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Fiber Stresses in Bows
Part 2 of 5

Let us examine a section of a 40 pound bow having all sections rectangular in shape, constant thickness and a width which tapers along a straight line from the handle to the bow tip. Suppose we take a section which is 30 inches from the end, where the width is one and three-eighth inches and the thickness is three-quarters of an inch. Substituting in equation (2) we find for the maximum fiber stress:

The fiber stress at elastic limit, given by The Forests Products Laboratory, for Yew is 10,100 pounds per square inch. The stress at rupture for this wood is given as 16,800 pounds per square inch. We thus see that in this bow the wood is being worked close to, but below the elastic limit.

If this bow had the same thickness and same width throughout its length, the fiber stress would become less and less as you approach the tips. The bow would bend in a form as shown in figure 6a of the paper referred to above. However if the width is tapered on a straight line from the handle to the tips, the decrease in the value of x is exactly offset by the decrease in the width w so that the fiber stress is the same throughout the length of the bow.

For example if we take a section which is 15 inches from the tip the thickness is still three-quarters of an inch, but the width is only one-half as much as before, i. e. 11/16 inches.

Substituting in equation (2) we get for the maximum fiber stress:

This is the same value obtained for the section 30 inches from the tip. In like manner every other section will have the same maximum fiber stress.

Let us now determine the maximum fiber stress in a yew bow manufactured by a well known manufacturer of archery tackle. This bow weighs forty pounds and is seventy inches long. On examining a section 30 inches from one end of the bow and locating the neutral plane of bending by the method described in the preceding publication, we find that the bow has a thickness of one inch, a width of one and one-eighth inches and a shape such that the neutral plane is .55 inches from the belly and .45 inches from the back.

As shown in the preceding publication, the moment of force due to extension is wfb³/3, where b is the distance from the back to the neutral plane. The total moment about the neutral plane is twice this value. The moment due to the load F is Fx, where x is the distance from the end to the section. Therefore f equals

 3Fx/2wb³ (3)

Substituting in this equation we get:

The maximum fiber stress will be on the belly side where the distance from the neutral plane to the belly is .5 5 inches. Since the maximum fiber stress is f times this distance we have: 17,600 X .5 5 equals 9,660 pounds per square inch for the maximum fiber stress.