general law has been established for the position of the live load for a maxi mum stress at any particular point. According to equation 21 the stress at any point of the arch ring varies with (1) the tangential thrust, T; (2) the true pole distance, H; (3) the intercept ac; and (4) the depth of the arch ring. Each of these quantities varies according to a different law; and it will he very difficult, if not impossible, to state a general law for the position of the live load which will give a maximum stress at any point. Practice varies considerably as to the positions of the live load employed in testing the stability of an arch. The less exacting engineers test the arch for only two posi tions of the live load, viz.: over half the span and over the whole span; while others test for four positions, viz.: over one quarter of the span, one half, three quarters, and the whole span; and still others test for a live load over two fifths, one half, three fifths, and the whole span. Professor Wm. Cain suggests* that probably the fol lowing positions are the best: live load over three tenths of the span, half of the span, six tenths of the span, and the whole span.
To fully check the design of an arch, a table similar to the first four columns of Table 95 should be made out for each of the above positions of the live load; and then a table similar to the latter part of Table 95 should be made out showing the maximum stresses for any of the positions.
The temperature stresses in an arch ring having fixed ends may be quite high, and should therefore be carefully considered. To compute the temper ature stresses, conceive that the arch is without weight and exactly fits between the skewbacks, without stress anywhere, at a certain mean temperature. Let 1 = the span of the neutral line; e = the expansion of concrete per unit of length per 1° Fahr.; to= the difference in degrees Fahrenheit between the mean temper ature and the actual temperature of the arch ring.
Then the total change in length of the span of the neutral line is 1 e t°. As the abutments resist this change, a horizontal force and also a bending moment will be developed at each abutment. Conceive that the bending moment is resisted by a horizontal force, Q, applied at some distance, q, above each springing line, and that these forces act inward for a rise of temperature and outward for a fall; and also conceive that at each springing two horizontal forces, each equal to Q, act opposite to each other. The first Q and one of the latter form a couple whose bending moment at the abutment is Q . q, and the remaining Q at the springing resists the horizontal
thrust (or pull) at the abutment.
We may regard the arch as being without weight and acted upon by the couples and by the horizontal thrusts, and that we are to find the resulting stresses in the arch ring. Since the arch has fixed ends, the three equations of condition, equations 4, 5, and 6 (page 674) must be satisfied.
If the upper Q at each end be conceived as acting along the line
Fig. 233, i.e., if q = dk, then the bending moment at any point of the arch ring due to temperature changes is Q . ak. The bending moment at any point of the arch ring due to the external loads is H H. ac. Hence, by analogy, we see that Q . ak may replace H. ac in equations 4, 5, and 6, page 674; and consequently the equa tions of conditicn for temperature stresses become The line k,k,,: has been so located that F ak = o, and also that ak . x = o; and therefore, if the yet unknown force Q acts along
equations 22 and 23 are thereby satisfied, i.e., the first two of the equations of conditions (§ 1300) are satisfied.
To satisfy the third condition, notice that a rise.of temper ature tends to increase and a fall to decrease the span; and hence the forces Q at each abutment must be just sufficient to resist this ten dency, and must act toward the center of the span to counteract a rise of temperature and from the center to counteract a fall. In § 1332 the change of span was shown to be 1 e t°; and by equation e, page 673, the differential change of span is M y ds. Hence EI Substituting the value of M from § 1333, and taking the summation for one half of the arch ring we get and by transposition E in equation 27 is to be taken in accordance with the character of the concrete in the arch ring; and in the example in hand we will assume a 1 : 2 : 4 concrete, and take E = 1,500,000 lb. per sq. in. or (1,500,000 X 144) lb. per sq. ft. (see § 478 and 493). 1 is the span of the neutral line, and is known. In the example in hand, 1 = 50 ft. Different observers find values of e varying from 0.000,004,3 to 0.000,008,0 per 1° Fahr., although the more reliable results are between 0.000,004,3 and 0.000,006,5.* We will use 0.000,005,4, the value obtained by Professor W. D. Pence.f I _ 4s is the reciprocal of 4s _ I, which was computed in determining the stresses due to external loads. For the example in hand, I - 4s is equal to the reciprocal of the mean of the quantities in the last column of Table 93, page 680; or 1 - J8 = 1 _ 2.3275. The value of ak.y is given in Table 94, page 683, and is equal to 192.72.