This section is from the book "Cyclopedia Of Architecture, Carpentry, And Building", by James C. et al. Also available from Amazon: Cyclopedia Of Architecture, Carpentry And Building.

This will be illustrated by a numerical example. A beam having a span of 18 feet supports one side of a 6-inch slab 8 feet wide which carries a live load of 200 pounds per square foot. In addition, a special piece of machinery, weighing 2,400 pounds, is located on the slab so near the middle of the beam that we shall consider it to be a concentrated load at the center of the beam. The floor area carried by the beam is 18 feet by 4 feet = 72 square feet. Adding 3 inches to the 6 inches thickness of the slab as an allowance for the weight of the beam, we have 9X12 = 108 pounds per square foot for the dead weight of the floor. With a factor of 2 for dead load, this equals 216. Using a factor of 4 on the live load (200), we have 800 pounds per square foot. Then the ultimate load on the beam, due to these sources, is (216 + 800)

72 = 73,152 pounds. So far as its effect on moment is concerned, the concentrated load of 2,400 pounds at the center would have the same effect as 4,800 pounds uniformly distributed. As it is a piece of vibrating machinery, we shall use a factor of six (0), and thus have an ultimate effect of 6 X 4,800 = 28,800 pounds. Adding this to 73,152, we have 101,952 pounds as the equivalent, ultimate, uniformly distributed load. Then,

M0 = 1/8 Wo l = 1/8 X 101,952 X 216 = 2,752,704.

Fig. 102. Reinforced Beam.

In order to reduce as much as possible the size and weight of this beam, we shall use 1:2:4 concrete, and therefore apply Equation 24:

2,752,704 = 565 bd2; bd2 = 4,872.

If b =16 inches, d? = 304.5, and d = 17.5 inches.

A still better combination would be a deeper and narrower beam with 6 = 12 inches, and d = 20.15 inches. With this combination, the required area of the steel will equal:

A = .0121 X bd = .0121 X 12 X 20.15 = 2.93 square inches.

This can be supplied by eight bars 5/8 inch square.

The total ultimate load as determined above, is 101,952 pounds.

One-half of this gives the maximum shear at the ends, or 50,976 pounds. Applying Equation 31, we have, since d - x = .85 d = 17 inches:

v = V/ b(d-x) = 50,976 /12x17 = 249 pounds per square inch.

As already discussed in previous cases, the ends of the beam must be reinforced against diagonal tension, since the above value of v is too great, even as an ultimate value, for such stress. Therefore the ends of the beam must be reinforced by turning the bars up, or by the use of stirrups. The beam must therefore be reinforced about as shown in Fig. 102. Although the concentrated center load in this case is comparatively too small to require any change in the design, it should not be forgotten that a concentrated load may cause the shear to change so rapidly that it might require special provision for it by means of stirrups in the center of the beam, where there is ordinarily no reinforcement which will assist shearing stresses.

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