must be carried up with sets of 10 or 12 inches in width. Abutments are very different from the piers, they sustain both the weight of half the extreme arches, and the horizontal thrust of those arches. Therefore, their thickness should be regulated to counteract both those, whose tendency is to cause the abut. ment to yield. Counterforts are often added to the abutments, where the horizontal thrust is great. When the bridge and road surface is above the natural surface, it becomes necessary to widen the access towards the end of the bridge, and also sustain the embankments, by building wing-walls from the bridge; this should receive a batter, and a suitable thickness to resist the action of the pressure. The style of architecture depends on the locality and general character of the structure itself. To find the Fall of Water in the Arches of a Bridge. ACCORDING TO C. HUTTON, L.L. &c. Principles. "1. A heavy body, that in first second of time has falled the height of a feet, has acquired such a velocity, that the moving uniformly with it, will in the next second of time move the length of 2 feet. "2. The spaces run through by falling bodies are proportional to one another as the squares of their last or acquired velocities. These two principles are proved by writers on mechanics. "3. Water forced out a large channel, through one or more smaller passages, will have the stream through those contracted in the ratio of 25 to 21. This is shown in the 36th problem of the 2 Book of Newton's Principia. "4. In any stream of water, the velocity is such as would be acquired by the fall of a body from a height above the surface of that stream. This is evident from the nature of motion. "5. The velocities of water through different passages of the same height, are reciprocally proportional to their breadths. For, at some time, the water must be delivered as fast as it comes; otherwise the bounds would be overflowed. At that time, the same quantity which in any time flows through a section in open channel, is delivered in equal time through the narrower passages; or the rectangle under the section of the narrow passages by the mean velocity, must be equal to the rectangle under the section of the open channel by its mean velocity. Therefore the velocity in open channel is to the velocity in the narrower passages, as the section of those passages is to the section of the open channel. But, the heights in both sections being equal, the sections are directly as the breadths. Consequently the velocities are reciprocally as the breadths. "6. In a running stream, the water above any obstacle put therein will rise to such a height, that by its fall the stream may be discharged as fast as it comes. For the same body of water, which flowed in the open channel, must pass through, made by the obstacles: and the narrower the passages, the swifter will be the volocity of the water but the swifter the velocity of the water, the greater is the height from which it hath descended: consequently the obstacles, which contracts the channel, cause the water to rise against them. But the rise will cease, when the water can run off as fast as it comes: and this must happen when, by the fall between the obstacles, they will require a volocity in a reciprocal proportion to that in the open channel, as the breadth of the open channel is the breadth of the narrow passages. "7. The quantity of the fall caused by an obstacle in a running stream, is measured by the difference between the height fallen from, to acquire the velocities in the narrow passages and open channel. For, just above the fall the velocity of the stream is such as would be acquired by a body falling a height higher than the surface of the water: and at the fall, the velocity of the stream is such, as would be acquired by the fall of a heavy body from a height more. elevated than the to of the falling stream: and consequently the real fall is less than the height. Now as the stream comes to the fall with a velocity belonging to a fall above its surface, consequently the height belonging to the velocity at the fall, must be diminished by the height belonging to the velocity with which the stream arrives at the fall." "Prob. In a channel of running water, whose breadth is contracted by one or more obstacles; the breadth of the channel, the mean velocity of the whole stream, and the breadth of the water-way between the obstacles, being given, to find the quantity of the fall occasioned by those obstacles. Let b = breadth of the channel in feet; v = mean velocity of the water in feet; c = breadth of the water-way between the obstacles." Now 25:21::c: c, the water-way contracted, by prin. 3. And cb::v :v, the velocity in the contracted way, prin. 5. Also (2a)': v2::a:, height fallen to gain the vel. v, 1 and 2 prin. And (2a): (v)2::a: (2b)2×2, ditto for the velocity v, prin. 1 and 2. Then - is the measure of the fall required, prin. 7, Or thus, [()-1]x is a rule for computing the fall. In this case a = 16.0899 feet; and 4a = 64.3596. Prob. 2. To compute the quantity of water that passes through a bridge or any section of a river. Procure a slender wooden rod, to one end of which at tach a weight sufficient to sink it, leaving the other end above the water. This being put into the stream whose velocity is required, the greater velocity at the surface will make it incline towards the direction in which the water runs ; when it has acquired an equilibrium, it mus move in an oblique position, the velocity being nearly equal to that of the water there. But as the velocity varies with the depth, the experiment should be tried in different parts of the cross-section, and a mean taken for the mean velocity. Having determined the mean velocity of the stream, the next thing to be done is to find the area of the cross-section, and a mean taken for the mean velocity. Having determined the mean velocity of the stream, the next thing to be done is to find the area of the cross-section, by taking a few depths at equal distances across the river. In finding the mean velocity, a stop-watch will be necessary to tell the time to seconds. There is an instrument for this purpose called a Patamometer, which is by far the best way to go to work. Example. Suppose the breadth of a river at m, to be 40 feet, and the depths at 6 different equidistant places, to be 3, 5, 7, 6, 4, 5, feet, 3+ 5 = 8, and 8÷ 2 = 4; then (4+5+7+6+4)÷5=26÷5=5}, and 5×40=208 square feet for the area of the cross-section at m. Now suppose the velocities taken at six different places, in the cross-section, to be 40, 46, 50, 56, 50, 42, feet per minute, the mean velocity will be (40+46+50+56+50+42)÷6=471⁄2 feet, for the mean velocity per minute. Then 47×208=13490 cubic feet, the quantity that flows through the section of the river in one minute. To find the velocity at any other section da; da: mn: velocity through mn: the velocity through da. On the Oblique Arch. Prob. 1.-Given the width of the arch-way, the angle of obliquity, and the distance between the two faces, to draw the plan of the arch-way. Draw the straight line A n, (Fig. 32,) and make An = the given width. Make the angle n A B = the complement of the angle of obliquity. Draw n B perpendicular to A n, and B C perpendicular to A B. Make BC = the distance between the faces, and Q D C parallel to A B Then is the parallelogram A Q D B the plan, A Q, and B D the two springing lines, A B and Q D the lengths of the faces. Prob. 2.-Given the width of the arch-way, the distances of obliquity, and the distance between the two faces, to draw the plan of the arch-way. (Fig. 32.) Draw the straight line As the width of the arch. Draw s B perpendicular to As, and the distance of obliquity, and A B. Erect the perpendicular Bo= the distance between the faces. Through o draw on parallel to A B, and A n parallel to B m. Then will A B, m n be the required plan. Prob. 3.-Having executed the piers of an oblique arch to the height of the springers, to draw the plan of the archway from the dimensions then taken. (Fig. 33.) Suppose you find A B the length of one of aces to be 20 feet, and B D the length of one of the abutments = 30 feet, and Ao the perpendicular width of the arch= 16 feet. Draw A B = 20 feet, and upon this describe the semicircle Ao B; from A as centre with 20 feet as radius describe an arch to cut the circle in o, join B o and produce it off to D making BD = 30 feet the length of the abutments; draw DC and A C parallel to B A and BD; then will the parallelogram A CDB be the required plan. Note. It is evident you may take other dimensions as well as the last, from which you may construct the plan. |