0.007. Answer: Upper limit: 3.390 Lower limit: 3.368  Subtopic 2 Stress Analysis STRESS AND STRENGTH Strength (S) Stress (𝛔) Property: due to structure, State: due to thermal or processing, and treatment mechanical loading Test data Calculated FACTOR OF SAFETY 𝐹𝑎𝑖𝑙𝑢𝑟𝑒 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝐹𝑆 = 𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑜𝑟 𝑑𝑒𝑠𝑖𝑔𝑛 𝑠𝑡𝑟𝑒𝑠𝑠 Consideration Lower FS Higher FS Effect of failure Minor inconvenience High financial loss or danger to life Type of load Static or well-defined forces Impact or variable forces Force analysis Known forces Uncertain or unpredictable forces Cost Cost-sensitive parts Reliability and safety prioritized Stable, non-corrosive, Service conditions Corrosive or high temperatures normal temperatures SAMPLE PROBLEM A rod loaded with 2000 lbf axial force undergoes stress of σ = F∕A. Given a material strength of 24 kpsi and a factor of safety of 3.0, determine the minimum diameter of the rod. SAMPLE PROBLEM A rod loaded with 2000 lbf axial force undergoes stress of σ = F∕A. Given a material strength of 24 kpsi and a factor of safety of 3.0, determine the minimum diameter of the rod. Solution: 𝑆 𝐹𝑆 = 𝜎 24000 𝑝𝑠𝑖 3= 𝜎 𝜎 = 8000 𝑝𝑠𝑖 𝐹 𝐹 𝜎= = 𝜋𝑑2 𝐴 4 2000 𝑙𝑏𝑓 8000 𝑝𝑠𝑖 = 𝜋𝑑2 4 𝑑 = 0.564 𝑖𝑛 STRESS ANALYSIS Normal stress Shear stress 𝐹 𝐹 𝜎= 𝜏= 𝐴⊥ 𝐴∥ STRESS ANALYSIS 𝑀𝑐 Bending stress 𝜎 = 𝐼 Solid circular section: 32𝑀 where 𝑀 = bending moment 𝜎= 𝑑 = diameter 𝜋𝑑 3 Hollow circular section: 32𝑀𝑑𝑜 𝜎= 𝜋 𝑑𝑜4 − 𝑑𝑖4 STRESS ANALYSIS 𝑇𝑟 Torsional shear stress 𝜏 = 𝐽 Solid circular section: where 𝑇 = torque 16𝑇 𝜏= 𝑑 = diameter 𝜋𝑑 3 2𝜋𝑇𝑁 𝑃= 60 Hollow circular section: 16𝑇𝑑𝑜 𝜏= 𝜋 𝑑𝑜4 − 𝑑𝑖4 SAMPLE PROBLEMS 1. An 80 kN pull is transmitted to the bars through a pin. If the bar yield strength is 200 MPa and the pin shear yield strength is 160 MPa, find the diameters of bar and pin. Consider factor of safety of 2. SAMPLE PROBLEMS 1. An 80 kN pull is transmitted to the bars through a pin. If the bar yield strength is 200 MPa and the pin shear yield strength is 160 MPa, find the diameters of bar and pin. Consider factor of safety of 2. Solution: a. Diameter of bars 𝑆𝑦 𝐹𝑆 = 𝜎 200 𝑀𝑃𝑎 2= 𝜎 𝜎 = 100 𝑀𝑃𝑎 𝐹 𝜎= 𝐴 80×103 𝑁 100 𝑀𝑃𝑎 = 𝜋 2 𝑑 4 𝑏 𝑑𝑏 = 32 𝑚𝑚 SAMPLE PROBLEMS 1. An 80 kN pull is transmitted to the bars through a pin. If the bar yield strength is 200 MPa and the pin shear yield strength is 160 MPa, find the diameters of bar and pin. Consider factor of safety of 2. b. Diameter of pin 𝑆𝑠𝑦 𝐹𝑆 = 𝜏 160 𝑀𝑃𝑎 2= 𝜏 𝜏 = 80 𝑀𝑃𝑎 𝑉 𝜏= 𝟐𝐴 80×103 𝑁 80 𝑀𝑃𝑎 = 𝜋 2 𝟐× 4 𝑑𝑝 𝑑𝑝 = 25.2 𝑚𝑚 SAMPLE PROBLEMS 2. The pump lever exerts forces of 25 kN and 35 kN as shown. Find the diameter of the shaft if the stress is not to exceed 100 MPa. SAMPLE PROBLEMS 2. The pump lever exerts forces of 25 kN and 35 kN as shown. Find the diameter of the shaft if the stress is not to exceed 100 MPa. Solution: ∑𝑀𝐴 = 0 𝑅𝐵 × 950 𝑚𝑚 − 35 𝑘𝑁 × 750 𝑚𝑚 − 25 𝑘𝑁 × 150 𝑚𝑚 = 0 𝑅𝐵 = 31.58 𝑘𝑁 = 31.58 × 103 𝑁 ∑𝐹𝑦 = 0 𝑅𝐴 + 𝑅𝐵 − 25𝑘𝑁 − 35𝑘𝑁 = 0 𝑅𝐴 = 28.42 × 103 𝑁 SAMPLE PROBLEMS 2. The pump lever exerts forces of 25 kN and 35 kN as shown. Find the diameter of the shaft if the stress is not to exceed 100 MPa. 𝑀 = 6.316 × 106 𝑁 𝑚𝑚 32𝑀 𝜎= 𝜋𝑑 3 32(6.316×106 𝑁 𝑚𝑚) 100 𝑀𝑃𝑎 = 𝜋𝑑 3 𝑑 = 86.3 𝑚𝑚 SAMPLE PROBLEMS 3. A shaft is transmitting 97.5 kW at 180 rpm. If the allowable shear stress in the material is 60 MPa, find the suitable diameter for the shaft. SAMPLE PROBLEMS 3. A shaft is transmitting 97.5 kW at 180 rpm. If the allowable shear stress in the material is 60 MPa, find the suitable diameter for the shaft. Solution: 2𝜋𝑇𝑁 𝑃= 60 2𝜋×𝑇×180 𝑟𝑝𝑚 97.5 × 103 𝑊 = 60 𝑇 = 5172 𝑁 𝑚 = 5172 × 103 𝑁 𝑚𝑚 16𝑇 𝜏= 𝜋𝑑 3 16(5172×103 𝑁 𝑚𝑚) 60 MPa = 𝜋𝑑 3 𝑑 = 76 𝑚𝑚 KNOWLEDGE CHECK 1. Calculate the allowable stress for an object with failure strength of 300 MPa given a factor of safety of 1.5. Answer: 200 MPa 2. Find the diameter of the pin with permissible shear stress of 75 MPa if a pull of F = 70 kN is transmitted across the plate. Answer: 24.38 mm KNOWLEDGE CHECK 3. Determine maximum moment of a 48-in. beam with a static load of 6000 pounds at the center. Answer: 72,000 lb-in 4. Find the torque that can be applied to a 1-in diameter solid shaft if the shearing stress is not to exceed 8000 psi. Answer: 1570 lb-in Q&A SESSION