GATE ME 2023 Set 1 — Question 15

MCQ+1 / -0.33MediumWork Holding & 3-2-1 LocationCNC & Work HoldingManufacturing Processes

Manufacturing Processes → CNC & Work Holding → Work Holding & 3-2-1 Location

Last updated 5 September 2026

Question

A cuboidal part has to be accurately positioned first, arresting six degrees of freedom and then clamped in a fixture, to be used for machining. Locating pins in the form of cylinders with hemi-spherical tips are to be placed on the fixture for positioning. Four different configurations of locating pins are proposed as shown. Which one of the options given is correct?

Correct answer (A) Configuration P1 arrests 6 degrees of freedom, while Configurations P2 and P4 are over-constrained and Configuration P3 is under-constrained. Solution A cuboid is located with the 3-2-1 principle . Three pins on the primary (bottom) face arrest 3 DOF: one translation and two rotations. Two pins on a secondary side face arrest 2 DOF, and one pin on a tertiary face arrests the last DOF. That is 6 pins in all. Counting the pins in the figure: P1: 3 pins under the base and 3 horizontal pins on the sides (2 on one face, 1 on the adjacent face). This is exactly 3-2-1, so all 6 DOF are arrested . P2: 4 pins under the base (plus the side pins). A plane needs only 3 points, so the base is over-constrained . P3: 3 pins under the base but only 2 side pins, 5 in total. One DOF stays free, so it is under-constrained . P4: 4 pins under the base and 4 side pins. This is over-constrained . So P1 arrests 6 DOF, P2 and P4 are over-constrained, and P3 is under-constrained. Answer: (A) Turn this into a strength. Explore AI-powered practice and doubt support with Success Tracker.

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Correct answer (A) Configuration P1 arrests 6 degrees of freedom, while Configurations P2 and P4 are over-constrained and Configuration P3 is under-constrained. Solution A cuboid is located with the 3-2-1 principle . Three pins on the primary (bottom) face arrest 3 DOF: one translation and two rotations. Two pins on a secondary side face arrest 2 DOF, and one pin on a tertiary face arrests the last DOF. That is 6 pins in all. Counting the pins in the figure: P1: 3 pins under the base and 3 horizontal pins on the sides (2 on one face, 1 on the adjacent face). This is exactly 3-2-1, so all 6 DOF are arrested . P2: 4 pins under the base (plus the side pins). A plane needs only 3 points, so the base is over-constrained . P3: 3 pins under the base but only 2 side pins, 5 in total. One DOF stays free, so it is under-constrained . P4: 4 pins under the base and 4 side pins. This is over-constrained . So P1 arrests 6 DOF, P2 and P4 are over-constrained, and P3 is under-constrained. Answer: (A)

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