Rational

joint, line, springing, lb, ft, crown, factor and stability

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Stability against Crushing. Since the center of pressure is always within the middle third, there is no tension in any joint, and therefore the unit crushing stress may be found by equation 2, page 611. It is not always possible by inspection to tell which joint has the maximum unit crushing stress; and therefore it is usually necessary to compute the stress at several joints. In the case in hand, the maximum pressure will be found for three joints—the crown, joint 5, and the springing.

At the crown, d = *1, and hence P = 2 W _ l; or, since W = 9,369 lb. and 1 = 1.25 ft., P = 14,990 lb. per sq. ft. = 103 lb. per sq. in.

At joint 5, W = the component of

R1 normal to the joint = 13,900 lb., 1 = 2.42 ft., and d = 0.40 ft.; and therefore i.e., P = 11,460 lb. per sq. ft. = 80 lb. per sq. in.

Similarly, at the springing W = 21,090 lb.,. 1 = 4.5 ft., and d = 0.16 ft.; and therefore P = 5,680 lb. per sq. ft. = 39 lb. per sq. in.

Except for

a particular kind of stone and a definite quality of masonry, it is impossible even to discuss the probable factor of safety; but it is certain that in this case the nominal factor is ex cessive (see f 582-84), while the real factor is still more so (see § 1200-02).

If the maximum pressure at the most compressed joint had been more than the safe bearing power of the masonry, it would have been necessary to increase the depth of the arch stones and repeat the entire process. Notice that the total pressure on the joints increases from the crown toward the springing, and that hence the depth of the arch stones also should increase in the same direction.

Stability against Sliding. To determine the degree of stability against sliding, notice that the angle between the resultant pressure on any joint and the joint is least at the springing joint; and hence the stability of this joint against sliding is less than that for any other. The nominal factor of safety is equal to the coef ficient of friction divided by tan (90° – 76°) = tan 14° = 0.25. An examination of Table 74 (page 464) shows that when the mortar is still wet the coefficient is at least 0.50; and hence the nominal factor for the joint in question is at least 2, and probably more, while the real factor is still greater. The nominal factor for joint 7 is at least 4.6, and that for joint 3 is about 6. There is little or no prob ability that an arch will be found to be stable for rotation and crush ing, and unstable for sliding. If such a condition should occur, the direction of the assumed joint could be changed to give stability.*

The actual joints should be as nearly perpendicular to the line of resistance as is consistent with simplicity of workmanship and with stability. For circular arches, it is ordinarily sufficient to make all the joints radial. In Fig. 196, page 629, the joints are radial to the intrados; but if they had been made radial to the extrados or to an intermediate curve, the stability against sliding, particularly at the springing joint, would have been a little greater.

Special Solution.

The special feature of the following method is that it enables one to draw a line of resistance without previously having computed the crown thrust, and also enables one to find a line of resistance which will pass through two prede termined points; and one of the most useful applications of this method is in determining the line of resistance for a segmental arch having a central angle so small as to make it obvious that the joint of rupture is at the springing line.

For example, assume that it is required to draw the line of resistance for the circular arch shown in Fig. 197. The span is 50 feet, the rise 10 feet, the depth of voussoir 2.5 feet, and the height of the earth above the summit of the arch ring is 10 feet. The angular distance of the springing from the crown is 43° 45'; and since the angle of rupture is nearly always more than 45°, it is safe to assume that the joint of rupture is at the springing. The problem then is to find a line of resistance that will pass through U (the upper end of the middle third of the crown joint) and through b (the lower end of the middle third of the springing joint).

The first step is to compute the external forces similarly as in 4 1205-09, which see.

Next construct a load line, as shown in the force diagram, Fig. 197, by laying off w, and h,, and w, and h,, etc., in succession, and drawing F,, F„ etc. Since the crown thrust has not yet been deter mined, the position of the pole is not known, and hence a trial posi tion must be assumed. Since the load is symmetrical, we may assume that the thrust at the crown is horizontal; and hence we may choose a pole at any point, say P', horizontally opposite 0. Draw lines from P' to the extremities of F,, F,, etc. Construct a trial equilibrium polygon by drawing through U a line parallel to the line P' 0, of the force diagram, and prolong it to an intersection with F,; from this point draw a line parallel to R„ and prolong it to an intersection with F„ etc., continuing to, an intersection, b', with the springing line prolonged.

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