\fdO'^ - Old Forge Coal Mines

STRENGTH OF MATERIALS. 227
Tzd'
{c) As above, r=>|/^=/iL = /|=^. Ans.
i)d
(600) Using formula 69, pd—^^tS, we have t = ^.
Using a factor of safety of 0,
^, 4/5 ^ epd 6X100X8 _, .
^^=^'^^^^ ir= 4X20,000
=-^^- ^"^-
(601) The graphic solution is shown in Fig. 52. The
uniform load is divided into 14 equal parts, and lines drawn
through the center of gravity of each part. These loads
are laid off on the line through the left reaction, the pole P
chosen, and the rays drawn. The polygon b c d e f a is then
drawn in the usual manner. The shear diagram is drawn
as shown. The maximum shear is either t T ox r v ^=- 540 lb.
The maximum moment is shown by the polygon to be at
/"r vertically above the point ?/, where the shear line crosses
the shear axis. The pole distance /*7 is 1,440 1b. to the
scale of forces, and the intercept/"^ is 14 inches to the scale
of distances. Hence, the bending moment is 20,160 in. -lb.
(602) From formula 74,
Any
^=-p*- = 20,160. 5. = 9,000;/= 8.
_,, / 20,160X8 _ ^-
=^^-^^'
^^^"'r= 9,000
But, — = ^^. . z=i —bd"^ for a rectangle.
c ^d 6
Hence, i^^' = 17.92, ot bd' = 107.52.
number of beams will fulfil this condition.
Assuming d=iQ', b— — -^ — = 3', nearly.
Assuming d=h\ b = — ^ — = 4.3'.
(603) Using the factor of safety of 10, in formula 71,
y~ 10 Id 108x2.5