Rural arithmetic by Thomas Augustus Orloff 1863-1935

Author:Thomas, Augustus Orloff, 1863-1935
Language: eng
Format: epub
Tags: Arithmetic
Publisher: New York, American Book Co
Published: 1916-03-25T05:00:00+00:00

43. Atf5.17acubicyard, how much will it cost to construct a cement foundation for a building 26 ft. long and 20 ft. wide with walls 8 in. thick and 7J ft. high?

44. With the cost of cement 15.17 per cubic yard, how much will it cost to construct a retaining wall 100 ft. long and 9 ft. high, the wall to be 60 in. wide at the bottom and 8 in. wide at the top ?

45. How muclv will it cost, at f 5.17 per cubic yard, to build a curb about a corner lot 75 ft. wide a-nd 150 ft. deep, the curb to be 16 in. deep and 6 in. wide ?

Solution. —75 ft. + 150 ft., = 225 ft., entire length of curb. (225 X li X J) cu. ft. = 150 cu. ft. 150 cu. ft. -I- 27 cu. ft. = 5|, number of cubic yards. Cost of 1 cu. yd. = $5.17. Cost of 5| cu. yd. = 5^ x «5.17 = $28.72.

46. Find the cost of constructing a curb in front of a lot 50 ft. wide, the curb to be 14 in. deep and 7 in. wide.

47. Find the cost of laying the cement floor of a pergola 36 ft. long and 12 ft. wide, the floor to be 6 in. thick.

48. Find the cost of constructing a series of 6 steps, each step 12 ft. long, 14 in. wide, and 8 in. thick.

49. How many cubic yards of concrete will it require to construct 1000 ft. of sewer, the inside diameter being 5 ft.,

^emittt^ the concrete 5J in. thick ?

f \^ Note. —The outer surface of the sewer pipe forms a cylin-

L-y- J drical space whose diameter is 5 ft. 11 in. or 5|i ft.; the inner \^ ^^ surface forms a cylindrical space whose diameter is 5 ft. ^^"^^ The difference between the volumes of these two cylinders (see p. 228) is the volume of tlie concrete (27 cu. ft. = 1 cu. yd.).

BUILDING PROBLEMS

50. Find the cost of constructing the foundation of a , public building according to the accompanying plans.

Note. — The small cross section shows that the lower Il2* P^^^ ®f *^® foundation is 3 ft. wide and 12 in. deep, and the upper part, 2 ft. wide and 8 in. deep. Solution. — Perimeter = 922 ft. (No deduction for corners.) Area of cross section = (3 x ||) sq. ft. + (2 x ^) sq. ft. = 4J sq. ft. (922 X ^)^27 = 147.975, number of cubic yards. 147.975 x»5.I7 = ^765.03.

171'

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