Fig. 55

Fig. 55.

For intermediate proportions the BM parallel to the short sides may be found by multiplying by L4

L4 + B4 that obtained when considering the slab as a simple beam of length B; where L = length of slab and B = breadth. The BM parallel to the long sides may be similarly found by multiplying by B4 the BM

L4+B4 obtained when considering the slab as a simple beam of length L.

Slabs are usually reinforced with rods at right angles to one another across length and breadth of slab. These reinforcements must be capable of meeting the BM as found with the aid of the above factors. The dimensions may be ascertained as set forth for beams, a strip of the slab 12 inches wide being taken for calculation. Suppose that a slab 8 feet x 12 feet is to support a load of 160 lbs. per square foot, which load includes the weight of the slab, and that the slab will be freely supported at its edges. BM across short wav of slab -

=wB2_x L4

8 L4+B4

= 160x82 x 124. foot-lbs.

8 124 + 84

= 12826 inch-lbs. BM across long way of slab - wL2 x B4 8 L4+B4

= 160x 122 x 84 foot-lbs. 8 12+84

= 5701 inch-lbs.

Assuming that a reinforcement of 1 per cent. is to be employed (see table above) d2 = 12826

Q2.8 bd2= 12826, but b= 12.

Armoured Or Reinforced Concrete Beams Part 3 88

92.8 x 12

Armoured Or Reinforced Concrete Beams Part 3 89

d=3.4, or say 3 1/2 inches

And area of reinforcement = 31/2x12 =.42 square inch per foot of width, say 4 - 3/8-inch rods to the foot, - that is, 3 inches apart.

Now, considering the reinforcement in the length of the slab d=3 1/8 inches (see Fig. 282), and representing the coefficient of bd2 by k, kbd2 = 5701.

Armoured Or Reinforced Concrete Beams Part 3 90

k x12 x (3 1/8)2 = 5701.

Armoured Or Reinforced Concrete Beams Part 3 91

k = 48.6,

On turning to the table given above it is seen that something less than 1/2 per cent. is required. 1/4-inch diameter rods 4 inches apart may, however, be used, which is equivalent to .39 per cent.

The full depth of the slab will be 3 1/2 inches plus the covering to the reinforcement, - or, say, a total of 4 1/2 or 5 inches (see Fig. 56).

Calculations of this sort will not often be necessary, as uniformity in size and arrangement of rods is to be aimed at as far as possible.

In either slabs or beams it is preferable to use a number of rods of fairly small section in place of a few large ones, in order that their effect may be evenly distributed, and also that the metal may be removed as far as possible from the neutral axis.

Tee Beams

Where floors are entirely moulded in situ the floor slab, together with the beam attached to it, form a beam of tee section, the portion of the slab on either side of the rib affording resistance to compression, while the rib itself chiefly acts in holding the tensile reinforcement. The form of the beam produced may be seen in Fig. 57.

As the slab is continuous on either side of the rib, the width that may be considered as forming part of the tee beam must be decided upon. In the case where a simple series of beams supported at their extremities is used the width of the flange of tee may be taken as extending to points centrally between beams; where, however, secondary beams supported on main beams are employed, the compressive stress induced in the horizontal member in carrying its load as a slab will act in the same direction as that induced in the flange of the main beam. Thus in the latter case the full width of slab cannot be considered as taking the compressive stress of the beam. It will be safe, however, in all cases to consider that the flange of the beam extends to a quarter of the span on either side of the rib.

Tee Beams 92

Fig. 56.

Tee Beams 93

Fig. 57.

The calculations for tee beams will be precisely the same as those for rectangular beams, provided that the neutral axis does not fall below the lower surface of the slab. Fig. 57 illustrates a case in which the neutral axis is above the lower surface of the slab. The formulŠ given on page 37 may then be applied, substituting B (or half width of bay) for b, and total depth D for d; while p, the percentage of reinforcement, must be taken on area B x D as shown dotted. This dotted area, in fact, represents a beam of equal strength to that of the tee beam, - that is, if the resistance of concrete in tension be neglected; however, in the case of the tee beam the use of an ample supply of reinforcement passing through the depth of the beam is most necessary.

The formulŠ become -

Tee Beams 94

Mr=2/3 Bhc (D - 2/3h)

It will be noticed that the width of the rib does not enter into the calculations. It should, however, be proportioned to the depth of the rib, and should be sufficient to allow the concrete to be well packed round the reinforcement.

Fig. 58 shows a tee beam with its stress area, in which the neutral axis falls below the lower surface of the slab. The above formulae are not strictly applicable to this case; but formulae deduced on the same principles are too complicated to warrant their adoption.

Compressional Reinforcement

It is sometimes desirable to use a metal reinforcement in the compressional portion of a beam as well as in the tensional portion, in order to gain a reduction in depth, while some systems always adopt it. Its use, however, in this position is not economical.

Compressional Reinforcement 95

Fig. 58.

The metal must be strained to the same extent as the surrounding concrete, and as the stress required to produced strain in steel equal to that in concrete = Es/Ec.c, the stress upon the reinforcement will be ten Ec times that upon the concrete. Thus if the surrounding concrete is stressed to 500 lbs. per square inch, the stress upon the steel will be 500 x 10 = 5000 lbs. per square inch; but as the steel must necessarily be embedded below the surface of the concrete, the actual stress will be less than this.