This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

D = 0.0000065 X 10 X l = 0.000065 l.

Now, since the rod could not shorten, it has a greater than normal length at the new temperature; that is, the fall in temperature has produced an effect equivalent to an elongation in the rod amounting to D, and hence a tensile stress. This tensile stress can be computed from the elongation D by means of formula 17. Thus,

S = E s; and since s, the unit-elongation, equals d/l = 0000165 l / l = 0000065.

S = 30,000,000 X .0000065 = 195.0 pounds per square inch. This is the value of the temperature stress; and the new unit-stress equals

10,000 + 195.0 = 10,195 pounds per square inch.

Notice that the unit temperature stresses are independent of the length of the rod and the area of its cross-section.

2. Suppose that the change of temperature in the preceding example is a rise instead of a fall. What are the values of the temperature stress due to the change, and of the new unit-stress in the rod?

The temperature stress is the same as in example 1, that is, 1,950 pounds per square inch; but the rise in temperature releases, as it were, the stress in the rod due to its being screwed up, and the final unit stress is

10,000 - 1,950 = 8,050 pounds per square inch.

1. The ends of a wrought-iron rod 1 inch in diameter are fastened to two heavy bodies which are to be drawn together, the temperature of the rod being 200 degrees when fastened to the objects. A fall of 120 degrees is observed not to move them. What is the temperature stress, and what is the pull exerted by the rod on each object?

Ans. | Temperature stress, 22,000 pounds per square inch. |

Pull, 17,280 pounds. |

97. Deflection of Beams. Sometimes it is desirable to know how much a given beam will deflect under a given load, or to design a beam which will not deflect more than a certain amount under a given load. In Table B, page 55, Part I, are given formulas for deflection in certain cases of beams and different kinds of loading.

In those formulas, d denotes deflection; I the moment of inertia of the cross-section of the beam with respect to the neutral axis, as in equation 0; and E the coefficient of elasticity of the material of the beam (for values, sec Art. 95).

In each case, the load should be expressed in pounds, the length in inches, and the moment of inertia in biquadratic inches; then the deflection will be in inches.

According to the formulas for d, the deflection of a beam varies inversely as the coefficient of its material (E) and the moment of inertia of its cross-section (I); also, in the first four and last two cases of the table, the deflection varies directly as the cube of the length (l3).

Example. What deflection is caused by a uniform load of 0,400 pounds (including weight of the beam) in a wooden beam on end supports, which is 12 feet long and (5 X 12 inches in cross-section ? (This is the safe load for the beam ; see example 1, Art. 05.)

The formula for this case (see Table B, page 55) is d = 5 Wl3 /384 EI .

Here W = 6,400 pounds ; l = 144 inches ; E = 1,800,000 pounds per square inch; and

I = 1/12 ba3 = 1/12 6 X 123= 864 inches4.

Hence the deflection is d = [ 5 X 6,400 X 1443 ] /[ 384 X 1,800,000 X 864] = 0.16 inch.

1. Compute the deflection of a timber built-in cantilever 8X8 inches which projects 8 feet from the wall and bears an end load of 900 pounds. (This is the safe load for the cantilever, see example 1, Art. 65.)

Ans. 0.43 inch.

2. Compute the deflection caused by a uniform load of 40,000 pounds on a 42-pound 15-inch steel I-beam which is 10 feet long and rests on end supports.

Ans. 0.28 inch.

98. Twist of Shafts. Let Fig. 57 represent a portion of a shaft, and suppose that the part represented lies wholly between two adjacent pulleys on a shaft to which twisting forces are applied (see Fig. 54). Imagine two radii ma and nb in the ends of the portion, they being parallel as shown when the shaft is not twisted. After the shaft is twisted they will not be parallel, ma having moved to ma', and nb to nb'. The angle between the two lines in their twisted positions (ma' and nb') is called the angle of twist, or angle of torsion, for the length /. If a'a" is parallel to ab, then the angle a"nb' equals the angle of torsion.

Fig. 57.

If the stresses in the portion of the shaft considered do not exceed the elastic limit, and if the twisting moment is the same for all sections of the portion, then the angle of torsion a (in degrees) can be computed from the following:

For solid circular shafts, a =584 Tl /E1 d4 = 36,800,000 Hl / E1 d4n. For hollow circular shafts, (19) a = 584 Tld /E1 d4-d14) =30,800,000 Hl /E1 (d4-d14) n

Here T, l, d, d1 H, and n have the same meanings as in Arts. 93 and 94, and should be expressed in the units there used. The letter E1 stands for a quantity called coefficient of elasticity for shear; it is analogous to the coefficient of elasticity for tension and compression (E), Art. 95. The values of E1 for a few materials average about as follows (roughly E1 = 2/5 E):

For Steel, | 11,000,000 pounds per square inch. |

For Wrought iron, | 10,000,000 " " " " |

For Cast iron, | 6,000,000 " " " " |

Example. What is the value of the angle of torsion of a steel shaft 60 feet long when transmitting 6,000 horse-power at 50 revolutions per minute, if the shaft is hollow and its outer and inner diameters are 16 and 8 inches respectively?

Here I = 720 inches; hence, substituting in the appropriate formula (19), we find that a= [30,800,000 X 6,000 X 720] /[11,000,000 X (164 - 84) 50] = 47 degrees.

Suppose that the first two pulleys in Fig. 5-1 are 12 feet apart; that the diameter of the shaft is 2 inches; and that P1 = 400 pounds, and a1 = 15 inches. If the shaft is of wrought iron, what is the value of the angle of torsion for the portion between the first two pulleys?

Ans. 3.15 degrees.

99. Non-elastic Deformation. The preceding formulas for elongation, deflection, and twist hold only so long as the greatest unit-stress does not exceed the elastic limit. There is no theory, and no formula, for non-elastic deformations, those corresponding to stresses which exceed the elastic limit. It is well known, however, that non-elastic deformations are not proportional to the forces producing them, but increase much faster than the loads. The value of the ultimate elongation of a rod or bar (that is, the amount of elongation at rupture), is quite well known for many materials. This elongation, for eight-inch specimens of various materials (see Art. 16), is :

For Cast iron, about | 1 per cent. | |

For Wrought iron (plates), | 12 - 15 per cent. | |

For " " (bars), | 20-25 " " . | |

For Structural steel, | 22-26 " " . |

Specimens of ductile materials (such as wrought iron and structural steel), when pulled to destruction, neck down, that is, diminish very considerably in cross-section at some place along the length of the specimen. The decrease in cross-sectional area is known as reduction of area, and its value for wrought iron and steel may be as much as 50 per cent.

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