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\fdO'^ - Old Forge Coal Mines

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RAILROAD CONSTRUCTION. 287<br />

120 = 12,231 pounds. Multiplying this weight by -/ / =<br />

8.56 feet, we have 12,221 X 8.50= 104,612 pounds, which,<br />

divided by o d, 16 feet, = 6,538 pounds = the pressure of<br />

the backing. This pressure to a scale of 4,000 pounds to<br />

the inch equals 1.63 inches. P, the center of pressure,<br />

is at ^ of the height of b d measured from d. At P erect a<br />

perpendicular to the back of the wall. Lay off on this perpendicular<br />

the distance /*//= 1.63 in. Draw /*/, making<br />

the angle ;//*/= 33° 41' = the angle of wall friction. At //<br />

draw a perpendicular to P n^ intersecting the line P t \n h.<br />

Draw h k, completing the parallelogram ;/ h k P\ h n, or its<br />

equal k J\ will represent the friction of the backing against<br />

the wall and the diagonal h P, which, to the same scale, =<br />

7,857 pounds, will be the resultant of the pressure of the<br />

backing and the friction. Produce P // to 5. The section<br />

a b d c oi the wall is a trapezoid. Its area is 84 square feet,<br />

which, multiplied by 154 pounds, the weight of rubble per<br />

cubic foot, gives 12,936 pounds, as the weight of the wall,<br />

which, to a scale of 4,000 pounds to the inch, = 3.23 inches.<br />

Find the center of gravity g of the section a b d c, diS explained<br />

in Art. 1488. Through g draw the vertical line<br />

g ty intersecting the prolongation of P h in /. Lay off from<br />

/, on g i, the distance / z\ 3.23 inches = weight of the wall,<br />

and on Is, the distance I in = 1.96 inches, the length of the<br />

resultant P h\ complete the parallelogram I in u v. The diagonal<br />

/ u is the resultant of the weight of the wall and the<br />

pressure. The distance c r from the toe c to the point where<br />

the resultant / u cuts this base is 3.6 feet, or nearly \ of the<br />

base c d. This guarantees abundant stability to the retaining<br />

wall.<br />

The distance o f\s obtained as follows : c d=S feet ; bat-<br />

ter of c a = 1 inch for each foot of length of ^ ^= 16 inches;<br />

hence, a o = S feet — 16 inches = 6 feet 8 inches; a b =<br />

2. 5 feet; therefore, ^/= 12.73 — (6 feet 8 inches — 2. 5 feet) =<br />

8 56 feet.<br />

(862)<br />

See Art. 1491.<br />

(863) See Art. 1491 and Figs. 409, 410, and 411

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