62. First Beam Formula. As shown in the preceding article, the resisting and bending moments for any section of a beam are equal; hence

SI/c = M, (6) all the symbols referring to the same section. This is the most important formula relating to beams, and will be called the " first beam formula."

The ratio I ÷ c is now quite generally called the section modulus. Observe that for a given beam it depends only on the dimensions of the cross-section, and not on the material or anything else. Since I is the product of four lengths (see article 51), I ÷ c is the product of three; and hence a section modulus can be expressed in units of volume. The cubic inch is practically always used; and in this connection it is written thus, inches3. See Table A, page 54, for values of the section moduli of a few simple sections.

63. Applications of the First Beam Formula. There are three principal applications of equation 6, which will now be explained and illustrated.

64. First Application. The dimensions of a beam and its manner of loading and support are given, and it is required to compute the greatest unit-tensile and compressive stresses in the beam.

This problem can be solved by means of equation 6, written in this form,

S = Mc/ I or M/ I + c (6)

Unless otherwise stated, we assume that the beams are uniform in cross-section, as they usually are; then the section modulus (I ÷ C) is the same for all sections, and S (the unit-fibre stress on the remotest fibre) varies just as M varies, and is therefore greatest where M is a maximum.* Hence, to compute the value of the greatest unit-fibre stress in a given case, substitute the values of the section modulus and the maximum bending moment in the preceding equation,and reduce.

If the neutral axis is equally distant from the highest and lowest fibres, then the greatest tensile and compressive unit-stresses are equal, and their value is S. If the neutral axis is unequally distant from the highest and lowest fibres, let c denote its distance from the nearer of the two, and S' the unit-fibre stress there. Then, since the unit-stresses in a cross-section are proportional to the distances from the neutral axis,

S'/s + c'/c, or S' = c'/c S.

If the remotest fibre is on the convex side of the beam, S is tensile and S' compressive; if the remotest fibre is on the concave side, S is compressive and S' tensile.

Examples. 1. A beam 10 feet long is supported at its ends, and sustains a load of 4,000 pounds two feet from the left end (Fig. 37, a). If the beam is 4 X 12 inches in cross-section (the long side vertical as usual), compute the maximum tensile and compressive unit-stresses.

The section modulus of a rectangle whose base and altitude are b and a respectively (see Table A, page 54), is 1/6 ba2; hence, for the beam under consideration, the modulus is

1/6 X 4 X 122 = 96 inches3,

To compute the maximum bending moment, we have, first, to find the dangerous section. This section is where the shear changes sign (see article 45); hence, we have to construct the shear diagram, or as much thereof as is needed to find where the change of sign occurs. Therefore we need the values of the reaction. Neglecting the weight of the beam, the moment equation with origin at C (Fig. 37, a) is

R1 X 10 - 4,000 X 8 = 0, or R1 = 3,200 pounds

* Note. Because S is greatest in the section where M is maximum, this section is usually called the " dangerous section" of the beam.

Then, constructing the shear diagram, we see (Fig. 37, b) that the change of sign of the shear (also the dangerous section) is at the load. The value of the bending moment there is

3,200 X 2 = 6,400 foot-pounds, or 6,400 X 12 = 76,800 inch-pounds.

Substituting in equation 6', we find that

S = 76,800/96 = 800 pounds per aquare inch

Fig. 37.

2, It is desired to take into account the weight of the beam in the preceding example, supposing the beam to be wooden. The volume of the beam is

[4 X 12/144] x 10 = 3 1/3 cubic feet; and supposing the timber to weigh 45 pounds per cubic foot, the beam weighs 150. pounds (insignificant compared to the load). The left reaction, therefore, is

3,200 + (¼ X 150) = 3,275; and the shear diagram looks like Fig. 37, c, the shear changing sign at the load as before. The weight of the beam to the left of the dangerous section is 30 pounds; hence the maximum bending moment equals

3,275 X 2 - 30 X 1 = 6,520 foot-pounds, or 6,520 X 12 = 78,240 inch-pounds.

Substituting in equation 6', we find that

S = 78,240/96 = 815 pounds per square inch.

The weight of the beam therefore increases the unit-stress produced by the load at the dangerous section by 15 pounds per square inch.

3. A T-bar (see Fig. 38) 8 feet long and supported at each end, bears a uniform load of 1,200 pounds. The moment of inertia of its cross-section with respect to the neutral axis being 2.42 inches*, compute the maximum tensile and compressive unit-stresses in the beam

Evidently the dangerous section is in the middle, and the value of the maximum bending moment (see Table B, page 55, Part I) is ⅛ Wl, W and I denoting the load and length respectively. Here

Fig. 38.

1/8 Wl = 1/8 X 1,200 X 8 = 1,200 foot-pounds, or 1,200 X 12 = 14,400 inch-pounds.

The section modulus equals 2.42 ÷ 2.28 = 1.06; hence

S = 14,400/1.06 = 13,585 pounds per square inch.

This is the unit-fibre stress on the lowest fibre at the middle section, and hence is tensile. On the highest fibre at the middle section the unit-stress is compressive, and equals (see page 62):

S' = c'/c S = 0.72/2.28 x 13,585 = 4,290 pounds per square inch.

DRAWING SHOWING FRACTURE OF VARIOUS WOODS UNDER BENDING STRESS.

Reproduced from "Strength and Properties of Materials " by W. G. Kirkaldy.

DRAWING SHOWING FRACTURE OF VARIOUS WOODS UNDER THRUSTING STRESS.

Reproduced from "Strength and Properties of Materials," by IF. G. Kirkaldy.