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ASRB NET Machine Design: practice questions

ASRB NET — Farm Machinery & Power / Land & Water Management Engineering / Agricultural Structures & Process Engineering · India · Agricultural Scientists Recruitment Board (ICAR)

Original practice questions written to the published syllabus · not past papers
9worked here, free
42written to this exam in this subject
13syllabus units

What this subject covers

Sizing 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

What ASRB NET asks from it

ASRB NET — Farm Machinery & Power

Design principles of farm machinery components, materials of soil-engaging tools, instrumentation (strain gauges, load cells, dynamometers) and overload safety devices.

Where the 42 questions fall

How our questions for ASRB NET spread across the units of Machine Design. Difficulty: 22 easy, 14 medium, 6 hard.

Overload safety devices: shear bolts and slip clutches7
Rolling and sliding bearings and bearing life5
Keys and couplings4
Measurement of force, torque and strain on machine elements4
Stress concentration and fatigue (Goodman and Soderberg)3
Knuckle and cotter joints3
Bolted joints3
Welded joints3
The design process, materials and factor of safety2
Shafts under combined bending and torsion2
Springs2
Belt and chain drives2
Spur gear design (Lewis and Buckingham)2

9 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Knuckle and cotter joints · medium

The hitch pin of a trailed implement is a clevis-and-pin (knuckle-type) joint. In service the pin bends and wears oval. Which statement about its design is correct?

  1. Only double shear matters, because pins never fail in bending
  2. A single-shear calculation gives the correct diameter because the drawbar is one plate, and bending can be neglected in a short pin
  3. The pin must be checked for bending and bearing as well as double shear, and it should be secured so it cannot work out
  4. A loose fit is preferred because it reduces bending stress
Answer and working

C. The pin must be checked for bending and bearing as well as double shear, and it should be secured so it cannot work out

In a clevis the pin acts as a short beam between the fork prongs, so bending often governs; bearing pressure causes the oval wear seen. Double shear is also checked. Clearance increases bending and impact, and the securing clip matters because a lost pin causes run-away implements.

Further reading: Bhandari, Design of Machine Elements

2 · Bolted joints · medium

Why does a high preload improve the fatigue life of a bolt in a joint under fluctuating external load?

  1. Preload makes the bolt stiffer than the clamped members, so the bolt safely carries a larger share of the load change
  2. Preload removes the stress concentration at the thread root
  3. The stiff clamped members absorb most of the load fluctuation, so the alternating stress in the bolt is small
  4. Preload raises the endurance limit of the bolt material
Answer and working

C. The stiff clamped members absorb most of the load fluctuation, so the alternating stress in the bolt is small

With preload, only the fraction C = kb/(kb + km) of the external load change reaches the bolt while the joint stays closed, so the bolt's stress amplitude is small even though its mean stress is high. Preload does not change material properties or thread geometry, and it does not change stiffnesses.

Further reading: Budynas & Nisbett, Shigley's Mechanical Engineering Design

3 · Springs · easy

Why is the spring index (D/d) of helical springs usually kept between about 4 and 12?

  1. An index above 12 always makes the Wahl factor greater than 2
  2. The index affects only the colour of the heat treatment
  3. Below 4 the spring has no stiffness at all, and above 12 its stiffness becomes infinite, so it cannot deflect under load
  4. Low indices are hard to coil and have high curvature stress; high indices tend to buckle and tangle
Answer and working

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

4 · Overload safety devices: shear bolts and slip clutches · easy

Why is a shear pin designed using the ultimate shear strength of its material rather than an allowable stress with a factor of safety?

  1. Because it is meant to break at a chosen torque, so its breaking strength must be predicted, not kept below
  2. Because a factor of safety would make the pin too small to fit the hub bore, so the drive could not be assembled
  3. Because ultimate strength is always lower than yield strength
  4. Because pins never carry fluctuating loads
Answer and working

A. Because it is meant to break at a chosen torque, so its breaking strength must be predicted, not kept below

A shear pin is a deliberate weak link: it must fail at the set torque to protect the drive. Using allowable stress with a factor of safety would make it stronger than intended, so it would not break when it should. Ultimate strength is higher than yield, not lower.

Further reading: Srivastava et al., Engineering Principles of Agricultural Machines (ASABE)

5 · Overload safety devices: shear bolts and slip clutches · medium

A farmer keeps breaking shear bolts on a post-hole digger in stony ground and replaces them with high-tensile bolts. A week later the gearbox fails. What is the correct advice?

  1. Weld the flange to the hub to remove the bolt
  2. Use only the specified shear bolt grade and size; if bolts break often, reduce the load or fit an auto-reset slip clutch
  3. Grease the shear bolt so it breaks less often
  4. Use an even stronger bolt so it never breaks, since the gearbox failure shows the driveline needs a stiffer, stronger connection
Answer and working

B. Use only the specified shear bolt grade and size; if bolts break often, reduce the load or fit an auto-reset slip clutch

The shear bolt is sized to fail below the gearbox's damaging torque. A stronger bolt moves the weak link to the gearbox, which is what happened. The fix is the correct bolt and, where overloads are frequent, a self-resetting device. Welding removes protection altogether.

Further reading: Srivastava et al., Engineering Principles of Agricultural Machines (ASABE)

6 · Measurement of force, torque and strain on machine elements · easy

Which transducer is widely used in tillage research to measure the horizontal (draft) and vertical forces on a soil-engaging tool?

  1. The extended octagonal ring transducer with strain gauges
  2. A tachogenerator
  3. A thermocouple
  4. A Bourdon tube pressure gauge
Answer and working

A. The extended octagonal ring transducer with strain gauges

The extended octagonal ring, fitted with strain gauges at nodes that respond selectively to horizontal and vertical force, separates draft and vertical load and can also give moment. A Bourdon tube measures fluid pressure, a thermocouple temperature and a tachogenerator speed.

Further reading: Srivastava et al., Engineering Principles of Agricultural Machines (ASABE)

7 · Springs · hard

A spring-tine cultivator tine spring is subject to a load that fluctuates between 300 N and 900 N at every stone and clod. Which approach is correct for its design?

  1. Find the mean and alternating shear stresses in the wire, apply the Wahl factor, and check against a fatigue criterion such as Goodman in torsion using the wire's torsional endurance strength
  2. Design only for 900 N with the static shear yield strength
  3. Ignore the curvature factor, because it applies only to static loading
  4. Design for the average load of 600 N with the static shear yield strength
Answer and working

A. Find the mean and alternating shear stresses in the wire, apply the Wahl factor, and check against a fatigue criterion such as Goodman in torsion using the wire's torsional endurance strength

Fluctuating loading needs a fatigue check. The mean load is 600 N and the alternating load 300 N; each gives a stress through tau = K 8PD/(pi d^3), with the Wahl factor (curvature matters most in fatigue). The pair is checked against Goodman or Soderberg in shear. Static checks alone miss fatigue.

Further reading: Budynas & Nisbett, Shigley's Mechanical Engineering Design

8 · Overload safety devices: shear bolts and slip clutches · hard

Shear pins on a forage harvester feed roll drive sometimes fail after long use at torques below their design breaking torque. What is the most likely mechanism, and what reduces it?

  1. Fatigue of the pin under repeated torque peaks, aggravated by clearance; a tight-fitting pin in hardened bushes and a controlled neck reduce it
  2. The ultimate shear strength rising with use; use a thicker pin
  3. Creep of the steel at harvest temperatures; use a lower-strength pin
  4. Corrosion alone; paint the pin
Answer and working

A. Fatigue of the pin under repeated torque peaks, aggravated by clearance; a tight-fitting pin in hardened bushes and a controlled neck reduce it

Repeated peaks below the breaking torque accumulate fatigue damage, and clearance adds impact and bending. A snug fit in hardened bushes and a consistent reduced section give more repeatable breaking. Creep does not occur in steel at field temperatures, and strength does not increase with use.

Further reading: Srivastava et al., Engineering Principles of Agricultural Machines (ASABE)

9 · Belt and chain drives · medium

A V-belt drive on a chaff cutter squeals and the belts glaze within weeks. The pulleys are aligned. Which cause is most likely and what is the fix?

  1. Belt dressing oil should be applied weekly
  2. The belts are too tight (over-tensioned); loosen until they flap
  3. The belts are slipping because of too little tension (or a worn groove); retension to the specified value and replace worn pulleys
  4. A flat belt should replace the V-belts to increase grip
Answer and working

C. The belts are slipping because of too little tension (or a worn groove); retension to the specified value and replace worn pulleys

Squeal and glazing are signs of slip, which heats and hardens the belt surface. Correct tension and sound grooves (a belt riding low in a worn groove loses the wedge action) stop it. Loose, flapping belts slip more, oil on belts is harmful, and a flat belt grips less than a V-belt.

Further reading: Bhandari, Design of Machine Elements

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Read the subject

Machine DesignThe full study manual, free ASRB NETFormat, sections and how it has changed

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Machine Design for other exams

GATE · IndiaICAR AIEEA-PG · IndiaICAR AICE-JRF/SRF · IndiaAgricultural and Biosystems Engineers Licensure Examination · PhilippinesNCEES PE Agricultural and Biological Engineering · USANCEES FE Other Disciplines · USA

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