Where d = the total depth (height), in inches, from top of top flange or chord to bottom of bottom flange or chord.

If girder or truss is of steel, use 1 3/5 instead of 1 7/8.

We see therefore that a beam of diminishing cross-section throughout is only about § as stiff, as one with uniform cross-section, as its amount of deflection will be one-half more than that of the latter. Both deflections are approximate only, however, as we see by comparing the amount for the uniform cross-section to that obtained from Formula (79). The deflection for varying cross-sections however can be assumed as nearly enough correct, as these are never diminished so much practically as we have assumed in theory. Now taking the case of a trussed beam.

Table XVIII Trussed Beams 100241

Fig. 146.

In Figure 146, let A B be one half of a trussed beam, let B C be the strut and A C the tie. We will consider the load concentrated at B. Now the first effect is to shorten A B by compression, let us say to D B.

Then, of course, A D will represent one half of the contraction in the whole beam A G. Now the end of rod A moving to D will, of course, let the point C down to E, if we make D E = A C.

But there will be an elongation in D E besides, due to the tension in it, which will let it down still further, say to F, if D F= A C + elongation in A C, of course the point B will move down too, but we can overlook this to avoid complication. We now have C F representing the amount of the deflection. To this should be added the amount of contraction of B C due to the compression in it. We can readily find C F.

We know that

Deflection Trussed Beam.

Deflection Trussed Beam

Now D F we know is = A C plus the elongation of A C due to the tension in it, which we can find from Formula (88). From same formula we find the amount of contraction in A G of which A D is one-half, subtracting this from A B or 1/2 leaves, of course, D B.

Now having found B F we substract from it B C, the length of which is known, and the balance is of course the deflection C F; to this we add the contraction of B C and obtain the total deflection of the whole trussed beam.

If the load had been a uniform load, instead of a concentrated one over the strut, there would be a deflection in that part of A G which would be acting as a continuous girder. But this deflection would take place between B and G and between B and A and would not affect the deflection of the whole trussed beam.

An example will make much of the foregoing more clear.

Example.

A trussed Georgia Pine beam is 16" deep and of 24 feet clear span; it bears 16" on each support and is trussed as shown in Figure 147. The beam carries a uniformly distributed load of 40800 pounds on the whole span including weight of beam and trussing. Of what size should the parts be?

Table XVIII Trussed Beams 100243Table XVIII Trussed Beams 100244

Fig. 147.

We draw the longitudinal neutral axes of each part, namely A B, B C and A C. The latter is so drawn that the neutral axis of the reaction, which is of course half way between end of girder and E (or 8" from E) will also pass through A.

In designing trusses this should always be borne in mind, that so far as possible all the neutral axes at each joint should go through the same point.

The beam A F virtually becomes a continuous girder, of two equal spans of 12 feet or 144" each, uniformly loaded with 20400 pounds each, and supported at three points A, B and F. From table XVII we know that the greatest bending moment is at B and

= u.l/8 = 20400.144/8 = 367200 pounds-inch.

The modulus of rupture for Georgia pine (Table IV) is (k/f)= 1200, therefore moment of resistance (r) from Formula (18) and Table I, section No. 2, r = b.d2/6 = 367200/1200 or b. d2 =1836 Now we know that d = 16, or d2= 256, therefore b = 1836/256 = 7,2 or say we need a beam 7 1/4" x 16" for the transverse strain. We must add to this however for the additional compression due to the trussing.

The amount of the load carried by strut C B, see Table XVII, is = 5/8. u from each side, or = 25500 on the strut B C, of which = 12750 from each side. If now we make at any scale a vertical line b c = half the load carried at point B or = 12750 in our case, and draw b a horizontally and a c parallel to A C, we find the strain in B A by measuring b a = (32300 pounds) or in A C by measuring a c = (34638 pounds) both measured at same scale as b c. We find, further, in passing around the triangle c b a c - (c b being the direction of the reaction at A), that b a is pushing towards A, therefore compression; and that a c is pulling away from A, therefore tension. Using the usual signs of + for compression, and - for tension, we have then:

Compression in Strut.

Compression in Beam.

AB

=

+

32300

pounds.

AC

=

+

34638

pounds.

BC

=

+

25500

pounds.

Had we used Table XVIII we should have had the same result for:

Compression in A B = 25500/2.AB/BC = + 32300 pounds and

Tension in A C= 25500/2.AC/BC = - 34638 pounds.

Now the safe resistance of Georgia pine to compression along fibres

(Table IV) is

(c/f) = 750 pounds.

If A B were very long, or the beam very shallow or very thin, we should still further reduce (c/f) by using Formulae (3), or (5). But we can readily see that the beam will not bend much by vertical flexure due to compression, nor will it deflect laterally very much, so we can safely allow the maximum safe stress per square inch, or 750 pounds, that is, consider A B a short column.

The necessary area to resist the compression, Formula (2) is: