Near the crown of the arch, the thrust is not so great, and will not wholly neutralize the tension due to moment. In order to with stand the various stresses, the rib must have a larger cross-section than would be required for moment alone. This means that the equation developed in Article 271, Pait III, 02 14 2 , can be utilized only in a roundabout way. For example, the moment at the abut ment was computed above as 149,490 inch-pounds. But a section 12 inches wide and 19.125 inches deep to the reinforcement, should with stand a moment of 272,300 inch-pounds with 0.43 per cent of steel. The above moment is only 55 per cent of this. Therefore 55 per cent of the steel, or .55 X .0043bd = .00236bd of steel, could safely carry the tension. But the actual ratio of steel adopted was .00245. As shown above, the steel at the abutment is unnecessary for transverse moment and for this condition of loading. Nearer the crown, the moment is less; but the relief to tension afforded by thrust is very much less, and the steel has much more to do. With other conditions of loading, it will also be different; and therefore the same amount of steel 4-inch bars spaced 12 inches, in both top and bottom, is used throughout.
43S. Maximum Thrust Due to this Loading. The thrust at any point of the arch is measured by the projection onto the tangent to the arch at that point, of the corresponding ray of the force diagram. Since the rays of the force diagram which are parallel to the segments of the equilibrium polygon are approximately tangent to the arch rib, it is approximately true to measure the thrust at any point by measuring the corresponding rays of the force diagram; but a more precise value may be found by drawing a line from one end of a ray parallel to the corresponding tangent, and projecting the ray on it. This method is particularly useful, since it measures at the same time the amount of the shear at that point, as will be explained in the next article. Since the increase in the thickness of the arch is compara tively slight from the crown to the abutment, in this particular problem, and since the amount of the thrust evidently increases very rapidly toward the abutment, the critical point, so far as the thrust is con cerned, is evidently at the abutment. Therefore we may draw from the lower end of load 20, a line parallel to the tangent to the curve at the abutment, and obtain the line wx, which scales 40,200 pounds, and which therefore measures the thrust at the abutment. Since the abutment is 21.25 inches thick, the area of a section one foot wide is 255 square inches. Dividing this into 40,200, we have a quotient of 15S as the unit-compression per square inch due to thrust. Adding the value of the compression at the intrados which is clue to moment, (151), to the compression just found for thrust, we have a total of 309 pounds per square inch at the abutment. This compression
does not allow for temperature stresses, which will be computed later, and which may exist simultaneously with the stresses due to moment and to thrust. • 439. Shear at Any Section. The shear at any section is meas ured by the projection onto the normal to the arch rib, of the corre sponding ray of the force diagram. It is seldom that the shear is a serious factor in the design of an arch. Whenever (as in the case just being worked out) the equilibrium polygon coincides approxi mately with the arch rib, the shear is very small. When the amount of the thrust is definitely computed, as determined above, the amount of the shear at the same point may be readily determined at the same time.
For example, the shear at the abutment is the projection of the ray onto the line which is parallel to the normal to the curve at B. This line, scaled at the rate of 5,000 pounds per inch, indi cates a shear of about 2,200 pounds. Dividing this by 225, the area of the section of the arch at that point, the unit-shear is less than 10 pounds per square inch, which of course may he neglected. The shear in any arch may be very easily tested by noting the portion of the special equilibrium polygon which makes the largest angle with the direction of the arch rib at any point; the larger the angle, the greater the shear. If the arch is tested at that point, and the shear is found to be insignificant, or well within the power even of plain con crete to carry, there is no need for further investigation.
On the other hand, there are minor stresses which occur in arches as well as other concrete structures, which must be provided for. These are caused by a possible excessive concentration of loading; possible structural weakness due to a poor quality of concrete in com paratively limited areas; the effect of slight settlement of the founda tions, etc. On account of these various stresses, which are more or less non-computable, it is the usual practice to insert bars, which are not only useful in taking up shear, but also tend to bind the whole structure together, make it act more nearly as a unit, and permit the structural weakness in local spots to be made up by the strength of the sounder portions of concrete around it; and therefore shear bars are put in, such as arc illustrated in Fig. 2-35. These bars are laced Letween the upper and lower sets of bars that run parallel with the axis of the arch. By this means, not only are the upper and lower sets of reinforcing bars tied together, but the bars have such a di rection that they can take up any shearing force which may, by any chance, be developed.