Art. 236. TO FIND THE SOLID CONTENT OF THE WALLS OF A

RECTANGULAR BUILDING, OF ANY GIVEN DIMENSIONS.

RULE. From the outside perimeter of the building, subtract four times the thickness of the walls; the remainder will be the length of the walls. Multiply the length by the height, and the product by the thickness; the result will be the solid content.

1. What is the number of cubic feet in the walls of a house 40 feet by 24, the walls being 16 inches in thickness, and 32 feet in height, no deduction being made for doors and windows?

40+40+24+24=128-5} = 122%, and 122 × 32 × 13: 52337 cubic feet.

=

2. What is the number of cubic feet in the walls of a school-house 75 feet by 54, and 40 feet in height, the walls being 20 inches in thickness to the height of 28 feet, and the remaining 12 feet 16 inches, making no deduction for doors and windows?

Art. 237. TO FIND THE NUMBER OF BRICKS WHICH WILL BE

REQUIRED TO BUILD A WALL OF ANY GIVEN DIMENSIONS.

RULE. Find the content of the wall in cubic feet, find also the number of bricks of the given dimensions that will be required to build a cubic foot, multiply the number of cubic feet in the wall by the number of bricks in a cubic foot, the product will be the required number of bricks. Bricks of the usual size are 8 inches in length, 4 inches in width, and 2 inches in thickness.

8X4X2=64, the number of cubic inches in one brick, and 17286427, the number of bricks in a cubic foot.

A deduction of about one tenth of the required number of bricks is usually made for the thickness of the mortar between the courses.

1. What number of bricks of the usual size will be required to build a garden wall 75 feet in length, 6 feet in height, and the width of 4 bricks in thickness?

75 X6 X 13=600 X 27=16200, the required number of bricks. 2. What number of bricks will be required to build the walls of the school-house, of which the dimensions were given in question 2d, Art. 236?

Art. 238. TO FIND THE CONTENTS OF CIRCULAR WALLS, SUCH

AS THE WALLS OF RESERVOIRS, CISTERNS, DRAINS, AND WELLS. The length of a circular wall is the circumference of a circle whose diameter is the thickness of the wall longer than the inside diameter.

1. What number of bricks will be required to build a circular cistern, 8 feet in depth, 6 feet 4 inches in diameter inside, the circular walls being 8 inches in thickness, and the bottom 4 inches?

6 ft. 4 in. 8 in. 7 ft. X 3.1416-21.9912 ft., the circumferor length of the wall. 21.9912 X 8: 175.9296 X= 117.2864 cubic ft. in the circular wall.

ence

=

6 ft. 4 in. +1 ft. 4 in. 7 ft. 8 in., the diameter of the bottom. And 7.666 X 3.1416 = 24.0835, and 12.04175 X 3.833 X 15.3853+, the number of cubic feet in the bottom. And 117.2864+ 15.3853132.6717 × 27=3582.1359+, the required number of

bricks.

Art. 239. TO FIND THE NUMBER OF PERCHES OF STONE IN A

WALL OF ANY GIVEN DIMENSIONS.

RULE. Find the number of cubic feet in the given wall, and divide by 24.75, the quotient will be the number of perches.

1. What number of perches of stone will be required for the walls of a cellar whose length is 33 feet, and breadth 18 feet, within the walls, height of walls 8 feet, thickness 1.5 feet?

Art. 240. TO FIND THE SOLIDITY OF ANY IRREGULAR BODY

WHOSE DIMENSIONS CANNOT BE ASCERTAINED.

RULE. Immerse the solid in some regular formed vessel, partly filled with water, and the contents of that part of the vessel filled by the rise of the water will be the solid content of the body immersed.

A school-boy, wishing to find the solid content of an irregular piece of granite, immersed it in a vessel 15 inches square, partly filled with water, and found that the water rose 4 inches; what was the solid content of the piece of granite?

Ans. 900 cubic inches.

Art. 241. The unit of measure for excavations, such as canals, cellars, wells, &c., is a cube whose side is 6 feet, called a square of earth, and contains 216 cubic feet. Excavations are also estimated by the cubic yard.

TO FIND THE NUMBER OF SQUARES OF EARTH IN ANY GIVEN EXCAVATION.

RULE. Find the content in cubic feet, and divide it by 216; the quotient will be the number of squares.

1. What number of squares of earth must be removed in digging a rectangular cellar 36 feet in length, and 2.1 feet in width; depth at the corners 8 feet, 7 feet, 6 feet, and 5 feet?

NOTE. The mean depth may be found by dividing the sum of the depths at the corners by 4; the quotient will be the mean depth.

8+7+6+5=26÷4=6.5 ft., the mean depth. 36 X 21 X 6.5=491421622.75 squares.

2. What number of cubic yards of earth must be removed in digging a well 24 feet in depth, 6 feet in diameter at the top, and 5 feet at the bottom?

Art. 242. Artificers compute the contents of their work by different units of measure. Painting, plastering, and paving, are estimated by the square yard. Slating and tiling are usually esti

mated by the square of 100 square feet. Glazing is estimated by the square foot. Plumbers' work is usually estimated by the pound or hundred weight.

Sheet-lead, used in roofing, &c., weighs from 7 to 12 lb. per square foot. Leaden pipes of inch bore weigh 10 lb.; of 1 inch bore, 12 lb.; of 14 inch bore, 16 lb.; of 14 inch bore, 18 lb.; of 13 inch bore, 21 lb.; and of 2 inches bore, 24 lb. per yard, in length.

1. What is the number of square yards of paving in a side-walk 66 feet in length, and 8 feet in width? Ans. 58 square yards. 2. How many squares of slating are there in the roof of a house 44 feet in length, and each side of the roof 24 feet in breadth? Ans. 21 squares.

GAUGING.

Art. 243. GAUGING IS THE METHOD OF FINDING THE CAPACITY

OF CASKS, AND OTHER VESSELS, IN GALLONS.

RULE. Take the length of the cask in inches, also the diameter at the bung and head; subtract the head diameter from the bung diameter, and note the difference.

If the staves of the cask are much curved, multiply the difference noted by 7; if they are but little curved, by .6; if they are of a medium curve, by .65; add the product to the head diameter; the sum is the mean diameter, which changes the form of the cask to a cylinder.

Find the square of the mean diameter and multiply it by .7854; the product is the mean area; then multiply this mean area by the length of the cask; the product is the capacity of the cask in cubic inches.

Divide this product by 231, the quotient will be the number of wine gallons; or divide it by 284, the quotient will be the number of beer gallons, that the cask will contain.

The length of the cask is usually taken with callipers, and from 1 to 2 inches is deducted for the thickness of the heads, according to the size of the cask.

The head diameter should be taken within the chimes and close to the head, and from .2 to .5 of an inch, according to the size of the cask, must be added, on account of the diameter being greater inside the head.

1. What number of wine gallons will a cask contain, 36 inches in length, 20 inches in diameter at the bung, and 17 inches in diameter at the head; the staves being much curved?

Ans. 44.65+wine gallons.

2. What number of beer gallons will a cask contain, whose bung diameter is 32 inches, head diameter 27 inches, and length 45 inches; the staves being of a medium curve?

22*

STRENGTH OF MATERIALS.

Art. 244. MATERIALS are exposed to four different kinds

of strains.

1st. They may be torn asunder, as in the case of ropes, tiebeams, and stretchers. The strength of a body to resist this kind of strain is called its resistance to tension, or absolute strength.

2d. They may be crushed or compressed in the direction of their length, as in the case of columns and posts.

3d. They may be broken across, as in the case of beams and joists. The strength of a body to resist this kind of strain is called its lateral strength.

4th. They may be twisted or wrenched, as in the case of axles and screws.

Extensive and accurate experiments are necessary to determine the several measures of these strengths in the different materials.

TABLE I.

Showing the weight in pounds that will pull asunder a prism one inch square, of the following materials, according to the experiments of Mr. Rennie:

Pounds.

Pounds.

Cast steel,

113077

Swedish iron,

Cast iron,

19096

Cast copper,

19072 Do. 1 inch in diameter, 5026

TABLE II.

Showing the cohesive strength of a bar of iron of different kinds, one inch square, obtained from different experimentists, expressed in pounds.

Pounds.

Pounds.

Showing the lateral strength of the following materials, the bar being one foot long, and one inch square, expressed in pounds.

Art. 245. The strength of materials to resist a strain lengthwise is in proportion to the squares of their diameters, or similar sides.

1. If a good hemp rope, 1 inch in diameter, will resist a strain of 6400 pounds, what will be the resistance of a cable of the same materials, 3 inches in diameter?

3 X 3

=

9 X 6400=57600 lbs. Ans. 2. If the cohesive strength of a bar of English iron 1 inch square be 66900 pounds, what is the cohesive strength of a bar 4 inches square? Ans. 1070400 lbs.

Art. 246. The lateral strength of bars and beams of the same material, and of the same length, is in proportion to their width multiplied by the square of their depth.

In square bars and beams, the lateral strengths are in proportion to the cubes of the sides.

1. Suppose a bar of cast iron 1 inch square, and 10 feet in length, to bear a weight of 320 pounds, what number of pounds will a bar of the same length, and 3 inches square, support?

3X 3X3=27 X 320 8640 lbs. Ans.

2. If a joist 10 feet long, 2 inches in thickness, and 4 inches in depth, sustain a weight of 1280 pounds, what number of pounds will a joist of the same length sustain, which is 3 inches in thickness, and 9 inches in depth? Ans. 9720 lbs.

Art. 247. The strength of bars, joists, and beams, of the same width and depth, is diminished in proportion to their length.

1. There are two joists of the same depth and thickness; one is 15 feet in length, the other 12; the shorter joist will sustain a weight of 1500 pounds; what number of pounds will the longer joist sustain? Ans. 1000 lbs.

2. There are two beams of the same depth and thickness; one is 24 feet in length, and the other 18 feet; the longer beam will sustain a weight of 9000 pounds; what number of pounds will the shorter beam sustain? Ans. 12000 lbs.

Art. 248. The strength of a hollow cylinder is to that of a solid cylinder of the same length, and the same quantity of matter, as the greater diameter of the hollow cylinder is to the diameter of the solid cylinder; and the strength of hollow cylinders of the same length, weight, and materials, are as their greater diameters.

1. What is the comparative strength of a solid cast iron cylinder 12 feet in length, and 3 inches in diameter, with a cast iron hollow cylinder of the same weight and length, whose greater diameter is 6 inches?

Ans. The solid cylinder is one half the strength of the hollow cylinder.