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Fiber Stresses in Bows
Part 3 of 5

Taking another section 15 inches from the tip, we find the thickness is .60 inches, the width .95 inches and the neutral plane is .27 inches from the back and .33 inches from the belly.

Using these values in equation (3) we get:

formula05 (1K)

The maximum fiber stress will be on the belly side which is .33 inches from the neutral plane and will equal: 48,000 X .33 equal 15,900 pounds per square inch.

This is an increase of 68% over the stress at the other section and is beyond the elastic limit given for this wood. It is close to the value given for the modulus of rupture.

If the thickness had been left the same as at the other section, and the width reduced to ½ its former value (.562), the maximum stress would have been reduced to the same value as at the other section. We would have had on substituting in equation (3):

formula06 (1K)

And: 17,000 X .55 equals 9,660 pounds per square inch.

The bow would have had a better cast and would not have taken a permanent set.

If instead of the parabolic section for the belly, a rectangular section had been used, the maximum stress would have been reduced about 18% because the distance of the belly from the neutral plane would have been .45 inches instead of .55 inches. The bow would have had exactly the same weight. The full thickness of 1 inch might have been used, with an increase in the weight of the bow of 27%. In which case the maximum fiber stress would have been by equation (2):

formula07 (1K)

Even with this increase in bow weight, the maximum fiber stress is 1,530 pounds less than for the parabolic section.