ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersSizing the parts of farm machines so they do not fail: shafts, keys, joints, springs, bearings, drives and overload devices.
In the exams: GATE AG Section 2 Machine Design (belts, spur gears, shear pins recur); ICAR AICE subject 39 Units 1–2; NCEES PE machinery component design
ICAR AIEEA-PG — Agricultural Engineering & Technology
Units III and IV (theory of machines, strength of materials) and Unit V (farm machinery): definitions, theories of failure, stress concentration, key and coupling types, bearing life, spring formulas and short calculations.
How our questions for ICAR AIEEA-PG spread across the units of Machine Design. Difficulty: 39 easy, 38 medium, 10 hard.
| Belt and chain drives | 11 | |
| The design process, materials and factor of safety | 9 | |
| Keys and couplings | 9 | |
| Springs | 8 | |
| Spur gear design (Lewis and Buckingham) | 8 | |
| Stress concentration and fatigue (Goodman and Soderberg) | 7 | |
| Shafts under combined bending and torsion | 7 | |
| Knuckle and cotter joints | 6 | |
| Welded joints | 6 | |
| Rolling and sliding bearings and bearing life | 6 | |
| Overload safety devices: shear bolts and slip clutches | 5 | |
| Measurement of force, torque and strain on machine elements | 3 | |
| Bolted joints | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Steel with a tensile yield strength of 400 MPa is used for a part in pure shear. According to the distortion energy theory, what is its yield strength in shear?
D. 231 MPa
Distortion energy theory gives Ssy = Syt / sqrt(3) = 0.577 x 400 = 231 MPa. 200 MPa is the maximum shear stress theory value (0.5 Syt). 283 MPa is 400 / sqrt(2), which is not a failure theory result.
Further reading: Budynas & Nisbett, Shigley's Mechanical Engineering Design
A shaft is specified in AISI/SAE 1045 steel. What does the designation tell the designer?
B. It is a plain carbon steel with about 0.45 per cent carbon
In the AISI/SAE four-digit system, the first two digits (10) mean plain carbon steel and the last two give carbon content in hundredths of a per cent (45, so about 0.45 per cent). The number is not a strength, and sulphurised free-cutting grades are in the 11xx series.
Further reading: Budynas & Nisbett, Shigley's Mechanical Engineering Design
A ductile steel part with a sharp groove carries only a steady load. How is stress concentration normally treated in its design?
C. It is usually neglected
Under static load a ductile material yields locally at the notch and the load spreads to the surrounding material, so the peak stress does not cause failure; stress concentration is normally ignored. It must be applied for brittle materials and for all materials under fatigue loading. Kf is never larger than Kt. The reason is that local yielding at the notch redistributes the stress.
Further reading: Budynas & Nisbett, Shigley's Mechanical Engineering Design
Which material behaviour is correct for fatigue design?
A. Steels usually show an endurance limit, while aluminium alloys have no true endurance limit
The S-N curve of steel flattens beyond roughly 10^6 to 10^7 cycles, giving an endurance limit. Aluminium and most non-ferrous alloys keep falling, so a fatigue strength at a stated life (for example 5 x 10^8 cycles) is used. The 0.5 Sut estimate applies to wrought steel.
Further reading: Budynas & Nisbett, Shigley's Mechanical Engineering Design
A tractor PTO delivers 15 kW at 540 rpm to an implement. What torque does the PTO shaft carry?
A. 265 N m
T = 60 P / (2 pi n) = 60 x 15000 / (2 pi x 540) = 265 N m. Dividing power by rpm directly gives 27.8, and forgetting the factor 60 (treating n as rev/s) gives 4.42 N m. 531 N m drops the 2 from 2 pi n (60 x 15000/(pi x 540)).
Further reading: Bhandari, Design of Machine Elements
A 12 mm x 12 mm square key fixes a pulley to a 48 mm shaft transmitting 400 N m. Allowable stresses for the key are 60 MPa in shear and 100 MPa in crushing. What minimum key length is required?
A. 27.8 mm
Shear: T = tau w l d/2, so l = 2T/(tau w d) = 2 x 400000/(60 x 12 x 48) = 23.1 mm. Crushing: T = sigma c (h/2) l d/2, so l = 4T/(sigma c h d) = 4 x 400000/(100 x 12 x 48) = 27.8 mm. The key must satisfy both, so l = 27.8 mm. 23.1 mm checks shear only. 13.9 mm drops one of the halves (2T/(sigma c h d)); 55.6 mm enters the 24 mm radius in place of d.
Further reading: Bhandari, Design of Machine Elements
Why is a cotter given a small taper?
A. So that driving it in draws the parts tightly together and friction holds it in place
The taper converts the driving force into an axial pull that tightens the joint, and a small taper keeps the cotter from loosening under friction. It does not make the joint carry bending or permit rotation; a knuckle joint does the latter.
Further reading: Bhandari, Design of Machine Elements
A helical spring with 10 active coils, wire diameter 6 mm and mean coil diameter 48 mm carries 500 N. Take G = 80 GPa. What is its deflection?
B. 42.7 mm
delta = 8 P D^3 N/(G d^4) = 8 x 500 x 48^3 x 10/(80000 x 6^4) = 42.7 mm. Using 12 total coils (active plus two end coils) instead of 10 active coils gives 51.2 mm; the end coils do not deflect. Using the outer coil diameter (54 mm) gives 60.8 mm, and using 4 instead of 8 in the formula halves the answer to 21.3 mm.
Further reading: Bhandari, Design of Machine Elements
Why is the spring index (D/d) of helical springs usually kept between about 4 and 12?
D. Low indices are hard to coil and have high curvature stress; high indices tend to buckle and tangle
A small index means sharply curved wire: difficult to coil and a large Wahl correction. A large index gives a floppy spring prone to buckling and tangling in handling. Stiffness varies continuously with the index, and the Wahl factor approaches 1 as the index grows.
Further reading: Budynas & Nisbett, Shigley's Mechanical Engineering Design
A spur gear has 40 teeth and a pitch circle diameter of 160 mm. What is its module, and what is the usual textbook choice of face width?
D. 4 mm; about 40 mm
m = d/z = 160/40 = 4 mm. Face width is usually taken as about 10 m (range about 9.5 m to 12.5 m), so about 40 mm. 0.25 is z/d, the inverse.
Further reading: Bhandari, Design of Machine Elements
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