# Strength of

## beam, section, pounds, hence, modulus, moment and equation

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STRENGTH OF BEAMS—(CONCLUDED).

62. First Beam Formula. As shown in the preceding article, the resisting and bending moment's for any section of a beam are equal; hence 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 foragiven beam it depends only on the dimensions of the cross-section, and not on the material or any thing 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, inches'. 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 ex plained 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 call be solved by means of equation 6, written in this Rim, S= or (6') Unless otherwise stated, we assume that the beams are uniform in cross-section, as they usually are; then the section modulus (I=cy is the same for all sections, and S (the unit-fibre stress on 71 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 sectiun, modulus and the mazimuni hauling moment in the preceding equation, and reduce.

If the neutral axis is equally distant from the highest and low est 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 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, c' c' or S' = 7 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 hence, for the beam under consideration, the modulus is 1 x 4 X 12' = 96 inches'.

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 dia 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 R, X 10- 4,000 X 8 = 0, or It, = 3,200 pounds 72 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 0 = 800 pounds per squar© inch.

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 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 and the sheer diagram looks like Fig. 37, e, 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.

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