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THE ELASTIC ARCH 163. Analysis of Fixed Arch.—Keinforced concrete arches are commonly constructed as solid curved beams firmly fixed to the abutments. lnan alyzing them, it is assumed that the abutments are im movable and the ends of the arch firmly held in their original positions.

Let Fig. SS repre sent the left half of an arch, fixed in position at the end C–H, and carrying loads which produce thrusts and bending moments throughout the arch ring.

The arch may be considered as made up of a number of small divisions. Suppose ABCD to be one of these divisions, small enough so that its ends are practically parallel and its section area constant. The loads upon the arch bring a bending moment upon the division ABCD, which causes the end CD to take the posi tion EF.

Let 31 = bending moment on the division; s= length of division, AD = BC; e= distance from center of section to outside fiber; ds = elongation of fiber distant e from neutral axis; Unit stress upon fiber distant e from neutral axis; k = angle of distortion, COE; E = modulus of elasticity of material; of inertia of section; x and y=horizontal and vertical coordinates of center of section, 0, with reference to center of crown sec tion, .I.

If the crown of the arch be free to move, the deflection of ABCD into its new form ABET will bring the middle point of the arch ring J, to the new position K. Let dx and dy be the horizontal and ver tical coordinates of K with respect to J. Then from similar tri angles, JK,'OJ=dy'x=k, and As the end section GH is fixed in position, the sununation of all the angular distortions k, for the left half of the arch gives the dis tortion at the crown section. The sununation similarly of those for the right half must give the same result with opposite sign, or indicating the left and right sides of the arch by the subscripts L and R respectively, and indicating sununation by the sign 1, we have Substituting for these distortions, their values as found above, we have for a symmetrical arch: If the length of the divisions of the arch ring be made directly proportional to the corresponding values of the moment of inertia, /=constant, the terms in Equations (4), (5), and (6) are stant and may be eliminated, and we have, Fig. S9 represents a symmetrical arch divided into parts the

lengths of which are directly proportioned to the moments of inertia of the cross-sections at their middle points. s I=constant. Sup pose the arch to be cut at the crown and the separate halves supported by introducing the stresses acting through the crown section as exterior forces. These may be resolved into a horizontal thrust, a vertical shear and a bending moment.

thrust at crown; V,=vertical shear at crown; = bending moment at crown.

F, is considered to be positive when acting in the direction indicated by the arrows. Moments are taken as positive when they produce compression on the upper and tension on the lower side of the section.

Let moment on mid-section of any division in left half of arch; moment on mid-section of any division in right half of arch; at middle of any division in left half, caused. by external loads between that division and the crown section; ma=moment at mid-section of any division in right half, caused by external loads between that section and the crown section; x and y= coordinates of middle point of any division with respect to middle of crown section.

The bending moment at any section of a beam is equal to the moment at any other section plus the moments of the intermediate loads about the center of the section. Therefore, In analyzing an arch by this method, the arch is first divided into a number of parts in which s11 is a constant. The loads upon the divisions are then found and mL and Inn, computed for the several centers of division. The values of and 3Ie may then be found from Formulas (15), (16) and (17), after which the line of thrust may be drawn beginning with the known values of He and Vc at the crown. The moment .3J is due to the eccentricity of the thrust at the crown, and the point of application for may be found by divid ing by This gives the vertical distance of from the center of gravity of the crown section. For M, positive, is above, and for M, negative, is below the center of section.

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