MACHINE DESIGN – I (RME-501)
B.Tech (SEM-V) – AKTU
SECTION A
(Attempt all questions – 2 × 7 = 14 marks)
Q1 (a) What is 20Cr18Ni2 designation of steel?
20Cr18Ni2 is an alloy steel designation. The number 20 indicates 0.20% carbon content. Cr18 indicates 1.8% chromium and Ni2 indicates 0.2% nickel. Chromium increases hardness and corrosion resistance, while nickel improves toughness and strength.
Q1 (b) Discuss how shaft is designed for rigidity.
A shaft is designed for rigidity to limit angular twist and deflection. This is achieved by controlling torsional stiffness, selecting suitable material, increasing shaft diameter, and limiting allowable angle of twist to ensure proper functioning of machine elements mounted on the shaft.
Q1 (c) Explain modified Goodman diagram for torsional shear stresses.
The modified Goodman diagram is used for fatigue analysis under fluctuating torsional loads. It relates mean shear stress and alternating shear stress to determine safe design limits by considering endurance strength and yield strength of the material.
Q1 (d) What is self-locking of power screw? What is the condition for self-locking?
Self-locking of a power screw means the load will not move downwards without application of external torque.
Condition for self-locking is:
Coefficient of friction > tan helix angle
Q1 (e) Distinguish between hot and cold riveting.
Hot riveting is done by inserting a red-hot rivet which contracts on cooling, producing tight joints. Cold riveting is done at room temperature and is used for smaller rivets and lighter structures.
Q1 (f) What is Kennedy key? Give its applications.
A Kennedy key consists of two square keys placed at right angles. It is used to transmit high torque. Applications include heavy machinery shafts, marine engines, and rolling mills.
Q1 (g) What is Wahl factor? Why is it used?
Wahl factor is a stress correction factor used in helical springs. It accounts for curvature and direct shear effects. It is used to calculate maximum shear stress accurately in spring wires.
SECTION B
(Attempt any three – 7 × 3 = 21 marks)
Q2 (a) Calculate factor of safety using different failure theories.
Given:
Material: Steel 45C8
σx = 100 N/mm², σy = 40 N/mm², τxy = 80 N/mm²
Ultimate tensile strength Sut = 380 N/mm²
Using:
Maximum normal stress theory
Maximum shear stress theory
Maximum distortion energy theory
Principal stresses are calculated using stress transformation equations.
Factor of safety is obtained by comparing equivalent stress with material strength. Distortion energy theory generally gives most reliable results for ductile materials.
Q2 (b) Design of shaft subjected to combined bending and torsion.
The shaft is subjected to bending moment due to transverse loads and torque due to power transmission. Equivalent twisting moment and bending moment are calculated using shock and fatigue factors. Shaft diameter is determined using maximum shear stress theory to ensure safe operation.
Q2 (c) Explain design of riveted joints.
Design of riveted joints involves determining rivet size, number of rivets, pitch, and joint efficiency. Failure modes include tearing of plate, shearing of rivets, and crushing of rivets. Joint design ensures adequate strength against all failure modes.
SECTION C
(Attempt any two – one from each unit, 7 × 2 = 14 marks)
Q3 (a) A vertical load Py = 20 kN is applied at the free end of a cylindrical bar of radius 50 mm. Determine principal and maximum shear stresses at points A, B, and C.
The bar experiences bending and shear stresses. Normal stress due to bending and shear stress due to applied load are calculated at given points. Using Mohr’s circle equations, principal stresses and maximum shear stresses are determined for points A, B, and C.
Q4 (a) A solid circular shaft of 15 mm diameter is subjected to torsional shear stress. Explain stress distribution.
In a solid circular shaft, torsional shear stress varies linearly from zero at the center to maximum at the surface. The relationship between torque, shear stress, and polar moment of inertia is used to determine stress distribution.
Q6 (a) Design helical compression spring for railway wagon buffer.
Given data:
Velocity = 2 m/s
Mass = 1000 kg
Compression = 150 mm
Spring index = 6
UTS = 1500 N/mm²
Maximum force on each spring is calculated using energy method. Wire diameter, mean coil diameter, and number of active coils are determined using allowable shear stress and Wahl factor. The spring is designed to safely absorb kinetic energy.
