Estimates. calculate 900 shingles for each "square" (10 ft. by 10 ft.) to be covered. If the shingles are to be laid only 4 inches to the weather, 1 M. will be required for each square to be covered. 4. Estimate the shingles (to be laid 4 inches to the weather) for a double roof that is 36 feet long and whose rafters are 20 feet long. In giving the answer to this problem remember that lumber dealers will not divide a bunch of shingles. 5. Allowing 10% for waste, 1 thousand shingles laid 5 inches to the weather, will cover how many square feet? 6. Allowing 10% for waste and laying the shingles 5 inches to the weather, how many bunches must be purchased for a double roof that is 60 feet long and whose rafters are 26 feet long? SIDING. } NOTE.-Siding is about inch in thickness and 5 inches wide, but it is counted as though it were an inch thick and 6 inches wide; that is, a 16-foot piece of siding is said to contain 8 feet of lumber. When siding is laid "4 inches to the weather," what per cent of itself must be added to the number of square feet to be covered to denote the number of feet of lumber required? 7. Estimate the siding for a building that is 30 feet by 40 feet and 18 feet to the plates,* the roof running lengthwise of the building and its highest point being 9 feet higher than the plates. Consider the openings (doors and windows) sufficient to make up for defective lumber and waste in cutting. 8. Allowing 7% for waste, 1 thousand feet of siding laid "4 inches to the weather" will cover how many square feet of surface? 9. A sufficient amount of siding was purchased to com *The plates are the horizontal timbers upon which the rafters rest near their lower ends. The height to the plates is practically the height to the gables. The gables are triangular, and in this case are 30 ft. by 9 ft. Estimates. plete a job, provided the siding was laid "4 inches to the weather." It was laid "4 inches to the weather." What part of the amount purchased was unused? MATCHED FLOORING. NOTE.-Common matched flooring is, omitting the tongue, but little more than 5 inches in width, but since it is made from 6-inch boards, it is called 6-inch flooring and is counted as 1 foot of lumber for every 2 linear feet of flooring. To find the amount of flooring needed for a given surface add of itself to the number of feet to be covered. 10. Estimate the amount of 6-inch flooring needed for two floors throughout a building, main part, 24 by 36 with an ell 18 by 20. 11. Find the cost at $27 per M. of the flooring for the building described in Problem 10. 12. One thousand feet of 6-inch matched flooring will cover how many feet of surface? SHEATHING. NOTE.-A Somewhat low grade of "matched flooring" is often used for the sheathing of buildings; that is, for a first covering of the frame work. The siding covers the sheathing. 13. Estimate the sheathing of "matched flooring” for a building 18 feet by 24 feet and 15 feet to the plates, the roof running lengthwise of the building and the highest point being 6 feet higher than the plates. STOCK BOARDS. NOTE.-Barns are sometimes covered with stock boards 1 foot in width. These boards are put on vertically and the cracks between them are covered with battens. 14. Estimate the stock boards for covering a barn that is 40 feet by 60 feet and 24 feet to the plates, the roof running lengthwise of the barn and its highest point being 10 feet higher than the plates. Consider the openings sufficient to Estimates. make up for the waste in cutting the lumber and for defective boards. SIDEWALKS. 15. Find the cost of lumber for 240 feet of side walk 6 feet wide the walk to be made of 2-inch plank laid on three 2 by 6 stringers and the lumber to cost $17.50 per M. 16. Find the cost of lumber for 80 feet of walk, 3 feet wide, the walk to be made of 1-inch boards laid on two 2 by 4 stringers, the lumber to cost $16 per M. • BRICK AND MORTAR. NOTE.-For data for estimating brick, see page 244 of this book. 17. Estimate the brick for the outer walls of a building, the outside perimeter of which is 240 feet, the height of the walls to the gables being 29 feet and the thickness, 3 bricks or, including the mortar, about 13 inches. Make no allowance for openings or double corners" but assume that the material saved at these points will be sufficient to lay the gables. "" 18. To lay 1000 brick, about 13 bbl. lime and 13 bushels sand are required. Find the amount of lime and sand necessary to lay the brick in the outer walls of the building described in Problem 17. 19. Estimate the brick for the wall of a well 22 feet deep and 20 feet inside diameter, the brick to be laid without mortar and the wall to be 12 inches in thickness.* STONE. NOTE. For data for estimating stone, see page 243 of this book. 20. Estimate the stone (cords) for the foundation of the outer walls of the building described in Problem 17, the * For this estimate, it will be sufficiently accurate for practical purposes to use 3 as the ratio of diameter to circumference, to regard the length of the wall as equal to the circumference of a circle 21 feet in diameter, and to compute 25 bricks to the cubic foot. Estimates. foundation to be 5 feet high and 20 in. in thickness. Make no allowance for “double corners" or openings. The material saved at these points may be used in the expansion of the wall at the bottom. 21. Which is the cheaper, so far as material is concerned, a 12-inch (practically 13-inch) wall of brick or an 18-inch wall of stone, the brick costing $6.00 per M., and the stone $8.00 per cord? PLASTERING. 22. Find the cost at 28¢ per square yard, of plastering three rooms, walls and ceilings, the dimensions of the rooms being as follows: Room 1, 18 ft. by 20 ft.; room 2, 14 ft. by 16 ft.; room 3, 13 ft. by 19 ft.; height of each room, 10 ft. Make no allowance for openings. WATER. NOTE.—In computing capacity for water, it is customary to regard 7 gallons as equal to 1 cubic foot. Water weighs 624 lb. to the cubic foot. 23. When the water in the well described in Problem 19, is 12 feet deep, how many gallons of water in the well? 24. A tank 10 feet deep and 12 feet in diameter will contain how many gallons of water? 25. What is the weight of the water that will exactly fill the tank described in Problem 24? 26. Estimate the weight of water sufficient to fill a tank as large as your school room, to the depth of 3 feet. 27. Estimate the number of gallons of water sufficient to cover the floor of your school room to the depth of 1 inch. 28. What is the weight of water to the acre when there is a rain-fall of 1 inch? 29. A heavy shower raised the water in a cistern 8 feet long and 6 feet wide, 2 feet 6 inches. The water came from Estimates. a roof 30 feet long and 22 feet wide. The rain-fall was how many inches? 30. How many wells or cisterns 4 feet in diameter are equal in capacity to a well or cistern 20 feet in diameter, the depth being the same in each case? 31. Which would require the greater number of brick for walls, twenty-five 4-foot wells or one 20-foot well, the depth of the wells being the same in each case, but the 4-foot wells to have a 4-inch wall and the 20-foot well to have a 12-inch wall? WEIGHT OF IRON. NOTE-Iron is about 74 times as heavy as water, the same bulk being considered. 32. (a) How much does a cubic foot of iron weigh? (b) A cubic inch? 33. Estimate the weight of a bar of iron, 1 inch by 6 inches by 12 feet. 34. (a) Estimate the weight of a 4-inch iron ball. (b) Of a 2-inch iron ball. 35. (a) Estimate the weight of an iron rod 1 inch in diameter and 12 feet long. (b) Of an iron rod 2 inches in diameter and 12 feet long. 36. Estimate the weight of a sheet of boiler iron, 8 feet. square and of an inch thick. 37. Compare the weight of an iron ball 6 inches in diameter with the weight of one 2 inches in diameter. WEIGHT OF ICE. NOTE.-The specific gravity of a body that floats on water is less than 1; its weight is less than 62 lb. to the cubic foot. The fraction that indicates the part of a floating body that is below the surface of the water, also indicates its specific gravity. Thus, if three-fourths of a floating body is immersed, its specific gravity is 3, and its weight is 3 of 62 lb. to the cubic foot. |