WORLD FOR CIVIL

Subtopic 1.31 - Torque

Formally, a torque (also know as the Moment of Force) is a vector cross product, that is, given a distance vector R (from origin, as shown in diagram 1) and a force vector F, the torque is defined as: . This cross product is a vector and has a resultant magnitude (size) of , where theta is the angle between the distance vector and force vector. As diagram 2 shows, what this cross product effectively does is to take the component of the force vector perpendicular to the distance vector and then multiply it by the distance vector.

The direction of the torque vector is perpendicular to both the distance and force vectors, and is given by the right hand rule. There are a variety of ways of expressing the right hand rule. In this case it could be expressed this way. If vector R were a rod, pinned at the origin, notice that force component would cause R to start to rotate counterclockwise about the origin. If we now curl the fingers of our right hand in the same way as the rotation, our thumb will point the direction or the torque vector - in this case, out of the screen. (+ z direction.)

If this all seems confusing at this point, don't panic. We will use a somewhat less formal approach to determining torque acting on structures, as shown in the following example(s).

Example 1. In Diagram 3, a 10 foot long beam, pinned at point P, is shown with 100 lb. force acting upward on the beam. We wish to determine the torque on the beam due to the 100 lb. force.
The very first thing we must do when determining torque is to PICK A POINT. Torque is always calculated with respect to a point (or axis), and so before we can proceed we first choose a point (sometimes called the pivot point) to calculate torque about. In this example we will use point P, at the left end of the beam.

We next calculate the torque caused by the force by Torque(about P) = Force x perpendicular distance. That is, the torque caused by the 100 lb. force with respect to point P will be the 100 lb. force times the perpendicular distance from point P to the line of action of force (d = 10 ft.). By perpendicular distance, we mean, that there is only one line we can draw which begins at point P and intersects the line of action (direction) of the force at 90 degrees, and the length of this line is the perpendicular distance. In diagram 3a, it is the thin dark line with 'd' below it, and its value is 10 ft., so the torque due to the 100 lb. force with respect to point P will be: Torque = F x d = 100 lb. x 10 ft = 1000 ft-lb.

In diagram 3b, we have moved the 100 lb. force to halfway down the beam toward point P, and so the perpendicular distance is changed to 5 ft., and the torque due to the 100 lb. force with respect to point P now becomes: Torque = F x d = 100 lb. x 5 ft. = 500 ft-lb.

In diagram 3c, we have moved the 100 lb. force three quarters length down the beam toward point P, and so the perpendicular distance is now 2.5 ft., and the torque due to the 100 lb. force with respect to point P now becomes: Torque = F x d = 100 lb. x 2.5 ft. = 250 ft-lb.

Finally in diagram 3d, the 100 lb. force acts directly below point P, and so the perpendicular distance is 0 ft, and therefore there is zero torque due to the 100 lb. force with respect to point P. This does not mean the 100 lb. force does not push on point P, it does, however it produces no torque (no rotation) with respect to point P.

To see what happens if we do not have a force acting a 'nice' direction as in example 1 above, we will do a second example.

Example 2: In this example we use the same beam as above but now the 100 lb. force acts at an angle of 37^o with respect to the horizontal (below) as shown in diagram 4.
Once again we wish to calculate the torque produced by the 100 lb. force with respect to point P. We will first do this directly from our less formal definition of torque: Torque = Force (times) perpendicular distance.

Notice in Diagram 4a, we have extended the line of action of the force, and then we have started at point P and drawn a line from point P which intersects the force line at 90^o. This is the perpendicular distance 'd' from the pivot point to the line of the force. We can find the value of the perpendicular distance d by noting that the force, perpendicular distance d, and beam length form a right triangle with the beam length as the hypotenuse. Therefore from trigonometry, d = 10 sin 37^o = 6 ft, and the torque will be: Torque = 100 lb. x 6 ft = 600 ft-lb. (Positive torque, since if the beam were actually pinned at point P, the 100 lb. force would start the beam rotating in the counterclockwise direction.)

While this method of calculating the torque is fine and does work, it actually is more effective in problems, certainly in problems with a number of forces, to first break the force into it's equivalent x and y-components before calculating torque. This is the process we will normally use, and we have shown this process in Diagram 4b.

We first found the x and y-components of the 100 lb. force as shown in the diagram. Then the resultant torque about point P will be the sum of the torque produced by each component force. Notice the y-component force results in Torque = 100 lb. sin 37^o x 10 ft. = 600 ft-lb., while the x-component force results in Torque = 100 lb. cos 37^o x 0 ft. = 0 ft-lb since its line of action passes through point P, and thus its perpendicular distance is zero, resulting in zero torque. Summing the two torque we get a total of 600 ft-lb., just as we found in the first method.

As we continue with examples of statics problems, the applications of torque in solving problems will be shown in complete detail. While we have determined torque here in only 2-dimensions, the procedure is the same 3-dimensions, with the exception that there are 3-axis of rotation possible. Since the type of problems we will consider are mainly 2-dimensional, we will not go into calculating torque in 3-dimensions at this point.