\fdO'^ - Old Forge Coal Mines

212 STRENGTH OF MATERIALS.
Since 40^ of the plate is removed by the rivet holes, 60%
remains, cmd the actual thickness required is
_^^ 2X120X4 8^
.60 .00X55,000
(568) Using a factor of safety of G, in formula 68,
^ 6 3
Hence Hence, ^--5-- ^0,000
. _ 3/^ _ 3 X 6 X 200 _
" '^^ ' ^"'•
(569) Using formula 71, with a factor of safety of 10,
9,600,000/'-"' nr-AAAA^''"
^= 10/^ =960,000^
''«/ //^ _ ^.18/130 X 12 X 12 X 3
Hence, /= f -£-- = f - = .272".
^
960,000
960,000
Ans.
(570) From formula 70,
._ S^ or.-- ^^ - ^>OOOXt _4,000_
/*-;-4_/'°^^- ^^-^-2,800-2,000- 800-^- ^''^•
(571) See Fig. 46. (a) Upon the load line, the loads
0-1^ 1-2, and 2-3 are laid off equal, respectively, to
F^, i%, and F^ ; the pole P is chosen, and the rays drawn in
the usual manner; the pole distance //= 2,000 lb. The
equilibrium polygon is constructed by drawing a c, c d, d e,
and e f parallel to PO, Pi, P 2, and PS, respectively, and
finally drawing the closing line/" a to the starting point n.
P in is drawn parallel to the latter line, dividing the load
line into the reactions in = P^, and 3 vi = R^. The shear
axis in n is drawn through in, and the shear diagram
h I . . . . s' n m \s constructed in the usual manner. To
the scale of forces in = 1,440 lb., and 3 in = 2,160 lb. To
the scale of distances the maximum vertical intercept j' =
^'^=31.2 ft., which, multiplied by N, = 21.2 x 2,000==
62,400 ft. -lb. = 748,800 in. -lb. Ans.
{d) The shear at a point 30 ft. from the left support =
in = l,UO\h. Ans.
{c) The maximum shear = ns' — — 2,160 lb. Ans.