Page:American Journal of Mathematics Vol. 2 (1879).pdf/24

The direction of action of the stress whose intensity is $N_{1}$ is normal to its plane of action, which is a normal section of a fibre parallel to the axis of the beam. Now, since the applied bending forces are perpendicular in direction to the axis of the beam, no part of $N_{1}$ can result directly from the forces; that is, they have no component parallel to the fibres subjected to the normal stress $N_{1}.$

The stress, whose intensity is $N_{1},$ exists only, therefore, in consequence of the shearing, or tangential, stresses called into action by the slipping over each other of the fibres parallel to the axis of the beam, or in consequence of $T_{2}$ and $T_{3}.$ The expression for $N_{1}$ cannot therefore have a part independent of the quantities $T_{2}$ and $T_{3},$ except in the case (not of pure flexure) where the beam is subjected to the action of an external force acting in the direction of its own length. The function $\Psi \,(y,z)$ cannot, therefore, depend on the variables $y$ and $z$ unless they appear raised to the zero power; or, in other words, $\Psi \,(y,z)$ cannot exist except as a constant, since the integrating equation (10) was made in respect to $x$. But the case treated is that of pure flexure, in which no external force acts upon the beam in the direction of its own length, and in which, consequently, no part of $N_{1}$ can be independent of the tangential stresses $T_{2}$ and $T_{3};$ hence $\Psi \,(y,z)=0$ or $c,$ according as the origin of co-ordinates is at a section of no flexure or not.

Again, differentiate equation (10) in respect to $y,$ there results

${\tfrac {dN_{1}}{dy}}=-x\,\left[f''_{y}\,(y,z)+{\tfrac {d\,(F'_{x}\,(y,\,z))}{dy}}\right]+{\tfrac {d{\it {{\Psi }\,(y,\,z)}}}{dy}}$

....(11)

In this equation any value of $z$ may be assumed while $y$ is considered the only variable. Let such a value for $z$ be assumed that the equation will apply to the neutral surface. It will not destroy the force of the reasoning to suppose that surface plane, for if it is not plane the equation of its trace on the plane of normal section of the beam will be $z=f\,(y).$

Now, in the neutral surface $N_{1}=0,$ $T_{3}=0$ and $F'_{z}\,(y,z)=0$ since $T_{2}$ has there its maximum value. Consequently ${\tfrac {dN_{1}}{dy}}=0,$ $f''_{y}\,(y,z)=0,$

and

${\tfrac {d{\it {{\Psi }\,(y,\,z)}}}{dy}}\,dy=0.$

........(12)

Next, differentiate equation (10) in respect to $z,$ and there results

${\tfrac {dN_{1}}{dz}}=-x\,\left[{\tfrac {d\,(f'_{y}\,(y,\,z))}{dz}}+F''_{z}\,(y,z)\right]+{\tfrac {d{\it {{\Psi }\,(y,\,z)}}}{dz}}.$

....(13)