**BRIDGES**

**291**

abutment N. Let the load on the beam, including the weight of the beam, be divided into any number of equal or unequal parts. Let those loads which lie to the left of the section be called io v u 2 , w 3 , &c., and let the distances of their centres of gravity fro a the section be called l v 1 2> l a . Let those loads which are to the right of the. section be called ! ^l. 2 , w 3 , and let their distances from the section be called r v T 2> r s & c - J je k the bending moment at the section be

called M, then by the above definitionAn image should appear at this position in the text.A high-res raw scan of the page is available. To use it as-is, as a placeholder, edit this page and replace "{{missing image}}" with "{{raw image|Encyclopædia Britannica, Ninth Edition, v. 4.djvu/335}}". If it needs to be edited first ( e.g. cropped or rotated), you can do so by clicking on the image and following the guidance provided.
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Fig. 13.

This bending moment is resisted at the section by the elastic forces in the beam called into play by compression at the top and extension at the bottom, or, in other words, called into action by the stresses on the material. Limiting our consideration to those cross sections which have a vertical axis of symmetry, and to those materials which have a constant modulus of elasticity for tension and compression, it is easy to find a general expression for the moment of the elastic forces, which will hereafter be designated by the letter /x. Let the horizontal axis ZZ : (fig. 14) be a line along which the material is unstrained,—

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Fig. 14.

all the portions above this line being compressed, and all the portions below this line being extended. Conceive the sec tion divided into little elementary surfaces, the area of each being equal to the product of the little elementary length Ay into the little elementary length A*, the force exerted by the elasticity of this element will be proportional to its area AyAs and to its distance y from the axis 2Z 15 for the longitudinal extension or compression is directly propor tional to this distance ; then calling a, as before, the rate at which the stress increases with the increase of y, or in other words the intensity of the stress at the unit distance from ZZj, we shall have for the force exerted by each element the expression ayAyA?. Now, as the section is not moved as a whole along the beam in either direction, we must have 2yAyA,r ? 0, that is to say, the sum of the positive must be equal to the sum of the negative forces. Let the constant quantity AyA# be conceived as the weight w of a thin plate of the area AyA#, the moment of this weight relatively to the axis will be wy, but as we have the sum of the moments ^ivy ? ^ayAyAa; ? 0, the axis ZZ t round which these moments are taken must pass through the centre of gravity of the section. This axis is called the neutral axis of the beam (vide 18). The expression for the force exerted by each element being ay&.yb.x, the moment of this force, or, in other words, for the moment due to the stress, is given by the expression ay 2 AyA2 ; then for the moment p. of all the forces round ZZ we have—

where I is the moment of inertia of the surface of the cross section. This expression /M is Rankine s moment of resist ance to flexure ; it may also be called the moment of the elastic forces, or, as suggested by Professor Kennedy, the moment of stress.

The greatest intensity of stress p l occurs at the greatest distance from ZZ 1? or at the distance ^d if the depth of the beam be called d, and the axis ZZ X be equidistant from top and bottom. In this case we have—

ad Zp l Pl ~ -2, 01 d hence we may write for sections with two axes of symmetry, and for materials having equal strengths and constant moduli of elasticity under tension and compression—

3. , = 2 d Now, the general condition of equilibrium between the external forces and the elastic forces at the section is simply M = /A, hence when the beam does not break we have also—

Whenever the amount and distribution of load on a given girder is known the bending moment M is to be calculated from equation 1, and then equation 4 allows us to calculate >! ; or if we know the value of /, the ultimate strength of the material to resist tension and compression, this equa tion enables us to find what moment M : is required to break the beam at that section, and therefore what load distri buted in a given way the beam can bear ; thus we have—

2/1 In a beam of uniform strength the value of ~- will at GP every section be equal to the value of M I} the moment due to external forces at that section.

and compression, let f t and f c be the two strengths (per unit of cross section), but let the modulus of elasticity be assumed constant as before. Then, as above, the unstrained axis will be the axis passing through the centre of gravity of the section, and the intensity of stress at any distance y abo^e or below that axis will be ay ; let y e be the distance of the uppermost element of the cross section from the horizontal axis, and let y, be the distance of the lowermost element. .Then the greatest stress will occur at the top if ?/ c be greater than y t , but at the bottom if y t be greater than y c ; since, however, the material is not equally strong to resist tension and compression, it does not follow that it will give way where the stress is greatest, and the beams

will yield first at that edge where the ratio ? is greatest.