When the floors of these buildings are constructed with rolled-iron beams and brick arches, then the following (Art. 568, Transverse Strains) is the appropriate equation for the area of cross-section of the bottom flange of the girder:

a' = (140+y/3c) 700/700-l x c'mn/2 dk; (85.)

in which a' is in inches, and c, c , d, l, m, and n are in feet. Also, a' is the area required; y is the weight per yard of the rolled-iron beam of the floor; c, their distances from centres; c', the distance from centres at which the tubular girders are placed, or the breadth of floor carried by one girder; d, the depth of the girder; k, the effective resistance of the metal per inch in the flanges of the girder; and;m and n are the distances respectively from the two ends of the girder to the point at which the area of cross-section of the bottom flange is required. The rule may be thus described:

Rule LXIII. - Divide the weight per yard of the rolled-iron beams by 3 times their distance from centres; to the quotient add 140 and reserve the sum; deduct the length in feet from 700, and with the remainder as a divisor divide 700; multiply the quotient by the above reserved sum, and by the value of c'; divide the product by the product of twice the depth into the value of k, and the quotient multiplied by the values of m and of n will be the required area of cross-section of the bottom flange at the point in the length distant from the two ends equal to m and n respectively.

Example. - In a floor of 9-inch, 70-pound beams, 4 feet from centres, what ought to be the area of the bottom flange of a tubular girder 40 feet long between bearings, 2 feet 8 inches deep, and placed 17 feet from the walls or from other girders; the area of the flange to be ascertained at every 5 feet of the length: the value of k to be put at 9000? Here y = 70, c = 4, c' - 17, 1=40, and d= 2 2/3. Therefore, by the rule -

a' = (140+70/3x4) 700/700-40 x 17/2x2 2/3x9000 x mn;

a' - 145.8 1/3 x 1 .0606 x 0.000354l 2/3 x mn; a' = 0.05478 m n.

The values of m and n are -

At the middle................................................................

m

=

20;

n

=

20

5 feet from middle.......... ........

m

=

15;

n

=

25

10 " " " ..................

m

=

10;

n

=

30

15 " " " ...............................................

m

=

5;

n

=

35

These give -

At the middle.............................

a'

=

0.05478

x

20

x

20

=

21 .91

" 5 feet from middle...............

a'

=

0.05478

x

15

x

25

=

20.54

" 10 " " " .....

a'

=

0.05478

x

10

x

30

=

16.43

"15 " " " .........

a'

=

0.05478

x

5

x

35

=

9.59

These are the areas of uncut fibres at the points named, in the lower flange; the upper flange requires the same sizes.

189. - Tubular Iron Girders, for Floors of First-Class

Stores. - The equation proper for this is (Transverse Strains, Art. 570) -

a' = (320+y/3c) 700/700-l x c'mn/2 dk' (86.)

a rule the same in form as that of the previous article; hence it needs no particular exemplification.

Rule LXIII. of last article may be used for this case, simply by using the constant 320 in place of that of 140.

Cast-Iron Girders

190. - Cast-iron Girders: Inferior. - Rolled-iron beams have been so extensively introduced within a few years as to have superseded almost entirely the formerly much used cast-iron beam or girder. The tensile strength of cast iron is far inferior to that of wrought iron. This inferiority and the contingencies to which the metal is subject in casting render it very untrustworthy; it should not be used where rolled-iron beams can be procured. A very substantial girder to carry a brick wall is made by placing two or more rolled-iron beams side by side, and securing them together by bolts at mid-height of the web; placing thimbles or separators at each bolt. As there may be cases, however, in which cast-iron girders will be used, a few rules for them will here be given.

191. - Cast-Iron Girder: Load at Middle. - The form of cross-section given to this girder usually is as shown in

Fig. 47.

In the cross-section, the bottom flange is made to contain in area four times as much as the top flange. The strength will be in proportion to the area of the bottom flange, and to the height or depth of the girder at middle. Hence, to obtain the greater strength from a given amount of material, it is requisite to make the upright part, or the web, rather thin; yet, in order to prevent injurious strains in the casting while it is cooling, the parts should be nearly equal in thickness. The thickness of the three parts - web, top flange, and bottom flange - may be made in proportion as 5, 6, and 8.

Cast Iron Girders 74

Fig. 47.

For a weight at middle, the form of the web should be that of a triangle; the top flange forming two straight lines declining from the centre each way to the bottom flange at the ends, like the rafters of a roof to its tie-beam. From Transverse Strains, Art. 583, we have -

a' = Wal/4850 d, (87.),

which is a rule for the area in inches of the bottom flange, for a load at middle; the area of the top flange is to be equal to one fourth of that of the bottom flange. To secure this, make the width of the top flange equal to one third of the width of the bottom flange; the thickness of the former, as before directed, being made equal to 6/8 or 3/4 of the latter. The weight W is in pounds; the length l is in feet; and the depth d is in inches. The factor of safety a should be taken at not less than 3; better at 4 or 5.

The equation in words may be as follows:

Rule LXIV. - Multiply the weight by the length, and by the factor of safety; divide the product by 4850 times the depth at middle, and the quotient will be the area in inches of the bottom flange; divide this area by the width of the bottom flange, and the quotient will be its thickness. Of the top flange make its width equal one third that of the bottom flange, and its thickness equal to three quarters that of the latter. Make the thickness of the web equal to 5/8 that of the bottom flange.

Example. - What should be the dimensions of the cross-section of a cast-iron girder 20 feet long between bearings, and 24 inches high at middle, where 30,000 pounds is to be carried; the factor of safety being put at 5?

Here we have W = 30000; a = 5; l = 20; and d = 24; therefore, by the rule -

a' = 30000x5x20/4850x24 = 25 .773.

This is the area of the bottom flange. If the width of this flange be 12 inches, then 25.773 divided by 12 gives 2.15, or 2 1/8 full, as the thickness. One third of 12 equals 4, equals the width of the top flange; and 3/4 of 2 . 15 equals 1 . 61, or 1 5/8 - its thickness. The thickness of the web equals 5/8 x 2 . 15 = 1 .34 or 1 1/3 inches.

192. - Cast-iron Girder: Load Unformly Distributed.

The equation suitable to this is -

a' = Ual/9700 d' (88.)

a rule of like form with that of the last article; therefore, Rule LXIV. may be used for this case, simply by substituting 9700 for 4850.