Masonry structures by Spalding Frederick P. (Frederick Putnam) 1857-1923

Author:Spalding, Frederick P. (Frederick Putnam), 1857-1923
Language: eng
Format: epub, pdf
Tags: Masonry
Publisher: New York, John Wiley & sons, inc. ; [etc., etc.]
Published: 1921-03-25T05:00:00+00:00

and (42)

A' = 19980/8085 = 2.47 in. 2

Area of tension steel (43),

T 36680+19980

115. Tables. — The labor of computation may be materially lessened by the use of tables, which may be made in several ways, of which the following seem most convenient for use.

Table XI. Transposing the terms of formula (43) we have

r c 1

A - -4-A — f T"5T«

Js Js

This may be placed in the form,

A= Pl bd+^p, ....... (46)

Js

in which p\ is the ratio of steel for a beam with the same unit stresses and without compression steel.

Formula (45) may be put in the form

M = Rbd 2 +f' s A'(d-d'), or solving for A'

-f s (d-d'y

Values of R, pi and f' s , in terms of various values of f s , f c , and d'/d for n=15, are given in Table XI. This table may be used to find the areas of steel required when a beam of given dimensions must carry a bending moment too great to be resisted by tension reinforcement only.

Table XII. Combining (41), (42) and (45) we have

M = ^f c jkbd 2 -i-f s p f (l = d'/d)bd 2 , from which

= ±f c jk+f' s p'(l-d'/d)=G, .... (48)

in which G is constant for definite values of unit stresses and steel ratios. In Table XII, values of p' and p are given directly for various values of / c and G when n = 15 and /,= 16,000 lbs./in. 2

To use this table in design, it is only necessary to find G by dividing the bending moment M by bd 2 for the proposed beam and take the required ratios of steel directly from the table.

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