In a preceding paper we have defined, and given a method for locating the neutral plane of bending of a bow.

The fiber stress in any cross-section of a bow is proportional to the distance from the neutral plane of bending. The fiber stress is therefore greatest at the back or belly of the bow. This stress is usually referred to as the maximum fiber stress.

If the cross-section of the bow is symmetrical with respect to the neutral plane, the fiber stress at the back, which is one of stretch, will equal the fiber stress at the belly, which is one of compression.

If the cross-section of the bow is non-symmetrical with respect to the neutral plane, as discussed in the previous publication, the fiber stress at the back will not equal that at the belly. For example if the cross-section of the bow is such that the belly side of the neutral plane is a parabola and the back side of the neutral plane is a rectangle, the fiber stress at the belly surface will be approximately 30% higher than at the back.

If the cross-section of the bow were kept rectangular in shape on the belly side as well as for the back, the same weight bow could be obtained with a much smaller maximum fiber stress. It is, of course, desirable to work the wood to as high a stress as possible without fracture or serious permanent set.

We shall now determine the effect that the cross-section has upon the maximum fiber stress in a bow.

It has been shown in a preceding paper, (The Effect of Thickness and Width of a Bow on its Form of Bending) equation (3), that the total moment of force about the neutral plane of bending, at any section of a rectangular shaped bar, is equal to ft³v/12, where t equals thickness, w equals width, and f is a constant, which when multiplied by the distance from the neutral plane, gives the force per unit of area at the point.

It was also shown, equation (4), that ft³v/12 equals Fx, where F is the load on the end of the bar and x is the distance from the end of the bar to the section. Therefore:

f equals 12Fx/wt³ | (1) |

The maximum stress will be at the greatest distance from the neutral plane. In this case, where the cross-section of the bar has the shape of a rectangle, the neutral plane will be in the middle and the greatest distance from it will be t/2. The force per unit of area at the surface under tension and also at the surface under compression will be ft/2.

Making use of equation (1) we get for the maximum fiber stress:

ft/2 equals 6Fx/wt² | (2) |

It has been shown in some of the early publications that the tension in the string of most bows is approximately equal to the weight of the bow at full draw. Also at full draw, the string pulls in such a direction that the string tension or bow weight may be substituted for the load F in the above formula. This formula will give a rough idea of the maximum fiber stress in a bow.