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weight of the body in air, so is the specific gravity of water to the specific gravity of the body. NOTE. — A cubic foot of water weighs 1000 ounces.

1. A stone weighed 10 pounds, but in water only 64 pounds. Required the specific gravity.

Ans. 2608.6+. 2. Suppose a piece of elm weigh 15 pounds in air, and that a piece of copper, which weighs 18 pounds in air and 16 pounds in water, is affixed to it, and that the compound weighs 6 pounds in water. Required the specific gravity of the elm.

Ans. 600.

SECTION LXXXII.

STRENGTH OF MATERIALS.

The force with which a solid body resists an effort to separate its particles or destroy their aggregation can only become known by experiment.

There are four different ways in which the strength of a solid body may be exerted; first, by resisting a longitudinal tension; secondly by its resisting a force tending to break the body by a transverse strain; thirdly, in resisting compression, or a force tending to crush the body; and, fourthly, in resisting a force tending to wrench it asunder by torsion. We shall, however, only consider the strength of materials as affected by a transverse strain.

When a body suffers a transverse strain, the mechanical action which takes place among the particles is of a complicated nature. The resistance of a beam to a transverse strain is in a compound ratio of the strength of the individual fibres, the area of the cross section, the distance of the centre of gravity of the cross section from the points round which the beam turns in breaking.

The following are the facts and principles on which mechanics make their calculations.

1. A stick of oak one inch square and twelve inches long, when both ends are supported in a horizontal position, will sustain a weight of 600 pounds; and a bar of iron of the same dimensions will sustain 2190 pounds.

2. The strength of similar beams varies inversely as their lengths; that is, if a beam 10 feet long will support 1000 pounds, a similar beam 20 feet long would support only 500 pounds.

3. The strength of beams of the same length and depth is directly as their width ; that is, if there be two beams, each 20 feet long and 6 inches deep, and one of them is 6 inches wide and the other but 3 inches, the former will support twice the weight of the latter.

4. The strength of beams of the same length and width is as the squares of their depths; that is, if there be two beams, each of which is 20 feet long and 4 inches wide, but one is 6 inches deep and the other is 3 inches deep, their strength is as the squares of these numbers. Thus, 6 x 6 = 36; 3 x3=9; that is, the strength of the former is to the latter as 36 to 9. It will, therefore, sustain four times the weight of the latter. Thus, 36 = 9=4.

5. To compare the strength of two beams of the same length, but of different breadth and depth, we multiply their widths by the squares of their depths, and their products show their comparative strength. Thus, if we wish to ascertain how much stronger is a joist that is 2 inches wide and 8 inches deep, than one of the same length that is 4 inches square, we multiply 2 by the square of 8, and 4 by the square of 4; thus, 2

X 8 X 8= 128; 4 X 4 X 4 = 64; 128 = 64=2. Thus, we see that although the quantity of material in one joist is the same as in the other, yet the former will sustain twice the weight of the latter. Hence “deep joists” are much stronger than square ones, which have the same area of a transverse section.

6. To compare the strength of two beams of different lengths, widths, and depths, we multiply their widths by the squares of their depths, and divide their products by their lengths, and their quotients will show their comparative strength. Therefore, if we wish to ascertain how much stronger is a beam that is 20 feet long, 8 inches wide, and 10 inches deep, than one 10 feet long, 6 inches wide, and 5 inches deep, we adopt the following formulas :

- 8 X 10 X 10 6 x5x5

20

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15.

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The strength of the former, therefore, is to the latter as 40 to 15 ; that is, if the first beam would sustain a weight of 40cwt., the latter would sustain only 15cwt.

7. Having all the dimensions of one beam given, to find another, part of whose dimensions are known, that will sustain the same weight. We multiply the width of the given beam by the square of its depth, and divide this product by the length, and the result we call the reserved quotient ; then, if we have the length and breadth of the required beam given to find the depth, we multiply the reserved quotient by the length of the required beam, and divide the product by its width, and the quotient is the square of the depth of the required beam. If the length and depth of the required beam were given to find the width, we multiply the reserved quotient by the length of the required beam, and divide this product by the square of the depth of the required beam, and the quotient is the breadth. But if the width and depth of the required beam were given to find the length, we multiply the width of the required beam by the square of the depth, and divide this product by the reserved quotient, and he result is the length of the required beam.

8. A triangular beam will sustain twice the weight with its edge up that it will with its edge down. Hence split-rails have twice the strength with the narrow part upward, which they have with the narrow part downward.

9. In making the above calculations, we have not noticed the weight of the beam itself, and in short distances it is of but little consequence; but where a long beam is required, its weight is of importance in the calculation.

10. A beam supported at one end will sustain only one fourth part the weight which it would if supported at both ends.

11. The tendency to produce fracture in a beam by the application of a weight is greatest in the centre, and decreases towards the points of support; and this ratio varies as the square of half the length of the beam to the product of any two parts where the weight may be applied. Hence the tendency of a weight to break a bar 8 feet long, when applied to the centre, to that of the same weight, when applied 3 feet from one end, is as 4 X 4 = 16 to 3 x 5 = 15.

QUESTIONS TO BE PERFORMED BY THE PRECEDING RULES.

1. If a stick of oak 1 inch square and 12 inches long, when both ends are supported in a horizontal position, will sustain a weight of 600 pounds, how many pounds would a similar stick sustain, that was 36 inches long? Ans. 200 pounds.

2. If a beam 4 inches square and 12 feet long would support a weight of 1000 pounds, how many pounds would a similar beam support, that was 3 feet long? Ans. 4000 pounds.

3. If a beam 20ft. long, 4in. wide, and bin. deep, will sustain a weight of 2000lbs., how many pounds would a similar beam sustain, that was 3in. wide ?

Ans. 1500lbs. 4. If a beam 10 feet long, 4 inches wide, and 3 inches deep, will sustain a weight of 1000 pounds, how many pounds would a similar beam support, that was 6 inches deep?

Ans. 4000 pounds. 5. If a beam 10 feet long, 2 inches wide, and 4 inches deep, will sustain a weight of 1000 pounds, how many pounds would a beam, that is 10 feet long, 4 inches wide, and 6 inches deep, sustain?

Ans. 4500 pounds. 6. If a beam 2 feet long, 2 inches wide, and 3 inches deep, will sustain a weight of 4000 pounds, what will a beam, that is 4 feet long, 3 inches wide, and 6 inches deep, sustain ?

Ans. 12000 pounds. 7. If a beam that is 10 feet long, 4 inches wide, and 6 inches deep, will sustain a weight of 4 tons, what weight will a beam sustain, that is 20 feet long, 8 inches wide, and 10 inches deep ?

Ans. 113 tons. 8. If a beam 6 inches square and 8 feet long will support a weight of 2000 pounds, what weight will a beam 10 feet long and 10 inches square sustain ?

Ans. 740741 pounds. 9. If a beam 15 feet long and 5 inches square will sustain a weight of 1200 pounds, required the length of a beam, that is 8 inches square, that will sustain a weight of 2000 pounds.

Ans. 36125 feet. 10. If a beam 8 feet long and 7 inches square will sustain a weight of 3000 pounds, how many inches square must be the beam, that is 6 feet long, that will sustain 2000 pounds ?

Ans. 5.5+ inches. 11. If a bar of iron 10 feet long, 2 inches wide, and 3 inches deep, will sustain 10 tons, what must be the depth of a bar, that is 12 feet long and 3 inches wide, that will sustain 30 tons ?

Ans. 4.64+ inches. 12. If it require 1000 pounds to break a certain beam, that is 24 feet long, when placed in its centre, required the weight necessary to break it, when placed within 4 feet of one end of the beam?

Ans. 13. If a beam 6 inches square and 10 feet long will sustain a weight of 2000 pounds from its centre, what weight would a beam of the same material sustain that is 10 inches square and 12 feet long, if the weight were suspended 2 feet from the centre?

Ans.

SECTION LXXXIII.

ASTRONOMICAL PROBLEMS.

PROBLEM I. To find the dominical letter for any year in the present century, and also to find on what day of the week January will begin.

Rule. - To the given year add its fourth part, rejecting the fractions ; divide this sum by 7; if nothing remains, the dominical letter is A; but if there be a remainder, subtract it from 8, and the residue will show the dominical letter, reckoning 0 A, 2 B , 3= C, 4 D , 5=E, 6=F,7=G. These letters will also show on what day of the week January begins. For when A is the dominical letter, January begins on the Sabbath; when B is the dominical letter, January begins on Saturday; C begins it on Friday; D, on Thursday; E, on Wednesday; F, on Tuesday; G, on Monday.

Note. - If it be required to find the dominical letter for the last centu. ry, proceed as above; only, if there be a remainder after division, subtract it from 7, and the remainder shows the dominical letter, reckoning 1=A, 2= B, 3=C, 4=D, 5=E, 6=F, 0=G.

1. Required the dominical letter for 1835.
OPERATION.
4)1835 8—4=4=D= dominical letter.
458

As D is the dominical letter, January began 7)2293

on Thursday, and the fourth day was the 327-4 Sabbath. 2. Required the dominical letter for 1836. OPERATION. 4)1836 8-6=2= B and C= dominical letters.

459 7)2295 In leap years there are two dominical letters.

3996 The last letter, C, is for January and February,

"no and B for the remainder of the year. As C is the dominical letter, January began on Friday, and the third day was the Sabbath.

3. Required the dominical letter for 1841 ? Ans. C. 4. Required the dominical letter for 1899. Ans. A. 5. Required the dominical letters for 1896. Ans. D and E. 6. What is the dominical letter for 1786 ? Ans. A. 7. What is the dominical letter for 1837 ? Ans. A.

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