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The other formulas are the same as in the last problem, if 212 is used instead of 850; and for wrought iron, use × 950, or 238 instead of 212.

121. When the weight is uniformly distributed over the beam, take of that found by the preceding problem. Or in the formulas of last problem use 425 for 850.

EXERCISES.

1. A beam is 30 feet long, 8 inches deep, and 21 broad; what weight can it support? Ans. 1132 lbs. 2. What load uniformly distributed over a beam 16 feet long, 4 inches deep, and 2 broad, can it sustain?

Ans. 425 lbs. 3. A beam 20 feet long and 10 inches deep supports a load of 17000 lbs. ; what is its breadth? Ans. 16 inches. 4. A beam 24 feet long and 2 inches broad supports 1735 lbs. uniformly distributed; required its depth.

Ans. 7 inches.

122. PROBLEM X.-To find the weight which a solid cylinder of cast-iron, supported at both ends, can sustain at the middle.

'Multiply the cube of the diameter in inches by 500, and divide the product by the length in feet, then the quotient will be the weight.'

W = 500 d3 ÷ 7, where d — the diameter.

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123. When the weight is uniformly diffused, it will be double that given by the preceding rule; or,

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Since 8819= nearly, the weight sustained by a cylinder is nearly equal to of that sustained by a square beam of the same length, and whose depth is equal to the diameter of the cylinder.

EXERCISES.

1. What weight will a cylinder 10 feet long and 4 inches diameter support? Ans. 3200 lbs. 2. What weight will a uniformly loaded cylinder support, its length being 24 feet, and diameter 10 inches?

Ans. 41659. 3. What will be the diameter of a cylinder 20 feet long, which is capable of supporting 3134 lbs.? Ans. 5 inches. 4. What will be the limit of the length of a cylinder uniformly loaded by a weight of 100000 lbs., whose diameter is 12 inches? Ans. 17.28 feet.

124. PROBLEM XI.-To find the weight that a cylinder, fixed at one end, can sustain at the free end.

'Multiply the cube of the diameter in inches by 125, and divide the product by the length in feet, and the quotient will be the weight.'

W = 125 ď3÷l, or W = × 500 ď3 ÷ l. The weight is just the fourth of that found in article 122, and the formulas the same as in that article, if 125 = 500 is taken for 500.

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1. What weight will a cylinder 10 feet long and 4 inches diameter support at its free end?

Ans. 800 lbs. 2. What will be the diameter of a cylinder 20 feet long that can support 784 lbs. ? Ans. 5 inches. 3. What will be the length of a cylinder, which is 12 inches diameter, that supports 12500 lbs.? Ans. 17.28 feet.

125. PROBLEM XII.-To find the exterior diameter of a hollow cylinder of cast-iron, supported at both ends, to sustain a weight applied at the middle, the ratio of the interior and exterior diameters being given.

Let the ratio of the exterior to the interior diameter be that of 1 to n; then take the difference between 1 and the

fourth power of n, and multiply it by 500; find also the product of the length and the weight; divide the latter product by the former; then the quotient will be the cube of the diameter.".

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When the exterior diameter d is found, the interior diameter will be obtained by multiplying d by n. If d' the interior diameter, and t = the thickness, of the metal, then d'nd, and t }(d—d') = 4 (1-n) d.

EXAMPLE. The weight supported by a hollow cylinder is 32000 lbs., its length is 12 feet, and the ratio of the exterior and interior diameters is 10 to 1; what are its diameters?

d3 TW ÷ 500 (1 — n1) =

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768 768000

•999 999

Hence,

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= 868.9, and d = 3/868·9 = 9.5 inches.

d'nd 1 x 9.595 inch.

and t(d-d') = (9·5 — ·95) = × 8·554·27; ort = (1—n) d = 1 (ì —·1) × 9.5×·9×954-27

EXERCISES.

1. A hollow cylinder 10 feet long supports 2500 lbs., the ratio of its diameters is 2 to 1; what are the diameters?

Ans. 1.89 and 3·78, and thickness of metal 99 inch. 126. If the thickness of the metal is to be of the exterior diameter, then

ď3 — ·132 3/ lW.

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For the equation = (1-n)d becomes}d=1(1-n)d; hence 2-5-5n, and n ==·6, and 1 — n1 — 1 — 1296 =8704, 500 (1—n1) = 500 × 8704 435, and =.132.

2. A hollow cylinder 9 feet long is intended to support 15000 lbs., and the thickness of the metal is to be of the exterior diameter; required its diameters.

Ans. 6.6 and 3.96 inches.

127. PROBLEM XIII.-To find the deflexion of a rectangular beam of cast-iron, supported at the ends, and fully loaded in the middle.

'Divide the square of the length in feet by 50 times the depth in inches, and the quotient is the deflexion in inches.' Let D the deflexion,

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D= 1250 Dd, and d

50d'

12

50 D'

The cohesive strength of the beam is considered to be equivalent to 15300 lbs. on the square inch.

EXERCISES.

1. Find the deflexion of a beam 18 feet long and 12 inches deep. Ans. 54 inch. 2. What is the deflexion of a beam 30 feet long and 15 inches deep? Ans. 1.2 inches.

128. PROBLEM XIV. To find the deflexion of a rectangular beam of cast-iron, supported at both ends, when fully loaded, and the weight uniformly distributed over it.

'Divide the square of the length in feet by 40 times the depth in inches, and the quotient is the deflexion in inches."

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inches deep?

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Ans. 72 inch.

2. Find the deflexion of a beam 15 feet long and 51 inches deep. Ans. 1.07 inches.

129. PROBLEM XV.-To find the resistance to torsion of a square shaft of cast-iron.

'Find the continued product of 92-5, the number of degrees in the angle of flexure, and the fourth power of the side in inches; divide this product by the product of the length and the leverage of the power both in feet, and the quotient will be the resistance in pounds.'

Let a and

the number of degrees in the angle of flexure, s = the side of the shaft in inches,

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the length in feet,

L= the leverage of the power W in feet,
W the resistance in pounds;

then W92·5 as1 ÷ IL.

Hence,

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1. The side of a square shaft is 3 inches and its length 12 feet, and it is driven by a power acting with a leverage of 2 feet; what power may be applied to it, so as not to cause a flexure of more than 1° 6'? Ans. 343-4 lbs.

2. What must be the side of a square shaft of cast-iron 10 feet long to resist a power of 1500 lbs. acting with a leverage of 3 feet, with an angle of flexure of 14 degrees? Ans. 4.244.

130. PROBLEM XVI.-To find the resistance to torsion of a solid cylinder of cast-iron.

'Find the continued product of 55, the number of degrees in the angle of flexure, and the fourth power of the diameter in inches; divide this product by the product of the length and the leverage both in feet, and the quotient will be the resistance.'

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tain nearly of the torsion that a square beam is capable of, whose side is equal to the diameter of the cylinder.

EXERCISES.

1. A shaft is 35 feet long and 10 inches diameter; what power can it sustain, acting at a leverage of 2 feet with an angle of flexure of 1°? Ans. 7857 lbs.

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