In a double-acting cylinder, with a piston area of 3 in^2 and a rod area on the rod side of 2 in^2, what is the effective area on the rod side when retracting?

Prepare for the General Aandamp;P Test with study guides, flashcards, and multiple-choice questions. Every question includes hints and comprehensive explanations to ensure you are well-prepared. Get ready to succeed!

Multiple Choice

In a double-acting cylinder, with a piston area of 3 in^2 and a rod area on the rod side of 2 in^2, what is the effective area on the rod side when retracting?

Explanation:
When a double-acting cylinder retracts, the fluid on the rod side acts on the annular space around the rod, not on the full piston face. So the effective area for the rod side during retract is the bore area minus the rod area. With a bore area of 3 square inches and a rod cross-sectional area of 2 square inches, the annular area available for retracting is 3 − 2 = 1 square inch. This smaller area is why the retracting force (at the same pressure) is less than the extending force. In other words, the proper rod-side effective area during retract is 1 square inch. (If a value of 2 square inches is seen, that would be using the rod area by itself rather than the annular area.)

When a double-acting cylinder retracts, the fluid on the rod side acts on the annular space around the rod, not on the full piston face. So the effective area for the rod side during retract is the bore area minus the rod area.

With a bore area of 3 square inches and a rod cross-sectional area of 2 square inches, the annular area available for retracting is 3 − 2 = 1 square inch. This smaller area is why the retracting force (at the same pressure) is less than the extending force.

In other words, the proper rod-side effective area during retract is 1 square inch. (If a value of 2 square inches is seen, that would be using the rod area by itself rather than the annular area.)

Subscribe

Get the latest from Passetra

You can unsubscribe at any time. Read our privacy policy