\fdO'^ - Old Forge Coal Mines

STRENGTH OF MATERIALS. 221
(586) Here IF = 21,000; /= 10 ; 5, = 150,000; ^ =
6,250; / = 7.5 X 12 = 90*. For using formula 78, we have
.3183 IVf _ .3183 X 21,000 X 10 ~
_
S,
150,000
11^^11X81(^^20.7360.
g- 6250
--4455.
Therefore,
d= 1.41421^.4456 + i^.U56 (.4456 + 20.7300) =
1.4142-/. 4456 + 3.0722 = 2.65', or say 2f'.
(587) For this case, A = 3.1416 sq. in. ; /=
48'; 5, = 55,000; /= 10; /= .7854; ^=20,250.
Substituting these values in formula 76,
W=
S,A 55,000X3.1416
A^'\ .J. 3.1416 X 48'
.
/(x.f) 4, 20,250 X .7854/
5,500 X 3.1416
1.4551
Steam pressure = 60 lb. per sq. in.
rru c
4 X 12 =
f loo^a
gQ 1.4551X60
Hence d^ - 5>500X 3.1416 _
^^"""^'"^ - " ^^^' "^^'^^y*
.7854 X 1.4551 X 60
•
Then, area of piston = .7854 d* = — = — .
. r,c.. ^. ^ 5,500 X 3.1416
and d= |/252 = 15|-", nearly. Ans.
(588) {a) The strength of a beam varies directly as the
width and square of the depth and inversely as the length.
Hence, the ratio between the loads is
6X 8'.4x 12' — ^^ ,^ -, , A
• = 16 : 15, or Ijiy. Ans.
iQ—
—Jq—
{d) The deflections vary directly as the cube of the
lengths, and inversely as the breadths and cubes of the
depths.
Hence, the ratio between the deflections is