JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD
JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech II Year II Semester DESIGN OF MACHINE MEMBERS I Previous Question Paper (March - 2022 - R16)
Answer any five questions. All questions carry equal marks.
Main
A hollow shaft is required to transmit 600 kW at 110 r.p.m., the maximum torque being 20% greater than the mean. The shear stress is not to exceed 63 MPa and twist in a length of 3 meters not to exceed 1.4 degrees. Find the external diameter of the shaft, if the internal diameter to the external diameter is 3/8. Take modulus of rigidity as 84 GPa.
Answer
- Given Data: Power (P) = 600 kW = 600,000 W; Speed (N) = 110 r.p.m.; Max Torque (Tmax) = 1.2 × Mean Torque (Tmean); Allowable Shear Stress (τ) = 63 MPa = 63 N/mm2; Length (L) = 3 m = 3000 mm; Angle of twist (θ) = 1.4° = 1.4 × (π/180) radians = 0.0244 rad; Modulus of Rigidity (G) = 84 GPa = 84,000 N/mm2; Ratio (k = di/do) = 3/8 = 0.375.
- Mean Torque: Tmean = (P × 60 × 106) / (2 × π × N) = (600,000 × 60) / (2 × π × 110) = 52,087.6 N-m = 52.088 × 106 N-mm.
- Max Torque: Tmax = 1.2 × 52.088 × 106 = 62.506 × 106 N-mm.
- Design based on Strength: τ = (16 × Tmax × do) / (π × (do4 - di4)) = (16 × Tmax) / (π × do3 × (1 - k4)). Substituting values: 63 = (16 × 62.506 × 106) / (π × do3 × (1 - 0.3754)). Solving for do: do3 = (16 × 62.506 × 106) / (π × 63 × 0.9823) = 5,123,790. do = 172.4 mm.
- Design based on Rigidity: θ = (Tmax × L) / (G × J), where J = (π/32) × do4 × (1 - k4). 0.0244 = (62.506 × 106 × 3000) / (84,000 × (π/32) × do4 × (1 - 0.3754)). Solving for do: do4 = (62.506 × 106 × 3000 × 32) / (84,000 × π × 0.0244 × 0.9823) = 9.47 × 108. do = 175.7 mm.
- Conclusion: Selecting the larger value for safety, the external diameter do = 175.7 mm.
A shaft is subjected to a bending load of 3 kN, pure torque of 1000 N-m and an axial pulling force of 15 kN. Calculate the stresses at A and B.
Answer
- Given Data: Bending Load (W) = 3 kN = 3000 N; Torque (T) = 1000 N-m = 106 N-mm; Axial Force (P) = 15 kN = 15,000 N. Note: As the shaft dimensions (diameter) are not provided, the stresses are expressed in terms of the shaft diameter d.
- Axial Stress (σa): σa = P / Area = 15,000 / (π/4 × d2) = 19,098.6 / d2 N/mm2.
- Bending Stress (σb): Assuming a cantilever or simply supported beam configuration is required to calculate bending moment (M). Assuming a standard length L for the bending moment M = W × L. If L is not provided, M = 3000 × L. σb = (32 × M) / (π × d3).
- Torsional Shear Stress (τ): τ = (16 × T) / (π × d3) = (16 × 106) / (π × d3) = 5,092,958 / d3 N/mm2.
- Resultant Stresses: The principal stresses are calculated using the formula σ1,2 = (σb + σa)/2 ± √[((σb + σa)/2)2 + τ2]. The maximum shear stress is τmax = √[((σb + σa)/2)2 + τ2].
- Note: Without the specific shaft diameter and beam length, the final numerical values for stresses cannot be computed. The student must substitute the specific diameter 'd' and bending moment 'M' into these derived expressions.
What is the difference between failure due to static load and fatigue failure? Explain.
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A hot rolled steel shaft is subjected to a torsional moment that varies from +350 Nm to -115 Nm and an applied bending moment at a critical section varies from 445 Nm to 225 Nm. The shaft is of uniform cross-section. Determine the required shaft diameter. The material has an ultimate strength of 550 MPa and yield strength of 410 MPa. Take the endurance limit as half the ultimate strength, factor of safety of 2, size factor of 0.85 and a surface finish factor of 0.62.
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A double riveted lap joint with chain riveting is to be made for joining two plates 10 mm thick. The allowable stresses are: σt= 60 MPa; τ = 50 MPa and σc = 80 MPa. Find the rivet diameter, pitch of rivets and distance between rows of rivets. Also, find the efficiency of the joint.
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What is the difference between caulking and fullering? Explain with the help of neat sketches.
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Design a cotter joint to connect two mild steel rods for a pull of 30 kN. The maximum permissible stresses are 55N/mm2 in tension, 40N/mm2 in shear and 70N/mm2 in crushing. Draw a neat sketch of the joint.
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Briefly explain the procedure to design a shaft based on any two theories of failures.
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Discuss the function of a coupling. Give at least three practical applications.
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Design and draw a muff coupling to transmit 50 HP at 120 rpm. The shaft and key are made of the same material having allowable shear stress of 30N/mm2 and compressor stress of 80N/mm2. The flange is made, as cast Iron with allowable shear stress is 15N/mm2.
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A bracket carries a load of 10 kN. Find the size of the weld if the allowable shear stress is not to exceed 80 MPa.
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A truck spring has 12 leaves, two of which are full length leaves. The spring supports are 1.05 m apart and the central band is 85 mm wide. The central load is to be 5.4 kN with permissible stress of 280 MPa. Determine the thickness and width of the steel spring leaves. The ratio of the total depth to the width of the spring is 3. Also, determine the deflection of the spring.
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A machinery shaft is subjected to torsion only. The bearings are 2.50 meters apart. The shaft transmits 190 kW at 220 rev/min. Allow shear stress of 45 MPa after an allowance for keyways. i) Calculate the shaft diameter for steady loading and ii) Calculate the shaft diameter if the load is suddenly applied with minor shocks.
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Write the applications of sleeve and cotter joint.
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Design a knuckle joint to transmit 150 kN. The design stresses may be taken as 75 MPa in tension, 60 MPa in shear and 150 MPa in compression.
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Explain the modified Goodman diagram for torsional shear stresses.
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