If the required torque of a trunnion nut is 320 inch-pounds and a 2-inch straight adapter is used on an 18-inch torque wrench, how many foot-pounds will be indicated?

To determine the indicated torque on the torque wrench in foot-pounds when using an adapter, it's important to understand how torque values change with the use of different tools and whether they are extension adapters or not.

In this case, we have a torque requirement of 320 inch-pounds at the trunnion nut, and a 2-inch adapter is used with an 18-inch torque wrench. When using an adapter, the torque applied is affected by the ratio of the lengths of the adapter and the wrench.

To convert from inch-pounds to foot-pounds, you can remember that 12 inch-pounds equals 1 foot-pound. Therefore, to convert 320 inch-pounds to foot-pounds, you divide by 12:

320 inch-pounds ÷ 12 = approximately 26.67 foot-pounds.

However, because of the leverage effect created by using the 2-inch adapter on the 18-inch wrench, the effective torque will actually be less than the calculated torque at the nut.

The effective torque can be calculated by applying the ratio of the lengths (or arm distances), adjusting the 320 inch-pounds downward due to the leverage effect. The formula can be understood as:

Effective Torque = (Torque Requirement)