STRESSES DuE TO DEAD AND LIVE LOADS. In Fig. 234, let GJ represent a portion of the arch, ab the neutral line, and ce the side of the true equilibrium polygon to the left of the point a, ac the vertical intercept between the neutral line and the equilibrium polygon, and ae is a perpendicular from a upon ce. Then ce is the line of action of the resultant, R, of all the external forces to left of the section ea, i.e., R is the resultant of the reaction at the left abutment and of all the loads between the left abutment and the section ea. The amount and the direction of R is given by the cor responding ray of the force diagram. Assume that two opposite forces, and each equal to R and parallel to that force, be applied at a. These forces will not disturb the equilibrium, and the single force R acting at c is then replaced by the couple R R, and a force R, act ing at a. The force R, may be decomposed into two compo nents—T tangent to the neutral line ab, and N normal to the neutral line. The couple R produces bending, the force T causes a shortening of the arch ring, and the force N produces shear in a normal section through a. The bending, the shortening, and the shear of the elastic arch are somewhat analogous to the tendency in the voussoir arch to overturn, to crush, and to slide.
To find the stresses in the arch ring, let ac = the intercept between the neutral line and the true equili brium polygon; b = the breadth of the unit section of the arch, i.e., b = 1 ft.; c = the distance of the most remote fiber from the neutral line; d = the depth of the arch ring; / = the unit fiber stress; H = the true pole N = the component parallel to the radius at any point of the neutral line of all the forces to one side of the point; T = the component parallel to the tangent at any point of the neutral line of all the forces to one side of the point; v = the unit shearing stress.
In Fig. 234, the moment of the couple is R. ae; but if H is the horizontal component of R, i.e., is the true pole distance, then by similar triangles R. ae = H. ac. The value of H was computed in
§ 1325, and as can be measured in Fig. 233, facing page 688; and there fore the bending moment at any point of the arch may be found. The maximum unit fiber stress in the section ae due to bending is lb is compression on the side next to R and tension on the side opposite.
The value of T can be found by resolving the ray in the force diagram that is parallel to the side of the equilibrium polygon ad jacent to the point a, into a component parallel to the tangent at a. The unit compressive stress due to the force tending to shorted the rib is: The force N is the component parallel to the radius at a; and the unit shearing stress, v, is: The total maximum fiber stress due to combined bending and shortening is: The first term of the right-hand side of equation 21 is always com pression. For the extrados, the last term is plus, i.e., compression, when the equilibrium polygon lies outside of the neutral line of the arch ring, and minus, i.e., tension, when inside; and for the intrados the stress is the reverse of that in the extrados. Ultimately the stress given by equation 21 must be combined with that due to a change of temperature of the arch ring.
In using equation 21, the intercepts ac should be measured' in the verticals through a„ a,, a,, etc., the points at which ds ± I is con stant, in accordance with the equations of condition, which also secures the greatest accuracy (see § 1316). The values of ac can be determined by taking the difference between ak and ck, which is more accurate than measuring the small quantity ac directly.
Equation 21 above is analogous to equation 2 (page 611) of the voussoir arch. -The former is limited to the tensile strength of the concrete, while the latter is limited to the condition that the line of pressure must remain within the middle third of the depth.