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Design of Beam Bridges 152

feet, slab, inches, pounds, load and width

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DESIGN OF BEAM BRIDGES 152. Slab Bridges.—When the span of a bridge is not more than 12 to 15 feet, the simple slab spanning the opening and resting upon the abutments at its ends is usually the most economical form to use. Under heavy loading, the economic limit of length may be only 10 to 12 feet, while for lighter loads, slabs 16 to 20 feet in length may be desirable. The design of a bridge slab will be illustrated by a numerical example.

Example 1.—Design a highway slab of 11 feet clear span, and width of 18 feet to carry a macadam road with the loading given in Section 149.

Solution.— Assume weight of road material= 80 pounds per square foot. Weight of slab = 145 pounds per square foot.

Total dead Load = 225 pounds per square foot.

Live load is auto truck with 14,000 pounds on each of two wheels 6 feet apart. From Section 150, effective width, c=.6S+1.7. As .65 is more than the distance apart of wheels, the loads would over lap, and we consider both loads distributed over 14.3 feet. The live load per foot of width is 28000.'14.3=1950 pounds. This load may be considered as applied over a length of 1.7 feet=20 inches. The effective length of the beam is distance between centers of bearings, or 1 foot more than the clear span. (11+1=12 feet.) Bending moments, M (live) = 1950/2 (72-5) = 65325 in.-lb.

M (impact) = 25 per cent of live = 16330 in.-lb.

M (dead) = 225X 12X 12X 12/8 = 48600 in.-lb.

Taking f = 650, n=15, Table VII (p. 163) gives R=108, 12d'=13025/10S=1206, and inches.

Maximum shear occurs when center of live load is 1.7/2 feet from support., in which case, Make d= 10 inches, then allowing concrete to extend 1.5 inches below steel, weight of beam is 12X 11.5X 150 144 =144 pounds per square foot, which is within the assumed weight.

Reinforcement.—The arca of steel required per foot of width, From Table XV (p. 199), we sec that -inch round bars spaced 5.5 inches apart, or i-inch square bars 5 inches apart will answer. For the latter the maximum unit bond stress is Fig. SO shows the slab in longitudinal section. For Lateral rein

forcement ;-inch round bars, 12 inches apart, are used. To prevent cracking due to negative moment. where the slab joins the abutments, ;-inch round bars 12 inches apart are placed in the ends of the slab at the top. Expansion joints, usually tar paper, are placed on the top of the abutment the slab, thus preventing the develop ment of negative moment and allowing for temperature changes.

153. T-Beam Bridges.—When the length of the bridge is too great for a simple slab, it is found economical to use girders to support the slab. If the head room is sufficient and the span not too great, T beam construction may be used. This consists of a series of 'I'-beams extending from abutment to abutment, girders being placed under the slab to form the stems of the T-bearns, and the slab being continuous over the girders for the width of the bridge.

Example 2.—Design a T-beam bridge with clear span of 24 feet, to carry a roadway 18 feet wide, using loadings as in Example 1.

Solution.—Allowing 12 inches for width of base of guard rail, the full width is 20 feet. Use five girders, spaced 4 feet on centers, the outside girders being 2 feet from end of beam (sec Fig. 81).

The live load is a single wheel load of 14,000 pounds distributed over a width .0X4+1.7=4.1 feet. The live Load per foot of width is 14000,4.1=3415 pounds. This may be considered as distributed over 2 feet of length. The slab is continuous and taking the moment of the concentrated load as four-fifths of the moment for a simply supported beam, we have 12 108=311, and inches.

The shear is a maximum when the load is placed next to the support, and assuming width of girder at 12 inches, ? (live) = 3415X 2.5/ 4 =2134 pounds.

? (impact) = 25 per cent of 3415 = 533 pounds.

? (dead) =180X 3 "2 = 270 pounds.

From Table XV (p. 199), --inch square bars spaced 5 inches apart will answer.

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