February 1932 by National Field Archery Association

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Sylvan Archer

VOL. 5., NO. 10

ALBANY, OREGON

Entered as second-class matter October 14, 1931, at the post office at Albany, Oregon, under the Act of March 3, 1879. Published monthly by Ye Sylvan Archer Publishing Co. 325 W. 2nd Street, Albany, Oregon

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CONTENTS

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The Neutral Plane of Bending of a Bow By Dr. C. N. Hickman

Light Bow in Deer Hunting

Archery on the Fashion Page.

Bow Sights Containing Glass By G. R. H

Bowstring Comments.

.10

Venison Features Archery Dinner.

.12

Junior Tournament.

.14

As Others See Us.

.15

Fresno Club Elects.

.16

1ft

February, 1932

Oze Neutral Plane of Bending of a Bow By C. N. Hickman, Ph. D. Long Island, N. Y.

When a symmetrical homogeneous member is stressed by bending, the elongation on one side is exactly equal to the compression on the other side. If a bow having a symetrical cross section is all made from either heart or sap-wood, the elongation along the back of the bow will in general equal the compression along the belly. In such a bow there is a thin section located midway between the belly and back which is neither stretched nor compressed. We may call this layer the neutral plane of bending. Many bows are cut from the tree so that the back is composed of sap wood and the rest of the bow is com posed of heart-wood. Other bows are backed with various materials. Any of these bows will have the neutral plane of bending moved back or for ward depending on the elasticity of the materials used. Practically all bows are non sym metrical in cross-section. The belly is usually much narrower than the back. This type of construction moves the neutral plane of bending toward the back of the bow. It is necessary to know the location of this neutral plane in order to de termine the maximum fiber stress in the bow. Since it is the object of the succeeding paper to show the effect of the cross-section on the fiber stresses of a bow, a method for locat ing the neutral plane of bending will be briefly outlined in this paper. Let us consider a bow all made from the same kind of wood having a cross section as shown in Fig. 7a. Let the width of the bow at the back equal w,

the thickness from belly to back equal T, the distance from the belly to the neutral plane of bending equal a, and the distance from the back to the neutral plane equal b. Let us further assume that the por tion back of the neutral plane Z Z' is rectangular in shape and that the portion in front of the neutral plane is represented by the conic section z equals \/c(a — y) (a parabola) ;When y equals 0, z equals w/2 equals yea This type of construction may be found in many bows. Of course the corners B B' are usually rounded and the transition from the parabolic curve to the straight portion is grad ual at N and N' Note: We are using the Z axis here in order to reserve the X axis to repre sent the length of the bow in sub sequent treatments. Let the force of compression at any point P be fy The moment of force about the Z Z' axis of a segment Dy is equal to 2zfy’Dy The total moment of force due to compression in therefore: The integral from y equals a to y equals o of 2fzy2dy. ----------„ Substituting the value of equals z yc(a — y) we have for the total moment of force due to compression: The integral from y equals a to y equals 0 of: 2f y2dy^/c (a-y) Integrating we .get: (32/105)fasyac —7 find the In like manner we- may moment of force due d— to 1. extension equals:

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The integral from y equals b to y equals 0 of: 2fzy2dy. But here as seen from Fig. 7a, z equals a constant equals w/2 Therefore the moment of force due to extension equals: The integral from y equals b to y equals 0 of wfy2dy. Integrating we get: wfb3/3 Since w equals 2\/ac, we have for the total moment of force due to ex tension, (2/3)fb3-^/ac The numerical value of the moment of force due to compression is equal to the numerical value of the moment of force due to extension. Therefore: (32/105) a3f \/ac equals (2/3) b3-\/ac Or a3 equals (35/16) b3 and a equals 1.30 b approximately. Since the thickness of the bow T is equal to a -L b we have a equals .565 T and b equals .435 T We therefore find that the distance from the belly of the bow to the plane of neutral bending is 30% greater than the corresponding distance to the back of the bow. If the same type of bow is backed with a layer of material having a thickness of t and a coefficient of elasticity equal to n times that of the .bow wood, the neutral plane of bend ing will be moved toward the back of the bow As before let a equal the distance from the belly of the bow to the neu tral plane and b equal the distance from the neutral plane to the back, including the applied layer of backing. The moment of force due to com pression will be the same as before: (32/105) a3fy/ac The moment of force due to exten sion will be different. If as before we let fy equal the force per unit of area within the wood

at any point, then the total m

tow 4 will be equal to: ’ The integral from yequalsb to y equals 0 of w f Integrating we get: | We also have a moment of [2 I due to the backing material. I Since the elasticity of the batfe I: material is n time as much as that/ the bow wood we have for the moms, of force due to the backing: The integral from y equals b toy equals b — t of wnfy3 Integrating we get: (wnf/3) [b3 — (b — t)1] thiti w equals 2y/ac Adding these two moments ri equating them to the moment due a compression we have: a3 equals (35/16) [nb3 — n (bt)3 4- (b — t)3] ' equals (35/16) [nb3 — (n — D (b - t)3] Let t equal mb Then: a3 equals (35/16)b3[n — (n - 'I (1 —■ m)3]

a equals b3-\/(35/16) [n —(n (1 — m)3] If n equals 2 and m equals ■ a equals b (1.30x1.083) equ3 a equals .58 T, b equals .41 The distance from the neuW^ of the bow to the back has duced by almost 5%It should be clearly unders‘ the neutral plane is moved t back of the bow only 1° fle which have higher elastic!d it' bow wood and which ' edillg elongation without elastic limit. ^tioii Now consider the crost bow such as shown m • . pla»e, the belly side of the neu^, sj a semicircle having: ra*u W the distance from the n «i the back of the bow is

February, 1932

of the bow at the back is w. Following the same procedure as before, the moment of force about the Z Z' axis due to compression is the integral from y equals a to y equals 0 of 2zy2dy. But z equals -\/(a-—y-) Substituting the value of z ana in tegrating we get: 3.1416fa4/8 The moment of force due to exten sion is as before wfb3/3. Substituting the value of w equals 2a The moment is 2fab3/3 Since the moment due to compres sion is equal to the moment due to extension we have: a*f 3.1416fa'/8 equals 2afbs/3 From which a equals 1.194 b With a bow having this type of cross-section the distance from the neutral plane to the belly of the bow is about 20% greater than the cor responding distance to the back of the bow. This type does not have the neutral plane as close to the back as as the type having the parabolic shaped belly. On the other hand a bow which has a belly with a hyper bolic shape will have its neutral plane closer to the back of the bow than either of these types just discussed. In case the section of the bow is

all of the same kind of wood, a simple method of locating the neutral plane of bending may be followed by the non-mathematician. Lay out the shape of the cross-sec tion on a heavy cardboard, using any convenient enlarged scale. Cut this cardboard section out with a pair of shears. Draw a line from the belly to the back which divides the section into two equal parts like the lines Y Y' in figures 7a and 7b. Punch a small pin through the cardboard section on this line at such a position that the section will remain in balance on the pin for any position. After a number of trials you will find a point where the cardboard section may be rotated on the pin to any position, and at which, it will be in balance. This point is known as the center of .gravity and is on the line corresponding to the plane of neutral bending. The distance from this point to the back therefore gives the distance of the neutral plane of bending from the back. If care is used in laying out the cross-section shape on the cardboard and in obtaining the proper balancing point, accurate results may be obtained. Y

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nS. 7a. Fig. 7b.