720 X 30 X .0000065 X2,400,000 X 6,529 //i - 881.
2,495,951 d) for center = 1S3 127.2 = 55.8 (y d) for abutments = 0 127.2 = 127.2.
The moment produced by the assumed change of temperature 30 degrees, is therefore as follows: At the center, 881 X 55,S = 49,160 inch-pounds.
At the abutments, SS1 X ( 127.2) = 112,063 inch-pounds.
It was computed above, that a moment of 149,490 ineh-pounds at the abutment would produce a unit-stress in the concrete of 151 pounds per square inch; therefore a unit-stress produced in the abutment by a moment of 112,063 pounds would be x 151 149,490 = 113 pounds per square inch. In cold weather the effect of the moment due to temperature would be to produce compression at the intrados at the abutment, and tension at the intrados at the crown; but it wa's found above, that the compression at the intrados at the abutment due to transverse moment and to thrust totaled 309 pounds per square inch. Adding the stress due to temperature, we have a total of 422 pounds per square inch. If the reduction of temperature below the temperature of construction was more than 30 degrees, the stresses would be increased in direct proportion. If the reduction of temperature was, sa3:, 40 degrees, instead of 30 degrees, the added stress due to temperature would be about 151 pounds per square inch. On the other hand, during warm weather, the effect of a rise in tem perature will be to relieve the strain; and the net compression or tension will be less than that which would be due to direct loading and to thrust. For the loading which has been computed, the moment due to loading at the crown is very small, and is in the opposite direc tion to the moment usually produced by a load over the entire arch. It is generally true that in cold weather the arch is stressed by the sum of the stresses due to transverse moment, thrust, and temperature; in warm weather the stresses usually tend to counteract each other. Cold weather is therefore a critical time for an arch, and the time when an excess of live load would be particularly dangerous.
441. Stresses Due to Rib Shortening. The compression in a rib results in shortening the arch rib very slightly; and this produces precisely the same effect in altering the moment as an equivalent fall in temperature. For example, in the above case, we have at the abut ment a thrust of 296 pounds per square inch; dividing this by E, the modulus of elasticity, 2,400,000, we have .0(101233, the proportional shortening; dividing this by .0000065, the coefficient of expansion, we find that the thrust clue to this rib shortening is the equivalent of a reduction of temperature of 19 degrees. Since we have found that a reduction of temperature of 30 degrees produced a unit-stress of 113 pounds per square inch, a virtual reduction of 19 degrees would pro duce a unit-stress of 72 pounds per square inch. Since such a stress
is always the same as that due to a reduction of temperature, and since this always has the effect of increasing the stresses for usual loading, such unit-stress must be added to the value found above; therefore; adding this 72 pounds per square inch to the total previously found (422), we have a unit-compression at the intrados at the abutment, of 494 pounds per square inch.
442. Testi.:g this Arch for Other Loading. A live load of 200 pounds per square foot over the entire arch would unquestionably increase the thrust over the entire arch, especially at the abutments. The stress due to shortening will of course be increased in proportion to the increase in the thrust. The stress due to moment cannot be accurately predicted. Of course such an examination and test for full loading should be made in the case of any arch to be constructed, and should be worked out precisely on the same principles and in general by identically the same method as was used above.
To test the arch for a concentrated loading such as would be produced by the passage of a road roller, or, in the case of a railroad bridge, by an especially heavy locomotive, the test must be made by assuming the position of that concentrated load which will test the arch most severely. Ordinarily this will be found when the concen trated load is at or near one of the quarter points of the arch. The only modification of this test over that given above in detail, is in the drawing of the load line, but the general method is identical.
443. Testing an Arch with Variable Moments of Inertia. It has already been indicated how the equations on which the arch theory is based may be simplified when the moment of inertia is constant.
The above problem was worked out on the basis that the moment of inertia varied in the ratio of In either case the solution is eon & siderably simplified. Arches are frequently designed where the ment of inertia varies according to some other law. The very quent practice is to increase the thickness of the arch toward the abut ds ment much more rapidly than the rule would call for, and thus dx increase the moment of inertia of the arch much more rapidly. In such a case, Equations 49 must be used; and the summations must be made up by computing for each unit-section the value of the moment of inertia for that point, and by measuring ds along the length of the arch rib. This means also that the sections of dead and live load, instead of having a constant width (as in the above problem), have a variable width, and the loads must be separately computed. While there is nothing especially difficult about such a solution, it involves considerably more work.