Another table could be made by substituting different values of h corresponding to different beams, and this would readily give for each beam the deflection by multiplying by the square of the span in inches.
If the fibre stress in the beam due to the loading was less than 16,000, the deflection would be obtained by•multiplying the result given in the table by the ratio of given fibre stress to 16,000.
The formula (2) applies directly to beams and channels only. If, therefore, a table based on this formula is made, and it is desired to use it for determining the deflection of unsymmetrical shapes, such as angles, tees, etc., the coefficients given must be divided by twice the distance of the neutral axis from extreme fibre, since both numerator and denominator of (1) has been multiplied by 2.
If a beam had a center load, its deflection could be obtained from this table by multiplying by I, this being the ratio of the deflection of a beam supported at the ends and loaded with a center load, to that of a similar beam with the same total load uniformly distributed.
In the table of safe loads it will be noted that a heavy black line divides the capacities specified. This is to denote the limit of span beyond which the deflection of the beam, if loaded to its full capacity, would be likely to cause the ceiling to crack. This limit of span can be determined from the formulae given above, as follows : A deflection of of the span can be safely allowed without causing cracks. Substituting a o for d, therefore, we have 1 X 8 X 16,000 360 384 E y and 1 48.3 y Making the substitutions of the value of y for different sized beams, gives limits agreeing with those in the Cambria Hand Book. The limits given in the Carnegie book are fixed arbitrarily at 20 times the depth of beam and some less than these.
This equation shows that if the table of properties is used to determine the capacity of a beam for a certain span which will be within the plaster limits of deflection, the product of the fibre strain and the span must be kept constant for a given depth of beam.
For example, if it is desired to know the fibre strain allowable for a 12-inch beam on an effective span of 30'4" (30 feet 0 inches) such that the plaster deflection will not be exceeded, we have The formula can be more quickly used by comparison with the limiting span given by the table of safe loads. In the above
case the limit of span for a 12-inch beam and a fibre strain of 16,000 lbs. is 24 feet; therefore the required f 24 30 x 16,000 = 12,800 Lateral Deflection of Beams. When beams are used for long spans, and the construction is such that no support against side deflection is given, the beam will not safely carry the full load indicated by the table, and the allowable fibre stress in top flange must be reduced. If such a beam were to carry a load giving a fibre stress of 16,000 lbs. per square inch, the actual fibre stress in top flange would be greater than this, as the deflection sideways would tend to distort the top flange and thus cause the additional stresses.
The length of beam which it is customary to consider capable of safely carrying the full calculated load without support against lateral deflection, is twenty times the flange width. The reason for thus fixing upon twenty times the flange width may be seen from the following In any consideration of a reduction of stress in a compression member due to bending caused by its unsupported length, it is customary to use Gordon's formula for the safe stress in columns. This formula is : 1 a r 2 For columns with fixed ends, a = 36,000. Now if we consider a 5-inch 9.75-lb. I, the moment of inertia about the neutral axis coincident with center line of web is I' = 1.23.
Since the moment of inertia of the web alone about this axis is inappreciable, the moment of inertia of each flange about this axis is = .62.
The area of the whole section = 2.87 sq. in.
Web = .86 Area of flanges = 2.01 sq. in.
Area of one flange = 1.00 Therefore .62 .79 The width of flange for 5-in. beam = b = 3.00 in.
b Therefore = — • Tests on full-sized columns show that columns of length less than ninety times the radius of gyration bend little if any under their load. It is, therefore, generally customary to disregard the effect of bending for lengths less than 90 radii. If in the above we multiply, we have : 90 = 23.7 6 The assumption that with full fibre stress of 16,000 lbs. beams should be supported at distances not greater than twenty times the flange width, brings the limit under that of 90 radii.