This section is from the book "Safe Building", by Louis De Coppet Berg. Also available from Amazon: Code Check: An Illustrated Guide to Building a Safe House.

That these strains just equal each other is due to the design of the truss and its uniform loading. Were the design of the parts less symmetrical, or the load not symmetrical, or the wind on one side only, the two strains would not be found to be equal.

Truss with abutment.

In Figure 246 whore we consider the wall as incapable of taking up the thrust, and the truss has to take care of it itself, it will be found that this horizontal member (now called P Y) is in tension.

The drawing of the strain diagram (Figure 248) presents no difficulty, but it will be seen that nearly all the strains on the truss are very much larger.

The actual thrust of the truss Figure 244 on the wall will be parallel to and equal to a o of Figure 245. To resist it there must be weight enough to deflect this line sufficiently not to overturn the wall. This subject was thoroughly treated in Chapter V (Arches), Vol. I, under the heading "Arch with Abutment."

The analysis of strains due to wind pressure, when taken separately on one side of the roof only, and at right angles to same, is treated the same as for dead loads, excepting that the load line is no longer vertical, and is drawn therefore parallel to the wind arrows (that is, at right angles to the main rafter on which wind blows). Where the roof is a broken one the wind load line is broken also, its different parts being proportioned to the amounts and parallel to the directions of the wind on the different parts.

It will be readily seen that the left and right reactions due to the wind cannot be equal, and the calculation of these reactions offers the only difficulty. Let us take a truss similar to the one shown in Figure 240 and overlooking the steady or dead load assume that the wind is blowing from the right hand side. Its effect will be as shown in Figure 249. The pressure on the apex or A B will be equal to half the length of rafter B G multiplied by the product of the distance between trusses and the wind pressure per square foot, as given in Table XLIV. It is indicated by the arrow A B. The pressure in the middle joint - indicated by arrow B C - will be this same (latter) product multiplied by the sum of half the length of B G plus half the length of E C. The pressure at the foot - or arrow CD - will be the same product multiplied by half the length of E C. The reactions at p and q which resist this wind pressure will evidently be in the opposite direction to the wind, or slanting, as shown. If, therefore, the truss and dead load is not heavy enough to sufficiently deflect these reactions, there would need to be holts or buttresses at p and q to keep the truss from sliding-off the wall. This will be explained at the end of the chapter in speaking of spires.

Truss without abutment.

Wind load-line normal to roof.

Wind reactions unequal.

We now have the amount and direction of wind pressures and direction of the reactions, but we do not know the amounts of the reactions, as they evidently cannot be equal.

Fig 249a.

If we imagine the neutral axis of the entire wind pressure, that is the entire wind pressure concentrated at the point 1 at the centre of the length of rafter, the effect, so far as reactions are concerned will be the same.

We now draw the horizontal line P Q connecting the feet of rafters, prolong the centralized wind pressure (or neutral axis) arrow 1 till it intersects P Q at 2, then will the reactions P and Q be inversely as the divisions of line P Q; that is, if P Q = l and P2 = m and 2 Q = n and the whole wind pressure = w, we should have

Amount of wind reactions.

p = w.n/l

(130) and q = w.m/l

(131)

Where p = the amount of (slanting) left reaction, in pounds, due to wind pressure.

Where q= the amount of (slanting) right reaction, in pounds, due to wind pressure.

Where l = the length, in inches, measured horizontally between centres of feet of truss.

Where m and n = respectively, in inches, the lengths into which the horizontal line is divided by the prolongation of the centre line (or neutral axis) of wind pressure, m being nearer p and n nearer 7.

The analogy between these formulas and Formulæ 14 to 17 should be noticed. The analyzing of the strains by means of the diagram Figure 249a will present no difficulty. We draw a b parallel and equal the pressure A B at the apex, be = 13 C pressure at central joint and cd = C D pressure at foot. Along this load line a d we now lay off in opposite directions the reactions, namely, d0= D 0 or reaction 7 and

0 n = 0 A or reaction p;. We first find the strains on the foot joint 0 A by drawing ahoa, then skip to the apex joint and draw a b g a; the rest presents no difficulty.

It will be noticed that the effect of the wind is to reverse the strains on two pieces. The upper part of the rafter on the side on which the wind is blowing is now in tension, whereas in Figure 240 it was in compression ; further the tie H O now becomes a thrust member opposing the wind, where before it was in tension.

Fig.250.

Fig.250a.

Had we drawn the wind on the left side of truss the calculation would have been similar. The direction of p and q would have been from right upwards to the left; p would have been the larger reaction and we should have found the strains same as in Figure 240, except on A G which would become tension and on HF (or 0 E) which would now become compression. It is not necessary to make more than one wind diagram, therefore, where both feet of the trusses are bolted down.

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