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Effect of Thickness and Width of a Bow on its Form of Bending
Part 2 of 5

In order to illustrate the principle used in determining the form of bending, we shall first treat the case of a rectangular bar clamped at one end and loaded with a force at the other end. Fig. 2a shows this bar. Fig. 2b shows a cross-section of the bar in the middle along its length and will be used as reference in the following treatment.

In Fig. 2b let l equal length, w equal width and t equal thickness of a bar clamped at one end and subjected to a force F at the other end.

Let P be a point which is a distance x from the loaded end and a distance y from the center of the cross-section o.

The force of compression at P will be proportional to y.

Let fy equal force per unit area at P.

(1) Then fywDy equals force for a segment having thickness Dy and full width of the bar w.

There is an extension force on the other side of the center-line at a point corresponding to P having the same value as the compression force.

Therefore the moment of force about o will be 2fy2w Dy.

The total reacting force will therefore be:

(2) The integral from y equals t/2 to y equals o of 2fy2 w dy

Integrating this expression we get:

(3) ft3w/12

But Fx equals moment due to the applied force, and since the applied force equals the reacting force,

(4) f w t3/12 equals Fx or f equals 12 Fx/wt3

Let Y equal Youngs modulus which is defined as the force unit area divided by the deflection per unit length.

If we consider an increment of length Dx and remember that the force per unit area at P is fy equals 12 Fxy/wt3:–

Then Y equals 12 Fxy/wt3 divided by the deflection at P for an increment of length Dx

Of the deflection at P equals 12 FxyDx/wt3Y

The deflection at the end of the bar is x y times as much as at P for the increment of length Dx.

Therefore the deflection at the end of the bar due to the bending of the length Dx equals 12Fx2Dx/wt3Y

Different bow shapes
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