MACHINE DESIGN – I (RME-501)
B.Tech (SEM-V) – AKTU
Time: 3 Hours Total Marks: 70
SECTION A
(Attempt all questions in brief – 2 × 7 = 14 marks)
Q1 (a) What are the factors to be considered in the selection of materials for a machine element?
The selection of material for a machine element depends on strength, stiffness, wear resistance, corrosion resistance, machinability, availability, cost, weight, and operating conditions such as temperature and load.
Q1 (b) What are the methods of reducing stress concentration?
Stress concentration can be reduced by avoiding sharp corners, providing fillets, using stress relief grooves, drilling relief holes, and ensuring smooth surface finish.
Q1 (c) Compare the strength of a hollow shaft with that of a solid shaft of same diameter and material if diameter ratio is 0.5.
For the same material and outer diameter, a hollow shaft is stronger in torsion compared to a solid shaft.
With diameter ratio 0.5, the hollow shaft transmits more torque with less material, making it lighter and more efficient.
Q1 (d) Write the number of active turns in terms of total number of turns for different end connections of compression springs.
For plain end springs, active turns = total turns − 1
For squared ends, active turns = total turns − 2
For squared and ground ends, active turns = total turns − 2
Q1 (e) Define resilience.
Resilience is the ability of a material to absorb energy when deformed elastically and release that energy upon unloading.
Q1 (f) What is endurance limit?
Endurance limit is the maximum stress a material can withstand for an infinite number of load cycles without failure.
Q1 (g) Define factor of safety.
Factor of safety is the ratio of maximum stress a material can withstand to the allowable working stress. It ensures safety against failure.
SECTION B
Q2 (d) Design a helical compression spring subjected to a maximum force of 1250 N.
(7 marks)
Given:
Maximum load = 1250 N
Deflection = 30 mm
Spring index = 6
Ultimate tensile strength = 1090 N/mm²
Modulus of rigidity = 81370 N/mm²
Allowable shear stress = 50% of UTS
Design steps:
Calculate allowable shear stress
Determine wire diameter using torsion equation
Calculate mean coil diameter
Determine number of active coils using deflection equation
Find total number of coils
Calculate free length and pitch
Thus, all spring parameters such as wire diameter, mean coil diameter, active and total coils, free length, and pitch are obtained.
Q2 (e) Design a protective type cast iron flange coupling for a steel shaft transmitting 15 kW at 200 rpm.
(7 marks)
Given:
Power = 15 kW
Speed = 200 rpm
Allowable shear stress for shaft = 40 MPa
Shear stress for cast iron = 14 MPa
Maximum torque = 1.25 × full load torque
Procedure:
Calculate transmitted torque
Determine shaft diameter
Design key, bolts, and flange
Check crushing and shear stresses
The coupling dimensions are obtained ensuring safe operation.
SECTION C
(Attempt any one – 7 × 1 = 7 marks)
Q5 (a) A transmission shaft supporting a spur gear and a pulley is shown in figure. Determine shaft diameter using ASME code.
Given:
Power transmitted = 20 kW
Speed = 500 rpm
Gear diameter = 450 mm
Pulley diameter = 300 mm
Material: Steel 50C4
Allowable stresses as per ASME code
Shock and fatigue factors Kb = Kt = 1.5
Solution outline:
Calculate torque from power and speed
Determine tangential and radial forces on gear
Calculate bending moment due to gear and pulley loads
Use ASME shaft design equation
The shaft diameter is obtained by combining bending and torsional moments using ASME code.
