Longitudinal Section Of A Folded Plate

where L is the longitudinal span (ft). Maximum bending stresses then may be determined by the beam formula / = ±M/S, where S is the section modulus. The positive sign indicates compression, and the negative sign tension.

For solid-web members, S = I/c, where I is the moment of inertia of the plate cross section and c is the distance from the neutral axis to the top or bottom of the plate. For trusses, the section modulus (in3) with respect to the top and bottom, respectively, is given by

where At = cross-sectional area of top chord, in2

Ab = cross-sectional area of bottom chord, in2 h = depth of truss, in

In the general case of a folded-plate structure, the stress in plate n at joint n, computed on the assumption of a free edge, will not be equal to the stress in plate n + 1 at joint n, similarly computed. Yet, if the two plates are connected along the fold line n, the stresses at the joint should be equal. To restore continuity, shears are applied to the longitudinal edges of the plates (Fig. 4.27d). The unbalanced stresses at each joint then may be adjusted by converging approximations, similar to moment distribution.

If the plates at a joint were of constant section throughout, the unbalanced stress could be distributed in proportion to the reciprocal of their areas. For a symmetrical girder, the unbalance should be distributed in proportion to the factor

where A = cross-sectional area, in2, of girder h = depth, in, of girder r = radius of gyration, in, of girder cross section

And for an unsymmetrical truss, the unbalanced stress at the top should be distributed in proportion to the factor

The unbalance at the bottom should be distributed in proportion to

H Ab

A carry-over factor of -V2 may be used for distribution to the adjoining edge of each plate. Thus the part of the unbalance assigned to one edge of a plate at a joint should be multiplied by - V2, and the product should be added to the stress at the other edge.

After the bending stresses have been adjusted by distribution, if the shears are needed, they may be computed from f + f

for true plates, and for trusses, from

where Tn = shear, kips, at joint n fn = bending stress, ksi, at joint n An = cross-sectional area, in2, of plate n

Usually, at a boundary edge, joint 0, the shear is zero. With this known, the shear at joint 1 can be computed from the preceding equations. Similarly, the shear can be found at successive joints. For a simply supported, uniformly loaded, folded plate, the shear stress fv (ksi) at any point on an edge n is approximately fv = k (1 - I) (4.161)

where x = distance, ft, from a support t = web thickness of plate, in L = longitudinal span, ft, of plate

As an illustration of the method, stresses will be computed for the folded-plate structure in Fig. 28a. It may be considered to consist of four inverted-V girders, each simply supported with a span of 120 ft. The plates are inclined at an angle of 45° with the horizontal. With a rise of 10 ft and horizontal projection a = 10 ft, each plate has a depth h = 14.14 ft. The structure is subjected to a uniform load w = 0.0353 ksf over its surface. The inclined plates will be designed as trusses. The boundaries, however, will be reinforced with a vertical member, plate 1. The structure is symmetrical about joint 5.

As indicated in Fig. 28a, 1-ft-wide strip is selected transversely across the structure at midspan. This strip is designed to transmit the uniform load w to the folds. It requires a vertical reaction of 0.0353 X 14.14/2 = 0.25 kip per ft along each joint (Fig. 28b). Because of symmetry, a typical joint then is subjected to a uniform load of 2 X 0.25 = 0.5 kip per

Concrete Folded Plate Sections
FIGURE 4.28 (a) Folded-plate roof. (b) Plate reactions for transverse span. (c) Loads at joints of typical interior transverse section. (d ) Forces at joint 4. (e) Forces at joint 3. f) Plate 4 acting as girder. (g) Loads at joints of outer transverse section. (h) Plate 2 acting as girder.

ft (Fig. 28c). At joint 1, the top of the vertical plate, however, the uniform load is 0.25 plus a load of 0.20 on plate 1, or 0.45 kip per ft (Fig. 28g).

The analysis may be broken into two parts, to take advantage of simplification permitted by symmetry. First. the stresses may be determined for a typical interior inverted-V girder. Then. the stresses may be computed for the boundary girders, including plate 1.

The typical interior girder consists of plates 4 and 5, with load of 0.5 kip per ft at joints 3, 4, and 5 (Fig. 28c). This load may be resolved into loads in the plane of the plates, as indicated in Fig. 28d and e. Thus a typical plate, say plate 4, is subjected to a uniform load of 0.707 kip per ft (Fig. 28/). Hence the maximum bending moment in this truss is

Renewable Energy 101

Renewable Energy 101

Renewable energy is energy that is generated from sunlight, rain, tides, geothermal heat and wind. These sources are naturally and constantly replenished, which is why they are deemed as renewable. The usage of renewable energy sources is very important when considering the sustainability of the existing energy usage of the world. While there is currently an abundance of non-renewable energy sources, such as nuclear fuels, these energy sources are depleting. In addition to being a non-renewable supply, the non-renewable energy sources release emissions into the air, which has an adverse effect on the environment.

Get My Free Ebook

Post a comment