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2. What must be the weight necessary to tear a cylinder of cast-iron 2 inches diameter of mean strength, for which C = 48000? Ans. 150797 lbs. 3. What must be the thickness of a spar of English oak, whose breadth is 2 inches, that will require 20 tons weight to tear it, the oak being of the strongest kind?

Ans. 1 inches. 4. Required the side of a square Norway spar, which is torn by a weight of 1.5 tons. Ans. 1.7 inches. 5. What must be the diameter of an iron wire, in order that it may just support a weight of 2000 lbs. ?

Ans. 0.178 inch.

98. PROBLEM II.-To find the deflexion of a beam fixed at one end and loaded at the other.

'Find the continued product of the tabular value of E, the breadth of the beam, and the cube of its depth, both in inches; find also the continued product of 32, the given weight in pounds, and the cube of the length in inches; divide the latter product by the former, and the quotient will be the deflexion in inches."

When the weight is uniformly distributed along the beam, the same rule applies, except that the factor 32 must be changed to 12.

The rule applies only to cases of small deflexion.

When the specific gravity differs from that given in the table, a proportion must be made, as in the last problem, namely, the tabular specific gravity is to that given, as the result found by the preceding rule to the true deflexion.

Let b, l, d, W the breadth, length, and depth of the beam and the weight, respectively, and D = the deflexion, then D=32 W3 Ebd3;

and when the load is uniformly distributed,

D12 Wi3 Ebď3.

EXAMPLE.-How much will a batten of larch 5 feet long, 1 inches broad, and 21 deep, be deflected by a weight of 15 lbs. suspended from its free extremity?

D=32 W13 ÷ Ebd3 — 32 × 15 × 603÷4200000 × 11 × (§)3

=

32×15×216000×2×8

4200000 × 3 × 125

16×1×72×64
700 × 100

16x10x216×64

2100×1×1000

= 1.05 inches.

Formulas may easily be obtained from the rule to find any one of the quantities l, b, d, W, D, or E; when the rest are

given; thus,

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32 13W

When the beam is square bd, and then 64 —

The constant, E, could be also found, for E=

EXERCISES.

DE

32 W 13

Dbd

1. A spar of Mar Forest fir is 10 feet long, 2 inches broad, and 3 deep, what will be its deflexion if a weight of 10 lbs. be suspended from its free end? Ans. 3 inches. 2. What will be the deflexion of the same spar, if it is supported by a spur at 4 feet distance from the wall?

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3. What will be the deflexion of the same batten as that in the first exercise, supposing the weight to be uniformly distributed along it? Ans. 1 inch. 4. What must be the side of a square ash batten which projects 4 feet from a wall, and which is deflected 13 inches by a weight of 200 lbs. ? Ans. 3 inches. 5. A bar of malleable iron 83 feet long, 2 inches deep, and 1 thick, supports a weight of 80 flexion?

lbs.; what is its de

Ans. 3 inches.

99. PROBLEM III. -To find the deflexion of a beam supported at both ends, and loaded with a weight at the middle.

'Find the continued product of the tabular value of E, the breadth, and cube of the depth, both in inches; find also the product of the cube of the length in inches into the

given weight in pounds; then divide the latter product by the former, and the quotient will be the deflexion in inches.'

100. When the beam is fixed at both ends, the deflexion, with an equal weight, will be only of the result found by the above rule.

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b= (4), d = 1 ~ DEb

DE

It is evident, from the above rule, that the deflexion in the former problem for the same beam and weight is 32 times greater than in this problem.

EXAMPLE. Find the deflexion of a beam of the best English oak, its length being 20 feet, breadth 4 inches, and depth 5 inches, when loaded with a weight of 1000 lbs. W13 1000 × 203 × 123

D

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1 x 8000 × 432

7000 × 1x125

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1. A joist of American birch 20 feet long, 3 inches thick, and 8 deep, is loaded with a weight of 3000 lbs. ; what is the depression? Ans. 4.7 inch. 2. A beam of iron 30 feet long, 3 inches thick, and 9 inches deep, is fixed at both ends, and loaded with 3 tons; what is the deflexion? Ans. 1.38 inches. 3. A square beam of the best English oak was deflected 1.53 inch by a weight of 1000 lbs; what was its side, its length being 20 feet? Ans. 6 inches. 4. A beam of red pine 33 feet 4 inches long, and 1 foot deep, was depressed in the middle 14 inches by a weight of 2000 lbs.; what was its breadth? Ans. 8 inches.

101. When the beam is uniformly loaded, and supported at both ends, the deflexion will be only of that found by the preceding problem.

102. When the beam is uniformly loaded and fixed at

both ends, the deflexion will be only of that found in the preceding problem, or of that found by the rule in the preceding article.

103. The deflexion of a beam, fixed at both ends, is only of the deflexion when its ends are only supported, whether it be loaded by a weight at the middle, or by a weight uniformly distributed over it.

The examples in the preceding problem are sufficient to illustrate the rules in articles 101 and 102.

The formula for the case in art. 101 is D =

and that for the case in art. 102 is D=

5 W 13

8 Ebd3

5 W/3

12 Ebd3 And from these the formulas for b, d, l, and W, are easily obtained.

104. PROBLEM IV. To find the ultimate deflexion of a beam before rupture, when supported at both ends.

'Multiply the tabular value of U by the depth of the beam in inches, and divide the square of the length also in inches by that product, and the quotient will be the ultimate deflexion."

Hence,

D= 12÷dU.

12=dDU, and d = 12 ÷ DU.

105. The same rule applies when the beam is fixed at both ends.

106. When the piece is fixed only at one end, the deflexion will be 8 times greater than that found by the above rule, or D = 87 ; dU.

EXAMPLE. Find the ultimate deflexion of a 2 inch plank of Dantzic oak, which is 25 feet long.

D=12÷dU = 3002 ÷ 2 × 724 =

62.165 feet 2 inches.

EXERCISES.

90000

1448

1. Find the ultimate deflexion of an ash plank 4 inches

thick and 40 feet long.

Ans. 145.8 inches.

2. A spar of ash, 2 inches deep and 6 feet long, is fixed at one end, and is broken by a weight applied at the free end; what was the ultimate deflexion? Ans. 52.5 inches. 3. If the ultimate deflexion of a plank of teak is 20 inches, its thickness being 2 inches; what is its length?

Ans. 180.9 inches. 4. The ultimate deflexion of a rod of larch 25 feet long is 40 inches; what was its depth? Ans. 5 inches.

107. PROBLEM V. To find the ultimate transverse strength of a rectangular beam, fixed at one end, and loaded at the other.

'Find the continued product of the tabular value of S, the breadth and square of the depth, both in inches, and divide this product by the length in inches, and the quotient will be the weight in pounds.'

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Wbd2S÷l.

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108. When the beam is uniformly loaded, the weight will be twice as great as that found by this rule.

Besides this rule, another, which is more accurate, is given in Barlow's Essay, which, in the examples given by him, leads to a result from to greater than that obtained by the above rule. The latter rule, however, is more simple, and therefore more convenient in practice, and for a permanent load, only of that obtained by this rule is

taken.

EXAMPLE. Find the ultimate transverse strength of a malleable bar of iron 10 feet long, 2 inches thick, and 3 deep.

Wbd2S ÷ 1 = 2 × 32 × 9000 ÷ 120=

2×9×9000

120

=1350 lbs.

EXERCISES.

1. What weight will break a spar of New England fir, its breadth being 2 inches, depth 3 inches, and length 5 feet?

Ans. 330 lbs.

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