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CHAPTER V.

STRESSES IN LATERAL SYSTEMS.

Introduction. The wind loads on bridges are carried to the abutments by the lateral systems. In a through truss bridge the lateral systems usually consist of the top lateral system, the bottom lateral system, the intermediate bents or sway bracing between the intermediate posts, and the portals in the planes of the end-posts as shown in Fig. 1, Chapter VIII. In shallow through truss bridges the sway bracing is sometimes omitted; in deck trusses the portals are replaced by sway bracing; while in low trusses the bottom lateral system only is used.

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Wind Loads.-The wind loads are usually given in specifications as a certain number of pounds per lineal foot of bridge or per square foot of exposed surface. The wind load is usually taken at 30 lb. per square foot of exposed surface when the live load is on the bridge and at 50 lb. per square foot of exposed surface when the bridge is unloaded.

The usual specification for highway bridges is: A wind load of 150 lb. per lineal foot of bridge on the unloaded chord to be treated as a dead load, and a wind load of 300 lb. per lineal foot of bridge on the loaded chord, 150 lb. of which is to be treated as a dead load and 150 lb. to be treated as a live load. In railroad bridges the dead load wind is usually taken the same as for highway bridges, while the live load wind is taken at 450 lb. to 600 lb. per lineal foot. For extracts from standard specifications, see Chapter IX.

STRESSES IN LATERAL SYSTEMS.-In the through Pratt truss bridge in Fig. I the wind joint loads on the upper chord are equal to U, while the joint loads on the lower chord are equal to L. Where sway bracing is used, part of the upper chord loads are transferred to the lower

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lateral system by the sway bracing. The exact amount thus transferred is statically indeterminate, but is usually assumed as U/2 at all joints having sway bracing. This load U/2 produces a vertical load, K, at each joint in the vertical trusses, which acts downward on the leeward and upward on the windward side of the bridge. Each portal transfers the load carried to the hip

Center line of Truss

--

FIG. 2.

joint by the upper lateral system, and the load at the hip joint to the abutments. This produces a tension V⚫sin 0 in the bottom chord on the leeward side and a compression V sin in the bottom chord on the windward side. In addition to the wind loads on the top chord that are transferred to the bottom lateral system, the wind load on the train of cars on steam and electric railway bridges, increases the loading on the vertical trusses on the leeward side and decreases the loading on the windward side of the bridge as shown in Fig. 2. This increase or decrease in vertical loading can be calculated by taking moments about the line of the pins in the lower chord. The wind load acting on the train is usually specified as applied six feet above the base of the rail.

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Skew Bridge. In a skew bridge the abutments are not at right angles to the center line of the bridge. This gives a warped portal, and ties which are not vertical unless the bridge is skewed one entire panel as is shown in Fig. 3, which is the common practice.

Initial Stresses.-In (a) Fig. 4 the diagonal lateral rods have an initial stress of 10,000 lb. in each rod. In (b) the lateral truss is loaded with loads of 12,000 lb. at joints B, C and D, producing stresses as shown. In (c) the combined stresses due to direct loads and the initial stresses are shown. The stresses in the chords and struts can now be calculated by algebraic resolution. The stresses are combined as follows: In panel B-C each rod has an initial stress of 10,000 lb., and in addition must transfer a wind shear of 6,000 lb. or an inclined stress of 9,000 lb. the 9,000 lb. or 4,500 lb. will be added to the initial stress in Bc, making the stress

Half of 14,500 lb.,

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while 4,500 lb. will be subtracted from the, initial stress in Cb, making the stress 5,500 lb. In panel A-B the initial stress in aB is entirely relieved by the direct stress, while the initial stress in Ab is increased by the direct stress remaining after the initial stress of 10,000 lb. was relieved in aB, or 17,000 lb., making the total stress in Ab = - 10,000 17,000 = - 27,000 lb., the same as if there had been no initial stress in the panel. This solution is based on the mathematical principle "That if a load may be carried from one point to another by more than one route, it will be divided between the routes in proportion to the rigidities of the routes." In the problem above the two routes are assumed to have the same rigidities.

PORTALS.-Portal bracing is placed at the ends of through bridges in the planes of the end-posts to transfer the wind loads from the upper lateral system to the abutments. The stresses in the sway bracing placed in the planes of the intermediate posts are calculated in the same manner as the stresses in portal bracing. Portal bracing should be designed so that the stresses will be statically determinate. Several different types of portals are shown in Fig. 5. Types (a), (b) and (d) are the types most used for highway bridges. The lower ends of the end-posts may be hinged (free to turn), or fixed. The criterion for determining whether the end-posts are fixed or not will be discussed in Chapter VI.

Case I. Stresses in Simple Portals: End-posts Hinged.-The deflections of the posts in the portals shown in Fig. 5 are assumed to be equal, and

H = H' ==

R/2

Taking moments about the foot of the windward post

V' = V = R.h/s

(1)

Having found the external forces, the stresses in the members may be found by either algebraic or graphic methods.

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Algebraic Solution. Portal (a).—To obtain the stress in member G-C, (a) Fig. 5, pass a section cutting G-F, E-F and G-C, and take moments of the external forces to the right of the section, about point F as a center.

But H =

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R/2, and (h — d) sin 0 = s.cos 0. Substituting these values in (2) we have

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(2)

(3)

Resolving at C and F we have, stress in E-F = 0, and also stresses E-E' and H-E' = 0. To obtain stress in G-D, pass section cutting H-G, H-E' and G-D, and take moments of the external forces to the left of the section, about point H as a center.

G-D = H•h/[(h — d) sin 0]

=

+ V.sec 0

(4)

To obtain stress in G-F, pass a section cutting G-F, E-F and G-C, and take moments of the external forces to the right of the section, about point C as a center

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To obtain stress in H-G, pass a section cutting H-G, H-E' and G-D, and take moments of the external forces to the left of the section, about the point D as a center.

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The stress in the windward post, A-F, is zero above and V below the foot of the knee brace C; the stress in the leeward post is zero above and V' below the foot of the knee brace D. The shear in the posts is H below the foot of the knee brace, and above the foot of the knee brace is given by the formula

SH.d/(h d)

= stress in H-G

(7)

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