The formula quoted is that drawn up by Tredgold, but known as Gordon's formula. It takes account of the least width (d) of a strut as one of the elements upon which the strength depends. Experiments, however, do not agree very closely with the results of calculation by this formula, and Rankine altered it by taking into account the least radius of gyration (r) instead of the least diameter or width, and this brought calculation and 1 experiment into close agreement. The in the Gordon formula is made

up of three elements

1 m

a

=

α

in which ma fixing modulus, say 1 for both ng

ends fixed, 2.5 for one end fixed and one rounded, 4 for both ends rounded, 16 for one end fixed and the other free; n = a shape modulus, say 11 for hollow cylindrical section, 1 for solid rectangular section, for solid cylindrical section, for cross, L, H or T section, to for built-up beams; q = a strength modulus, say 500 times the tensile working strength in tons per sq. in. Suppose the case to be a hollow cylindrical cast-iron pillar, 1 1 ends flat and fixed, the ordinary value for is ; but to show the confusion α 800

1 that exists, it may be pointed out that for the same thing is given by 400 Hurst, Molesworth, Wray, Campin, Horner, Lineham, Stoney, and The Builder; by Trautwine; by Rankine, Fiddler, Cambria Steel Co.,

1 600

1 800

3 800

Mitchell, and Notes in Building Construction; by Moos.

Proceeding, let us try a 41-in. × 4-in. × §-in. angle, and then the stresses produced will be equal to

= 3.90 tons compression and 2.70 tons per sq. in. tension, in which case, using Formula 460, the safe permissible stress is

Therefore this section will do nicely. We have purposely taken a wrong section previously to this in order to show the reader that in such cases as these the first or even the second and third approximations will not give the required results, when the calculations must be gone into again.

The horizontal ties would be 3-in. × 3-in. × 3-in. angles, and the cross braces 21 in. flats. A firm concrete base 14 ft. square × 4 ft. thick should be provided, and into it the angles should be bedded 2 ft. 6 in. deep. Details of the construction are shown in figs. 897, 898, 899, 900, and 901.

Another interesting case appears in fig. 901A.

It is required to find the dimensions of a suitable column AC. For this case we shall assume the wind pressure at 20 lbs. per sq. ft. We then proceed to resolve the wind pressure into 8 lbs. vertical and 3 lbs. horizontal on the roof itself. The roof bay 20 ft. x 43 ft. 860 sq. ft. and x 8

=

=

=

= 6880 lbs. 3 tons as the vertical component of the wind pressure, and 860 × 3 = 2580 lbs., say 1 ton horizontal component. We shall consider our column as fixed at A and hinged at C, being loaded at B. We must now obtain the point of contraflexure above ground-level, which is very often

assumed in practice as half AB, but is better found with accuracy by the formula

We then only take into consideration the wind pressure above Y, which is

=

FIG. 901.

So with a roof load of 1 ton horizontal we

2.7 tons each at Y, the point of contrathe foot of the knee braces

363 in.-tons on the leeward stanchion; that on the

windward one, being somewhat less, is not considered; so taking moments about points of contraflexure and dividing by the span we have—

=

=

the radius of gyration being, of course, taken in the direction of flexure. Now BM 363 in.-tons and M = modulus of section 62.58... S = BM 363 main comp. stress on the extreme fibres = M 62.58

=

=5.9 tons per sq. in., and

Designing a Cast-Iron Pillar by the Gordon Formula.-We have already treated the subject of cast-iron pillars in a previous chapter, giving therein certain formulæ whereby they could be designed.

Many, however, prefer to use Gordon's formulæ direct as being more accurate, although they are certainly more troublesome to use. To be able to dispose of the metal in the most economical way in a pillar of this description is not nearly so easy a matter as it first appears, and a little foundry experience is always valuable in this departure. In the first place, to cast a pillar less than in. thick would not be advisable, and the usual conditions on the design of castings previously set forth must be adhered to.

The usual foundry practice is to make the diameter about the height and the thickness the diameter. These rules, however, must not be taken as binding, because this practice will vary greatly. We proceed to investigate

a case.

Required

12

A cast-iron stanchion (or pillar, as they are usually called to distinguish them from steel stanchions) 15 ft. high for a load of 40 tons.

Investigating our equation, we have the constant = 1/320, being recognised as correct in most text-books for this particular case. Assuming one end fixed and one end free, we are taking a safe stress of 6 tons, which would be quite suitable for a dead load, while we are taking a 9-in, pillar at random to see what sort of a result it will give.

=

Now a 9-in. pillar in, thick has a total sectional area of 63.6 – 47·1 16.50, a pillar which would do, as it gives a fairly good distribution of metal. It is the duty of the designer, however, to repeat these calculations and see if, by using other diameters and thicknesses which, being within reasonable limits, would give safe results and a lighter pillar, because such castings usually cost so much per cwt.

Fig. 901в shows a suitable design for a C.I. pillar.

Lattice Bracing on Stanchions.-Although this work is not usually subject to more mathematical consideration than is given on the subject in this

chapter, yet the following method of accurately determining the design may

be of use to the student. Referring to fig. 901c,

and must not exceed eight times the least width of the segments connected. In practice it is usual to employ

(1) For 15-in. channels or sections built of 3-in. and 4-in. angles, 2-in. lattice bars having rivets.

(2) For 12-in. and 10-in. channels or sections built up of 3-in. angles, 21-in. lattice bars are used with 2-in. rivets.

(3) Likewise with 9-in. and 8-in. channels or sections of 2-in. angles, 2-in. bars and g rivets are used.

But in no case should any

FIG. 901B.

FIG. 901c.

lattice bar when single be thinner than, or, if double, than of the distance between the rivets connecting them to the main members, assuming the double bars to be riveted at their intersection.