## Pl3

where

The deflection at a distance xL from one support of the girder produced by concentrated loads P at distance kL from the support and an upward concentrated load P at a distance (1 - k)L from the support (antisymmetrical loading, Fig. 4.23) is given by

where

For convenience in analysis, the loading carried by the grid framing is converted into concentrated loads at the nodes. Suppose for example that a grid consists of two sets of parallel girders as in Fig. 4.19, and the load at interior node r is Pr. Then it is convenient to assume that one girder at the node is subjected to an unknown force Xr there, and the other girder therefore carries a force Pr - Xr at the node. With one set of girders detached from the other set, the deflections produced by these forces can be determined with the aid of Eqs. (4.134) to (4.141).

A simple example will be used to illustrate the application of the method of consistent deflections. Assume an orthogonal grid within a square boundary (Fig. 4.24a). There are n = 4 equal spaces of width h between girders. Columns are located at the corners A, B, C, and D. All girders have a span nh = 4h and are simply supported at their terminals, though continuous at interior nodes. To simplify the example. all girders are assumed to have equal and constant moment of inertia I. Interior nodes carry a concentrated load P. Exterior nodes, except corners, are subjected to a load P/2.

Because of symmetry, only five different nodes need be considered. These are numbered from 1 to 5 in Fig. 4.24a, for identification. By inspection, loads P at nodes 1 and 3 can be distributed equally to the girders spanning in the x and y directions. Thus, when the two sets of parallel girders are considered separated, girder 4-4 in the x direction carries a load of P/2 at midspan (Fig. 4.24b). Similarly, girder 5-5 in the y direction carries loads of P/2 at the quarter points (Fig. 4.24c).

Let X2be the load acting on girder 4-4 (x direction) at node 2 (Fig. 4.24b). Then P -X2 acts on girder 5-5 (y direction) at midspan (Fig. 4.24c). The reactions R of girders 4-4 and 5-5 are loads on the boundary girders (Fig. 4.24d).

Because of symmetry, X2 is the only unknown in this example. Only one equation is needed to determine it.

To obtain this equation. equate the vertical displacement of girder 4-4 (x direction) at node 2 to the vertical displacement of girder 5-5 (y direction) at node 2. The displacement of girder 4-4 equals its deflection plus the deflection of node 4 on BC. Similarly, the displacement of girder 5-5 equals its deflection plus the deflection of node 5 on AB or its equivalent BC.

When use is made of Eqs. (4.136) and (4.138), the deflection of girder 4-4 (x direction) at node 2 equals FIGURE 4.24 Square bay with orthogonal grid. (a) Loads distributed to joints. (b) Loads on midspan girder. (c) Loads on quarter-point girder. (d ) Loads on boundary girder.

n3 h3 EI

where 84 is the deflection of BC at node 4. By Eq. (4.137), [A, A] = (A8)(n/16). By Eq. (4.139), (1A 1/4) = V48. Hence ## Renewable Energy 101

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