Jawaharlal Nehru Technological University Hyderabad
Jawaharlal Nehru Technological University Hyderabad B.Tech III DESIGN OF MACHINE MEMBERS - I Previous Question Paper (August - 2022 - R13)
Answer any five questions
All questions carry equal marks
Note: Assume suitable data, if necessary.Main
Distinguish between the three types of fits.
Answer
Fits are classified based on the relative motion and tightness between mating parts:
- Clearance Fit: The internal member is always smaller than the external member, ensuring a positive clearance. This allows for relative motion between parts, such as in a sliding fit or running fit.
- Interference Fit: The internal member is always larger than the external member, creating a negative clearance. This results in a permanent or semi-permanent assembly where parts are held together by friction, such as in press fits or shrink fits.
- Transition Fit: The tolerance zones of the mating parts overlap, resulting in either a clearance or an interference depending on the actual sizes within the tolerance range. This is used for parts requiring accurate location but not necessarily relative motion, such as push fits.
Find the diameter of shaft required to transmit 60 kW at 150 rpm, if the maximum torque is likely to exceed the mean torque by 25% for a maximum permissible torsional shear stress of 60 N/mm2. Also find the angle of twist for a length of 2.5 meters. Take G = 80 GPa.
Answer
Given: Power (P) = 60 kW = 60,000 W; Speed (N) = 150 rpm; Max Torque = 1.25 × Mean Torque; Shear Stress (τ) = 60 N/mm2; Length (L) = 2.5 m = 2500 mm; Modulus of Rigidity (G) = 80 GPa = 80,000 N/mm2.
1. Calculate Mean Torque (Tmean):
2. Calculate Maximum Torque (Tmax):
3. Determine Shaft Diameter (d) using the Torsion Equation:
4. Calculate Angle of Twist (θ) in radians:
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. (Using Goodman’s Line).
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What are bolts of uniform strength?
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A circular steel bar 50 mm diameter and 200 mm long is welded perpendicularly to a steel plate to form a cantilever to be loaded with 5 kN at the free end. Determine the size of the weld, assuming the allowable stress in the weld as 100 MPa.
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A through bolt is used to connect two plates with a gasket in between. The combined stiffness of the parts held together by the bolt is three times the stiffness of the bolt. The bolted assembly is subjected to an external tensile force, which fluctuates from 0 to 5 kN. The initial preload in the bolt is 4.5 kN. The bolt has M8 coarse threads with a stress area of 36.6 mm2. The bolt is made of plain carbon steel 45C8 (Sut = 630 N/mm2 and Syt = 380 N/mm2). The fatigue stress concentration factor is 3. Determine the factor of safety used in the design for a reliability of 50%.
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Design a sleeve and cotter joint to resist a tensile load of 60 kN. All parts of the joints are made of the same material with the following allowable stresses: σt = 60 MPa, τ = 70 MPa, σc = 125 MPa. Make a dimensioned sketch of the joint.
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What are the functions of key? How is keyway made?
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A flat key is used to connect a pulley to a 45 mm diameter shaft. The standard cross section of the key is 14 × 9 mm. The key is made of commercial steel (Syt = Syc = 230 N/mm2) and the factor of safety is 3. Determine the length of the key on the basis of shear and compression considerations, if 15 kW power at 360 rpm is transmitted through the keyed joint. Assume (Ssy = 0.5 Syc).
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What are the advantages and disadvantages of bushed-pin flexible coupling?
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A propeller shaft is required to transmit 50 kW power at 600 rpm. It is a hollow shaft, having an inside diameter 0.8 times of the outside diameter. It is made of steel (Syt = 380 N/mm2) and the factor of safety is 4. Calculate the inside and outside diameters of the shaft. Assume (Ssy = 0.5 Syt).
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Derive an expression for bending stress in torsional helical spring.
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Design a compression helical spring to carry a load of 500 N with a deflection of 25 mm. The spring index may be taken as 8. Assume the following values of the spring material: Permissible shear stress = 350 MPa; Modulus of rigidity = 84 kN/mm2; Wahl’s factor = (4C-1)/(4C-4) + 0.615/C, where C is the spring index.
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