JNTU
JNTU B.Tech III PRINCIPLES OF MACHINE DESIGN Previous Question Paper (March 2021 - R13)
Answer any five questions. All questions carry equal marks.
Main
Consider the following states of biaxial stress: (a) 𝜎1 = 30 MPa, 𝜎2 = 0 MPa, (b) 𝜎1 = 30 MPa, 𝜎2 = -15 MPa, (c) 𝜎1 = 30 MPa, 𝜎2 = -30 MPa, (d) 𝜎1 = 30 MPa, 𝜎2 = 15 MPa, (e) pure shear, 𝜏 = 30 MPa. With the aid of a 𝜎1 - 𝜎2 plot, list these stress states in order of increasing likelihood of causing failure according to (i) the maximum-normal-stress theory, (ii) the maximum-shear-stress theory, and (iii) the maximum-distortion-energy theory. Assume Yield strength as 80 MPa.
Answer
To evaluate the likelihood of failure for the given stress states, we compare them against the yield strength 𝜎y = 80 MPa using the failure criteria boundaries plotted on a 𝜎1-𝜎2 plane.
The stress states are: (a) (30, 0), (b) (30, -15), (c) (30, -30), (d) (30, 15), and (e) pure shear (𝜎1=30, 𝜎2=-30). Note that (e) is identical to (c).
(i) Maximum-Normal-Stress Theory (Rankine): Failure occurs if |𝜎1| ≥ 𝜎y or |𝜎2| ≥ 𝜎y. Here, all 𝜎1 and 𝜎2 values are ≤ 30 MPa, which is well below 80 MPa. All states are equally safe (likelihood = 0).
(ii) Maximum-Shear-Stress Theory (Tresca): Failure occurs if the maximum shear stress 𝜏max ≥ 𝜎y/2 = 40 MPa. 𝜏max = max(|𝜎1-𝜎2|/2, |𝜎1|/2, |𝜎2|/2).
Calculated 𝜏max values:
(a) 15 MPa, (b) 22.5 MPa, (c) 30 MPa, (d) 15 MPa, (e) 30 MPa.
Order of increasing likelihood: (a) = (d) < (b) < (c) = (e).
(iii) Maximum-Distortion-Energy Theory (von Mises): Failure occurs if 𝜎1² - 𝜎1𝜎2 + 𝜎2² ≥ 𝜎y² = 6400.
Calculated values (𝜎1² - 𝜎1𝜎2 + 𝜎2²):
(a) 30² - 0 + 0 = 900
(b) 30² - (30)(-15) + (-15)² = 900 + 450 + 225 = 1575
(c) 30² - (30)(-30) + (-30)² = 900 + 900 + 900 = 2700
(d) 30² - (30)(15) + 15² = 900 - 450 + 225 = 675
(e) Same as (c) = 2700.
Order of increasing likelihood: (d) < (a) < (b) < (c) = (e).
What is meant by endurance strength of a material? How does the size, surface condition of a component and type of load affect such strength?
Answer
Endurance strength (or fatigue limit) is the maximum stress amplitude that a material can withstand for an infinite number of stress cycles without failure. It is determined by testing polished specimens under completely reversed loading.
The endurance strength is modified by several factors:
- Size factor: Larger components have a higher probability of containing internal defects or flaws, which reduces the endurance strength compared to a small test specimen.
- Surface condition: Surface finish significantly affects fatigue life. Rough surfaces act as stress concentrators; therefore, polished surfaces yield higher endurance strength than ground, machined, or forged surfaces.
- Type of load: Endurance strength varies with the mode of loading. Rotating bending, axial loading, and torsional loading produce different stress distributions, with axial loading typically showing lower endurance strength than bending.
Calculate the fatigue strength for the specimen made of 30C8 steel with endurance strength 280 MPa and ultimate strength 600 MPa for a life of 200,000 cycles of stress reversals. What would be the life of a specimen, if the fluctuating reversible stress is taken as 420 MPa?
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A welded connection as shown in figure 1 is subjected to an eccentric force of 60 kN in the plane of the welds. Determine the size of the welds, if the permissible shear stress for the weld is 100 MPa. All the dimensions are in mm only.
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Design and draw a cotter joint to support a load varying from 30 kN in compression to 30 kN in tension. The material used is carbon steel for which the following allowable stresses may be used. The load is applied statically. Tensile stress = compressive stress = 50 MPa; shear stress = 35 MPa and crushing stress = 90 MPa.
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Determine the percentage increase in horse power capacity made possible in changing from a flat belt drive to a V-belt drive. The diameter of the flat belt pulley is same as the pitch diameter of grooved pulley. Speed of rotation is also same. The coefficient of friction for flat belt as well as V-belts is 0.3. Groove angle is 60°. Belts are of the same material and same cross sectional area. In each case the angle of wrap is 150°.
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A chain drive with pitch 25.4 mm rotates at 750 rpm. The number of teeth of driving sprocket is 23. Determine the average speed, the maximum and the minimum speed.
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A belt 100 mm wide and 10 mm thick is transmitting power at 1000 m/min. The net driving tension is 1.8 times the tension on the slack side. If the safe permissible stress on the belt section in 1.6 MPa, calculate the maximum power that can be transmitted at this speed. Assume density of the leather as 1000 kg/m³.
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The pitch circle diameters of the pinion and gear are 100 mm and 300 mm respectively. The pinion is made of plain carbon steel 40C8 while gear is made of grey cast iron FG 300. The pinion receives 5 kW power at 500 rpm through its shafts. The service factor and factor of safety can be taken as 1.5 each. The face width of the gear is 10 times module. Calculate module, the number of teeth on the pinion and gear, Beam strength and Wear strength.
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A 30 BC 04 deep groove ball bearing is subjected to the following load cycle:
| Fraction of Cycle | Radial Load (kN) | Axial Load (kN) | RPM | Load Condition |
|---|---|---|---|---|
| 0.2 | 5 | 2.5 | 700 | Steady |
| 0.5 | 6 | 1.5 | 1000 | Light Shock |
| 0.3 | 3 | 2 | 1500 | Steady |
The inner ring rotates. Determine the equivalent load, the rating life and mean life.
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