Adding this to the above we should have the total deflection δ = 0,655+ 0,0125 = 0,6675

This would be the amount we should have to "camber" up the team, or say 3/4".

The safe deflection not to crack plastering, would be (Formula 28) δ = L. 0,03 = 24.0,03 = 0,72

So that our trussed beam is amply stiff.

Table XIX gives all the necessary data in regard to the use of Tables XX, XXI, XXII, XXIII, XXIV and XXV. These tables give all the necessary information in regard to all architectural sections which are rolled. Where, after the name of the company rolling the section there are several letters, it means that practically the same section is rolled by several companies. It should be remarked that except in the case of the simplest kind of beam work, it is cheaper to frame up plate girders, or trusses, of angles, tees, etc., as there is a strong pool in the rolling of I-beams and channel sections, which keeps the price of these two sections unreasonably high, in proportion to other rolled sections. Steel beams and sections are sold as cheap as iron, (they are really cheaper to manufacture), and where their uniformity can be relied on, should be used in preference, as they are much stronger and also a trifle stiffen As a rule, however, the uniformity of steel in beams and other rolled sections cannot be relied on.

One example of an iron beam will make the application of the Tables to transverse strains clear, and help to review the subject, before taking up the graphical method of calculating transverse strains.

Explanation of Tables XIX to XXV.

I-sections not economical.

Example.

A wrought-iron I-beam of 25-foot clear span, carries a uniform load of 500 pounds per fool including weight of beam; also a concentrated load of 1000 pounds 10 feet from the rigid hand support. The beam is not supported sideways. What size beam should be used?

The total uniform load u = 500.25 = 12500 pounds of which one-half or G250 pounds will go to each reaction; of the 1000 pounds load

180/300 or 3/5 will go to the nearer support q (Formula 15), therefore q = 6250 + 3/5. 1000 = 6850 Similarly we should have (Formula 14) p = 6250 + 2/5. 1000 = 6650 As a check the sum of the two loads should = 13500, and we have, in effect:

6850 + 6650 = 13500 To find the point of greatest bending moment begin at q pass to load 1000, and we will have passed over ten feet of uniform load or 5000 pounds, add to this the 1000 pounds making 6000 pounds, and we still are 850 pounds short of the reaction, we pass on therefore towards p one foot, which leaves 350 pounds more, and pass on another 7/10 of a foot (to A) which very closely makes the amount. The point of greatest bending moment therefore is at A, say 1' 8" to the left of the weight, or 140" from q: As a check begin at p and we must pass along 160" or 13' 4" of uniform load before reaching the point A, at 500 pounds a foot this would make 13 1/3.500. = 6666 or close enough to amount of reaction p for all practical purposes.

Table XVIII Trussed Beams 100246

Fig. 148.

The uniform load per inch will be 500/12 =41 2/3 pounds.

Now the bending moment at A will be, taking the right-hand side (Formula 24) mA = 6850.140 - 41 2/3. 140.70 - 1000.20 = 530 667 pounds-inch. As a check take the left-hand side (Formula 23) mA = 6650.160 - 412/3.160.80

= 530614 pounds-inch, or near enough alike for all practical purposes.

Now the safe modulus of rupture for wrought-iron (Table IV) is

(k/f)= 12000 pounds, therefore the required moment of resistance r from Formula (18) r= 530667/12000 = 44,2

Looking at the Table XX we find the nearest moment of resistance to be 46,8 or we should use the 12" - 120 pounds per yard I-beam.

But the beam is unsupported sideways. The width of top flange is b = 5 1/2". We now use Formula (78) to find out how much extra strength we require.

In inserting value for y, we use the second column of Table XVI, as the beam is, of course, of uniform cross-section throughout, and have y= 0,0192. In place of w we can insert the actual value r of the beam, and see what proportion of it is left to resist the transverse strength, after the lateral flexure is attended to, or r1 = r/1+0,0192.252/5 1/22 = r/1+0,3966 or r1 = 46,8/1,3966 = 33,6 or the beam would not be strong enough. The next size would be the 12 1/4" - 125 pounds per yard beam, but as the 15" - 125 pounds per yard beam would cost no more and be much stronger we will try that. Its width of flange is b = 5" and moment of resistance r = 57,93. Inserting these values in (Formula 78) and using r in place of w we have r1 = 57,93/1+0,0192.252 /52= 57,93/1,48

= 39,14 The required moment of resistance was r = 44,2 so that this is still short of the mark, and we should have to use the next section or the 15" - 150 pounds per yard beam. The moment of resistance of this beam is r = 69,8 its width of flange the same as before, therefore: r1 =69,8/1,48 = 47,1

Or this beam would be a trifle too strong even if unsupported sideways. We need not bother with deflection, for the length of beam is only 1 2/3 times the span, and besides not even § of the actual transverse strength of the beam is required to resist the vertical strains, and, of course, the deflection would be diminished accordingly. The column in Table XX headed "Transverse Value," gives the safe uniform load, in pounds, if divided by the span in feet, for beams supported sideways. Of course the result should correspond with Table XV, except that the uniform load will be expressed in pounds here, while it is expressed in tons of 2000 pounds each in that table. For Tables XXI, XXII, XXIII, XXIV and XXV the use of the "Transverse Value" is similar, and as more fully explained in Table XIX.