\fdO'^ - Old Forge Coal Mines

STRENGTH OF MATERIALS. 217
F^ = 2,000 lb. ; ^-3 =40 Xl2 = 480 lb., is the uniform load
between F^ and F^\ 3-4- = 2,000 X 1.3 = 2, GOO lb., is F„ and
j^.5 = 40 X 10 = 400 lb., is the uniform load between F^ and
the right support. The pole P is chosen and the rays drawn.
Since the uniform load is very small compared with F^ and /%,
it will be sufficiently accurate to consider the three portions
of it concentrated at their respective centers of gravity
X, j\ and ^. Drawing the equilibrium polygon parallel to
the rays, we obtain the moment diagram a Ji b k c I d a.
From /*, drawing P in parallel to the closing line a d, we
obtain the reactions^ m andrnS equal, respectively, to 2,930
and 2,870 lb. Ans. The shear axis ;;/ x, and the shear diagram
r s t u V tt ;«, are drawn in the usual manner. The
greatest shear is m, 2,930 lb. The shear line cuts the shear
axis under F^. Hence, the maximum moment is under /'"",. By
measurement, £'^is64', andP^r is 5,000 lb. ; hence, the maximum
bending moment is 64 X 5,000 = 320,000 in. -lb. Ans.
(579) From the table of Bending Moments, the great-
est bending moment of such a beam is — ^— , or, in this case,
o
w X 240'
8
By formula 74,
_ wx_240]_ _ 5/ _ 45^000
280
8 fc 4 12 ^ 2
^, f 45,000 X 280 X 8 • ^^ no lu u r
Therefore, zu = — — = -f 72.92 lb. per mch of
'
240 X 240 X 4
^
X
length = 72.92 X 12 = 875 lb. per foot of length. Ans.
(580) From the table of Bending Moments, the maximum
bending moment is
From formula 74,
/=,^(^*-