Hence I/C = 105,600/16,000 = 6.6 inches3.

That is, an I-beam is needed whose section modulus is a little larger than 6.6, to provide strength for its own weight.

To select a size, we need a descriptive table of I-beams, such as is published in handbooks on structural steel.

Below is an abridged copy of such a table. (The last two columns contain information for use later.) The figure illustrates a cross-section of an I-beam, and shows the axes referred to in the table.

It will be noticed that two sizes are given for each depth; these are the lightest and heaviest of each size that are made, but intermediate sizes can be secured. In column 5 we find 7.3 as the next larger section modulus than the one required (6 6); and this corresponds to a 12¼-pound 6-inch I-beam, which is probably the proper size. To ascertain whether the excess (7.3-6.6 = 0.70) in the section modulus is sufficient to provide for the weight of the beam, we might proceed as in example 1. In this case, however, the excess is quite large, and the beam selected is doubtless safe.

## Table C. Properties Of Standard I-Beams Section of beam, showing axes 1-1 and 2-2.

 1 2 3 4 5 6 Depth of Beam, in inches. Weight per foot, in pounds. Area of cross-section, in square inches. axis 1- 1. Section modulus, axis 1 - 1. Moment of inertia, axis 2 - 2. 3 5.50 1.63 2.5 1.7 0.46 3 7.50 2.21 2.9 1.9 .60 4 7.50 2.21 6.0 3.0 .77 4 10.50 3.09 7.1 3.6 1.01 5 9.75 2.87 12.1 4.8 1.23 5 14.75 4.34 15.1 6.1 1.70 6 12.25 3.61 21.8 7.3 1.85 6 17.25 5.07 26.2 8.7 2.36 7 15.00 4.42 36.2 10.4 2,67 7 20.00 5.88 42.2 .12.1 3.24 8 18.00 5.33 56.9 14.2 3.78 8 25.25 7.43 68.0 17.0 4.71 9 21.00 6.31 84.9 18.9 5.16 9 35.00 10.29 111.8 24.8 7.31 10 25.00 7.37 122.1 24.4 6.89 10 40.00 11.76 158.7 31.7 9.50 12 31.50 9.26 215.8 36.0 9.50 12 40.00 11.76 245.9 41.0 10.95 15 42.00 12.48 441.8 58.9 14.62 15 60.00 17.65 538.6 71.8 18.17 18 55.00 15.93 795.6 88.4 21.19 18 70.00 20.59 921.2 102.4 24.62 20 G5.00 19.08 1,169.5 117.0 27.86 20 75.00 22.06 1,268.8 126.9 30.25 24 80.00 23.32 2,087.2 173.9 42.86 24 100.00 29.41 2,379.6 198.3 48.55

## Examples For Practice

1. Determine the size of a wooden beam which can safely sustain a middle load of 2,000 pounds, if the beam rests on end supports 16 feet apart, and its working strength is 1,000 pounds per square inch. Assume width 6 inches.

Ans. 0 X 10 inches.

2. What sized steel I-beam is needed to sustain safely a uniform load of 200,000 pounds, if it rests on end supports 10 feet apart, and its working strength is 10,000 pounds per square inch?

Ans. 95-pound 21-inch.

3. What sized steel I-beam is needed to sustain safely the loading of Fig. 10, if its working strength is 16,000 pounds per square inch ?

Ans. 14.75-pound 5-inch.

67. Laws of Strength of Beams. The strength of a beam is measured by the bending moment that it can safely withstand; or, since bending and resisting moments are equal, by its safe resisting moment (SI ÷ c ). Hence the safe strength of a beam varies (1) directly as the working fibre strength of its material, and (2) directly as the section modulus of its cross-section. For beams rectangular in cross-section (as wooden beams), the section modulus is 1/6 ba2, b and a denoting the breadth and altitude of the rectangle. Hence the strength of such beams varies also directly as the breadth, and as the square of the depth. Thus, doubling the breadth of the section for a rectangular beam doubles the strength, but doubling the depth quadruples the strength.

The safe load that a beam can sustain varies directly as its resisting moment, and depends on the way in which the load is distributed and how the beam is supported. Thus, in the first four and last two cases of the table on page 55,

 M = Pl, hence P = SI ÷ lc, M = ½ Wl, " w = 2SI ÷ lc, M = ¼ pl " p = 4SI ÷ lc, M = 1/8 wl, " w = 8SI ÷ lc, M = 1/8 pl, " p - 8SI ÷ lc, M = 1/12 wl " w = 12SI ÷ lc,

Therefore the safe load in all cases varies inversely with the length; and for the different cases the safe loads are as 1, 2, 4, 8, 8, and 12 respectively.

Example. What is the ratio of the strengths of a plank 2 X 10 inches when placed edgewise and when placed flatwise on its supports ?

When placed edgewise, the section modulus of the plank is 1/6 X 2 X 102 = 331/3, and when placed flatwise it is 1/6 X 10 X 22 = 62/3; hence its strengths in the two positions are as 331/3 to 62/3 respectively, or as 5 to 1.

## Example For Practice

What is the ratio of the safe loads for two beams of wood, one being 10 feet long, 3 X 12 inches in section, and having its load in the middle; and the other 8 feet long and 2x8 inches in section, with its load uniformly distributed.

Ans. As 135 to 100. 68. Modulus of Rupture. If a beam is loaded to destruction, and the value of the bending moment for the rupture stage is computed and substituted for M in the formula SI ÷ c = M, then the value of S computed from the equation is the modulus of rupture for the material of the beam. Many experiments have been performed to ascertain the moduli of rupture for different materials and for different grades of the same material. The fol-owing are fair values, all in pounds per square inch:

## Table D. Moduli Of Rupture

 Timber: Spruce ... 4.000 - 7,000, average 5,000 Hemlock ... 3,500 7,000, " 4,500 White pine ... 5,500 10,500, " 8,000 Long-leaf pine. 10,000 16,000, " 12,500 Short-leaf pine. . 8,000 14,000, " 10,000 Douglas spruce. 4,000 12,000, " 8,000 White oak.. 7,500 18,500, " 13,000 Red oak... 9,000 15,000, " 11,500 Sandstone ... 400 - 1,200, " Limestone .... 400 1,000. Granite... 800 1,400. One and one-half to two and one-quarter times its ultimate tensile strength. Hard steel: Varies from 100,000 to 150,000

Wrought iron and structural steels have no modulus of rupture, as specimens of those materials will "bend double," but not break. The modulus of rupture of a material is used principally as a basis for determining its working strength. The factor of safety of a loaded beam is computed by dividing the modulus of rupture of its material by the greatest unit-fibre stress in the beam.

69. The Resisting Shear. The shearing stress on a cross-section of a loaded beam is not a uniform stress; that is, it is not uniformly distributed over the section. In fact the intensity or unit-stress is actually zero on the highest and lowest fibres of a cross-section, and is greatest, in such beams as are used in practice, on fibres at the neutral axis. In the following article we explain how to find the maximum value in two cases-cases which are practically important.

70. Second Beam Formula. Let Ss denote the average value of the unit-shearing stress on a cross-section of a loaded beam, and A the area of the cross-section. Then the value of the whole shearing stress on the section is:

Resisting shear = Ss A.

Since the resisting shear and the external shear at any section of a beam are equal (see Art. 59),

SSA = Y. (7)

This is called the "second beam formula" It is used to investigate and to design for shear in beams.

In beams uniform in cross-section, A is constant, and Ss is greatest in the section for which V is greatest. Hence the greatest unit-shearing stress in a loaded beam is at the neutral axis of the section at which the external shear is a maximum. There is a formula for computing this maximum value in any case, but it is not simple, and we give a simpler method for computing the value in the two practically important cases: