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?

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?

• A. 3 square inches
• B. 1 square inch
• C. 2 square inches
• D. 4 square inches

Answer: (C)

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.)