632 moleswokth’s pocket-book
Differential and Integral Calculus—
continued.
yH 1
dxn = nxn~1dx; /x“dx =——r-
n + 1
Simple Example of Maxima and Minima.
, d u
u is a minimum or maximum when — = 0.
d x
W
In formula 8 = (L» - æ2) (see page 171); find the
/joint at which the strain is a maximum. .
W
v is the same in all cases, and therefore a constant;
2DL
substitute a for it; and u for S; or u = a L x — ax*;
, du ,1 — 1 2 — 1 T „
then — = aLx — 2ax = aL - 2ax.
d x
Make (or a L — 2 a æ) = 0;
dx s
then 2 ax = -a L; or 2x = L; or x = -■; or the point of
maximum or minimum value is at half the span.
. „ . . u .
When the second differential coefficient ——- is negative,
ctæ’
u is a maximum; but if positive, u is a minimum.
— 2 ax therefore w is a maximum.
dx*
Example of the Application of the Calculus to
Qirdeks.
In any girder (whether straight, curved, continuous, or dis-
continuous) if the bending moment at any point be expressed
as a function of a variable x, the normal shearing force at
that point will be expressed by the differential coefficient of
the function.
Thus, if the bending moment at any point be expressed by
M — W (2 ax - atf); the normal shearing force at that point
will be = W (2 a — 2 x) — 2 W (a — æ).
d <v