THE THEORY OF CONJUGATE PRESSURES. I6l
similar to that in Case L, and readily understood from the diagram, the
direction of R is found, and it will be noticed that it makes the same
angle, Ö, with the direction of Q, as the resultant in Case I., but on the
opposite side.
Case III., with which we are mainly concerned, is a combination of the
conditions obtaining in the preceding instances and may be solved from
them. For it is possible to take two subsidiary intensities such that the
principal intensity, q, is equal to their sum and the principal intensity, p,
to their difference, thus—
1 2 2
„ 1+P Q-p
P 2 2
Dealing with these subsidiary intensifies in pairs, the problem résolves
itself into finding, first, the resultant of two like intensities, each equal to
^, as in Case I. ; secondly, the resultant of two unlike intensities, each
equal to as in Case II. ; and, lastly, the combined resultant of these
two.
In fig. 84, set off O X=^-^, perpendicular to the plane A C, to
represent the resultant intensity due to two like equal intensities of that
amount. Next set off OY=^-^ at an angle X O Y = 2 0,to represent the
resultant of two unlike equal intensities. Completing the parallelogram,
O Z = r will be the resultant of these component intensities in direction and
magnitude.
The same result may be demonstrated by a slightly modified diagram,
which lends itself to a clearer analysis of the range of stress.
In fig. 85 draw O II at right angles to the plane A O, from the point of
intersection O, and set off O M=^^^.
Produce the line of action of the
stress Q to L, taking the point L such that OML is an isosceles triangle
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