Page:American Journal of Mathematics Vol. 2 (1879).pdf/26

beam is subjected at the section considered; or, in other words, it is simply proportional to the bending moment.

This gives at once a method of expressing $N_{1}$ in terms of the bending moment of the external forces, and it will be sometimes convenient hereafter to use it.

Hereafter, also, unless otherwise stated, $N,$ instead of $N_{1},$ will be written for the general value of the intensity of the normal stress parallel to the axis.

Let $n$ and $M_{1}$ represent the values of $N$ and the external bending moment respectively for any given section, and $M$ the general value of the external bending moment, then, by the principle just stated,

(17)

$N=n{\tfrac {M}{M_{1}}}$

This is a perfectly general expression whatever may be the position of the origin of co-ordinates.

It will now be necessary to return to the discussion of the general form of equation (16),

(18)

$N=-x\left[f_{y}^{\prime }(y,z)+F_{z}^{\prime }(y,z)\right]+c$

taken in connection with equations (8) and (9).

The functions $f(y,z)$ and $F(y,z)$ are perfectly arbitrary; hence it is sufficient for equilibrium to assign any laws whatever for the variations of the intensities $T_{2}$ and $T_{3},$ and when $T_{2}$ and $T_{3}$ are known $N$ at once results from equation (18). There are not, therefore, a sufficient number of equations founded on the principles of statics to insure a solution of the problem. The “Principle of Least Resistance,” however, furnishes the wanting condition. Now whatever may be the laws governing the quantities $N,T_{2}$ and $T_{3}$ there are two conditions which must be fulfilled, i. e. the moment of the internal stresses in any section must be equal to the moment of the external forces for the same section, and the total shearing stress in any normal section must equal the sum of the external forces acting on one side of that section. But the second of these conditions is really involved in the first, as will now be shown.

Let $f(z^{\prime },y^{\prime })=0$ be the equation of the perimeter of a normal section of the beam, and $A=\iint {dzdy}$ its area. Then, remembering that the coefficient of elasticity for tension is assumed equal to that for compression, the equation